r k bansal - A Textbook of Fluid Mechanics and hydraulic machines. 9-laxmi.pdf
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About This Presentation
Book
Slide Content
A Textbook of
FLUID MECHANICS
HYDRAULIC MACHINES
SL Uns
FLUID MECHANICS
HYDRAULIC MACHINES
SL Uns
A TEXTBOOK OF FLUID MECHANICS
AND
HYDRAULIC MACHINES
AND
HYDRAULIC MACHINES
WIN A CASH AWARD OF Rs. 500.00
I bas come to our notice that some booksellers are fraudulently selling fake/duplicate
copies of some of our fast sling ites
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‘cling ites, The Hologram displays a unique 2D sult-evel, mu caour effect rom diferent
angles: It has the following three levels of lt graphics merged together. The background artwork
Sem o be under’ or chia the Hologram and gives the illusion of depth unlike the fake Hologram
fon the fakeduplicate books.
P
Present, ony some es have et the Hologram his cate A Text Book of Fi
‘Machines (2010 edition, priced at Rs. 450.00) has gt the Hologram.
If ged say yee finado ds torno lab ey bak oe ll wot
ho LP Hologram, Nao requested to writ tous at Ms LAXMI PUBLICATIONS PVT. LTD.
113, Golden House, Daryagan), New Delhi-110002, giving the name and adress oh bookseller
rom where hole purchased is bok, together wit the photocopy ofthe cer and the 2a page on
hh e price ofthe book and name o the printer are printed. He/She willbe seta cash award of
Re 50000.
How to decide ifthe book is genuine or fake?
1 The above information may or may not. bo printed.
2 The counter tion of the back may have na LP Hologram or it has, it wil be without
{he lusionary depth as described above.
What is the harm in purchasing duplicate books ?
— Poor quality of paper and printing which alot your eye.
Normal o authors who are scholars and have put ther bard labour in writing the bok,
thus depriving them of her talent right
a offence. Legal action shall o ton aie th)
rer found gully of such an offence in any way
[Warning +Saling or buying pirated books is
books) and student) or m
I bas come to our notice that some booksellers are fraudulently selling fake/duplicate
copies of some of our fast sling ites
In our sincere efforts to provide you with our genuine books and to protect you against those
counterfeit hooks, Lax Publications (LP) has put a Hologram onthe cover of some of its fast
‘cling ites, The Hologram displays a unique 2D sult-evel, mu caour effect rom diferent
angles: It has the following three levels of lt graphics merged together. The background artwork
Sem o be under’ or chia the Hologram and gives the illusion of depth unlike the fake Hologram
fon the fakeduplicate books.
P
Present, ony some es have et the Hologram his cate A Text Book of Fi
‘Machines (2010 edition, priced at Rs. 450.00) has gt the Hologram.
If ged say yee finado ds torno lab ey bak oe ll wot
ho LP Hologram, Nao requested to writ tous at Ms LAXMI PUBLICATIONS PVT. LTD.
113, Golden House, Daryagan), New Delhi-110002, giving the name and adress oh bookseller
rom where hole purchased is bok, together wit the photocopy ofthe cer and the 2a page on
hh e price ofthe book and name o the printer are printed. He/She willbe seta cash award of
Re 50000.
How to decide ifthe book is genuine or fake?
1 The above information may or may not. bo printed.
2 The counter tion of the back may have na LP Hologram or it has, it wil be without
{he lusionary depth as described above.
What is the harm in purchasing duplicate books ?
— Poor quality of paper and printing which alot your eye.
Normal o authors who are scholars and have put ther bard labour in writing the bok,
thus depriving them of her talent right
a offence. Legal action shall o ton aie th)
rer found gully of such an offence in any way
[Warning +Saling or buying pirated books is
books) and student) or m
A TEXTBOOK OF
FLUID MECHANICS
AND
HYDRAULIC MACHINES
da SL Units)
(Eng. Services), AMLE: (India)!
By
Dr. RK. BANSAL
B, Se. Engg. (Mech), M.Tech, Hons. (LL, Del)
Ph.D, MLE. (India)
Formerly
Professor, Department of Mechanical Engineering
Dean (UG. Studies), Delhi College of Engineering, Delhi
LAXMI PUBLICATIONS (P) LTD
BANGALORE » CHENNAI = COCHIN = GUWAHATI. + HYDERABAD
JALANDHAR = KOLKATA + LUCKNOW + MUMBAI = RANCHI
NEW DELHI
FLUID MECHANICS
AND
HYDRAULIC MACHINES
da SL Units)
(Eng. Services), AMLE: (India)!
By
Dr. RK. BANSAL
B, Se. Engg. (Mech), M.Tech, Hons. (LL, Del)
Ph.D, MLE. (India)
Formerly
Professor, Department of Mechanical Engineering
Dean (UG. Studies), Delhi College of Engineering, Delhi
LAXMI PUBLICATIONS (P) LTD
BANGALORE » CHENNAI = COCHIN = GUWAHATI. + HYDERABAD
JALANDHAR = KOLKATA + LUCKNOW + MUMBAI = RANCHI
NEW DELHI
Published by
LAXMI PUBLICATIONS (P) LTD
113, Golden House, Daryaganj,
‘New Delbi-110002
Phone : OL1-43 53 25 00
ax : 011.43 53 25 28
wow laxmipublicationscom
Infuelamnipubliatins com
Author + Dr. R.K. Bansal
Compiled by + Smt. Nirmal Bi
O All rights reserved with Author and the Publishers. No part of this
‘publication may be reproduced, stored in a retrieval aystem, or transmitted
in any form or by any means, electronic, mechanical, photocopying,
‘recording otherwise without the prior written permission ofthe publisher
Price Rs. 495.00 Only First Edition: Sept 1983.
Ninth Edition 2005
Reprint ; 2006, 2007, 2008, 2009
"Revised Ninth Edition : 2010
OFFICES
080-26 61 16 61 © Chennai 044-24 34 47 25
0184-287 70 04, 406 19.03 © Guwahati 0361-25436 68,251 38 81
(040-24 65 23 33 © Jalandhar 0181222 12 73
© Kolkata” 099.22 27 43 84 © Lucknow 0522220 95 76
© Mumbai 022.24 91 54 15, 24 92 78 69 © Ranchi
EFM-0509-495-FLUID MECHANICS & HM-BAN =
petit a Suse Coese, New Dela Printed ot: Rey India LA, Mud
LAXMI PUBLICATIONS (P) LTD
113, Golden House, Daryaganj,
‘New Delbi-110002
Phone : OL1-43 53 25 00
ax : 011.43 53 25 28
wow laxmipublicationscom
Infuelamnipubliatins com
Author + Dr. R.K. Bansal
Compiled by + Smt. Nirmal Bi
O All rights reserved with Author and the Publishers. No part of this
‘publication may be reproduced, stored in a retrieval aystem, or transmitted
in any form or by any means, electronic, mechanical, photocopying,
‘recording otherwise without the prior written permission ofthe publisher
Price Rs. 495.00 Only First Edition: Sept 1983.
Ninth Edition 2005
Reprint ; 2006, 2007, 2008, 2009
"Revised Ninth Edition : 2010
OFFICES
080-26 61 16 61 © Chennai 044-24 34 47 25
0184-287 70 04, 406 19.03 © Guwahati 0361-25436 68,251 38 81
(040-24 65 23 33 © Jalandhar 0181222 12 73
© Kolkata” 099.22 27 43 84 © Lucknow 0522220 95 76
© Mumbai 022.24 91 54 15, 24 92 78 69 © Ranchi
EFM-0509-495-FLUID MECHANICS & HM-BAN =
petit a Suse Coese, New Dela Printed ot: Rey India LA, Mud
Dedicated.
The loving memory
of
my daughter, Babli
The loving memory
of
my daughter, Babli
PREFACE TO THE NINTH EDITION
The popularity of the eighth edition and roprints of the book A Textbook of Fluid
Mechanics and Hydraulic Machines emongst the students and the teschors of the
rites ofthe country, has prompted the bringing ut of the ninth edition ofthe
book so soon. The ninth edition has been thoroughly rovised and brought up-to-date. A large
number of problems from different BE. degroo examinations of Indian Universities and other
‘examining bodies such as Institution of Enginoors and U.P-SC. upto Summer 2002
‘examinations have been selected and have been solved at proper places on this edition. Most
of these problems have been worked out in Si units, All of the text along with existing
problems have been converted into S.L. Units
Tn the ninth edition, a new chapter entitled Ideal Flow (or Potential Flow) has been added.
Potential Now has been included in most of Indian Universities. This chapter has been written.
ina simple and easy-to-follow language so that even an average student can grasp the subject
matter by selEstudy. Also a few new topics such as “Liquide in Relative Equilibrium” and
¡pe Network” have been added in tis edition. The topic of Pipe Network has been included
the chapter of Flow Through Pipes. The pipe network is mostly used in city water supply
system, Laboratory supply system or house hold supply of water and gas.
"The objective type multiple-choice questions are often asked in the various competitive
‘examinations. Hence a large number of objetivo type questions with answers have been
added in the end ofthe book.
With these addition, itis hoped that the book will be quite useful for the students of
ierent branches of Engineering at various Engineering Institutions.
1oxpross my sincoro thanks to my colleagues, friends, students and the toachors of difer-
‘ent Indian Universities for their valuable suggestions and recommending the book oftheir
students
‘Suggestions forthe improvement of this book ave most welcome and would be incorporated
in tho next edition with view to make the book more uscfal.
= Author
The popularity of the eighth edition and roprints of the book A Textbook of Fluid
Mechanics and Hydraulic Machines emongst the students and the teschors of the
rites ofthe country, has prompted the bringing ut of the ninth edition ofthe
book so soon. The ninth edition has been thoroughly rovised and brought up-to-date. A large
number of problems from different BE. degroo examinations of Indian Universities and other
‘examining bodies such as Institution of Enginoors and U.P-SC. upto Summer 2002
‘examinations have been selected and have been solved at proper places on this edition. Most
of these problems have been worked out in Si units, All of the text along with existing
problems have been converted into S.L. Units
Tn the ninth edition, a new chapter entitled Ideal Flow (or Potential Flow) has been added.
Potential Now has been included in most of Indian Universities. This chapter has been written.
ina simple and easy-to-follow language so that even an average student can grasp the subject
matter by selEstudy. Also a few new topics such as “Liquide in Relative Equilibrium” and
¡pe Network” have been added in tis edition. The topic of Pipe Network has been included
the chapter of Flow Through Pipes. The pipe network is mostly used in city water supply
system, Laboratory supply system or house hold supply of water and gas.
"The objective type multiple-choice questions are often asked in the various competitive
‘examinations. Hence a large number of objetivo type questions with answers have been
added in the end ofthe book.
With these addition, itis hoped that the book will be quite useful for the students of
ierent branches of Engineering at various Engineering Institutions.
1oxpross my sincoro thanks to my colleagues, friends, students and the toachors of difer-
‘ent Indian Universities for their valuable suggestions and recommending the book oftheir
students
‘Suggestions forthe improvement of this book ave most welcome and would be incorporated
in tho next edition with view to make the book more uscfal.
= Author
PREFACE TO THE FIRST EDITION
‘Lam glad to present the book entitled, A Textbook of Fluid Mechanics and Hydraulic
Machines to the engincering students of mechanical, civil, electrical, aeronautical and
chomical and also tothe students preparing for tho new schome of Section B of AMIE.
Examination of Institution of Engineers (India), Tho course contents have been planned in
such a way that the general requirements e all engineering students are falild
During my long experience of teaching this subject to undergraduate and postgraduate
engineering students for the past 16 years, Ihave observed that the students face difficulty in
understanding clearly the basic principles, fundamental concopts and theory without
‘adequate solved problems along with the text. To meet this very basic requirement to the
Students, a large number of the questions taken from the examinations of the various
‘Universities of India and from other professional and competitive examinations (such as
Institution of Engineers and U.PS.C. Engineering Service Examination) have been solved
along with the text in MKS, and S.L, units,
‘The book is written in a simple and easy-to-follow language, so that even an average ctu
dents can grasp the subject by selfstudy. At the end of each chapter highlights, theoretical
‘questions and many unsolved numerical problems with answer are given for the students to
solve them,
Y am thankful to my colleagues, fiends and students who encouraged me to write this
‘ook. Ham grateful to Institution of Engineers (India), various Universities ofl ‘those
‘authorities whose work have been consulted and gave me a great help in preparing the book.
1 express my appreciation and gratefulness to my publisher. Shri RIK. Gupta (as
Mechanical Engineer for his most cooperativo, painstaking otitudo and untiring efforts for
bringing out Ihe book in a short period
Mes, Nirmal Bansal desorvos special credit as she not only provided an ideal atmosphere
at homo for book writing but also gave inspiration and valuable suggestions
‘Though every care has boen taken in checking the manuscripts and proof reading, yet
claiming perfection is very difficult. hall be very grateful to the readers and users of this
ook for pointing any mistakes that might have cropt in. Suggestions for improvement aro
most welcome ond would be inearporated in the next edition with a view to make the book
more sofa,
= Author
‘Lam glad to present the book entitled, A Textbook of Fluid Mechanics and Hydraulic
Machines to the engincering students of mechanical, civil, electrical, aeronautical and
chomical and also tothe students preparing for tho new schome of Section B of AMIE.
Examination of Institution of Engineers (India), Tho course contents have been planned in
such a way that the general requirements e all engineering students are falild
During my long experience of teaching this subject to undergraduate and postgraduate
engineering students for the past 16 years, Ihave observed that the students face difficulty in
understanding clearly the basic principles, fundamental concopts and theory without
‘adequate solved problems along with the text. To meet this very basic requirement to the
Students, a large number of the questions taken from the examinations of the various
‘Universities of India and from other professional and competitive examinations (such as
Institution of Engineers and U.PS.C. Engineering Service Examination) have been solved
along with the text in MKS, and S.L, units,
‘The book is written in a simple and easy-to-follow language, so that even an average ctu
dents can grasp the subject by selfstudy. At the end of each chapter highlights, theoretical
‘questions and many unsolved numerical problems with answer are given for the students to
solve them,
Y am thankful to my colleagues, fiends and students who encouraged me to write this
‘ook. Ham grateful to Institution of Engineers (India), various Universities ofl ‘those
‘authorities whose work have been consulted and gave me a great help in preparing the book.
1 express my appreciation and gratefulness to my publisher. Shri RIK. Gupta (as
Mechanical Engineer for his most cooperativo, painstaking otitudo and untiring efforts for
bringing out Ihe book in a short period
Mes, Nirmal Bansal desorvos special credit as she not only provided an ideal atmosphere
at homo for book writing but also gave inspiration and valuable suggestions
‘Though every care has boen taken in checking the manuscripts and proof reading, yet
claiming perfection is very difficult. hall be very grateful to the readers and users of this
ook for pointing any mistakes that might have cropt in. Suggestions for improvement aro
most welcome ond would be inearporated in the next edition with a view to make the book
more sofa,
= Author
CONTENTS
Chapter Pages
Chapter 1. Properties of Fluide fen
1.1. Introduction 1
12, Properties of Fluide A
121. Density or Mass Density 1
122 Specific Weight or Weight Density 1
123. Specifi Volume 2
124. Specie Gravity 2
Solved Problems 11-12 2
1.8, Viscosity a
131. Unite of Viscosity 3
132. Kinematic Visual 5
133. Newton's Law of Viscosity 5
184. Variation of Viscnity with Temperature 6
185. Types of Fluids 5
Solved Problems 19-110 8
1.4. Thermodynamic Properties n
LAL Dimension of R e 15
142. othermal Process 18
143. Adiabaie Process 18
TAA Universal Gas Constant 1
Solved Problems 120-122 1
1.8. Compressiblity and Bulk Modulus -
Solved Problems 125-124
1.6. Surface Tension and Capilaity
161. Surface Tension on Liquid Droplet 2
162. Surface Tension on a Hollow Bubble 2
183. Surface Tension on a Liquid Jet E
Solved Problems 126127 E]
184. Capilar 5
Solved Problems 128-132 25
1.7. Vapour Pressure and Cavitation 29
Heights 30
Exercise 30
Chapter 2. Pressure and Its Measurement 35-68
21. Fluid Pressure at a Point a as
22, Pascal Law a
23. Pressure Variation in a Fluid at Rest 36
Solved Problems 2.1027 E
oy
Chapter Pages
Chapter 1. Properties of Fluide fen
1.1. Introduction 1
12, Properties of Fluide A
121. Density or Mass Density 1
122 Specific Weight or Weight Density 1
123. Specifi Volume 2
124. Specie Gravity 2
Solved Problems 11-12 2
1.8, Viscosity a
131. Unite of Viscosity 3
132. Kinematic Visual 5
133. Newton's Law of Viscosity 5
184. Variation of Viscnity with Temperature 6
185. Types of Fluids 5
Solved Problems 19-110 8
1.4. Thermodynamic Properties n
LAL Dimension of R e 15
142. othermal Process 18
143. Adiabaie Process 18
TAA Universal Gas Constant 1
Solved Problems 120-122 1
1.8. Compressiblity and Bulk Modulus -
Solved Problems 125-124
1.6. Surface Tension and Capilaity
161. Surface Tension on Liquid Droplet 2
162. Surface Tension on a Hollow Bubble 2
183. Surface Tension on a Liquid Jet E
Solved Problems 126127 E]
184. Capilar 5
Solved Problems 128-132 25
1.7. Vapour Pressure and Cavitation 29
Heights 30
Exercise 30
Chapter 2. Pressure and Its Measurement 35-68
21. Fluid Pressure at a Point a as
22, Pascal Law a
23. Pressure Variation in a Fluid at Rest 36
Solved Problems 2.1027 E
oy
o
2.4. Absolute, Gauge, Atmospheric and Vacuum.
Pressures a
‘Solved Problem 28 2
125. Measurement of Presse se
251. Manometers : Y
252. Mechanical Gauges ® 4
26. Simple Manometers 43
281. Piesmmeter “
262, Utube Manometer 43
Solved Problems 20-213 à “
263. Single Column Manometer is
Solved Problem 214 50
2. Differential Manometers 50
271. Ustube Differential Manometer 50
Solved Problems 215-217 E
2.72, Inverted U-tube Dierential Manometer E
Solved Problems 218221 5
28, Pressure ata Point in Compresihe Fluid = 56
251. Isothermal Prowse 56
282. Adiabate Process 56
283, Temperature at any Point in
Compressible Fluid 58
254. Temperature LapseRate (L) 59
Solved Problems 222-226 E
Highlights se
Exercise
Chapter 3. Hydrostatic Forcos on Surfaces 69-190
31. Introduction 60
32. Total Pressure and Centre of Pressure E
33. Vertical Plane Surface Sub merged in Liquid ”
Solved Problems 3.13.12 2
34. Horizontal Plane Surface Sub merged in Liquid Ss
Solved Problem 3.13 88
35. Inclined Plane Surface Sub-merged in Liquid E
Solved Probleme 3.1(0) 3.21 E 83
36. Curved Surface Sub-merged in Liquid = E
Solved Problems 322331 E
37. Total Pressure and Centre of Pressure 0
Lack Gates 107
Solved Problems 2222.20 109
3.8. Pressure Distribution in a Liquid Subjected to
Constant HorizontaVVertical Acceleration ue
381. Liquid Containers Subject to Constant
Horizontal Aceleration me
Solved Problems 3343.36 1
382. Liquid Containers Subjected to Constant
Vertical Acceleration 120
Solved Problems 297-336 me
Highlights mi
Exercise 185
2.4. Absolute, Gauge, Atmospheric and Vacuum.
Pressures a
‘Solved Problem 28 2
125. Measurement of Presse se
251. Manometers : Y
252. Mechanical Gauges ® 4
26. Simple Manometers 43
281. Piesmmeter “
262, Utube Manometer 43
Solved Problems 20-213 à “
263. Single Column Manometer is
Solved Problem 214 50
2. Differential Manometers 50
271. Ustube Differential Manometer 50
Solved Problems 215-217 E
2.72, Inverted U-tube Dierential Manometer E
Solved Problems 218221 5
28, Pressure ata Point in Compresihe Fluid = 56
251. Isothermal Prowse 56
282. Adiabate Process 56
283, Temperature at any Point in
Compressible Fluid 58
254. Temperature LapseRate (L) 59
Solved Problems 222-226 E
Highlights se
Exercise
Chapter 3. Hydrostatic Forcos on Surfaces 69-190
31. Introduction 60
32. Total Pressure and Centre of Pressure E
33. Vertical Plane Surface Sub merged in Liquid ”
Solved Problems 3.13.12 2
34. Horizontal Plane Surface Sub merged in Liquid Ss
Solved Problem 3.13 88
35. Inclined Plane Surface Sub-merged in Liquid E
Solved Probleme 3.1(0) 3.21 E 83
36. Curved Surface Sub-merged in Liquid = E
Solved Problems 322331 E
37. Total Pressure and Centre of Pressure 0
Lack Gates 107
Solved Problems 2222.20 109
3.8. Pressure Distribution in a Liquid Subjected to
Constant HorizontaVVertical Acceleration ue
381. Liquid Containers Subject to Constant
Horizontal Aceleration me
Solved Problems 3343.36 1
382. Liquid Containers Subjected to Constant
Vertical Acceleration 120
Solved Problems 297-336 me
Highlights mi
Exercise 185
Gi
Chapter 4. Buoyancy and Floatation ace
A1. Introduction m
42, Buoyaney m
43. Centre of Buoyaney a m
Solved Problems 41-46 m
Ad. Meta-contre a 136
15. Metacrntrie Height 4 136
4.6. Analytical Method for Meta Centre Height u m
Solved Problems 474.11 138
4.7. Conditions of Equlibrivm of a Fasting and
Sulbmerged Bodies us
ATA Stability ofa Sab-merged Body 18
472, Stability ofa Floating Body : 1
Solved Problems 412-418 Ma
4.8, Experimental Method of Determination of
Metarcentrie Heigat 154
Solved Problems 410-120 185
49. Oscillation (Rolling) of a Floating Body 156
Solved Problems 421.422 158
Highlights 150
Exercise 160
Chapter 5. Kinematics of Flow and Ideal Flow 169-258
A KINEMATICS OF FLOW
51. Introduction 163
52. Methods of Deseibing Fluid Motion. 163
53. Types of Fuld Flow m
531. Steady and Unsteady Flows 163
532. Uniform and Noneunform Flows 164
533. Laminar and Turbulent Flows 164
534, Compressible and incompressible lows 1. m
535. Rotational and Irotational Flows 165
536. One, two and Three Dimensional Flows 165
5.4. Rate of Flow or Dischargo (Q) 165
55. Continuity Equation 2 165
Solved Problems 5.155 166
5.6. Continuity Equation in Taree Dimensions m
581. Continuity Equation in Cylindrical
Polar Coordinates m
Solved Problema 3.54 Ys
57. Velocity and Acceleration m
571. _ Lacal Aceteration and Convective
Acceleration 1m
Solved Problems 56-59 1m
58. Velocity Potential Funcion and Stream Function 18
S81. Velocity Potential Function. m
582, Stream Funetion m
583. Equipotental Line 183
584, Line of Constant Stream Function 1
Chapter 4. Buoyancy and Floatation ace
A1. Introduction m
42, Buoyaney m
43. Centre of Buoyaney a m
Solved Problems 41-46 m
Ad. Meta-contre a 136
15. Metacrntrie Height 4 136
4.6. Analytical Method for Meta Centre Height u m
Solved Problems 474.11 138
4.7. Conditions of Equlibrivm of a Fasting and
Sulbmerged Bodies us
ATA Stability ofa Sab-merged Body 18
472, Stability ofa Floating Body : 1
Solved Problems 412-418 Ma
4.8, Experimental Method of Determination of
Metarcentrie Heigat 154
Solved Problems 410-120 185
49. Oscillation (Rolling) of a Floating Body 156
Solved Problems 421.422 158
Highlights 150
Exercise 160
Chapter 5. Kinematics of Flow and Ideal Flow 169-258
A KINEMATICS OF FLOW
51. Introduction 163
52. Methods of Deseibing Fluid Motion. 163
53. Types of Fuld Flow m
531. Steady and Unsteady Flows 163
532. Uniform and Noneunform Flows 164
533. Laminar and Turbulent Flows 164
534, Compressible and incompressible lows 1. m
535. Rotational and Irotational Flows 165
536. One, two and Three Dimensional Flows 165
5.4. Rate of Flow or Dischargo (Q) 165
55. Continuity Equation 2 165
Solved Problems 5.155 166
5.6. Continuity Equation in Taree Dimensions m
581. Continuity Equation in Cylindrical
Polar Coordinates m
Solved Problema 3.54 Ys
57. Velocity and Acceleration m
571. _ Lacal Aceteration and Convective
Acceleration 1m
Solved Problems 56-59 1m
58. Velocity Potential Funcion and Stream Function 18
S81. Velocity Potential Function. m
582, Stream Funetion m
583. Equipotental Line 183
584, Line of Constant Stream Function 1
oa
585. Flow Net
585. Relation between Stream Function and
Velocity Potential Funetion
Solved Problems 810517
59. Types of Motion
594. Linear Translation
592, — Linear Deformation
593. Angular Deformation
‘or Shear Deformation
594,
595,
Solved Problems 518-519
5.10. Vortex Flow
5101. | Force Vortex Flow
510.2. Free Vortex Flow
5.10.8. Equation of Motion for Vortex Flow
5.10, Equation of Foreed Vortex Flow
Solved Problems 520-625
5.10.5. Closed Cylindrical Vesels
Solved Problems 8265.31
5.10. Equation of Free Vortex Flow
Solved Problem 5.32
Ah IDEAL FLOW (POTENTIAL FLOW)
511. Introduction
5.12. Important Cases of Potential Flow
54, Uniform Flow
513.1. Uniform Flow Parallel to x Axis
5.182. Uniform Potential Flow Parallel to y-Axis
514. Source Flow
515. Sink Flow
Solved Problems 533-535
5.16. Free Vortex Flow
SAT. Super Imposed Flow
5171. Source and Sink Pair
Solved Problems 5:26—6.37
5.172. Dones
Solved Problem 5:38
5173. À Plane Soure in Uniform Flow
low Past a HalBods)
Solved Problems 619-641
5174. À Source and Sink Pirin Uniform Flow
‘low Past a Rankine Oval Body)
Solved Problem 5.12
A Double in a Uniform Plow
(ose Post a Cirevlar Cylinder)
Solved Problems 543544
sans
m
104
164
191
191
191
m
192
192
192
193
m
104
195
197
202
202
209
210
mo
au
an
En
En]
En
216
216
219
En
En
=
Ed
En
ma
237
au
Ed
260
252
En
585. Flow Net
585. Relation between Stream Function and
Velocity Potential Funetion
Solved Problems 810517
59. Types of Motion
594. Linear Translation
592, — Linear Deformation
593. Angular Deformation
‘or Shear Deformation
594,
595,
Solved Problems 518-519
5.10. Vortex Flow
5101. | Force Vortex Flow
510.2. Free Vortex Flow
5.10.8. Equation of Motion for Vortex Flow
5.10, Equation of Foreed Vortex Flow
Solved Problems 520-625
5.10.5. Closed Cylindrical Vesels
Solved Problems 8265.31
5.10. Equation of Free Vortex Flow
Solved Problem 5.32
Ah IDEAL FLOW (POTENTIAL FLOW)
511. Introduction
5.12. Important Cases of Potential Flow
54, Uniform Flow
513.1. Uniform Flow Parallel to x Axis
5.182. Uniform Potential Flow Parallel to y-Axis
514. Source Flow
515. Sink Flow
Solved Problems 533-535
5.16. Free Vortex Flow
SAT. Super Imposed Flow
5171. Source and Sink Pair
Solved Problems 5:26—6.37
5.172. Dones
Solved Problem 5:38
5173. À Plane Soure in Uniform Flow
low Past a HalBods)
Solved Problems 619-641
5174. À Source and Sink Pirin Uniform Flow
‘low Past a Rankine Oval Body)
Solved Problem 5.12
A Double in a Uniform Plow
(ose Post a Cirevlar Cylinder)
Solved Problems 543544
sans
m
104
164
191
191
191
m
192
192
192
193
m
104
195
197
202
202
209
210
mo
au
an
En
En]
En
216
216
219
En
En
=
Ed
En
ma
237
au
Ed
260
252
En
Gi
Chapter 6. Dynamics of Fluid Flow 259-016
6.1. Introduction 250
62, Equations of Motion 259
3. Bulee's Equation of Motion 260
8.4. Bernoull's Equation from Euler's Equation au
65. Assumptions E 261
Solved Problems 61-66. 21
6. Bermoull Equation fr Real Fluid 265,
Solved Problems 67-69 266
6.7. Practical Applications ol Bernpullh Equation 208
671. Venturimeter 268
Solved Problema 610-621 20
1872. Oribe Meter or Orifice Plate au
Solved Problems 622 623 23
673. Piste 2
Solved Problems 624-628 288
68. The Momentum Equation 288
Solved Problems 620-635 29
169. Moment of Momentum Equation 2 298
Solved Problems 0366.37 298
6.10. Free Liquid Jota EN
‘Soived Problems 638641 308
Highlights 307
Exercise soo
Chapter. Orificos and Mouth aros
7. Introduction su
72, Closißcationa of Oricon sur
73. Flow Through on Ori a7
TA. Hydraulic Coefficients 318
TAA. Collien of Velocity (C,) : se
142, Coefüien of Contraction (C.) sis
743. Go-lfiient of Discharge (C,) En
Solved Probleme 7.1.12 En
75. Experimental Determination of Hydraulic
Concent 7 320
751. Determination of Co-fcient of Discharge(C,) m
752. Determination of Co-lfiient af Velocity (C,) sa
153, _ Determination of Coefficient of
Contraction (0) am
Solved Problems 73.10 sn
76. Flow Through Large Orifice : 62
164, Discharge Tarough Large
Rectangular Orifice . ses
Solved Problems 2217.13 3
7.7. Discharge Through Fully Sub-merged Orifice 330
‘Solved Problems 214-715 En
78. Discharge Through Partially Sub-merged Orifice E
‘Solved Problem 7.16 En
Chapter 6. Dynamics of Fluid Flow 259-016
6.1. Introduction 250
62, Equations of Motion 259
3. Bulee's Equation of Motion 260
8.4. Bernoull's Equation from Euler's Equation au
65. Assumptions E 261
Solved Problems 61-66. 21
6. Bermoull Equation fr Real Fluid 265,
Solved Problems 67-69 266
6.7. Practical Applications ol Bernpullh Equation 208
671. Venturimeter 268
Solved Problema 610-621 20
1872. Oribe Meter or Orifice Plate au
Solved Problems 622 623 23
673. Piste 2
Solved Problems 624-628 288
68. The Momentum Equation 288
Solved Problems 620-635 29
169. Moment of Momentum Equation 2 298
Solved Problems 0366.37 298
6.10. Free Liquid Jota EN
‘Soived Problems 638641 308
Highlights 307
Exercise soo
Chapter. Orificos and Mouth aros
7. Introduction su
72, Closißcationa of Oricon sur
73. Flow Through on Ori a7
TA. Hydraulic Coefficients 318
TAA. Collien of Velocity (C,) : se
142, Coefüien of Contraction (C.) sis
743. Go-lfiient of Discharge (C,) En
Solved Probleme 7.1.12 En
75. Experimental Determination of Hydraulic
Concent 7 320
751. Determination of Co-fcient of Discharge(C,) m
752. Determination of Co-lfiient af Velocity (C,) sa
153, _ Determination of Coefficient of
Contraction (0) am
Solved Problems 73.10 sn
76. Flow Through Large Orifice : 62
164, Discharge Tarough Large
Rectangular Orifice . ses
Solved Problems 2217.13 3
7.7. Discharge Through Fully Sub-merged Orifice 330
‘Solved Problems 214-715 En
78. Discharge Through Partially Sub-merged Orifice E
‘Solved Problem 7.16 En
iv)
73. Time of Emptying a Tank Through an Orifice
st te Bottom
Solved Problems 7177.18
7.10. Time of Emptying a Hemisphercal Tank
Solved Problema 7197.21
7.11. Time of Emptying a Cireuar Horizontal Tank
Solved Problems 720-724
7.12, Classification of Mouthpisces
7,13. Flow Through an External Gylindrcal Mouthpiece
Solved Problems 121-725
7.14. Flow Through a Convergent Divergent Mouthpiece
Solved Problems 726-728
7.15, Flow Through Internal or Re-entrant on.
Bordes Mouthpiece
Solved Problem 7.20
Highlights
Exercise
Chapter 8. Notohes and Weirs
81. Introduction
#2, Cinsifcation of Notenos and Woirs
83. Discharge Over a Rectangular Notch or Weir
Solved Problema 6183
8.4. Discharge Over a Tiangular Notch or Weir
Solved problems 8:18.
85. Advantages of Triangular Notch or
Weir over Rectangular Notch or Weir
8.6, Discharge Over a Trapezoidal Noteh or Weir
Solved Problem 87
Discharge Over a Stepped Notch
‘Solved Problem 5
18:8. Etet on Discharge Over a Notch or Weir
Due to Error in the Measurement of Head
881. For Rectangular Weir or Notch
852. For Trangular Weir or Notch
Solved Problems 898.11
89, (a) Time Required to Empty a Reservoir or a
‘Tank with a Rectangular Weir or Notch
(0) Time Required to Empty a Reservoir ora
‘Tank with a Telangular Weir or Notch
Solved Problems &12—8.14
8.10. Velocity of Approach
Solved Problems 8158.19
8.11. Empirical Formule for Discharge Over
Rectangular Weir
Solved Problems 820-822
8.12. Collett Weir or Note
Solved Problems 8258.24
8.13. Dischargo Over a Broad Crest Weir
ET
ss
2
336
E]
E]
au
zu
Er
Er
sus
sr
so
350
se
365-086
355,
355,
356
356
E]
269
si
EN
302
262
ET
364
sa
364
305
206
27
368
En)
zw
am
m
316
E
318
73. Time of Emptying a Tank Through an Orifice
st te Bottom
Solved Problems 7177.18
7.10. Time of Emptying a Hemisphercal Tank
Solved Problema 7197.21
7.11. Time of Emptying a Cireuar Horizontal Tank
Solved Problems 720-724
7.12, Classification of Mouthpisces
7,13. Flow Through an External Gylindrcal Mouthpiece
Solved Problems 121-725
7.14. Flow Through a Convergent Divergent Mouthpiece
Solved Problems 726-728
7.15, Flow Through Internal or Re-entrant on.
Bordes Mouthpiece
Solved Problem 7.20
Highlights
Exercise
Chapter 8. Notohes and Weirs
81. Introduction
#2, Cinsifcation of Notenos and Woirs
83. Discharge Over a Rectangular Notch or Weir
Solved Problema 6183
8.4. Discharge Over a Tiangular Notch or Weir
Solved problems 8:18.
85. Advantages of Triangular Notch or
Weir over Rectangular Notch or Weir
8.6, Discharge Over a Trapezoidal Noteh or Weir
Solved Problem 87
Discharge Over a Stepped Notch
‘Solved Problem 5
18:8. Etet on Discharge Over a Notch or Weir
Due to Error in the Measurement of Head
881. For Rectangular Weir or Notch
852. For Trangular Weir or Notch
Solved Problems 898.11
89, (a) Time Required to Empty a Reservoir or a
‘Tank with a Rectangular Weir or Notch
(0) Time Required to Empty a Reservoir ora
‘Tank with a Telangular Weir or Notch
Solved Problems &12—8.14
8.10. Velocity of Approach
Solved Problems 8158.19
8.11. Empirical Formule for Discharge Over
Rectangular Weir
Solved Problems 820-822
8.12. Collett Weir or Note
Solved Problems 8258.24
8.13. Dischargo Over a Broad Crest Weir
ET
ss
2
336
E]
E]
au
zu
Er
Er
sus
sr
so
350
se
365-086
355,
355,
356
356
E]
269
si
EN
302
262
ET
364
sa
364
305
206
27
368
En)
zw
am
m
316
E
318
tw)
814. Discharge Over Norrow-Crested Weir En
BAS. Discharge Over an Ogee Weir En
BAG. Discharge Over Submerged or Drowned Weir \. 370
Solved Problems 828827 380
Highlights " EN
Exercise = 383
Chapter 9. Viseous Flow 387-432
94. Introduction je Ed
92. Flow of Viscous Fluid Through Circular Pipe 387
Solved Problems 91-96 2 u
93. Flow of Viscous Fluid between Two Parallel Plates a
Solved Problems 99-912 400
94. Kinetic Energy Correction and Momentum
Correction Facters a
‘Solved Problem 9.13 404
9.5. Power Absorbed in Visa Flow = 407
951. Visonus Resistance of Journal Bearings 407
Solved Problems 9.149,18 408
952. Visonus Resistance of Fost-tep Bearing au
Solved Problems 9.19-9.20 an
953. _ Viren Resistance of Collar Bearing a a2
Solved Problems 921-022 = a
96, Lou of Head Due to Friction in Viscous Flow as
Solved Problems 923-024 as
9.2. Movement el Piston in Dashrpot an
‘Solved Problem 9.25 48
98. Methods of Determination of Coicient of Viscosity as
951. Capillary Tube Method ag
952, Falling Sphere Resistance Method 420
953, Rotating Cylinder Method aan
8.54. Orifee Type Viscometor a
Solved Problems 9269.32 3
Highlights am
Exercise as
Chapter 10, Turbulent Flow 433-464
101. Introduction a
102. Reynolds Experiment : 133
103. Frictional Loss in Pipo Flow u au
1031. Expression for Los of Head Due
to Friction in Pipes a
103.2. Exprestion fur Coeficient of Friction
in Terms of Shear Stress 46
10.4. Shear Stress in Turbulent Flow : 47
1041. Reynolds Expression for Turbulent
Shear Stress a
10.42. Prandtl Mixing Length Thoory for
‘Turbulent Shear Stress s ans
814. Discharge Over Norrow-Crested Weir En
BAS. Discharge Over an Ogee Weir En
BAG. Discharge Over Submerged or Drowned Weir \. 370
Solved Problems 828827 380
Highlights " EN
Exercise = 383
Chapter 9. Viseous Flow 387-432
94. Introduction je Ed
92. Flow of Viscous Fluid Through Circular Pipe 387
Solved Problems 91-96 2 u
93. Flow of Viscous Fluid between Two Parallel Plates a
Solved Problems 99-912 400
94. Kinetic Energy Correction and Momentum
Correction Facters a
‘Solved Problem 9.13 404
9.5. Power Absorbed in Visa Flow = 407
951. Visonus Resistance of Journal Bearings 407
Solved Problems 9.149,18 408
952. Visonus Resistance of Fost-tep Bearing au
Solved Problems 9.19-9.20 an
953. _ Viren Resistance of Collar Bearing a a2
Solved Problems 921-022 = a
96, Lou of Head Due to Friction in Viscous Flow as
Solved Problems 923-024 as
9.2. Movement el Piston in Dashrpot an
‘Solved Problem 9.25 48
98. Methods of Determination of Coicient of Viscosity as
951. Capillary Tube Method ag
952, Falling Sphere Resistance Method 420
953, Rotating Cylinder Method aan
8.54. Orifee Type Viscometor a
Solved Problems 9269.32 3
Highlights am
Exercise as
Chapter 10, Turbulent Flow 433-464
101. Introduction a
102. Reynolds Experiment : 133
103. Frictional Loss in Pipo Flow u au
1031. Expression for Los of Head Due
to Friction in Pipes a
103.2. Exprestion fur Coeficient of Friction
in Terms of Shear Stress 46
10.4. Shear Stress in Turbulent Flow : 47
1041. Reynolds Expression for Turbulent
Shear Stress a
10.42. Prandtl Mixing Length Thoory for
‘Turbulent Shear Stress s ans
wi)
10.5. Velocity Distribution in Turbulent Flow in Pipes as
1051. Hydrodynamically Smooth and Rough
Boundaries 0
1052. Velocity Distribution for Turbulent Plow
in Smooth Pipes - sn
10:53. Velocity Distribution for Turbulent Flow
in Rough Pipes c se
Solved Problems 101-104 à 2
105.4. Velocity Distribution for Turbulent Flow
in Terms ol Average Velocity as
Solved Problems 105-106 us
1055, Velocity Distribution for Turbolent Flow
in Smooth Pipes by Power Law a 450
106. Resistance af Smooth and Rough Pipes 450,
‘Solved Problems 10-7101 43
Highlights a
Exercise 462
ter 11. Flow Through Pipes 465-558
11. Introduction as
112. Lace of Energy in Pipes z 465,
113. Lose of Energy (or head) Due to Frietion e 465,
Solved Problems 111-117 : 487
11.4. Minor Energy (Head) Losses an
11.41. Loss of Head Due to Sudden Enlargement am
HAZ, Loss of Head Due to Sudden Contention
Solved Problems 11.11.14 nl
11443. Lost of Head at the Entrance ofa Pipo ase
114. Loss of Head atthe Exit of Pipe ae
11455, Loss of Head Due to on Obstruction
ina Pipe . 492
1146, Loss of Head Due to Bend in Pipe 485,
1147. Loss of Head in Various Pipe Ptings ass
Solved Problema 11.15.1121 ass
115, Hydraulie Gradient and Total Energy Line : a
11,54. Hydrouie Gredient Line 2 491
11,52. Total Enengy Line 3 a
Solved Problem 11.29-11.26. a
11.6. Flow Through Syphon 408
Solved Problems 11.27—11.29 498
112. Flow Tarough Pips in Series or Flow Through
Compound Pipes 502
Solved Problems 1130-11 30À ' 503
118, Bguivalent Pipe 507
Solved Problem 1131 EN
11.0. Flow Through Parallel Pipes 508
Solved Problems 11.39-11.41 EN
11.10. Flow Through Branched Pipes y su
Solved Problems 11.49-11.44 > so
10.5. Velocity Distribution in Turbulent Flow in Pipes as
1051. Hydrodynamically Smooth and Rough
Boundaries 0
1052. Velocity Distribution for Turbulent Plow
in Smooth Pipes - sn
10:53. Velocity Distribution for Turbulent Flow
in Rough Pipes c se
Solved Problems 101-104 à 2
105.4. Velocity Distribution for Turbulent Flow
in Terms ol Average Velocity as
Solved Problems 105-106 us
1055, Velocity Distribution for Turbolent Flow
in Smooth Pipes by Power Law a 450
106. Resistance af Smooth and Rough Pipes 450,
‘Solved Problems 10-7101 43
Highlights a
Exercise 462
ter 11. Flow Through Pipes 465-558
11. Introduction as
112. Lace of Energy in Pipes z 465,
113. Lose of Energy (or head) Due to Frietion e 465,
Solved Problems 111-117 : 487
11.4. Minor Energy (Head) Losses an
11.41. Loss of Head Due to Sudden Enlargement am
HAZ, Loss of Head Due to Sudden Contention
Solved Problems 11.11.14 nl
11443. Lost of Head at the Entrance ofa Pipo ase
114. Loss of Head atthe Exit of Pipe ae
11455, Loss of Head Due to on Obstruction
ina Pipe . 492
1146, Loss of Head Due to Bend in Pipe 485,
1147. Loss of Head in Various Pipe Ptings ass
Solved Problema 11.15.1121 ass
115, Hydraulie Gradient and Total Energy Line : a
11,54. Hydrouie Gredient Line 2 491
11,52. Total Enengy Line 3 a
Solved Problem 11.29-11.26. a
11.6. Flow Through Syphon 408
Solved Problems 11.27—11.29 498
112. Flow Tarough Pips in Series or Flow Through
Compound Pipes 502
Solved Problems 1130-11 30À ' 503
118, Bguivalent Pipe 507
Solved Problem 1131 EN
11.0. Flow Through Parallel Pipes 508
Solved Problems 11.39-11.41 EN
11.10. Flow Through Branched Pipes y su
Solved Problems 11.49-11.44 > so
aid
11.11, Power Transmission Through Pipes
1111. Condition for Maxim
"Transmission of Power
11.112. Maximum Efficiency of Transmission
of Power
Solved Problems 11.45-11.47
11.12, Flow Theoogh Norsles
1112.1. Power Transmitted Through Nozzle
11422. Condition for Maximum Power
‘Teansmited Through Neto
11.123. Diameter of Nol fr Maximum
‘Transmission of Power Through Nora
Solved Problems 1148-1151
11.18, Water Hammer in Piper
1138.1. Gradual Closure of Valve
11132. Sudden Clovue of Valve and Pipe i Rigid
11.183. Sudden Closure o Valve and Pipe i Elastic
11194. Time Taken by Pressure Wave to Travel
from the Valve tothe Tank and from
‘Tank tothe Valve
Salved Problema 1.62—11.55.
11.14. Pipe Network
ALL. Hardy Cross Method
Solved Problem 1156
‘Chapter 12. Dimensional and Model Analysis
12. Introduction
122. Secondary or Derived Quantities
‘Solved Prohlem 12.1
123. Dimensional Homogeneity
124. Methods of Dimensional Analysis
12.41. Rayleigh Method.
Solved Problems 12.2127.
124.2, Buckingham x Theorem
12.43. Method of Selecting Repeating Variables
12.44. Procedure fr Solving Problems by
Buckinghan’s x-Theorem
Salved Problems 12,812.14
125. Model Analyst
126. Similtude-Types of Similaiios
12:7. Types of Frees Acting in Moving Fluid
128, Dimensionless Numbers
1251. Reynolds Number (R)
1282. Proudes Number (E)
1288. Euler's Number (E.
1284. Weber Number 0)
1285, Mach’ Number (MN
E)
so
EN
sn
535
E
E
En
En
EN
En
Er
ss
ss
55
EN
En
552
En
559-610
EN
EN
E)
EN
E
se
56
Es
566
568
518
En
50
EN
ser
582
582
582
562
11.11, Power Transmission Through Pipes
1111. Condition for Maxim
"Transmission of Power
11.112. Maximum Efficiency of Transmission
of Power
Solved Problems 11.45-11.47
11.12, Flow Theoogh Norsles
1112.1. Power Transmitted Through Nozzle
11422. Condition for Maximum Power
‘Teansmited Through Neto
11.123. Diameter of Nol fr Maximum
‘Transmission of Power Through Nora
Solved Problems 1148-1151
11.18, Water Hammer in Piper
1138.1. Gradual Closure of Valve
11132. Sudden Clovue of Valve and Pipe i Rigid
11.183. Sudden Closure o Valve and Pipe i Elastic
11194. Time Taken by Pressure Wave to Travel
from the Valve tothe Tank and from
‘Tank tothe Valve
Salved Problema 1.62—11.55.
11.14. Pipe Network
ALL. Hardy Cross Method
Solved Problem 1156
‘Chapter 12. Dimensional and Model Analysis
12. Introduction
122. Secondary or Derived Quantities
‘Solved Prohlem 12.1
123. Dimensional Homogeneity
124. Methods of Dimensional Analysis
12.41. Rayleigh Method.
Solved Problems 12.2127.
124.2, Buckingham x Theorem
12.43. Method of Selecting Repeating Variables
12.44. Procedure fr Solving Problems by
Buckinghan’s x-Theorem
Salved Problems 12,812.14
125. Model Analyst
126. Similtude-Types of Similaiios
12:7. Types of Frees Acting in Moving Fluid
128, Dimensionless Numbers
1251. Reynolds Number (R)
1282. Proudes Number (E)
1288. Euler's Number (E.
1284. Weber Number 0)
1285, Mach’ Number (MN
E)
so
EN
sn
535
E
E
En
En
EN
En
Er
ss
ss
55
EN
En
552
En
559-610
EN
EN
E)
EN
E
se
56
Es
566
568
518
En
50
EN
ser
582
582
582
562
Go
129, Model Laws or Similarity Laws 5
129.1. Reynolds Model Law E
Solved Problems 12.16-12.16. a EN
1292. Froude Model Law rg
Solved Problems 12.19-12.27 Es]
1293. Balers Model Lave a 595,
1294, Weber Model Law 596
1295. Mach Model Law E]
Solved Problem 12.28 Ed
12.10. Model Testing of Parilly Sub-merged Bodies EN
Solved Probleme 1229-1242 : soo
12.01. Closifenion of Models EN
12.1.1. Undistorted Models En
12.112. Distorted Models a 605
12.113. Scale Ratios for Distorted Models 60
Solved Problem 12.33 os
Highlights 606
Exercise or
Chapter 13, Boundary Layer Flow 611-056
134. Introduction su
122. Definitions E]
13:21. Laminar Boundary Layer 5 su
1322. Turbulent Boundary Layer Er
19:23. Laminar Sub-layer sl
1324, Boundary Layer Thickness (8) sr
1325. Displacement Thickness (89) sl
19:26. Momentum Thickness (0) = as
1827. Energy Thickness (5 sis
Solved Problems 131-122 as
133. Drag Force on a Flat Plate Due to Boundary Layer si
1321. Local Co-eMcient of Drag [Col ox
13.32. Average Co-ecient of Drag (Col EN
13,33. Boundary Conditions for the
Velocity Profiles ox
Solved Problems 12.13.12 om
134. Turbulent Boundary Layer on a Flat Plate ES
Solved Problem 13.19 ss
135, Analysis of Turbulent Boundary Layer. 2 sa
13.6. Total Drag on a Flat Plate Due to Laminar and
‘Turbulent Boundary Layer sa
Solved Problems 13.14-13.17 : En
13:7. Separation of Boundary Layer ss
1971. Effet of Pressure Gradient on
Boundary Layer Separation : EN
12.7.2, Location of Separation Point so
Solved Problem 12.18 soo
129, Model Laws or Similarity Laws 5
129.1. Reynolds Model Law E
Solved Problems 12.16-12.16. a EN
1292. Froude Model Law rg
Solved Problems 12.19-12.27 Es]
1293. Balers Model Lave a 595,
1294, Weber Model Law 596
1295. Mach Model Law E]
Solved Problem 12.28 Ed
12.10. Model Testing of Parilly Sub-merged Bodies EN
Solved Probleme 1229-1242 : soo
12.01. Closifenion of Models EN
12.1.1. Undistorted Models En
12.112. Distorted Models a 605
12.113. Scale Ratios for Distorted Models 60
Solved Problem 12.33 os
Highlights 606
Exercise or
Chapter 13, Boundary Layer Flow 611-056
134. Introduction su
122. Definitions E]
13:21. Laminar Boundary Layer 5 su
1322. Turbulent Boundary Layer Er
19:23. Laminar Sub-layer sl
1324, Boundary Layer Thickness (8) sr
1325. Displacement Thickness (89) sl
19:26. Momentum Thickness (0) = as
1827. Energy Thickness (5 sis
Solved Problems 131-122 as
133. Drag Force on a Flat Plate Due to Boundary Layer si
1321. Local Co-eMcient of Drag [Col ox
13.32. Average Co-ecient of Drag (Col EN
13,33. Boundary Conditions for the
Velocity Profiles ox
Solved Problems 12.13.12 om
134. Turbulent Boundary Layer on a Flat Plate ES
Solved Problem 13.19 ss
135, Analysis of Turbulent Boundary Layer. 2 sa
13.6. Total Drag on a Flat Plate Due to Laminar and
‘Turbulent Boundary Layer sa
Solved Problems 13.14-13.17 : En
13:7. Separation of Boundary Layer ss
1971. Effet of Pressure Gradient on
Boundary Layer Separation : EN
12.7.2, Location of Separation Point so
Solved Problem 12.18 soo
aio)
187.3. Methods of Preventing the Separation
of Boundary Layer su
Highlights 51
Exercise 5 653,
Chapter 14. Forces on Subsmenged Bodi 57-692
Mal. Introduction 657
142. Force Exerted by a Flowing Fluid on
a Stationary Body 657
121. Drop E 558
1422, LR 658
148, Expression für Drag and Li. E
1431. Dimension Analysis of Drag and Li. 5
Solved Problems 141-1415 En]
14:32, Pressure Drag and Friction Drag so
1433. Streamlined Body om
1434. Blut Body i. on
14, Drag ona Sphere on
Solved Problem 1416 En]
145. Terminal Velocity ta Body. om
Solved Probleme 14.17-14.20 sm
146, Dragon a Cylinder A on
147. Development of Lit on a Cireulor Cylinder = m
1671. Flow of Heal Fluid Over Stationary
Cylinder 66
147.2, Flow Pattern Around the Cylinder
when a Constant Circulation fis
Imparte to the Cylinder EN
14.73. Expreston fr LIA Porco Acting on
Rotating Cylinder 7 oo
14.74. Drag Force Acting on a Rotating Cyinder 2. sz
1475. Expression for LR Co oMciont for
Rotating Cylinder se
14.76, Location af Stagnation Pints fr a
Rotating Cylinder in a Uniform Flow field 683
14.24, Magnus Effect z 83,
Solved Problems 14.21-14.29 E 3
148. Development of lft on an Ai ss
1481. Steady-state ofa Flying Objet 7
Solved Problema 1424-1425 E
Highlights Es]
Exercise EN
Chapter 15. Compressible Flow 693-736
15.1. Introduction . sa
132. Thermodynamie Relations 2 1
1521. Equation of State E EN
15:22. Expansion and Compression of Perfect Gas EN
187.3. Methods of Preventing the Separation
of Boundary Layer su
Highlights 51
Exercise 5 653,
Chapter 14. Forces on Subsmenged Bodi 57-692
Mal. Introduction 657
142. Force Exerted by a Flowing Fluid on
a Stationary Body 657
121. Drop E 558
1422, LR 658
148, Expression für Drag and Li. E
1431. Dimension Analysis of Drag and Li. 5
Solved Problems 141-1415 En]
14:32, Pressure Drag and Friction Drag so
1433. Streamlined Body om
1434. Blut Body i. on
14, Drag ona Sphere on
Solved Problem 1416 En]
145. Terminal Velocity ta Body. om
Solved Probleme 14.17-14.20 sm
146, Dragon a Cylinder A on
147. Development of Lit on a Cireulor Cylinder = m
1671. Flow of Heal Fluid Over Stationary
Cylinder 66
147.2, Flow Pattern Around the Cylinder
when a Constant Circulation fis
Imparte to the Cylinder EN
14.73. Expreston fr LIA Porco Acting on
Rotating Cylinder 7 oo
14.74. Drag Force Acting on a Rotating Cyinder 2. sz
1475. Expression for LR Co oMciont for
Rotating Cylinder se
14.76, Location af Stagnation Pints fr a
Rotating Cylinder in a Uniform Flow field 683
14.24, Magnus Effect z 83,
Solved Problems 14.21-14.29 E 3
148. Development of lft on an Ai ss
1481. Steady-state ofa Flying Objet 7
Solved Problema 1424-1425 E
Highlights Es]
Exercise EN
Chapter 15. Compressible Flow 693-736
15.1. Introduction . sa
132. Thermodynamie Relations 2 1
1521. Equation of State E EN
15:22. Expansion and Compression of Perfect Gas EN
ox
153, Basie Equations of Compre
1831. "Continuity Equation
1632. Bernoull’s Equation
Solved Problems 161-153
153.3. Momentum Equations z
18.4. Velocity of Sound or Pressure Wave in a Fluid
15.4.1. Expression for Velocity of Sound
Wave in a Flora. o
15.42. Velocity of Sound in Terms of
Bull Modulus
15:43. Velocity of Sound for Iothermal Pecera
15.44. Velocity of Sound for Adíabati Process
155. Mach Number. E
‘Solved Problems 194-167 2
15.6. Propagation of Pressure Waves (or Disturbances}
in a Compresaible Fluid
15.61. Mach Angle
1582. Zone of Aetion
1563. Zone of Silence 5
Solved Problems 155—15.10 a
‘Stagnation Properties
167.1. Expression fur Stagnation Pressure 1.)
15.72. Expression fr Stagnation Density (.)
15,73. Expression for Stagnation Temperature (,)
Solved Problems 15:11—15.12 -
15.8. Area Velocity Relationship for Compresible Flow
15:0. Flow of Compressible Fluid Through Orifice and
Nous Fitted o a Large Tank
159.4, Value of m or AL for Maximum Value
le Flow
m
of Mons Rate of Flow
159.2, Value of Vy for Maximum Rate of Flow
of Fluid |
1593, Maximum Rate of Flow of Fluid Through
Nome
1594, Variation of Mass Rate of Flow of Compress
Fluid with Pressure ratio (2)
15.95, Velocity at Outlet of Nozle for Maximum
Rate of Flow it Equal to Sonie Velocity
Solved Probleme 1513-1615
15.10. Moss Rate of Flow of Compressible Fluid Through
Solved Problem 16.16
15.11. Pitot Static Tube in a Compressible Flow
Solved Problem 16-1
Highlights
Exercise
os
os
695,
on
m
m
u
m
705
705
705
106
208
zoo
En
o
mo
m
mu
m
m
716
ns
no
m
m
m
m
m
ms
330
mi
m
m
153, Basie Equations of Compre
1831. "Continuity Equation
1632. Bernoull’s Equation
Solved Problems 161-153
153.3. Momentum Equations z
18.4. Velocity of Sound or Pressure Wave in a Fluid
15.4.1. Expression for Velocity of Sound
Wave in a Flora. o
15.42. Velocity of Sound in Terms of
Bull Modulus
15:43. Velocity of Sound for Iothermal Pecera
15.44. Velocity of Sound for Adíabati Process
155. Mach Number. E
‘Solved Problems 194-167 2
15.6. Propagation of Pressure Waves (or Disturbances}
in a Compresaible Fluid
15.61. Mach Angle
1582. Zone of Aetion
1563. Zone of Silence 5
Solved Problems 155—15.10 a
‘Stagnation Properties
167.1. Expression fur Stagnation Pressure 1.)
15.72. Expression fr Stagnation Density (.)
15,73. Expression for Stagnation Temperature (,)
Solved Problems 15:11—15.12 -
15.8. Area Velocity Relationship for Compresible Flow
15:0. Flow of Compressible Fluid Through Orifice and
Nous Fitted o a Large Tank
159.4, Value of m or AL for Maximum Value
le Flow
m
of Mons Rate of Flow
159.2, Value of Vy for Maximum Rate of Flow
of Fluid |
1593, Maximum Rate of Flow of Fluid Through
Nome
1594, Variation of Mass Rate of Flow of Compress
Fluid with Pressure ratio (2)
15.95, Velocity at Outlet of Nozle for Maximum
Rate of Flow it Equal to Sonie Velocity
Solved Probleme 1513-1615
15.10. Moss Rate of Flow of Compressible Fluid Through
Solved Problem 16.16
15.11. Pitot Static Tube in a Compressible Flow
Solved Problem 16-1
Highlights
Exercise
os
os
695,
on
m
m
u
m
705
705
705
106
208
zoo
En
o
mo
m
mu
m
m
716
ns
no
m
m
m
m
m
ms
330
mi
m
m
Chapter 16. Flow in Open Channels
162.
162,
163.
165.
166.
167.
168.
Introduction
Clawfcaion of low in Channels
16.2. Steady Flow and Unsteady Flow
1622, Uniform Flow and Non uniform Flow
1625. Laminar Flow and Turbulent Flow
1624. Sulb<rtical, Critieal and Super-Critical
Flow
Discharge Through Open Channel by Chezy’s
Formula
Solved Problems 16.1—16.7
Empirical Formulas for the Value of Chexy's
Constant
Solved Problems 165-1612
Most Exonomical Section of Channels
1651. Most Bomomical Rectangular Channel
Salved Probleme 16.14-16.16.
165.2. Most Eemomical Trapezidal Channel
Solved Problems 16.16.1023.
10:53, Bost Side Slope for Most Economico!
rapezsidal Section
Solved Problems 1623-1624
1654. Flow Through Circular Channel
Solved Problems 1625-169,
165.5. Most Bemonical Circular Section
Solved Problems 1630-1632
‘Non-Uniform Flow through Open Channels
Specific Energy and Specific Energy Curve
16:71. Critical Depth dh)
16.72. Critical Velocity (Vo
16.73. Minimum Specific Energy in Terme of
Critical Depth
Solved Problema 1633-—16.35
1674. Critical Flow
16.75. Streaming Flow or Subrertieal Flow or
‘Teanquil Flow
1626. Super-Ceitical Flow or Shooting Flow or
‘Torrential Flow
16:7. Alternate Depths
1678, Condition for Maximum Discharge fora
Given Value of Specie Energy
Solved Problems 1630-1037
Hydraulie Jump or Standing Wave
16.81. Expresion fr Depth of Hydraulic Jump
1682. Expression for Los of Energy Due lo
Hydravlie Jump
108.3, Expression for Depth of Hydraulie Jump
in Terms or Upstream Froude Number
707-802
m
787
187
137
ES
ms
162.
162,
163.
165.
166.
167.
168.
Introduction
Clawfcaion of low in Channels
16.2. Steady Flow and Unsteady Flow
1622, Uniform Flow and Non uniform Flow
1625. Laminar Flow and Turbulent Flow
1624. Sulb<rtical, Critieal and Super-Critical
Flow
Discharge Through Open Channel by Chezy’s
Formula
Solved Problems 16.1—16.7
Empirical Formulas for the Value of Chexy's
Constant
Solved Problems 165-1612
Most Exonomical Section of Channels
1651. Most Bomomical Rectangular Channel
Salved Probleme 16.14-16.16.
165.2. Most Eemomical Trapezidal Channel
Solved Problems 16.16.1023.
10:53, Bost Side Slope for Most Economico!
rapezsidal Section
Solved Problems 1623-1624
1654. Flow Through Circular Channel
Solved Problems 1625-169,
165.5. Most Bemonical Circular Section
Solved Problems 1630-1632
‘Non-Uniform Flow through Open Channels
Specific Energy and Specific Energy Curve
16:71. Critical Depth dh)
16.72. Critical Velocity (Vo
16.73. Minimum Specific Energy in Terme of
Critical Depth
Solved Problema 1633-—16.35
1674. Critical Flow
16.75. Streaming Flow or Subrertieal Flow or
‘Teanquil Flow
1626. Super-Ceitical Flow or Shooting Flow or
‘Torrential Flow
16:7. Alternate Depths
1678, Condition for Maximum Discharge fora
Given Value of Specie Energy
Solved Problems 1630-1037
Hydraulie Jump or Standing Wave
16.81. Expresion fr Depth of Hydraulic Jump
1682. Expression for Los of Energy Due lo
Hydravlie Jump
108.3, Expression for Depth of Hydraulie Jump
in Terms or Upstream Froude Number
707-802
m
787
187
137
ES
ms
Chapter
ma.
ma.
ma
175.
xy
16.8.4. Length of Hydraulie Jump
Solved Probleme 16.281642
Gradually Varied Flow (GV.F)
169.1. Equation of Gradually Varied Flow
Solved Problems 1643-1644
16.92. Back Water Curve and Aff
169.3. Expression for the Length of Back
Water Curve
Solved Problem 1645
Highlights
Beerise
17. Impact of Jets and Jet Propulsion
Introduction
Force Exerted by th Jet on Stationary
Vertical Plate
1721. Force Exerted by a Jet on Stationary
Inclined Flat Plate
1722. Force Bxerted by a Jet on Stationary
Curved Plate
Solved Problems 171-176
oreo Bxerted by a Jot an a Hinged Plate
Solved Problems 17.717. 10 (a)
Force Bxerted by a Jet an Moving Plates
17.41. Foroe on Flat Vertical Plate Moving
in the Direction of Jet.
17.42, Force onthe Inclined Plate Moving in
{ie Direction of the Jet
Solved Probleme 1711-113
17.43, Kores om the Curved Plat when the
Plate is Moving inthe Direction of Jet
Solved Problems 17.14—1717
17.44. Force Exerted by a Jet of Water on an
Uneymmetrical Moving Curved Plate when
Jet Strikes Tangential at one ofthe Tips
‘Solved Problems 1718-1123
17.45, Fore Bxerted by a Jet of Water on a
Series ol Vanes
1748. Force Exerted on a Series of
Jet Propulsion
17.51.” Jet Propulsion ofa Tank with an Orifice
Solved Problema 17.27-17.26
17.62. Jet Propulion of Ships
Solved Problems 17291729
Highlights
Exercise
187
m
700
zo
792
293
m
ms
296
709
Den
son
sos
sos
ss
Eu
509
so
EN
sis
sas
EN
EN
EN
E
EN
a
su
E
so
sa
sa
sa
su
so
850
ma.
ma.
ma
175.
xy
16.8.4. Length of Hydraulie Jump
Solved Probleme 16.281642
Gradually Varied Flow (GV.F)
169.1. Equation of Gradually Varied Flow
Solved Problems 1643-1644
16.92. Back Water Curve and Aff
169.3. Expression for the Length of Back
Water Curve
Solved Problem 1645
Highlights
Beerise
17. Impact of Jets and Jet Propulsion
Introduction
Force Exerted by th Jet on Stationary
Vertical Plate
1721. Force Exerted by a Jet on Stationary
Inclined Flat Plate
1722. Force Bxerted by a Jet on Stationary
Curved Plate
Solved Problems 171-176
oreo Bxerted by a Jot an a Hinged Plate
Solved Problems 17.717. 10 (a)
Force Bxerted by a Jet an Moving Plates
17.41. Foroe on Flat Vertical Plate Moving
in the Direction of Jet.
17.42, Force onthe Inclined Plate Moving in
{ie Direction of the Jet
Solved Probleme 1711-113
17.43, Kores om the Curved Plat when the
Plate is Moving inthe Direction of Jet
Solved Problems 17.14—1717
17.44. Force Exerted by a Jet of Water on an
Uneymmetrical Moving Curved Plate when
Jet Strikes Tangential at one ofthe Tips
‘Solved Problems 1718-1123
17.45, Fore Bxerted by a Jet of Water on a
Series ol Vanes
1748. Force Exerted on a Series of
Jet Propulsion
17.51.” Jet Propulsion ofa Tank with an Orifice
Solved Problema 17.27-17.26
17.62. Jet Propulion of Ships
Solved Problems 17291729
Highlights
Exercise
187
m
700
zo
792
293
m
ms
296
709
Den
son
sos
sos
ss
Eu
509
so
EN
sis
sas
EN
EN
EN
E
EN
a
su
E
so
sa
sa
sa
su
so
850
oxi
Chapter 18, Hydraulic Machines—Turbines ss
18.1. Introduction ss
182. Turbines E]
183. General Layout ofa Hydreletrie Power Plant 853
184. Definitions of Heads and Eficincis of « Turbine E
185. Classification of Hodraulie Turbines 5 E
166. Pelton Wheel (or Turbine) 2 857
186.1, Velocity Triangles and Work Done for
Pelton Wheel E)
186.2. Point ta be Remembered for Pelton Wheel EN
Solved Problema 18,116.10 $62
1868. Design of Pelton Wheel sm
Solved Problems 18.11-18.19 : sm
Radi Flow Reaction Turbines : m
1871. Main Parte ola Radial Flow
Reaction Turbine m
1872. Inward Radial How Turbine EA
18.73. Degree of Reaction so
18.24. Definitions En
Solved Problems 16.14 18.20 En
18:75, Outward Radial Flow Reaction Turbine 1 02
Solved Problems 1821-1822 = EJ
188, Francis Turbine ss
18,51. Important Relations for Francis Turbines ss
Solved Problema 18.29-18.26 Ed
169. Axial Flow Reaction Turbine 903
189.1, Some Important Pont for Propeller
‘Kaplan Turbine) 2 905
Solved Problems 18.27-18.38 905
18.10. Drat-Tube sis
1810.1. Types of Dra Tubes 95
18.102. Deal Tube Theory 916
18.103. ficiency of Drat-Tube 16
Solved Problems 183 (0118.36 a our
18.11. Specie Speed m
1811.1. Derivation ofthe Specie Speed 920
18.112. Significance of Specific Speed oa
Solved Problema 1846-1841 m
18.12. Unit Quantities e m
1812.1. Unit Speed 3 om
18.122. Unit Discharge E
18.123. Unit Power z os
18.124. Use of Unit Quantities, Qu P,) EN
Solved Problems 18.1 (0) 18.48 1
18.13. Characteristie Curves of Hydraulic Turbines 2 933,
18.5.1. Main Characteriaie Curves or
Constant Head Curves > 93
18.132. Operating Characteristic Curves or
Constant Speed Curves ou
Chapter 18, Hydraulic Machines—Turbines ss
18.1. Introduction ss
182. Turbines E]
183. General Layout ofa Hydreletrie Power Plant 853
184. Definitions of Heads and Eficincis of « Turbine E
185. Classification of Hodraulie Turbines 5 E
166. Pelton Wheel (or Turbine) 2 857
186.1, Velocity Triangles and Work Done for
Pelton Wheel E)
186.2. Point ta be Remembered for Pelton Wheel EN
Solved Problema 18,116.10 $62
1868. Design of Pelton Wheel sm
Solved Problems 18.11-18.19 : sm
Radi Flow Reaction Turbines : m
1871. Main Parte ola Radial Flow
Reaction Turbine m
1872. Inward Radial How Turbine EA
18.73. Degree of Reaction so
18.24. Definitions En
Solved Problems 16.14 18.20 En
18:75, Outward Radial Flow Reaction Turbine 1 02
Solved Problems 1821-1822 = EJ
188, Francis Turbine ss
18,51. Important Relations for Francis Turbines ss
Solved Problema 18.29-18.26 Ed
169. Axial Flow Reaction Turbine 903
189.1, Some Important Pont for Propeller
‘Kaplan Turbine) 2 905
Solved Problems 18.27-18.38 905
18.10. Drat-Tube sis
1810.1. Types of Dra Tubes 95
18.102. Deal Tube Theory 916
18.103. ficiency of Drat-Tube 16
Solved Problems 183 (0118.36 a our
18.11. Specie Speed m
1811.1. Derivation ofthe Specie Speed 920
18.112. Significance of Specific Speed oa
Solved Problema 1846-1841 m
18.12. Unit Quantities e m
1812.1. Unit Speed 3 om
18.122. Unit Discharge E
18.123. Unit Power z os
18.124. Use of Unit Quantities, Qu P,) EN
Solved Problems 18.1 (0) 18.48 1
18.13. Characteristie Curves of Hydraulic Turbines 2 933,
18.5.1. Main Characteriaie Curves or
Constant Head Curves > 93
18.132. Operating Characteristic Curves or
Constant Speed Curves ou
io)
18.133. Constant Eficieney Curves or Muschel
‘Curves or lo-Eiciency Curves 9
18.14. Governing of Turbines 935,
Tiighlights 937
Exercite 7 939
Chapter 19. Centrifogal Pumps 945-992
Introduction ous
in Part ofa Centrifugal Pump ous
Work Done by the Centrifugal Pump
(or by Imphlee) on Water # our
194. Definition of Heads and Efiences ofa
Centrifugal Pump ss
Solved Problema 19.1—19.12 951
195. Minimum Speed for Starting a Centrifugal Pump 965,
Solved Problems 19.13—19.15 966
196, Multistage Gentetugal Pumps E vos
19:61. Multistage Centrifugal Pumps
for High Heads 68
196.2, Multistage Conteifagal Pumps for
High Diseharge 969
Solved Problems 19,16—19.17 969
19:7. Specific Speed ofa Centrifugal Pump (No) on
187.1. Expression for Specie Speed for a Pump m
19:8. Model Testing of Centefugal Pumps m
Solved Problema 19.18--19.22
199. Priming of Centrifugal Pump EN
19.10. Characteristic Carver of Centrifugal Pumps E 9
19.101. Main Characteriatie Curves E
19.102. Operating Characterisic Curves ps E
19.103. Constant Bficieney Curves 2 919
19.11, Cavitation 980
19.111. Precaution Against Cavitation so
19.112. Effets of Cavitation oi
19.113. Hydraulie Machines Subjected to Cavitation et
19.114. Cavitation in Turbinen 2 981
19.115. Cavitation in Centrifogal Pumps = 981
Salve Problem 19.23 982
19.12. Moximum Suetion Lit (or Suction Height) 985,
19:13. Net Positive Suction Head (NPSH) 985,
19.14. Cavitation in Contefugal Pump. 285,
Solved Problem 19:21 985
Highlights . 987
Exercise 6
Chapter 20. Reciproeating Pumps 990-1040
20.1. Introduction 980,
202. Main Parte of a Reciprocating Pump 99
203. Working of Reciprocating Pump EN
18.133. Constant Eficieney Curves or Muschel
‘Curves or lo-Eiciency Curves 9
18.14. Governing of Turbines 935,
Tiighlights 937
Exercite 7 939
Chapter 19. Centrifogal Pumps 945-992
Introduction ous
in Part ofa Centrifugal Pump ous
Work Done by the Centrifugal Pump
(or by Imphlee) on Water # our
194. Definition of Heads and Efiences ofa
Centrifugal Pump ss
Solved Problema 19.1—19.12 951
195. Minimum Speed for Starting a Centrifugal Pump 965,
Solved Problems 19.13—19.15 966
196, Multistage Gentetugal Pumps E vos
19:61. Multistage Centrifugal Pumps
for High Heads 68
196.2, Multistage Conteifagal Pumps for
High Diseharge 969
Solved Problems 19,16—19.17 969
19:7. Specific Speed ofa Centrifugal Pump (No) on
187.1. Expression for Specie Speed for a Pump m
19:8. Model Testing of Centefugal Pumps m
Solved Problema 19.18--19.22
199. Priming of Centrifugal Pump EN
19.10. Characteristic Carver of Centrifugal Pumps E 9
19.101. Main Characteriatie Curves E
19.102. Operating Characterisic Curves ps E
19.103. Constant Bficieney Curves 2 919
19.11, Cavitation 980
19.111. Precaution Against Cavitation so
19.112. Effets of Cavitation oi
19.113. Hydraulie Machines Subjected to Cavitation et
19.114. Cavitation in Turbinen 2 981
19.115. Cavitation in Centrifogal Pumps = 981
Salve Problem 19.23 982
19.12. Moximum Suetion Lit (or Suction Height) 985,
19:13. Net Positive Suction Head (NPSH) 985,
19.14. Cavitation in Contefugal Pump. 285,
Solved Problem 19:21 985
Highlights . 987
Exercise 6
Chapter 20. Reciproeating Pumps 990-1040
20.1. Introduction 980,
202. Main Parte of a Reciprocating Pump 99
203. Working of Reciprocating Pump EN
Gx)
203.1. Discharge Through a Reciprocating Pump
203.2. Work Done hy Reciprocating Pump
203.3. Discharge, Work Done and Power
Required to Drive a Doubleacting Pump
204. Slip of Reciprocating Pump
204.1. Negative Slip of the Reciprocating Pump
203. Classification of Reciprocating Pumps
‘Solved Problems 201-202
206. Variation of Velocity and Acceleration
in the Suction and Delivery Pipes Due to
Acceleration ofthe Piston
1207. fect of Variation of Velocity on Friction
inthe Section and Delivery Pipes
‘Solved Problem 20:3
208. Indicator Diagram
208.1. Kea! Indicator Diagram
208.2. Effect of Acceleration in Suction and
Delivery Pipes on Indicator Diagram.
Solved Problems 204-209
208.3. ft of Friction in Suetion and Delivery
Pipes on Indicator Diagram
208.4. ft of Acceleration and Friction in
Suction and Delivery Pipes on Indicator
Diagram
Solved Problems 20.10-20 12
2085. Maximum Speed of a Reciprocating Pump
Solved Problem 20.13
208. Air Vessels
Solved Problems 20.11-20.18
20.10. Comparison between Centrifugal Pumps and
Reciprocating Pumps
Higalighte
Exercise
Chapter 21. Fluid System
21.1 Introduction
212, The Hydraulic Press
2121. Mechanical Advantage
2122. Leverage o the Hydraulic Press
2123. Actual Heavy Hydraulic Preso
Solved Problems 211-215
21.3. The Hydraulic Accumulator
213.1. Capacity of Hydraulic Accumulator
Solved Problems 21.6—21.11
2132. Differential Hydraulic Accumulator
21.4. The Hydraulic Intensiier
Solved Problems 21.12-21.18
213, The Hydraulic Ram
Solved Problems 21.14-21.16
En
995,
995,
996
997
997
997
o
1001
1001
100
100
1000
1004
1012
1013
1015
1019
1020
io
1080
1037
1037
1038
1041-1070
mu
Fi
1002
10
1042
1048
1045
1046
1047
1051
1051
105
1083
1055
203.1. Discharge Through a Reciprocating Pump
203.2. Work Done hy Reciprocating Pump
203.3. Discharge, Work Done and Power
Required to Drive a Doubleacting Pump
204. Slip of Reciprocating Pump
204.1. Negative Slip of the Reciprocating Pump
203. Classification of Reciprocating Pumps
‘Solved Problems 201-202
206. Variation of Velocity and Acceleration
in the Suction and Delivery Pipes Due to
Acceleration ofthe Piston
1207. fect of Variation of Velocity on Friction
inthe Section and Delivery Pipes
‘Solved Problem 20:3
208. Indicator Diagram
208.1. Kea! Indicator Diagram
208.2. Effect of Acceleration in Suction and
Delivery Pipes on Indicator Diagram.
Solved Problems 204-209
208.3. ft of Friction in Suetion and Delivery
Pipes on Indicator Diagram
208.4. ft of Acceleration and Friction in
Suction and Delivery Pipes on Indicator
Diagram
Solved Problems 20.10-20 12
2085. Maximum Speed of a Reciprocating Pump
Solved Problem 20.13
208. Air Vessels
Solved Problems 20.11-20.18
20.10. Comparison between Centrifugal Pumps and
Reciprocating Pumps
Higalighte
Exercise
Chapter 21. Fluid System
21.1 Introduction
212, The Hydraulic Press
2121. Mechanical Advantage
2122. Leverage o the Hydraulic Press
2123. Actual Heavy Hydraulic Preso
Solved Problems 211-215
21.3. The Hydraulic Accumulator
213.1. Capacity of Hydraulic Accumulator
Solved Problems 21.6—21.11
2132. Differential Hydraulic Accumulator
21.4. The Hydraulic Intensiier
Solved Problems 21.12-21.18
213, The Hydraulic Ram
Solved Problems 21.14-21.16
En
995,
995,
996
997
997
997
o
1001
1001
100
100
1000
1004
1012
1013
1015
1019
1020
io
1080
1037
1037
1038
1041-1070
mu
Fi
1002
10
1042
1048
1045
1046
1047
1051
1051
105
1083
1055
216, The Hydraulic Li
216.1. Direct Acting Hydravlie Lit
216.2, Suapended Hydraulic Lite
Solved Problems 21.16-21.17
217. The Hydraulic Crane
‘Solved Problems 2119-2120
218. The Fluid or Hydraulic Coupling
218. The Hydraulic Torque Converter
21.10. The Air Lit Pump
BEAL. The Gear Wheel Pump
hts
Objective Type Questions
Appendix
Subject Index
1066
1057
1057
1058
1060
1060
1083
1084
1066
1086
1087
1068
1071-1004
1095-1096
10971102
216.1. Direct Acting Hydravlie Lit
216.2, Suapended Hydraulic Lite
Solved Problems 21.16-21.17
217. The Hydraulic Crane
‘Solved Problems 2119-2120
218. The Fluid or Hydraulic Coupling
218. The Hydraulic Torque Converter
21.10. The Air Lit Pump
BEAL. The Gear Wheel Pump
hts
Objective Type Questions
Appendix
Subject Index
1066
1057
1057
1058
1060
1060
1083
1084
1066
1086
1087
1068
1071-1004
1095-1096
10971102
|
PROPERTIES OF
BLOIDS
» 1.1 INTRODUCTION
Fluid mechanics is hat branch of science which deals withthe behaviour ofthe Mids (guid or
gases) a rest as wll in motion. Thus tis brane of scones deals withthe sti, kinematies and
dynamic aspects of Ads, The study of ui at rest is called fluid tates. The study of ui in
‘motion, where pressure force re not considered Is called fluid Kinematics and if the pressure forces
are als considered for he Nuk In motion, that branch of sienes called Mud dynamics.
> 1.2. PROPERTIES OF FLUIDS
1.2.1 Density or Mass Density. Density or mass density ofa fd is defined as the ratio of the
asso ido ts volume. Ths mass per unit volume of a ui alle density ts denote bythe
‘Symbol (ho) The un of mass density Ho S unit isk por cubie metro, ie. Kg”. Te density of
Tiguids may be considered as constant while that of gases changes wit hc variation of pressure and
temperature.
‘Mathematically, mass density I wea as
Mas of uid
Volume ofa
‘The valve of density of water is 1 pme or 1000 Kg
1.22, Specific Weight or Weight Density. Specific weight or weight density ofa ui i he
xo between the weight of a Hai! to its volume, Thos weight per unit volume of à fit Hs called
‘weight density and ts denoted bythe symbol
Weight of fd _ (Mass of Nui)» Acceleration duc to grav
‘Thus mathematical wa
Volume of fd Volume of u
Mass of id,
Volume of id
wR E Ata]
. * Volume of tid
w=pe am
PROPERTIES OF
BLOIDS
» 1.1 INTRODUCTION
Fluid mechanics is hat branch of science which deals withthe behaviour ofthe Mids (guid or
gases) a rest as wll in motion. Thus tis brane of scones deals withthe sti, kinematies and
dynamic aspects of Ads, The study of ui at rest is called fluid tates. The study of ui in
‘motion, where pressure force re not considered Is called fluid Kinematics and if the pressure forces
are als considered for he Nuk In motion, that branch of sienes called Mud dynamics.
> 1.2. PROPERTIES OF FLUIDS
1.2.1 Density or Mass Density. Density or mass density ofa fd is defined as the ratio of the
asso ido ts volume. Ths mass per unit volume of a ui alle density ts denote bythe
‘Symbol (ho) The un of mass density Ho S unit isk por cubie metro, ie. Kg”. Te density of
Tiguids may be considered as constant while that of gases changes wit hc variation of pressure and
temperature.
‘Mathematically, mass density I wea as
Mas of uid
Volume ofa
‘The valve of density of water is 1 pme or 1000 Kg
1.22, Specific Weight or Weight Density. Specific weight or weight density ofa ui i he
xo between the weight of a Hai! to its volume, Thos weight per unit volume of à fit Hs called
‘weight density and ts denoted bythe symbol
Weight of fd _ (Mass of Nui)» Acceleration duc to grav
‘Thus mathematical wa
Volume of fd Volume of u
Mass of id,
Volume of id
wR E Ata]
. * Volume of tid
w=pe am
2 Fluid Mechanics
“The value of specii weight or weight density +) for waters 9.81 1000 Newoa/n* in ST units
2.3 Specific Volume. Specific volume ofa uid is defined a he volume ofa fluid occupied
y 3 unit mass or volume por unit miss of a fuk i call specific volume. Mathematically es
express as
Speciie volume
Volume of id
‘Thus specific volume i the reciprocal of mass density. I expressed as mg. Is commonly
applied to ga
1.2.4. Specific Gravity. Specific gravity is defined as the rato ofthe weight density (or density)
of à tui tothe weight density (or density of a standard vid Fr liquids, te standard ids taken
Water and for gases, the standard lids take at. Specific raviy i also called relative density. tis
iment gam and ye
y (densiy) o liquid
‘Wid Lao
Weight density (ens) of gas
‘Weight density (den) of a
Weight density of wate
$1000 9.81 Nim?
So Density of water
5% 1000 kg/m am
IF the specie gravity ofa Mud is known, then Ihe density f the id wi he equal w specie
gravity of fad multiplica y the density of water. For example, the speci grat of mercury 136,
hence density of mercury = 13.6% 1000 = 13600 kg/m
Mathematically, Sor iquids) =
Sur pases)
“Thus weight density of quid
‘The density ofa quid
Problem 1.1. Calculote she specific weight, density and specific gravity of one lie of liquid
hich weighs 7
Solution. Given
(9 Specie weight (u)
Gi) Density (0)
Density of guid
ren rare Density of water
02135. Ans.
“The value of specii weight or weight density +) for waters 9.81 1000 Newoa/n* in ST units
2.3 Specific Volume. Specific volume ofa uid is defined a he volume ofa fluid occupied
y 3 unit mass or volume por unit miss of a fuk i call specific volume. Mathematically es
express as
Speciie volume
Volume of id
‘Thus specific volume i the reciprocal of mass density. I expressed as mg. Is commonly
applied to ga
1.2.4. Specific Gravity. Specific gravity is defined as the rato ofthe weight density (or density)
of à tui tothe weight density (or density of a standard vid Fr liquids, te standard ids taken
Water and for gases, the standard lids take at. Specific raviy i also called relative density. tis
iment gam and ye
y (densiy) o liquid
‘Wid Lao
Weight density (ens) of gas
‘Weight density (den) of a
Weight density of wate
$1000 9.81 Nim?
So Density of water
5% 1000 kg/m am
IF the specie gravity ofa Mud is known, then Ihe density f the id wi he equal w specie
gravity of fad multiplica y the density of water. For example, the speci grat of mercury 136,
hence density of mercury = 13.6% 1000 = 13600 kg/m
Mathematically, Sor iquids) =
Sur pases)
“Thus weight density of quid
‘The density ofa quid
Problem 1.1. Calculote she specific weight, density and specific gravity of one lie of liquid
hich weighs 7
Solution. Given
(9 Specie weight (u)
Gi) Density (0)
Density of guid
ren rare Density of water
02135. Ans.
Properties of Fiids 3)
Problem 1.2. Cafowate the density, specific weight and weight of one litre of pero of specific
prog =07
Soluton. Given: Volume = Le 11000 cn) !2 0001 n°
sate 5-07
D De 4
Ung eaten (118)
Dentro) 2 5x 1000 kl? = 07 x 1000 70 Km Ans
i Specife weigh (»)
Using equation (LD, w= p< 2 700x981 Nin! = 6867 Nim Ans
(iy Weight (w)
Weight
Votume
w w
rer ¿Ho
= We 6567 «0.001 = 6867 N Ans.
We knon that specie weigh =
> 13 viscosity
Viscosity defined aso property of fluid which offers resistance 10 the movement of one layer
of fluid over another adjacent ayer of th Mud, When wo layer ofa fuk, a distance “y” apart move
one over he other at different veloc say a and u + du as shown Fig. 11, the viscosity together
‘vith relative velocity causes a shear ses acting tc the ld layers.
The op layer cases a sear stens onthe
eject ower ayer wile de ower ge cores Lun
{iter son Le aa oy ya seat Ea uc
Sus le proportional o e rel change of ve: t La
locity with respect to y. It is denoted by symbol x Len
+ Crau t {VELOCITY PROFRE
er vun gerri
A Pig. Veoh carlton nera solid boundary.
e ES 12
>
ere pal m) se constant o proportionality and is ko a the of uf dynam isos
f
bar a e pc lo
From equation (1.2), we have = an am
(5)
‘Thus viscosity is also defined asthe scar suessreulred o produce uni at of shea st
1.3.1 Units of Viscosity. The units of viscosity s obtained by puting the dimensions of the
anis in equation (13)
Problem 1.2. Cafowate the density, specific weight and weight of one litre of pero of specific
prog =07
Soluton. Given: Volume = Le 11000 cn) !2 0001 n°
sate 5-07
D De 4
Ung eaten (118)
Dentro) 2 5x 1000 kl? = 07 x 1000 70 Km Ans
i Specife weigh (»)
Using equation (LD, w= p< 2 700x981 Nin! = 6867 Nim Ans
(iy Weight (w)
Weight
Votume
w w
rer ¿Ho
= We 6567 «0.001 = 6867 N Ans.
We knon that specie weigh =
> 13 viscosity
Viscosity defined aso property of fluid which offers resistance 10 the movement of one layer
of fluid over another adjacent ayer of th Mud, When wo layer ofa fuk, a distance “y” apart move
one over he other at different veloc say a and u + du as shown Fig. 11, the viscosity together
‘vith relative velocity causes a shear ses acting tc the ld layers.
The op layer cases a sear stens onthe
eject ower ayer wile de ower ge cores Lun
{iter son Le aa oy ya seat Ea uc
Sus le proportional o e rel change of ve: t La
locity with respect to y. It is denoted by symbol x Len
+ Crau t {VELOCITY PROFRE
er vun gerri
A Pig. Veoh carlton nera solid boundary.
e ES 12
>
ere pal m) se constant o proportionality and is ko a the of uf dynam isos
f
bar a e pc lo
From equation (1.2), we have = an am
(5)
‘Thus viscosity is also defined asthe scar suessreulred o produce uni at of shea st
1.3.1 Units of Viscosity. The units of viscosity s obtained by puting the dimensions of the
anis in equation (13)
4 Fluid Mechanics
Shear ses Tore A
Semmering * (Lem), T
Change oraisanes | Time J* Lengin
Force/(Lengt*_ Force x Time
FE en:
Time
Im MKS system, fore is repesemed hy kgf and length hy metro (m), in CGS system, force is
represeme by dye and length hy em and in ST system fore is represented hy Newton (N) and length
hy mote m).
er
MKS ait of viscosity
CGS unit oF viscosity ones
Inthe above expression Nim als Known a Pascal which is represen by Da, Hence Nin? Pa
seal
‘St unit of viscosity = Nein? = Pas,
Newton soe _ Ne
Sl uni of viscosty Mure,
‘The unit of viscosity in COS i alo called Pois whieh scqualto DE.
“The numerical conversion of th uit of viscosity from MKS unit to CGS unit is given Blow
one kpfsec _ 981N-see 1
Ir
9.81 Newton}
‘But one Newton
oe mus one (2) cea)
Dm 10 BH
ke
yesos
00x 100 em"
See pie)
“Thus for solving numeries problems, 4 vscosity gite in pls it mast he dvd y 98.110 got
A equivale numerical value in MKS.
one kghsce | 981NS
EN à PE poise 10 poise
But = 98. poise
Shear ses Tore A
Semmering * (Lem), T
Change oraisanes | Time J* Lengin
Force/(Lengt*_ Force x Time
FE en:
Time
Im MKS system, fore is repesemed hy kgf and length hy metro (m), in CGS system, force is
represeme by dye and length hy em and in ST system fore is represented hy Newton (N) and length
hy mote m).
er
MKS ait of viscosity
CGS unit oF viscosity ones
Inthe above expression Nim als Known a Pascal which is represen by Da, Hence Nin? Pa
seal
‘St unit of viscosity = Nein? = Pas,
Newton soe _ Ne
Sl uni of viscosty Mure,
‘The unit of viscosity in COS i alo called Pois whieh scqualto DE.
“The numerical conversion of th uit of viscosity from MKS unit to CGS unit is given Blow
one kpfsec _ 981N-see 1
Ir
9.81 Newton}
‘But one Newton
oe mus one (2) cea)
Dm 10 BH
ke
yesos
00x 100 em"
See pie)
“Thus for solving numeries problems, 4 vscosity gite in pls it mast he dvd y 98.110 got
A equivale numerical value in MKS.
one kghsce | 981NS
EN à PE poise 10 poise
But = 98. poise
Properties of Flids 5)
Alternate Method. One poise = DIES (ere)
eyo
à One poe po
1 bet ke 38
x En 1 HE à to pois.
1000 1 16 sm sm © OP
¡ote In S nis econ i represen hy “and ol by sec
{ip Ilse I given In po, mus be vided by 10 o ge quale! mur vals Sum,
Sortie a nto sony as ernie usd whee
{P= Genie, P= Poise]
“Te visos of water at 20€ 4.01 pose o 1.9 cei,
1.3.2 Kinematic Viscosity. Hi defined ash aio Petwccn the dynamic viscosity and density
of fluid. eis denoted hy the Greck symbol (u) Called "nu. Thus, mathematialy,
Viscosity _ u
u as
Density” p
‘The units of veau Viscosi ls obtained as
Unis of Force x Time Force x Time
Vaso ange MS M
CR Teng” Lem
Length
ace TERS Time Face = Mass x Ace
(Tine) =
Mas Ee
Ti
In MKS and SI, the unit of kiematc viscosity is mere ec or mee while in COS unis iis
write as em. In CGS uns, kinematie viscosity i also known a tke,
‘Thus, one stoke
‘Centstoks means
3. Newton's Law of Viscosity, tte thatthe scr ses (9) om Mid clement ayer
directly proportional to the rte of Shcar stan. The constant of proportionality is call the co
cit of viscosiy. Mathematical, is expressed as given by equation (1.2) oras
de
se
Alternate Method. One poise = DIES (ere)
eyo
à One poe po
1 bet ke 38
x En 1 HE à to pois.
1000 1 16 sm sm © OP
¡ote In S nis econ i represen hy “and ol by sec
{ip Ilse I given In po, mus be vided by 10 o ge quale! mur vals Sum,
Sortie a nto sony as ernie usd whee
{P= Genie, P= Poise]
“Te visos of water at 20€ 4.01 pose o 1.9 cei,
1.3.2 Kinematic Viscosity. Hi defined ash aio Petwccn the dynamic viscosity and density
of fluid. eis denoted hy the Greck symbol (u) Called "nu. Thus, mathematialy,
Viscosity _ u
u as
Density” p
‘The units of veau Viscosi ls obtained as
Unis of Force x Time Force x Time
Vaso ange MS M
CR Teng” Lem
Length
ace TERS Time Face = Mass x Ace
(Tine) =
Mas Ee
Ti
In MKS and SI, the unit of kiematc viscosity is mere ec or mee while in COS unis iis
write as em. In CGS uns, kinematie viscosity i also known a tke,
‘Thus, one stoke
‘Centstoks means
3. Newton's Law of Viscosity, tte thatthe scr ses (9) om Mid clement ayer
directly proportional to the rte of Shcar stan. The constant of proportionality is call the co
cit of viscosiy. Mathematical, is expressed as given by equation (1.2) oras
de
se
(6 Fd Mechanics
Fluids which obey the above relation are known as Neytonlan Aids and the Mids which do not
obey the above relation ae called Non-Newtonlan fads.
134 Vad re. Temperature affect the viscosiy, The
viscosity o liquids decreass with the increns of temperature while the viscose of gases increases
with the Increase of temperature. This is due o reason thatthe viseous forces In a ld are due to
cohesive forces and molecular momentum transfer. In liquds, the cohesive forces predominates
the molecular momentum transfer, duc to closely packed molceules and with the increase in
temperature, the enhesve forces deeecascs with the resul of deercasing viscosity. But ln ease of
ases the cohesive forces are small and molecular momentum transfor predominates. WIN the
Increase in temperature, molecular momentum (runter increases and ence viscosity increases. The
relation between viscosity and temperature or quis and gases are:
toon tas (rt) am
where 2 Viseosty of Tiuid at, in poise
Al = Viscosity of quid at 0°C In poise
B= Constants for he quid
For water us = 1.79 x 10 * pois, a = 0.03368 and B = 0.000221.
[Equation (1.4) shows that with te Increase of temperature, the viscosiy decrees
Gi) For a gas, Batt ar Be am
where for alr, = 0.000017, u = 0:000000056, 9 = 0.1189 x 10°.
[Equation LAB shows that with ih increase of temperature, Ihe viscosity increases
1.3.5, Types of Fluids. The Als may be classified into th following five types:
1. Ideal ui, 2, Real uid,
3. Newronian fui 4. Non-Newionian foi, and
5. Kcal past fui
1 eat Fluid. A ui, which is incompressible and is
having no siscosity is know as an ca Mal, dca iis
Only am imaginary uid as all the Mois, which exist, have
some viscosity
2. Real Fluid. A fluid, which possesses viscosity, is
‘known as tel foi, All the fli, in actual practice, ae eal
fluids
‘3. Newtonian Flold. A ral fui, in whieh the sbear
ress is directly proportional tothe rate of shear stain (or
ely pa, own a à New aı mc enorm (8)
% NomNewtonlan Fluid. A real id, in which the
tea stress is no proportional o the rate of shear stain (or Figs 12 Types of fas
velocity gradient), known 2 a Noo-Newtonian ui,
5. Ideal Plas Fluid. tui, in which shear stress is more than the yield value and tear
ses is proportional oth fate of ser tan (or velocity pradiem),i known as del plate ud.
Problem 1.3. (fe velocity distribution over a plate Is given by u
velocity in mere per second at a distance y metre above the plate, determine the shear stress at
Y= Dandy = 0.13 m. Take dynamic viscoso of fai at £63 pate.
Lys nene vice
Fluids which obey the above relation are known as Neytonlan Aids and the Mids which do not
obey the above relation ae called Non-Newtonlan fads.
134 Vad re. Temperature affect the viscosiy, The
viscosity o liquids decreass with the increns of temperature while the viscose of gases increases
with the Increase of temperature. This is due o reason thatthe viseous forces In a ld are due to
cohesive forces and molecular momentum transfer. In liquds, the cohesive forces predominates
the molecular momentum transfer, duc to closely packed molceules and with the increase in
temperature, the enhesve forces deeecascs with the resul of deercasing viscosity. But ln ease of
ases the cohesive forces are small and molecular momentum transfor predominates. WIN the
Increase in temperature, molecular momentum (runter increases and ence viscosity increases. The
relation between viscosity and temperature or quis and gases are:
toon tas (rt) am
where 2 Viseosty of Tiuid at, in poise
Al = Viscosity of quid at 0°C In poise
B= Constants for he quid
For water us = 1.79 x 10 * pois, a = 0.03368 and B = 0.000221.
[Equation (1.4) shows that with te Increase of temperature, the viscosiy decrees
Gi) For a gas, Batt ar Be am
where for alr, = 0.000017, u = 0:000000056, 9 = 0.1189 x 10°.
[Equation LAB shows that with ih increase of temperature, Ihe viscosity increases
1.3.5, Types of Fluids. The Als may be classified into th following five types:
1. Ideal ui, 2, Real uid,
3. Newronian fui 4. Non-Newionian foi, and
5. Kcal past fui
1 eat Fluid. A ui, which is incompressible and is
having no siscosity is know as an ca Mal, dca iis
Only am imaginary uid as all the Mois, which exist, have
some viscosity
2. Real Fluid. A fluid, which possesses viscosity, is
‘known as tel foi, All the fli, in actual practice, ae eal
fluids
‘3. Newtonian Flold. A ral fui, in whieh the sbear
ress is directly proportional tothe rate of shear stain (or
ely pa, own a à New aı mc enorm (8)
% NomNewtonlan Fluid. A real id, in which the
tea stress is no proportional o the rate of shear stain (or Figs 12 Types of fas
velocity gradient), known 2 a Noo-Newtonian ui,
5. Ideal Plas Fluid. tui, in which shear stress is more than the yield value and tear
ses is proportional oth fate of ser tan (or velocity pradiem),i known as del plate ud.
Problem 1.3. (fe velocity distribution over a plate Is given by u
velocity in mere per second at a distance y metre above the plate, determine the shear stress at
Y= Dandy = 0.13 m. Take dynamic viscoso of fai at £63 pate.
Lys nene vice
Properties of Fluids 7]
2,
Solution, Gin A
6 2
au) E
CPE En
(
a #) (4) -
ACL TRE AP) aos
cof = 8.63 poise E 51 nis = 0.863 Nin?
de
Now shar ss given hy equation (12) as =p DE
D Shear sues at y = Dis piven by
Fal = 0.863 40.667 = 05756 Nim. Ans.
(Gi) Shear res ty = 0.1 mis given by
al
Problem 1.4 A plate 0.025 mm disant rom a fixed pat, moves a 60 ends and requires a force of
2N per unit area Le, 2 No 1 malnuala is speed. Determine ih fluid viscosity between the ple.
Solution. Given
Distance between plates, dy= 025 mm
2.025% 10m
60 emis = 06 mis
0.863 90.367 = 0.3167 Nim’. Ans.
Velocity of upper plas,
Fig. 13
Force on upper pat,
“Tiss the valu of shear sos de €
[Let Mid viscosity etc te plates
Using the equation (1.2), we have
‘where du = Change of velocity = «= 0= 4 = 0.60 mis
= Change of distance = 025 4 10° m
Pare pr ont aa = 20 N
Do eu m ES
or © MT os po
439. 10° x 10 poke = 833 x 10% pul, Ans.
Problem 1.5 A fat late of rra 1.5 x 10 mm? is pale win speed of Od ms relative to
ater plat Incatd ata strc of 0.15 me from. ind he force and pones required o mada
‘he sped ifthe fu separating them shin icon a poire
2,
Solution, Gin A
6 2
au) E
CPE En
(
a #) (4) -
ACL TRE AP) aos
cof = 8.63 poise E 51 nis = 0.863 Nin?
de
Now shar ss given hy equation (12) as =p DE
D Shear sues at y = Dis piven by
Fal = 0.863 40.667 = 05756 Nim. Ans.
(Gi) Shear res ty = 0.1 mis given by
al
Problem 1.4 A plate 0.025 mm disant rom a fixed pat, moves a 60 ends and requires a force of
2N per unit area Le, 2 No 1 malnuala is speed. Determine ih fluid viscosity between the ple.
Solution. Given
Distance between plates, dy= 025 mm
2.025% 10m
60 emis = 06 mis
0.863 90.367 = 0.3167 Nim’. Ans.
Velocity of upper plas,
Fig. 13
Force on upper pat,
“Tiss the valu of shear sos de €
[Let Mid viscosity etc te plates
Using the equation (1.2), we have
‘where du = Change of velocity = «= 0= 4 = 0.60 mis
= Change of distance = 025 4 10° m
Pare pr ont aa = 20 N
Do eu m ES
or © MT os po
439. 10° x 10 poke = 833 x 10% pul, Ans.
Problem 1.5 A fat late of rra 1.5 x 10 mm? is pale win speed of Od ms relative to
ater plat Incatd ata strc of 0.15 me from. ind he force and pones required o mada
‘he sped ifthe fu separating them shin icon a poire
8 Fluid Mechanics
Solution. Given
ca of he plats 15% 10 mu
‘Speed of pte relative to another late, de =04 mls
Distance between the plates, d= 0.15 mm = 0.15 10% m
uns
15 n°
Viscoshy =I poise
1
10 * 15x10 ”
© Shear force, Be tx ater = 26646 % LS = 400 N. Ans
D Power” required o move the plat athe speed 04 misec
= Ex à = 400 4 DA = 160 W.
Problem 4.8. Determine the intensity of shear ofan où having vices = I poise. The ot à used
{for lubricating he clearance bern a shaft of diameter 10 em audi journal bearing, Th clearance
ds 1.5 mn and he shaft rotates at 130 pun.
du
Usine cquton (12) we ave SE
Solution. Given
Dia. of sha,
Distance hewsen shaft and jouma Dering
de 1Smm=15% 10%
Spesd of shat, Sopa
“Tangential pci of af is given by
RDN Rx O1x 150
ww
= 0.785 mis
du
Using pen (12 r=
where du = change of velocity betweea shaft and beating = u 0 = u
1, 0785
10 15x10
Problem 4.7 Calcule the demie viscosity of an où, which i wed fr labrication between à
square plate of sce 0.8 m X08 m and an inclined plane with angle of telnation 50° ax shown in
Fig Ih. Te weight of the square plate i 300 N ad slides down the inclined plane with uniform
ec of 05 m/s The hikes a où fi 1.5 mm. eS
Solution. Give
Arca of plat, A208 x08 = 064 m?
Angle of plane, 8230
Weight of plat. w=300N
Velocity of plc, ms
Towers Pre Nm
Su Wee Nine Wat)
Solution. Given
ca of he plats 15% 10 mu
‘Speed of pte relative to another late, de =04 mls
Distance between the plates, d= 0.15 mm = 0.15 10% m
uns
15 n°
Viscoshy =I poise
1
10 * 15x10 ”
© Shear force, Be tx ater = 26646 % LS = 400 N. Ans
D Power” required o move the plat athe speed 04 misec
= Ex à = 400 4 DA = 160 W.
Problem 4.8. Determine the intensity of shear ofan où having vices = I poise. The ot à used
{for lubricating he clearance bern a shaft of diameter 10 em audi journal bearing, Th clearance
ds 1.5 mn and he shaft rotates at 130 pun.
du
Usine cquton (12) we ave SE
Solution. Given
Dia. of sha,
Distance hewsen shaft and jouma Dering
de 1Smm=15% 10%
Spesd of shat, Sopa
“Tangential pci of af is given by
RDN Rx O1x 150
ww
= 0.785 mis
du
Using pen (12 r=
where du = change of velocity betweea shaft and beating = u 0 = u
1, 0785
10 15x10
Problem 4.7 Calcule the demie viscosity of an où, which i wed fr labrication between à
square plate of sce 0.8 m X08 m and an inclined plane with angle of telnation 50° ax shown in
Fig Ih. Te weight of the square plate i 300 N ad slides down the inclined plane with uniform
ec of 05 m/s The hikes a où fi 1.5 mm. eS
Solution. Give
Arca of plat, A208 x08 = 064 m?
Angle of plane, 8230
Weight of plat. w=300N
Velocity of plc, ms
Towers Pre Nm
Su Wee Nine Wat)
Properties of Flids 9)
Thickness of oil fa, Id LS mm= 1.5 10m
Lette iscosy of ui between plat and ¡clica plane i
Component weight W, along the plane = W os 60° = 300 cos 60
‘Thus the shear ree, om the boom surlace ofthe plate = 150 N
F
150 Nig?
coh = ISO on
Now using equation (12), we have
and shea ses,
du
un
where de = change of velocity = a 0 w= 03 mis
ate sio
2 150, 03
CS
15021510" a
= = LIT Nan = LA7 x 10 = 117 poise. Ans.
Problem 1
seiko ef viscosity 14 eine, Clear the hear sires in oil upper plates mored with a veloc
Of 25 ms
Solution. Given
Distance hetweon plats, dy = 1.25 em = 0.9125 m
ave
Viscosity, = tapes MN wim?
y We M poise = 1,
Velociy of per plat, w= 2.5 mee.
du
Shear ss i given by equation (1.2) as, € ap 2
ven by equation (1.2) a8, 12 À
where di Change of velocity Between plates = w=
0.0125 m.
ns”
Problem 1.9. The space beneen no square fla parallel paresis filed wir ol. Fac side ofthe
late i 0 em. The thickness of the a fim is 12.5 mm, The upper plat, which moves at 2.5 metre per
See requires a free of 98.1 N to main Ihe speed. Determine
(0) the amie viscosity of the on poise, and
(i) the knemrie viscosity ofthe lbn sake de specific grain ofthe oi is 095
Solution. Given
Bach side of a square plate = 60cm =0.60 m
ana, A2 06 %06 2 036 m
Thickness of ol fm, dy= 125 mm = 12 x 10 m
Velocity of upper plie, w= 25 mic
Tio horizontal plates are pled 1.25 em apart, the pace between them being filled
Thickness of oil fa, Id LS mm= 1.5 10m
Lette iscosy of ui between plat and ¡clica plane i
Component weight W, along the plane = W os 60° = 300 cos 60
‘Thus the shear ree, om the boom surlace ofthe plate = 150 N
F
150 Nig?
coh = ISO on
Now using equation (12), we have
and shea ses,
du
un
where de = change of velocity = a 0 w= 03 mis
ate sio
2 150, 03
CS
15021510" a
= = LIT Nan = LA7 x 10 = 117 poise. Ans.
Problem 1
seiko ef viscosity 14 eine, Clear the hear sires in oil upper plates mored with a veloc
Of 25 ms
Solution. Given
Distance hetweon plats, dy = 1.25 em = 0.9125 m
ave
Viscosity, = tapes MN wim?
y We M poise = 1,
Velociy of per plat, w= 2.5 mee.
du
Shear ss i given by equation (1.2) as, € ap 2
ven by equation (1.2) a8, 12 À
where di Change of velocity Between plates = w=
0.0125 m.
ns”
Problem 1.9. The space beneen no square fla parallel paresis filed wir ol. Fac side ofthe
late i 0 em. The thickness of the a fim is 12.5 mm, The upper plat, which moves at 2.5 metre per
See requires a free of 98.1 N to main Ihe speed. Determine
(0) the amie viscosity of the on poise, and
(i) the knemrie viscosity ofthe lbn sake de specific grain ofthe oi is 095
Solution. Given
Bach side of a square plate = 60cm =0.60 m
ana, A2 06 %06 2 036 m
Thickness of ol fm, dy= 125 mm = 12 x 10 m
Velocity of upper plie, w= 25 mic
Tio horizontal plates are pled 1.25 em apart, the pace between them being filled
10 Fluid Mechanics
‘Change of velocky between pates, du = 2.8 misee
Fores required on upper pls, F= 98.1 N
are FL 98IN
Shoat os, in
Ama "A" 036m
(0 Let à Dynamic viscosity of oi!
du 98 as
Using equation (12) HH op MBL yy 2S
eee ay 036 "* 135.107
sat 12510 Ne
PL Sy
CRE
263510 = 13.635 poise. Am.
(iD Sp. gr. of oi, = 095
Let = Kinematie viscosity of oi
Using equation (1.14),
Mass density fo % 1000 = 095 x 1000 = 950 kya?
m 1305555)
poe 550
= 1436 stokes. Ans. Ce en = stoke)
Problem 1.10. Find te Kinematic ici ofan ol having deny 98 gh. The shea sess a
a point in lis 02452 Min and veloc raden a tha pola is 0.2 per second
Solution. Given
001435 mee = 001435 x 10° nhs
Using the relation,
Mass density, = 981 ptm”
Shear sess, 2482 Nin?
du
Velocity gratin, SE 02s
ve 2
du
sing the equation (LD. HE or 02452 = x02
ous A
1226 Nim?
0200
Kinematic viscosity Vs given by
2 TS
pom
128 10 © x 10 em? = 0.25 10° mis
12.5 em = 125 stoke. Ans. Cs emis soko)
Problem 1.11 Determine the specific gravity of aid having viscosity 0.0 poise and Kinematic
viscorty 0.035 stokes,
Solution.
‘Viscosity. = 0.08 poise =
‘Change of velocky between pates, du = 2.8 misee
Fores required on upper pls, F= 98.1 N
are FL 98IN
Shoat os, in
Ama "A" 036m
(0 Let à Dynamic viscosity of oi!
du 98 as
Using equation (12) HH op MBL yy 2S
eee ay 036 "* 135.107
sat 12510 Ne
PL Sy
CRE
263510 = 13.635 poise. Am.
(iD Sp. gr. of oi, = 095
Let = Kinematie viscosity of oi
Using equation (1.14),
Mass density fo % 1000 = 095 x 1000 = 950 kya?
m 1305555)
poe 550
= 1436 stokes. Ans. Ce en = stoke)
Problem 1.10. Find te Kinematic ici ofan ol having deny 98 gh. The shea sess a
a point in lis 02452 Min and veloc raden a tha pola is 0.2 per second
Solution. Given
001435 mee = 001435 x 10° nhs
Using the relation,
Mass density, = 981 ptm”
Shear sess, 2482 Nin?
du
Velocity gratin, SE 02s
ve 2
du
sing the equation (LD. HE or 02452 = x02
ous A
1226 Nim?
0200
Kinematic viscosity Vs given by
2 TS
pom
128 10 © x 10 em? = 0.25 10° mis
12.5 em = 125 stoke. Ans. Cs emis soko)
Problem 1.11 Determine the specific gravity of aid having viscosity 0.0 poise and Kinematic
viscorty 0.035 stokes,
Solution.
‘Viscosity. = 0.08 poise =
Properties of Fluids 11
035 stokes
035 en Le Stoke
= 0035 x 10 “a
Kinemate visosiy,
en's)
N y
se get 0.035 x 10
P
Using the relation v
eos,
ETS
Sp. of liquid 4285 = LAR. Ans.
= Density oF water ©
Problem 1.12 Determine he viscosity ofa liquid having kinematie viscosity 6 stokes and specific
gravity 19.
Solution. Gien
‘Kinematic viscosiy
Sp. of quid
stokes = 6 em = 6 10 Ÿ x
>
e op
Now sperorangat 2 DEEE
we 2 Density of water
x on
or
à van ion se
u
Res
no
14x 10= 11.40 pose. Ans
Problem 1.19 The velocity distribution fr Row over a la plate is given by u = à y = in which
ic the veloci in metre per second at a distance y metre above the plate, Determine the shear tress
ty = 0.15 m Take dynamic viscosity of ud as RG poise
Solution. ven
Ac y= oul,
Viscosity,
035 stokes
035 en Le Stoke
= 0035 x 10 “a
Kinemate visosiy,
en's)
N y
se get 0.035 x 10
P
Using the relation v
eos,
ETS
Sp. of liquid 4285 = LAR. Ans.
= Density oF water ©
Problem 1.12 Determine he viscosity ofa liquid having kinematie viscosity 6 stokes and specific
gravity 19.
Solution. Gien
‘Kinematic viscosiy
Sp. of quid
stokes = 6 em = 6 10 Ÿ x
>
e op
Now sperorangat 2 DEEE
we 2 Density of water
x on
or
à van ion se
u
Res
no
14x 10= 11.40 pose. Ans
Problem 1.19 The velocity distribution fr Row over a la plate is given by u = à y = in which
ic the veloci in metre per second at a distance y metre above the plate, Determine the shear tress
ty = 0.15 m Take dynamic viscosity of ud as RG poise
Solution. ven
Ac y= oul,
Viscosity,
12 Flukd Mechanics
ES
a" 0
Problem 1.14 The dynamic viscosity of an o, sed for lubrication between a shaft and sleeve is
pois. The shafts of diameter 0:4 and rotates ot 190 rpm. Calculate the power los in the bearing
for a sleve length of 90 mim The thickness ofthe où fl i 1S mm
{Using equation (12)
PTE
Solution nen 30m
Viena 6 pote
ENS gg Ns A
Lion Oe :
ia of an, bettm DL.
‚Speed of shaf Na rpm sde +
s 90 mm = 90 x 10° m “
Ticknes of, Smm= 15x10’m Fig ts
ADN RXDAXI90
“Tangent vloiyof sat ADN EDAD. 8 y
, r 60 oO
de
Using te ation de
I dy
win 3.98.5
ere
10x 8. à 1592 Nm?
ET]
“Tiss shear tes on shat
‘Sha fore onthe shaft, = Shear ies X Ar
S92 «MDX L= 1592 XX AID x 10°
D os
ee 2 2 18005 x LE 3601 Nm
“Torque on the shat,
ENT. _ 2619013601 rr We Ans
Poser ost
©
Problem 1.15. ie velocity profile o aid over aplaeisporablle wih the vertex 20 cm from
‘the plate, where the veloc is 120 emer. Calculate the velocity gradients and shear siresse at à
distance of . 10 an 20cm from the ple, ifthe viscosity of he Au is 85 pote.
Solution. Given
Distance of verex from plate = 20 cm
Velocity at vertex, 20 eme F
Viscosity,
Fig. 16
EN
+ Power in Sante TO eT +
ES
a" 0
Problem 1.14 The dynamic viscosity of an o, sed for lubrication between a shaft and sleeve is
pois. The shafts of diameter 0:4 and rotates ot 190 rpm. Calculate the power los in the bearing
for a sleve length of 90 mim The thickness ofthe où fl i 1S mm
{Using equation (12)
PTE
Solution nen 30m
Viena 6 pote
ENS gg Ns A
Lion Oe :
ia of an, bettm DL.
‚Speed of shaf Na rpm sde +
s 90 mm = 90 x 10° m “
Ticknes of, Smm= 15x10’m Fig ts
ADN RXDAXI90
“Tangent vloiyof sat ADN EDAD. 8 y
, r 60 oO
de
Using te ation de
I dy
win 3.98.5
ere
10x 8. à 1592 Nm?
ET]
“Tiss shear tes on shat
‘Sha fore onthe shaft, = Shear ies X Ar
S92 «MDX L= 1592 XX AID x 10°
D os
ee 2 2 18005 x LE 3601 Nm
“Torque on the shat,
ENT. _ 2619013601 rr We Ans
Poser ost
©
Problem 1.15. ie velocity profile o aid over aplaeisporablle wih the vertex 20 cm from
‘the plate, where the veloc is 120 emer. Calculate the velocity gradients and shear siresse at à
distance of . 10 an 20cm from the ple, ifthe viscosity of he Au is 85 pote.
Solution. Given
Distance of verex from plate = 20 cm
Velocity at vertex, 20 eme F
Viscosity,
Fig. 16
EN
+ Power in Sante TO eT +
Properties of Fluids 15]
“The velocity profiles given parabolic and equtio of velocity profiles
pee
where a,b and ar constants. Tir values are determined from boundary conditions as
(@ my=0,
0) ay= 20 em, u = 120 emisce
te
fe) aty 2 200m, Lo,
om =
Suhsitin
owndary condition a) im equation (we get
‘Boundary condition (2) on substitution in (gives
120 = (20)? + BO) = 4000 + 200
Boundary condition (¢ on substation in equation () pies
Au days
ay
or De2xox mehr
Solving equations (i) and (i) for a and b
From equation di). b= ~ 40a
Substutng this valve in equation i), we get
120 = 4000 + 20 x (- 400) = 400 - 8004
4000
m3
-03
oo 10
40%(-03)= 12.0
Substituting he valcs ofa band in equation (D,
Or +.
Velocity Gradient
$03 «294 129 Oy + 12
aw
Velos gran ($4) 06 0 + 12= 126. Aw.
06% 10412
64 122 6h. Ans
aty=20em,
06420 4 122-1241
20 Am.
Shear Stresses
Shear sess i given by, opt
ti)
wi)
u
“The velocity profiles given parabolic and equtio of velocity profiles
pee
where a,b and ar constants. Tir values are determined from boundary conditions as
(@ my=0,
0) ay= 20 em, u = 120 emisce
te
fe) aty 2 200m, Lo,
om =
Suhsitin
owndary condition a) im equation (we get
‘Boundary condition (2) on substitution in (gives
120 = (20)? + BO) = 4000 + 200
Boundary condition (¢ on substation in equation () pies
Au days
ay
or De2xox mehr
Solving equations (i) and (i) for a and b
From equation di). b= ~ 40a
Substutng this valve in equation i), we get
120 = 4000 + 20 x (- 400) = 400 - 8004
4000
m3
-03
oo 10
40%(-03)= 12.0
Substituting he valcs ofa band in equation (D,
Or +.
Velocity Gradient
$03 «294 129 Oy + 12
aw
Velos gran ($4) 06 0 + 12= 126. Aw.
06% 10412
64 122 6h. Ans
aty=20em,
06420 4 122-1241
20 Am.
Shear Stresses
Shear sess i given by, opt
ti)
wi)
u
102 Nin
(in Stew suesaty=10, [e
i Sue EA
as x
(Gi) Shar sess at 2 20, te)
Problem 1.16 A Newton lid silted in he elearance between a luft and a concentric see.
The sleeve attains. sperdof SO em, hen a force of 40 i applied tothe sleeve parallel tthe shat.
Determine the speed if a force of 200 Nis applied
Solution. Given: Speed of eve, » 50 cms
sen rc pers
pen fon is when fs, Fy = 200
de
Using raton can
dy
LA
a
Onn
wt
>
A eu (eA, Wand y are constant)
EE
10. 200
Subt vals, ws gor 0, 200
si #50
ae EM 0 5 20 ak. Am
E
Problem 1.17 AS cm diameter vertical elinder rotates concentrically inside another esinder of
diameter 1510 om. Both esinders are 23 cm high The space been the esindes i filled with a
liquid whose viscosity is unknown. Ifa torque of 12.0 Non is required 10 rotate the inner elinder as
100 rpm, determine the vscosty of the fluid
Solution. ives
Diameter of eylidor 15 em = 0.15 m
Dia. of outer cylinder = 15.10 m= 0.151 m
Length of cyladers, L= 25 em = 025 m
Torque, T=120Nm
(in Stew suesaty=10, [e
i Sue EA
as x
(Gi) Shar sess at 2 20, te)
Problem 1.16 A Newton lid silted in he elearance between a luft and a concentric see.
The sleeve attains. sperdof SO em, hen a force of 40 i applied tothe sleeve parallel tthe shat.
Determine the speed if a force of 200 Nis applied
Solution. Given: Speed of eve, » 50 cms
sen rc pers
pen fon is when fs, Fy = 200
de
Using raton can
dy
LA
a
Onn
wt
>
A eu (eA, Wand y are constant)
EE
10. 200
Subt vals, ws gor 0, 200
si #50
ae EM 0 5 20 ak. Am
E
Problem 1.17 AS cm diameter vertical elinder rotates concentrically inside another esinder of
diameter 1510 om. Both esinders are 23 cm high The space been the esindes i filled with a
liquid whose viscosity is unknown. Ifa torque of 12.0 Non is required 10 rotate the inner elinder as
100 rpm, determine the vscosty of the fluid
Solution. ives
Diameter of eylidor 15 em = 0.15 m
Dia. of outer cylinder = 15.10 m= 0.151 m
Length of cyladers, L= 25 em = 025 m
Torque, T=120Nm
Properties of Fluids 15]
Fr.
the viscosity
“Tangential velocity ofeylider,u SPN = ROIS 100
2854 m
60 e
Surface are of einer, RD x T= RX O15 X 025 = 178m?
du
Now using ration de
. ue
whore d= w= 00 7858 mis
y= E op
wast
000$
Shoat force, hear ans x Arca EIA yng
s
»
Torque, rx?
de 2
15
?
= 0864 N sim?
TAS L178 AS
= 0.461 « 10 = 864 pose. Ans.
Problem 1.18 Two large plane surfaces arr 24 cm apart. The space been the surfaces is fled
swith glycerine, What force te required to droga very thin plate of surface area 05 square metre
between she 10 large plane surjces ata speed of 6 mis I
(1) the thin plate iin ih middle of the wo plane surfaces, and
(i) the thin plate is a a distance of 08 cm from one of the plane surfaces ? Take the mane
scot of eccrine = 8.10 x 10" Nm.
Solution. Given
Distance between two lye surfaces = 24 cm
‘Area of hi pate, A=05 08 1200
‘Velocity of tha plats, u= 06 m/s zn +r
Viscosity of gesting 10% 10 °N sin? es
{Cave L When the thin plate iin the middle ofthe two plane
surfaces [Roferto Fig 1. (0)
La
Fi 2 Shear force on the upp side ofthe io plat
E = Shea force onthe lower sido ofthe thi pate
Fa Total force required to drag the pla
Faber,
“The shear stes (5) om the upper id o he iin plate I given hy equation,
Fig. 17 (0)
Thea
Fr.
the viscosity
“Tangential velocity ofeylider,u SPN = ROIS 100
2854 m
60 e
Surface are of einer, RD x T= RX O15 X 025 = 178m?
du
Now using ration de
. ue
whore d= w= 00 7858 mis
y= E op
wast
000$
Shoat force, hear ans x Arca EIA yng
s
»
Torque, rx?
de 2
15
?
= 0864 N sim?
TAS L178 AS
= 0.461 « 10 = 864 pose. Ans.
Problem 1.18 Two large plane surfaces arr 24 cm apart. The space been the surfaces is fled
swith glycerine, What force te required to droga very thin plate of surface area 05 square metre
between she 10 large plane surjces ata speed of 6 mis I
(1) the thin plate iin ih middle of the wo plane surfaces, and
(i) the thin plate is a a distance of 08 cm from one of the plane surfaces ? Take the mane
scot of eccrine = 8.10 x 10" Nm.
Solution. Given
Distance between two lye surfaces = 24 cm
‘Area of hi pate, A=05 08 1200
‘Velocity of tha plats, u= 06 m/s zn +r
Viscosity of gesting 10% 10 °N sin? es
{Cave L When the thin plate iin the middle ofthe two plane
surfaces [Roferto Fig 1. (0)
La
Fi 2 Shear force on the upp side ofthe io plat
E = Shea force onthe lower sido ofthe thi pate
Fa Total force required to drag the pla
Faber,
“The shear stes (5) om the upper id o he iin plate I given hy equation,
Fig. 17 (0)
Thea
16 Fluid Mechanics
‘were du = Relative velocity betwoco thi plate and upper lage plane surface
06 mice
12 em = 0.012 m (plats ia thin one and hence ice of plate i melected)
10.10? (26) as
m
Now shear foes,
=1,xA=405x0,
Silly hear tes (1) om the lower sido ofthe thin plat given by
105107 [06 | 405 Nie?
a
Shear free, Bau x A= 405205 =2025N
Toral force, +, 408. Ans.
‘Cave I. When the thin plate at a distance of. em from onc of
fe plane surfaces [Rete Fig. 1.7 ——
Let the thin plate is ata distance 0.8 cm from the lower plane
sui ten
“Then distance ofthe plat rom the upper plane surface aan
=24-08= 16cm= 016m
(eplectiog thickness of the plate)
“The sbeat force on the upper sie ofthe thin plate,
Dore
au a
+ (ef) amos 65
“The shear force on ths lower sde ofthe thin pat,
"al force required = F +
Problem 1.19 A serial gap 2.2 em wide of inne etes contains a aid of viscosity 2.0 N n°
and specie gravity 09. A metallic plate 1.2 m x 1.2 m 02 em is 10 be Ud up witha constant
Velocity 070.1 msc, though the gap. Y le plat is nh middle he gap, find e force required
The weight ofthe plate is SON.
Solution. Given
‘With of gap
Sq, ge of fia
em, viscose = 20 N sh
‘were du = Relative velocity betwoco thi plate and upper lage plane surface
06 mice
12 em = 0.012 m (plats ia thin one and hence ice of plate i melected)
10.10? (26) as
m
Now shear foes,
=1,xA=405x0,
Silly hear tes (1) om the lower sido ofthe thin plat given by
105107 [06 | 405 Nie?
a
Shear free, Bau x A= 405205 =2025N
Toral force, +, 408. Ans.
‘Cave I. When the thin plate at a distance of. em from onc of
fe plane surfaces [Rete Fig. 1.7 ——
Let the thin plate is ata distance 0.8 cm from the lower plane
sui ten
“Then distance ofthe plat rom the upper plane surface aan
=24-08= 16cm= 016m
(eplectiog thickness of the plate)
“The sbeat force on the upper sie ofthe thin plate,
Dore
au a
+ (ef) amos 65
“The shear force on ths lower sde ofthe thin pat,
"al force required = F +
Problem 1.19 A serial gap 2.2 em wide of inne etes contains a aid of viscosity 2.0 N n°
and specie gravity 09. A metallic plate 1.2 m x 1.2 m 02 em is 10 be Ud up witha constant
Velocity 070.1 msc, though the gap. Y le plat is nh middle he gap, find e force required
The weight ofthe plate is SON.
Solution. Given
‘With of gap
Sq, ge of fia
em, viscose = 20 N sh
Properties of Fluids 17]
2 Weight density of fia
UE ] |
ia
macaco "Oden T
ae lin ge
VE Eve mo ep. de et
rés nes
(sten nu)
:
(22-0
em 01m.
7
[Now the shear ore on the et def te mel plat.
= Shar rose x Arca
Ce Ara 121200)
=052N
Similarly, Ue shear force on the right side ofthe metallic plate,
2 Toul sear fores 244304324422 864 N,
Im thisease the weight of plats (which acting vertically downward) and upward thrusts also 0 he
taken into account
“The upward thes
Weigh of fu displaced
(Weigh density of Mid) Volume of fa displaced
9.81 x 900 x.00288 N
(Volume of ld displaced = Volume of plat
= 2843.
"The not force ating i the downward direction duc to weight ofthe plate and upward thrust
= Weight of plate - Upwaed thrust = 40 ~ 25.43 = 14.57 N
the plate up
= Total shea force + 14.57=864 + 14.57 = 10097 No Ans
0288)
2 Total force required 1
> 1.4 THERMODYNAMIC PROPERTIES
Fluds consist o liquids or gases. But gases ae compresibe fluds and hence thermadyoamic
Properties play an important role, With the change of pressure and temperature, te pases undergo
2 Weight density of fia
UE ] |
ia
macaco "Oden T
ae lin ge
VE Eve mo ep. de et
rés nes
(sten nu)
:
(22-0
em 01m.
7
[Now the shear ore on the et def te mel plat.
= Shar rose x Arca
Ce Ara 121200)
=052N
Similarly, Ue shear force on the right side ofthe metallic plate,
2 Toul sear fores 244304324422 864 N,
Im thisease the weight of plats (which acting vertically downward) and upward thrusts also 0 he
taken into account
“The upward thes
Weigh of fu displaced
(Weigh density of Mid) Volume of fa displaced
9.81 x 900 x.00288 N
(Volume of ld displaced = Volume of plat
= 2843.
"The not force ating i the downward direction duc to weight ofthe plate and upward thrust
= Weight of plate - Upwaed thrust = 40 ~ 25.43 = 14.57 N
the plate up
= Total shea force + 14.57=864 + 14.57 = 10097 No Ans
0288)
2 Total force required 1
> 1.4 THERMODYNAMIC PROPERTIES
Fluds consist o liquids or gases. But gases ae compresibe fluds and hence thermadyoamic
Properties play an important role, With the change of pressure and temperature, te pases undergo
18 Fluid Mechanics
large variation in density. The relationship between pressure (absolut), specific volume and
temperature (absolute) ofa gas is given bythe equation of state as
pr
mor Lar as
P
where p = Absolute pressure of agus in Nim?
1
pect ola
D
Gas cons
Absolute temperature in °K
p= Density ofa gus
1 Dimension of R. The pas consta. depends upon the pacas pas. The dinension of A
$s obtained from equation (1.5)
=
eo
Din MES unis q hott’ kate
[Boule = Nm]
Na
ke
1.4.2. Isothermal Process. Ifthe change in density occurs at constant temperature, thee the
proces is called isothermal and relationship between pressure (7) and density (p) I given by
je au
2
2 à Consant ao
°
1.4.3 Adiabatic Process. inc change in density occurs with o eat exchange 10 and rom the
iS he proces is called adíabai. And Hf no cas generated within the gas due 1o fection, the
‘lationship between pressure and density given by
D 2 Constant an
>
where = Ratio of specific heut of à as a constant pressure and constar volum
IA for ai
large variation in density. The relationship between pressure (absolut), specific volume and
temperature (absolute) ofa gas is given bythe equation of state as
pr
mor Lar as
P
where p = Absolute pressure of agus in Nim?
1
pect ola
D
Gas cons
Absolute temperature in °K
p= Density ofa gus
1 Dimension of R. The pas consta. depends upon the pacas pas. The dinension of A
$s obtained from equation (1.5)
=
eo
Din MES unis q hott’ kate
[Boule = Nm]
Na
ke
1.4.2. Isothermal Process. Ifthe change in density occurs at constant temperature, thee the
proces is called isothermal and relationship between pressure (7) and density (p) I given by
je au
2
2 à Consant ao
°
1.4.3 Adiabatic Process. inc change in density occurs with o eat exchange 10 and rom the
iS he proces is called adíabai. And Hf no cas generated within the gas due 1o fection, the
‘lationship between pressure and density given by
D 2 Constant an
>
where = Ratio of specific heut of à as a constant pressure and constar volum
IA for ai
Properties of Fuids 19]
14 Universal Gas Constant
La mo Mas of gas in kg
= Volume of as of mass
= Absolut pressure
T= Absolute temperature
‘Ten, we bave pv nr
as
‘where = Gas constant
Equation (13) can be made universal, aplicable to al gases ts expressed in mole-basts
La = Number of moles in volume fa qu
= Volume ofthe pas
Mass ofthe as molecles
M ans of a hydrogen am
me Mass of gas kg
‘Then, we have nxM=m
Suhsttating he valve fm in equation (1.8), we got
pV = nic MT as
bg
‘The product M x Ri called universal gas constant and is equal 0 848 PCR in MKS un
ls ie = Keno K
and 8314 Mep-nole Kin St units.
‘One kilogram moe deine as tb product of one kilogram mass ofthe gas and is molecular weight.
Problem 1.20 A gos wright 16 Ni“ ar 25°C and at an absolute pressure of 0,25 Nun’. Deer
mine the gas const and density ofthe gos
Solution. Gien
Weight density, w= 16 Nm
Temperature, 1228°C
T=27321=213+25=288k
12 025 Nim (abs) = 029 x 10 Nin
Pes density Is obtained as
(Using relation e
w 16 A
2S = 1.68 kp’, Ans.
ss id
(Using estos (1.9, B=
ET
Nm
a - = sass Am an,
Dar 13
Problem 1.21. A cylinder of 0.6 m’ in volume contins air ot 50°C and 0.3 Nin? absolute
pressure The ai is compresed 100.3 m, Find () pressure inside the elinder assuming isothermal
process and (i) pressure and temperature assuming adiabatic process Tale k= 14.
Solution. Given
Initia volume,
ATEN
14 Universal Gas Constant
La mo Mas of gas in kg
= Volume of as of mass
= Absolut pressure
T= Absolute temperature
‘Ten, we bave pv nr
as
‘where = Gas constant
Equation (13) can be made universal, aplicable to al gases ts expressed in mole-basts
La = Number of moles in volume fa qu
= Volume ofthe pas
Mass ofthe as molecles
M ans of a hydrogen am
me Mass of gas kg
‘Then, we have nxM=m
Suhsttating he valve fm in equation (1.8), we got
pV = nic MT as
bg
‘The product M x Ri called universal gas constant and is equal 0 848 PCR in MKS un
ls ie = Keno K
and 8314 Mep-nole Kin St units.
‘One kilogram moe deine as tb product of one kilogram mass ofthe gas and is molecular weight.
Problem 1.20 A gos wright 16 Ni“ ar 25°C and at an absolute pressure of 0,25 Nun’. Deer
mine the gas const and density ofthe gos
Solution. Gien
Weight density, w= 16 Nm
Temperature, 1228°C
T=27321=213+25=288k
12 025 Nim (abs) = 029 x 10 Nin
Pes density Is obtained as
(Using relation e
w 16 A
2S = 1.68 kp’, Ans.
ss id
(Using estos (1.9, B=
ET
Nm
a - = sass Am an,
Dar 13
Problem 1.21. A cylinder of 0.6 m’ in volume contins air ot 50°C and 0.3 Nin? absolute
pressure The ai is compresed 100.3 m, Find () pressure inside the elinder assuming isothermal
process and (i) pressure and temperature assuming adiabatic process Tale k= 14.
Solution. Given
Initia volume,
ATEN
20 Fluid Mechanics
Temperature
Pressure
Final volume
© Lothermal
Using equation (1.6),
i Aaiabate process +
Using equation (1.7),
CON alt
30% 10" = 30x10
wo le sn 1082!
791 x 10° Nim? = 0.791 Nim. Ans.
For temperature, using equation (1.5), we get
or Lo Ris also constan]
OS
Problem 1.22. Colculaeıhe pressure exerted by 5 Ag o nitrogen as at a temperature ef 10°C Y
‘he volume 1.0.4 m’. Molecular weight of nitrogen ls 28. Assume, del ga laws are applicable.
Solution. «
Mass of nregen
Temperature,
Volume of nitrogen,
Molecular weight
Using equation (1.9), we have pV = ax Mac RT
Temperature
Pressure
Final volume
© Lothermal
Using equation (1.6),
i Aaiabate process +
Using equation (1.7),
CON alt
30% 10" = 30x10
wo le sn 1082!
791 x 10° Nim? = 0.791 Nim. Ans.
For temperature, using equation (1.5), we get
or Lo Ris also constan]
OS
Problem 1.22. Colculaeıhe pressure exerted by 5 Ag o nitrogen as at a temperature ef 10°C Y
‘he volume 1.0.4 m’. Molecular weight of nitrogen ls 28. Assume, del ga laws are applicable.
Solution. «
Mass of nregen
Temperature,
Volume of nitrogen,
Molecular weight
Using equation (1.9), we have pV = ax Mac RT
Properties of Fluids 21
‘where AR = Universal gas constant = 8314 N
EE maK
sd one Kp-mole = (gas) x Molecular weight (as) 28
au Xm
Atos trogn = DE 2 2959 Mm
E ik
“The gas laws for ige sp = MR, where R= Charctei gas constant
a x04 = 52909 1 283
PA gan. nh
08
> 15 COMPRESS!
TY AND BULK MODULUS
CCompressibility isthe reciprocal ofthe bulk modulus of = >
lis, K which is defined as the ratio of compressive sess z
E | Re »
LE Volume of arena eter i
Presa of gwen ot Y L
Le pe lead pe he om of pu ren
NT Fig
The neopreno =p kata?
Den m volume rv
a
olmo 0
v
vse ann ts volum dsc vi nero rau
Balk moss ES
a 41.10)
Commessiny -4 „u
Relationship between Bulk Modulus (K) and Pressure (p) for a Gas
‘The relaionsbip between bulk modulus of else (K) and pressure for a gas fortwo diferent
processes of compression are as
(For Isothermal Process Equation (1.6) pves te relationship between pressure () an density
(prota gasas
‘where AR = Universal gas constant = 8314 N
EE maK
sd one Kp-mole = (gas) x Molecular weight (as) 28
au Xm
Atos trogn = DE 2 2959 Mm
E ik
“The gas laws for ige sp = MR, where R= Charctei gas constant
a x04 = 52909 1 283
PA gan. nh
08
> 15 COMPRESS!
TY AND BULK MODULUS
CCompressibility isthe reciprocal ofthe bulk modulus of = >
lis, K which is defined as the ratio of compressive sess z
E | Re »
LE Volume of arena eter i
Presa of gwen ot Y L
Le pe lead pe he om of pu ren
NT Fig
The neopreno =p kata?
Den m volume rv
a
olmo 0
v
vse ann ts volum dsc vi nero rau
Balk moss ES
a 41.10)
Commessiny -4 „u
Relationship between Bulk Modulus (K) and Pressure (p) for a Gas
‘The relaionsbip between bulk modulus of else (K) and pressure for a gas fortwo diferent
processes of compression are as
(For Isothermal Process Equation (1.6) pves te relationship between pressure () an density
(prota gasas
22 Fluid Mechanics
1
o PY = Const {eves}
°
Pitre ns uan, wege and Y bth ate vai)
En
MOIN Er où pv Va oo pe
PAV Vpn 0 oc ptt == dp or p=
Substutng this vale in equation (110), we get
Kap am
(6 For Adiabatic Process. Using equation (17) fr adiabaie process
Constant or p = Constan
Diterenaing, we gt pay) à dp) =O
or PXRCR dvs ¥ dp a0
oe Play + dp =0 {Canceling obo iss}
vo
ar par dp or p= Nr
nes fom equation (1.10, we Rave
Kup am
where K = Bul
‘modulus and £ = Ratio of specific bess,
Problem 1.23, Deiemine the bulk modulus of elasticity of igi ifthe pressure ofthe guid is
increase rom 70 Nm’ 10 130 Nem The volume ofthe liquid decreases by 0.15 percent
Solution, Gen
nial pressure = 170 New’
Fina pressure 130 Ne?
2 dp= Increase in presure = 130-70
Dear in volume 015%
. EAN
vo
Bulk modulos Ks ven by equation (1-0) as
410" Nem? Ans
Problem 1.24 Whore bulk modulus of elasticity ofa liquid which i compressed i a elinder
rom a volume of 00125 m’ at 80 Nien? pressure to a volume of 0.0124 mat 150 Nem pressure ?
Solution. Give
Initial volume, ¥=00125 mt
inal volume 200128 m?
Decrease in volume, Y 0125-0124 = 0001 m?
1
o PY = Const {eves}
°
Pitre ns uan, wege and Y bth ate vai)
En
MOIN Er où pv Va oo pe
PAV Vpn 0 oc ptt == dp or p=
Substutng this vale in equation (110), we get
Kap am
(6 For Adiabatic Process. Using equation (17) fr adiabaie process
Constant or p = Constan
Diterenaing, we gt pay) à dp) =O
or PXRCR dvs ¥ dp a0
oe Play + dp =0 {Canceling obo iss}
vo
ar par dp or p= Nr
nes fom equation (1.10, we Rave
Kup am
where K = Bul
‘modulus and £ = Ratio of specific bess,
Problem 1.23, Deiemine the bulk modulus of elasticity of igi ifthe pressure ofthe guid is
increase rom 70 Nm’ 10 130 Nem The volume ofthe liquid decreases by 0.15 percent
Solution, Gen
nial pressure = 170 New’
Fina pressure 130 Ne?
2 dp= Increase in presure = 130-70
Dear in volume 015%
. EAN
vo
Bulk modulos Ks ven by equation (1-0) as
410" Nem? Ans
Problem 1.24 Whore bulk modulus of elasticity ofa liquid which i compressed i a elinder
rom a volume of 00125 m’ at 80 Nien? pressure to a volume of 0.0124 mat 150 Nem pressure ?
Solution. Give
Initial volume, ¥=00125 mt
inal volume 200128 m?
Decrease in volume, Y 0125-0124 = 0001 m?
Properties of Fluids 23]
. a ot
V os
ina pre = 80 Neo?
Final prone = 150 Nem?
Increase in pressure, dp = (150 - 80) = 70 Nic
Bulk modulos i iven by equation (1.10) as
TO 210x128 Nem
ra © AUS NE
von
= 475 x 10! Nam Ans.
» 1.6 SURFACE TENSION AND CAPILLARITY
Surface tension is defined a the tensile force acting onthe surface o a guid in contact wi as
or on the surface between two immiscible Uguids such thatthe contact surface bebaves ike a
membrane under tension. The magnitude of this foros er unit ent ofthe fee surface wil have the
‘same value asthe surco eneray per un ra. It is denoted by Greek letter (ale sima). In MKS
‘ls, is expressed as Kin while a Stunts as Ni.
The phenomenon of surface tension is explained by
Tig. 1.10. Consider three molecules A, B, C ofa liquid in a
mas of liquid. The molecule A is aacte in all dictions
quay bythe surtounding molecules ofthe liquid, Thus the
resultant force acting on the molecule A is zero, But the
molscule B, which is stuned near the fee surface, is acted
upon by upward and dowaward forces which are unbalanced.
‘Thos a oot resultant force on molecule 2 is acting In the
downward direcion. The molecule C, situated on the tree
Surface of qui, does experience resultant downward fee.
All the molecules on the tee surface experience à downward
forge Thus the fre surface ofthe qui acts Uke a very thin fm under tension ofthe surface of he
liquid act as though isan elastic membrane under tension
1.6.1. Surface Tension on Liquid Droplet. Consider a smal spbetcal droplet of a iqud of
aus + On the entire surface ofthe droplet, the tensile force due wo surface teasion wil be atin.
Let = Surface tension ofthe gid
= Pressure intensity inside the droplet (in xcess of the outside pressure intensity)
44 = Dis. of droit
Let the dople is ut into two halves. The forces acting om ons hal (ay et Ra) wil Be
(tensile foros due surface tension acting around the circumference of th cu portion as shown
in Fig. 1.11 0) and this sequal to
6x Chroumforence
=0xxd
Fig. 110. Surface tension
. a ot
V os
ina pre = 80 Neo?
Final prone = 150 Nem?
Increase in pressure, dp = (150 - 80) = 70 Nic
Bulk modulos i iven by equation (1.10) as
TO 210x128 Nem
ra © AUS NE
von
= 475 x 10! Nam Ans.
» 1.6 SURFACE TENSION AND CAPILLARITY
Surface tension is defined a the tensile force acting onthe surface o a guid in contact wi as
or on the surface between two immiscible Uguids such thatthe contact surface bebaves ike a
membrane under tension. The magnitude of this foros er unit ent ofthe fee surface wil have the
‘same value asthe surco eneray per un ra. It is denoted by Greek letter (ale sima). In MKS
‘ls, is expressed as Kin while a Stunts as Ni.
The phenomenon of surface tension is explained by
Tig. 1.10. Consider three molecules A, B, C ofa liquid in a
mas of liquid. The molecule A is aacte in all dictions
quay bythe surtounding molecules ofthe liquid, Thus the
resultant force acting on the molecule A is zero, But the
molscule B, which is stuned near the fee surface, is acted
upon by upward and dowaward forces which are unbalanced.
‘Thos a oot resultant force on molecule 2 is acting In the
downward direcion. The molecule C, situated on the tree
Surface of qui, does experience resultant downward fee.
All the molecules on the tee surface experience à downward
forge Thus the fre surface ofthe qui acts Uke a very thin fm under tension ofthe surface of he
liquid act as though isan elastic membrane under tension
1.6.1. Surface Tension on Liquid Droplet. Consider a smal spbetcal droplet of a iqud of
aus + On the entire surface ofthe droplet, the tensile force due wo surface teasion wil be atin.
Let = Surface tension ofthe gid
= Pressure intensity inside the droplet (in xcess of the outside pressure intensity)
44 = Dis. of droit
Let the dople is ut into two halves. The forces acting om ons hal (ay et Ra) wil Be
(tensile foros due surface tension acting around the circumference of th cu portion as shown
in Fig. 1.11 0) and this sequal to
6x Chroumforence
=0xxd
Fig. 110. Surface tension
24. Fluid Mechanics
(pressure toe om teuren Ea? =p das sown in
Tig. LAN (0. These two forces will he equal and opposte
under cquliium conditions, Le. > A
pee Peoxnd
oxnd 40 à
or = as
Leser
4 (PRESSURE FORCES.
[Equation (1.14) shows that withthe decrease of diameter Fig. LAN. Force on droplet.
lof the drole, pressure tensity Inside the droplet increases.
1.6.2. Surface Tension on a Hollow Bubble. A hollow bubble ike a soap bubble ai has two
faces in contact with af, one inde and other outside. Thus two surfaces ate subiste to surface
tension ln such ess, we have
ox gars
ELE as
np 4
4
on on a Liquid Jet. Consider guid jet of dlameter “a and length Las
a >
1.6.3 Surface Tens
sown in Fig. 112
Let p = Pressure intensity inside the quid jet above the outside pressure
{0 = Surface tension ofthe bavi +
(Consider the equilibrium of the semi Jt, we have |
1x area of semi jet
Force dus 10 pressure
pxLxd ‘ °
Force duc 10 surface tension 2 6 2
Equating the frees, we have
Dx Lxd =Ox2L |
ox,
xd
is o
ig. 112 Forces on liqid jet
Problem 1.25 Ihe surface tension of water in contact with ar 20°C 1 0.0725 Nin. The pressure
Inside a droplet of water ts to be 0.02 Nem? greater than the ouside pressure. Calcuot the diameter
ofthe drople of wer
Solution. Given
Surface tension 000725 Nin
Presse intensity, in excess of outside presune I.
= 002 Nam?
oor x ot A.
Let = di. ofthe droplet
(pressure toe om teuren Ea? =p das sown in
Tig. LAN (0. These two forces will he equal and opposte
under cquliium conditions, Le. > A
pee Peoxnd
oxnd 40 à
or = as
Leser
4 (PRESSURE FORCES.
[Equation (1.14) shows that withthe decrease of diameter Fig. LAN. Force on droplet.
lof the drole, pressure tensity Inside the droplet increases.
1.6.2. Surface Tension on a Hollow Bubble. A hollow bubble ike a soap bubble ai has two
faces in contact with af, one inde and other outside. Thus two surfaces ate subiste to surface
tension ln such ess, we have
ox gars
ELE as
np 4
4
on on a Liquid Jet. Consider guid jet of dlameter “a and length Las
a >
1.6.3 Surface Tens
sown in Fig. 112
Let p = Pressure intensity inside the quid jet above the outside pressure
{0 = Surface tension ofthe bavi +
(Consider the equilibrium of the semi Jt, we have |
1x area of semi jet
Force dus 10 pressure
pxLxd ‘ °
Force duc 10 surface tension 2 6 2
Equating the frees, we have
Dx Lxd =Ox2L |
ox,
xd
is o
ig. 112 Forces on liqid jet
Problem 1.25 Ihe surface tension of water in contact with ar 20°C 1 0.0725 Nin. The pressure
Inside a droplet of water ts to be 0.02 Nem? greater than the ouside pressure. Calcuot the diameter
ofthe drople of wer
Solution. Given
Surface tension 000725 Nin
Presse intensity, in excess of outside presune I.
= 002 Nam?
oor x ot A.
Let = di. ofthe droplet
Properties of Fluids 25]
so sxoons
Using equation (1.18, we gap = $F or 0.00 101. 4200728
1 cquation (1.19, we gp = 10 ss
= = OS à 00145 m = 00145 x 1000 = 1.45 mm. Ans.
TS
Problem 1.26. Find the surface tension in a soap bubble of 40 mm diameter when the inside
pressure is 25 Nin? above atmospheric pressure
Solution. Given
Dia. of bubb = 40 mm = 40% 10 'm
Pressure in exces of outside, = 2.5 Nam?
Fora sap bubble, using equation (.15), we gt
$8 ee asa 8h,
4 mxo
2514010"
210 im = 0.0128 Ni, Ans.
Problem 1.27 The pressure ouside the droplet of water of diameter 0:08 mix 10.82 Ne? (ut
mospheric pressure). Calewate the pressure within the drople if surfoce tension is given as
0.0725 Nim of water
Solution. Gien
Dia. of do 42004 mm = 01x10 ’m
Pressure outside the droplet = 10.32 Ne? = 10.32 x 10" Nim?
Surface endo, = 00728 Nim
The preso ide dope, in exes of outside pressure given y union 114)
A 49 _ 4200725 pu nn = SON. 28 Nem
PET ¡0
Prose inside te dpi =p + Presse ou te dp
0.725 + 10.32 = 11.045 N/em?. Ans.
164 Capiliaity. Copa is tn a à puma oc al of gid sra in a
sal tubo relative to the adjacent general love f quid whom th tube is hd vertical in he liquid.
The rise of liquid surface de known as Capilar rise while th Fl oF he quid surface I known 36
(apa depression. vs expressed in terms of m or mm of gui. Is m
Value depends upon the specific weiph of the quid, diameter of the x
tube and suce tension ofthe gui
Expression for Capillary Rise, Consier a glass tbe of small
diameter "opened at otk ends a is inserted in liquid, sa water
“The qu wil ise in the tube above the level ofthe gui,
Let h= height o the quid inte tb. Under a state equilibrio,
‘the Weight of gui of heights balanced by the force atthe surco of
‘te quid in the be, Butte force atthe surface ofthe quid ln the
Aube is dueto sua tension Fig. 113. Capillary rise
Let = Surface tension liquid
Angle of contact between Liquid and glass tute.
“The weight of ui of help nt tube = (Area of tube) p X €
so sxoons
Using equation (1.18, we gap = $F or 0.00 101. 4200728
1 cquation (1.19, we gp = 10 ss
= = OS à 00145 m = 00145 x 1000 = 1.45 mm. Ans.
TS
Problem 1.26. Find the surface tension in a soap bubble of 40 mm diameter when the inside
pressure is 25 Nin? above atmospheric pressure
Solution. Given
Dia. of bubb = 40 mm = 40% 10 'm
Pressure in exces of outside, = 2.5 Nam?
Fora sap bubble, using equation (.15), we gt
$8 ee asa 8h,
4 mxo
2514010"
210 im = 0.0128 Ni, Ans.
Problem 1.27 The pressure ouside the droplet of water of diameter 0:08 mix 10.82 Ne? (ut
mospheric pressure). Calewate the pressure within the drople if surfoce tension is given as
0.0725 Nim of water
Solution. Gien
Dia. of do 42004 mm = 01x10 ’m
Pressure outside the droplet = 10.32 Ne? = 10.32 x 10" Nim?
Surface endo, = 00728 Nim
The preso ide dope, in exes of outside pressure given y union 114)
A 49 _ 4200725 pu nn = SON. 28 Nem
PET ¡0
Prose inside te dpi =p + Presse ou te dp
0.725 + 10.32 = 11.045 N/em?. Ans.
164 Capiliaity. Copa is tn a à puma oc al of gid sra in a
sal tubo relative to the adjacent general love f quid whom th tube is hd vertical in he liquid.
The rise of liquid surface de known as Capilar rise while th Fl oF he quid surface I known 36
(apa depression. vs expressed in terms of m or mm of gui. Is m
Value depends upon the specific weiph of the quid, diameter of the x
tube and suce tension ofthe gui
Expression for Capillary Rise, Consier a glass tbe of small
diameter "opened at otk ends a is inserted in liquid, sa water
“The qu wil ise in the tube above the level ofthe gui,
Let h= height o the quid inte tb. Under a state equilibrio,
‘the Weight of gui of heights balanced by the force atthe surco of
‘te quid in the be, Butte force atthe surface ofthe quid ln the
Aube is dueto sua tension Fig. 113. Capillary rise
Let = Surface tension liquid
Angle of contact between Liquid and glass tute.
“The weight of ui of help nt tube = (Area of tube) p X €
26 Fluid Mechanics
Exp am
where p= Density of iid
Vertical componen ofthe surface tensile free
=(0 x Circumference) x cos 0
2 6 xr4x 0 as
For equiliium, oquating (1.17) and (1:18), we get
EP chxpxg=oxndxcos
a ° 8
xrdxe D _ 4000
prexd
or » 1.19)
CA
ioe.
‘The vale of between water and clean glass tbe is spprosimatly equal o zero and hence cos Dis
qual tunity. Then ie of waters given by
a
Prexd
Expression for Capillry Fall. Ihe glass tube is ded in mercury, de level of mercury inthe abe
Al be lower than the general level ofthe outside guid a show In Fi. LA.
Let h = Height of depression la nb.
‘Then in equilib, two forces are acting onthe mercury Inside the tube. First one is due to surface
tension ating in the downvad direction and I equal 0 6 Rad x co 0
Second fore is dus to hydrostatic force acting upward and i equal to ment of pressure ata
pth x Area
02
nr Eda pexnx ds p= pet
L 7 te p= pan)
Euaing the two, we pet
oxrdiccosd=peix E
x À
a nz test aa e
Ped
Vale o for mercury and glas tbe is 128 Fra
Fe.
Problem 1.28 Calcule the capillary rise in glass ube of 2.5 mm diameter when immersed
serial a) water and (b) mercury. Take surface tensions 6 0.0725 Nm for water and 0= 0.52 Nin
Jos mercury in contac wth air. The Spec gray for mercury is given as 13.6 and angle of comact
130%
Solution. Given
Dia. of ube, a
Surface tension, tor water =
for mercury
Sp. ge of mercury
5 mm = 25: 10m
0725 Nim.
Exp am
where p= Density of iid
Vertical componen ofthe surface tensile free
=(0 x Circumference) x cos 0
2 6 xr4x 0 as
For equiliium, oquating (1.17) and (1:18), we get
EP chxpxg=oxndxcos
a ° 8
xrdxe D _ 4000
prexd
or » 1.19)
CA
ioe.
‘The vale of between water and clean glass tbe is spprosimatly equal o zero and hence cos Dis
qual tunity. Then ie of waters given by
a
Prexd
Expression for Capillry Fall. Ihe glass tube is ded in mercury, de level of mercury inthe abe
Al be lower than the general level ofthe outside guid a show In Fi. LA.
Let h = Height of depression la nb.
‘Then in equilib, two forces are acting onthe mercury Inside the tube. First one is due to surface
tension ating in the downvad direction and I equal 0 6 Rad x co 0
Second fore is dus to hydrostatic force acting upward and i equal to ment of pressure ata
pth x Area
02
nr Eda pexnx ds p= pet
L 7 te p= pan)
Euaing the two, we pet
oxrdiccosd=peix E
x À
a nz test aa e
Ped
Vale o for mercury and glas tbe is 128 Fra
Fe.
Problem 1.28 Calcule the capillary rise in glass ube of 2.5 mm diameter when immersed
serial a) water and (b) mercury. Take surface tensions 6 0.0725 Nm for water and 0= 0.52 Nin
Jos mercury in contac wth air. The Spec gray for mercury is given as 13.6 and angle of comact
130%
Solution. Given
Dia. of ube, a
Surface tension, tor water =
for mercury
Sp. ge of mercury
5 mm = 25: 10m
0725 Nim.
Properties of Fuids 27]
2 Density 3.6 x 1000 ker,
da Capillary rise for water (@ 0°)
do 400725
Using equation (120), we get n= 40. AX
ing equaon (120), We BN ETE NECE
Em = LIB em. Ans,
(6) For mercury
Angle of comet etwcen mercury and glas ahs, 8 = 130%
400080 AXOS2 Kenn 130°
Pra xd” TIO x98! x25 ¥10"
= 004 m = 04 cm. Ans.
‘The negativo sign indicate he capilar depresion.
Using equation (121), we got
Problem 1.29 Coleulete she capilar fact in milimetres ina glas tbe of mm diameter, hen
immersed in (i) vtr, ond (i) mercury. The temperamre ofthe quid e 20°C and the vales of the
surface tension of water and mercury at 20°C i contact with ir are 0.073575 Nim and 0.51 Nim
spatial. The angle o contact for waters eva and thal for mercury is 130" Take density of water
(20°C as equal to 98 Kg
Solution. Given
Dia. of tubs 44 mm=4x 10m
Te capilar efect (apilar is or depression) given by equation (120) as
cost
he
rend
where @= surface tension in N
= angle of contact, and = density
(© Capillary effet for water
(6 = 0073575 Nim, 8 = 0°
(= 998 Lgim at 20°C
AK OOT3STSX cos"
a n= PEOR 5119 Sm = 7.81 mm. Ans.
EISEN
(iD Capillary effet for mercury
6 =051 Nim, 8= 130° and
= sp. BF 1000 = 136% 1000 = 13600 Keim!
sa a PISO 10m = 2.46 mum, Ans.
13600981 4107
“The negative sign indicates the capllary depression,
Problem 1.30 The capillary rise inthe glass tbe is wot to execed 0.2 mm of water. Determine its
minimum sie, given ha suce tension for water in contact with air = 0.0725 Nin.
Solution. Gien
CCopitory ic, 202mm =02% 10m
Surface tension 6 = 00725 Nim
2 Density 3.6 x 1000 ker,
da Capillary rise for water (@ 0°)
do 400725
Using equation (120), we get n= 40. AX
ing equaon (120), We BN ETE NECE
Em = LIB em. Ans,
(6) For mercury
Angle of comet etwcen mercury and glas ahs, 8 = 130%
400080 AXOS2 Kenn 130°
Pra xd” TIO x98! x25 ¥10"
= 004 m = 04 cm. Ans.
‘The negativo sign indicate he capilar depresion.
Using equation (121), we got
Problem 1.29 Coleulete she capilar fact in milimetres ina glas tbe of mm diameter, hen
immersed in (i) vtr, ond (i) mercury. The temperamre ofthe quid e 20°C and the vales of the
surface tension of water and mercury at 20°C i contact with ir are 0.073575 Nim and 0.51 Nim
spatial. The angle o contact for waters eva and thal for mercury is 130" Take density of water
(20°C as equal to 98 Kg
Solution. Given
Dia. of tubs 44 mm=4x 10m
Te capilar efect (apilar is or depression) given by equation (120) as
cost
he
rend
where @= surface tension in N
= angle of contact, and = density
(© Capillary effet for water
(6 = 0073575 Nim, 8 = 0°
(= 998 Lgim at 20°C
AK OOT3STSX cos"
a n= PEOR 5119 Sm = 7.81 mm. Ans.
EISEN
(iD Capillary effet for mercury
6 =051 Nim, 8= 130° and
= sp. BF 1000 = 136% 1000 = 13600 Keim!
sa a PISO 10m = 2.46 mum, Ans.
13600981 4107
“The negative sign indicates the capllary depression,
Problem 1.30 The capillary rise inthe glass tbe is wot to execed 0.2 mm of water. Determine its
minimum sie, given ha suce tension for water in contact with air = 0.0725 Nin.
Solution. Gien
CCopitory ic, 202mm =02% 10m
Surface tension 6 = 00725 Nim
28 Fluid Mechanics
Let. oF tbe
‘The angle @ for water
Density (p) for water
Using equation (1.20), we get
= 1000 kya?
so 4 _4xa0rs
TEO TE]
4x00725
DOTE]
“Thus minimum diameter ofthe tue should o 148 em.
Problem 1.91. Find out the minimum sie of glass be tha canbe used 10 measure water level
‘the capillary rise the tbe a 1 e restricted 10 2 mm. Consider surco tension of water in contact
with air as 0.073575 Nm.
»
148 m = 148 am. Ans.
Solution. Gien
Capilary ise. 0 mm = 20 x 10m
Surface tension = 0.073875 Nim
Leta. of tubs
“The angle @ for water y
The density or water, 000 Mg
Using equation (1.20), we get
AX0N73S75.
or 20% 10 3. LISIS
1000 x 981 xd
+ = 0015 m= 15 cm. Ans
de os.
IWoxosıx 2%
‘Thus minimum diameter ofthe tbe shouldbe LS cm.
Problem 1.82 An ol of sscosty $ poise ls used for lubrication between shaf and sleeve. The
diameter of the shaft 0. m and rotates at 200 rpm. Caleulate the power os in ol for a sleeve
enge of 100mm. The thickness of oil fi i 1.0 mme
Solution. Given
Viscosity,
Dia. of soa,
Speed of sha,
Sleeve length,
“Thickness of oi fen,
‘Tangential velocity of shat a= 5235 mis
Using the rtation, tu
Let. oF tbe
‘The angle @ for water
Density (p) for water
Using equation (1.20), we get
= 1000 kya?
so 4 _4xa0rs
TEO TE]
4x00725
DOTE]
“Thus minimum diameter ofthe tue should o 148 em.
Problem 1.91. Find out the minimum sie of glass be tha canbe used 10 measure water level
‘the capillary rise the tbe a 1 e restricted 10 2 mm. Consider surco tension of water in contact
with air as 0.073575 Nm.
»
148 m = 148 am. Ans.
Solution. Gien
Capilary ise. 0 mm = 20 x 10m
Surface tension = 0.073875 Nim
Leta. of tubs
“The angle @ for water y
The density or water, 000 Mg
Using equation (1.20), we get
AX0N73S75.
or 20% 10 3. LISIS
1000 x 981 xd
+ = 0015 m= 15 cm. Ans
de os.
IWoxosıx 2%
‘Thus minimum diameter ofthe tbe shouldbe LS cm.
Problem 1.82 An ol of sscosty $ poise ls used for lubrication between shaf and sleeve. The
diameter of the shaft 0. m and rotates at 200 rpm. Caleulate the power os in ol for a sleeve
enge of 100mm. The thickness of oil fi i 1.0 mme
Solution. Given
Viscosity,
Dia. of soa,
Speed of sha,
Sleeve length,
“Thickness of oi fen,
‘Tangential velocity of shat a= 5235 mis
Using the rtation, tu
Properties of Fluids 29,
where, de = Change of velocity = u = 235 mis
sty = Change of distance = 1= 1x 10 a
055235
ET
This isthe shear seres om the shaft
À Shea force om the shaft, P= Shoat at 4 Arca = 26175 RD XL Cs Area 2 RD XD)
= 26175 x05 X01 = 41095 N
26175 Nin?
E Power? lost = Toco Was = 7x ZEN w
a
“
» 1.7. VAPOUR PRESSURE AND CAVITATION
A change fom the liquid state 0 the gaseous tate is known as vaporization. The vaporization
(hich depends upon te prevailing pressure and temperature condition) occus because of cominuous
‘escaping of the molecules throuph the re quid surface
Consider a liquid (ay water) which is confinc in à closed vessel Let the temperature of liquid is
20°C and pressure is atmospheric, This quid will vaporise a 100-C. When vaporization takes place,
the molecu escapes from the fre surface of the qui. These vapour molecules get accumulated in
te space between the fe quid surface and top of the vessel. These aecumolated vapours exert a
presane on the Hg surface, This pressure is nown as vapour pressure ofthe qui or this i the
Pressure at whieh the lige is convert ito vapour
‘Again consider te same liquid a 20°C at amiospere pressure I the closed vessel If the pressure
hove the quid surface I reduced hy some means, the hong temperature wil aso reduce. I the
pressure is reduced 10 suc an extent tat becomes equal to or ess than the vapor pressure, the
Poing ofthe liquid wil star, though the temperature of the liquid Is 20°C. Thus a quid may boi
ven at ondinary temperature, he pressure shove the Liquid surfe is reduced so 15 o Be equal or
Jess ihn the vapour pressure of the liquid tht temperature
‘Now considera Mowing liquid in system. hc pressure at any pot in his Mowing quid hecomes.
equal to or less than the Vapour pressure, the vapoezation of the guid stats. The bubbles of these
vapours are cani hy the owing liquid into the region of high pressure where the collapse, giving
is to high impact pressure, The pressure developed by the collapsing bubbles isso high thatthe
‘mata fom the adjoining houndaris ets ere and eavitics ar formed on them. This phonomenon
Js known as enviados.
Hence the cavitation ls te phenomenon of formation of vapour bubbles of a flowing ul in a
gion whee the pressure of he quid falls low the vapour pressure and sudden collapsing of these
‘vapour bubbles in region of higher pressure, When the vapour bubbles collapse a very high pressure
creta, The metal surfaces, above which the gud is lowing, is subjected 1 hes igh pressure,
‘which cause pitin action onthe surface, Thus cavities ue formed on the metallic surface and ence
le name is coito.
= power leu ofS. Une ao IT watson ZT yw. The angular etc =
@ wo = ”
where, de = Change of velocity = u = 235 mis
sty = Change of distance = 1= 1x 10 a
055235
ET
This isthe shear seres om the shaft
À Shea force om the shaft, P= Shoat at 4 Arca = 26175 RD XL Cs Area 2 RD XD)
= 26175 x05 X01 = 41095 N
26175 Nin?
E Power? lost = Toco Was = 7x ZEN w
a
“
» 1.7. VAPOUR PRESSURE AND CAVITATION
A change fom the liquid state 0 the gaseous tate is known as vaporization. The vaporization
(hich depends upon te prevailing pressure and temperature condition) occus because of cominuous
‘escaping of the molecules throuph the re quid surface
Consider a liquid (ay water) which is confinc in à closed vessel Let the temperature of liquid is
20°C and pressure is atmospheric, This quid will vaporise a 100-C. When vaporization takes place,
the molecu escapes from the fre surface of the qui. These vapour molecules get accumulated in
te space between the fe quid surface and top of the vessel. These aecumolated vapours exert a
presane on the Hg surface, This pressure is nown as vapour pressure ofthe qui or this i the
Pressure at whieh the lige is convert ito vapour
‘Again consider te same liquid a 20°C at amiospere pressure I the closed vessel If the pressure
hove the quid surface I reduced hy some means, the hong temperature wil aso reduce. I the
pressure is reduced 10 suc an extent tat becomes equal to or ess than the vapor pressure, the
Poing ofthe liquid wil star, though the temperature of the liquid Is 20°C. Thus a quid may boi
ven at ondinary temperature, he pressure shove the Liquid surfe is reduced so 15 o Be equal or
Jess ihn the vapour pressure of the liquid tht temperature
‘Now considera Mowing liquid in system. hc pressure at any pot in his Mowing quid hecomes.
equal to or less than the Vapour pressure, the vapoezation of the guid stats. The bubbles of these
vapours are cani hy the owing liquid into the region of high pressure where the collapse, giving
is to high impact pressure, The pressure developed by the collapsing bubbles isso high thatthe
‘mata fom the adjoining houndaris ets ere and eavitics ar formed on them. This phonomenon
Js known as enviados.
Hence the cavitation ls te phenomenon of formation of vapour bubbles of a flowing ul in a
gion whee the pressure of he quid falls low the vapour pressure and sudden collapsing of these
‘vapour bubbles in region of higher pressure, When the vapour bubbles collapse a very high pressure
creta, The metal surfaces, above which the gud is lowing, is subjected 1 hes igh pressure,
‘which cause pitin action onthe surface, Thus cavities ue formed on the metallic surface and ence
le name is coito.
= power leu ofS. Une ao IT watson ZT yw. The angular etc =
@ wo = ”
Fluid Mechanics
HIGHLIGHTS
“The weight os or soci weight is equal o eight per unt volume, It is also equal,
Srecif volume isthe rcprcal of mass density
es pp oe vey at am =
Kinematic visa is given by v= À
ovicginen bya À
Poise and stokes arth ni visos an Kinematic wsosy respectively
"o conven th eit of vey from pois o MRS un pos ble ve y 8. dt conver
pote to Stun, te pis shoud Ye divided by 10 Sut ol vcs Nono Pa wee Nm
ora poet as, he equa of tates À = RT
po «1 £
ere R = es constant and for air 29
Em 7 ak
pS
For is pis 2 » Con wes rata pros, e = const
Del modulus o lacy fe given as Km
Compesaily the cipal of tak modus of ec ora E
pra (9) einen he isd and oui of tigi op igen p= ZT
so
tocata, p= 82
Foraligudiet, p= 29.
ou et >
A soo
Capilla iso Al fa iui given by = ELO,
“Te val of or water taken equal er and or meaty sua 128°
(A) THEORETICAL PROBLEMS
Define te fooing id properties
Dersiy. weight deny, specie volume and specie gravy of Mid.
Piernas two =) Ligue and gases, (1) Rel ld und del Mi, (i) Specie weight and
What iste dicen between dynamic veosy and kinematic ico ? Sate Weir vais of
HIGHLIGHTS
“The weight os or soci weight is equal o eight per unt volume, It is also equal,
Srecif volume isthe rcprcal of mass density
es pp oe vey at am =
Kinematic visa is given by v= À
ovicginen bya À
Poise and stokes arth ni visos an Kinematic wsosy respectively
"o conven th eit of vey from pois o MRS un pos ble ve y 8. dt conver
pote to Stun, te pis shoud Ye divided by 10 Sut ol vcs Nono Pa wee Nm
ora poet as, he equa of tates À = RT
po «1 £
ere R = es constant and for air 29
Em 7 ak
pS
For is pis 2 » Con wes rata pros, e = const
Del modulus o lacy fe given as Km
Compesaily the cipal of tak modus of ec ora E
pra (9) einen he isd and oui of tigi op igen p= ZT
so
tocata, p= 82
Foraligudiet, p= 29.
ou et >
A soo
Capilla iso Al fa iui given by = ELO,
“Te val of or water taken equal er and or meaty sua 128°
(A) THEORETICAL PROBLEMS
Define te fooing id properties
Dersiy. weight deny, specie volume and specie gravy of Mid.
Piernas two =) Ligue and gases, (1) Rel ld und del Mi, (i) Specie weight and
What iste dicen between dynamic veosy and kinematic ico ? Sate Weir vais of
Properties of Fluids 31
A. Explin em : 4) Dyaae vicos, and (i) Kisema visos Give toe meins
5. Sue the Newtons bw of viscosity and give examples of is aplication.
4% Emincite Newton aw of wc. Explain the portance of vod ie id moon. What the
fect of temperate co of ater and that oa?
1 Deine Newnan and Non Newton i
1 Wa do you understand by ns) Ista poss) Ada proces, andi) Univers as
3. Deine compres. Prove a compresily fra pret gs undergoing thermal comprcson à
wt for pet as din cul comparon s
10. Define sra tenso, Prove hate relationship been sure een and res nce done of
A1. spain ptenoeton of capillary. ban an apesin for caily ise 0 a igi
12 a) Dis between Leal Huis and ral is Ellos pora o compress la ld
Bow.
(0) Define ih ems: demi, sc volume, pecan, vaca presu, compressible and
compense Tus (GP. Vania, Bhopal $ 2002)
13, Define and opio Newon' a of viscoso
1, Conver 1 em dynam visos in pos
18. Why doc Be vice a gs creates ithe nes perle while Ut ys dre
sith increas in emperate?
16, (a) How does vscosy 0 a id vary wih empero 7
(0) Che oups where sere lemon effets Psy pone wie, UNT. Hyderabad 5 2002)
17, {Develo the expression forte relation Between gauge presse side a droplet of quid a the
suce tenon
(4 Esp he towing
Newtonian and Non Newtonian Mids, vapcar presi, and compres
RGP. Bhopal 5 2001)
(8) NUMERICAL PROBLEMS
“AD Ni 978 Sp 0978)
esky ten oo over a pe gen by w= y =P, e pl
same dynamic sony a pie agp Universi) Vans. 089 Ni
A ae 0095 mu dit for aed pe, move a 50 em ad requires foc of 1471 No
rain is speed Detemin he fi isa Between e pes inthe poise. (Ans: 7.357 10 "|
4 Determine the testy of Star of ano having vos) 12 ps a suse fr ation In he
toutes 200 cp Tan. 125.56 Nit
Two plaesare place a dicane of 0.15 mm apar. The ower ple ne while ne upper pie having
sara ares 1 pal 03 mí. Fin he fr nd power require o mais ped Y e
Aid sparing ther having sci 1.5 pie Ans 300,858 WI
6 Ano lao tcs 1.5 mun ls so for lication between a qua pl 09 2009 mand at
Ihe plane wi a uno veloc 002 ev. Fil the nai visos of he oi. (Ans 1242 pole
A. Explin em : 4) Dyaae vicos, and (i) Kisema visos Give toe meins
5. Sue the Newtons bw of viscosity and give examples of is aplication.
4% Emincite Newton aw of wc. Explain the portance of vod ie id moon. What the
fect of temperate co of ater and that oa?
1 Deine Newnan and Non Newton i
1 Wa do you understand by ns) Ista poss) Ada proces, andi) Univers as
3. Deine compres. Prove a compresily fra pret gs undergoing thermal comprcson à
wt for pet as din cul comparon s
10. Define sra tenso, Prove hate relationship been sure een and res nce done of
A1. spain ptenoeton of capillary. ban an apesin for caily ise 0 a igi
12 a) Dis between Leal Huis and ral is Ellos pora o compress la ld
Bow.
(0) Define ih ems: demi, sc volume, pecan, vaca presu, compressible and
compense Tus (GP. Vania, Bhopal $ 2002)
13, Define and opio Newon' a of viscoso
1, Conver 1 em dynam visos in pos
18. Why doc Be vice a gs creates ithe nes perle while Ut ys dre
sith increas in emperate?
16, (a) How does vscosy 0 a id vary wih empero 7
(0) Che oups where sere lemon effets Psy pone wie, UNT. Hyderabad 5 2002)
17, {Develo the expression forte relation Between gauge presse side a droplet of quid a the
suce tenon
(4 Esp he towing
Newtonian and Non Newtonian Mids, vapcar presi, and compres
RGP. Bhopal 5 2001)
(8) NUMERICAL PROBLEMS
“AD Ni 978 Sp 0978)
esky ten oo over a pe gen by w= y =P, e pl
same dynamic sony a pie agp Universi) Vans. 089 Ni
A ae 0095 mu dit for aed pe, move a 50 em ad requires foc of 1471 No
rain is speed Detemin he fi isa Between e pes inthe poise. (Ans: 7.357 10 "|
4 Determine the testy of Star of ano having vos) 12 ps a suse fr ation In he
toutes 200 cp Tan. 125.56 Nit
Two plaesare place a dicane of 0.15 mm apar. The ower ple ne while ne upper pie having
sara ares 1 pal 03 mí. Fin he fr nd power require o mais ped Y e
Aid sparing ther having sci 1.5 pie Ans 300,858 WI
6 Ano lao tcs 1.5 mun ls so for lication between a qua pl 09 2009 mand at
Ihe plane wi a uno veloc 002 ev. Fil the nai visos of he oi. (Ans 1242 pole
32 Fluid Mechanics
“The mas deny of sl 2684 hg er cb mee ad ico voy 6.30% 10 “aqua ms
pe end. Calculo sea ses a pot. ams 02 Nm]
8. Find Knemate soy aan having density DK" when at cer point ince he shear
cia AR [men toa
a ta i a iri
us
Toast Sees mst en eae nn
BE tte Sn an a
13. The seo due over plate gen by = À 3 in which she ve nme na
a om at pars nase SRE mn mate
ne
0 de ae RE
ue
GE pd he pil ne
pede pe criada
D pes seen
RSR the an nn nee
a mp nat e y e a ste)
pp Nit moe pi
D Rene ee one in ene
ias
A ee eee
loc weight of stwgea à 28 x A (ans 978 New
19. The pres 0 a guide rat from 60 Mie? 100 Nm and volume decreas by 02 pe cen,
Determine he bal modas a elt. ane 2 10" New
20. Determine he Dal modus uf late oa od which compre in aelinder om a one ot
0009 mu 0 Non pesar lo a volume 0.085 mat 290 Nam pense, [Ans 38% 10 Men]
21, The sara tension a mater contact with ara Ci ven ar 0.0716 Ni, The press fee &
pl of water it be ODLET Now peter than he ous prosas calculo the mse of the
drop of wae {ans 198 mu)
22, Ft aras tenio ina sop babble of 30m amet when te Inside pressure 952 Nh above
mosso Ans, 060738 Nm]
23, The le tension of wate in coat wih as given 00728 Nim The presu outside dept of
vee o dame 02 mm moe (10322), Ce he press win te dr o
wate {an 1.77 New)
“The mas deny of sl 2684 hg er cb mee ad ico voy 6.30% 10 “aqua ms
pe end. Calculo sea ses a pot. ams 02 Nm]
8. Find Knemate soy aan having density DK" when at cer point ince he shear
cia AR [men toa
a ta i a iri
us
Toast Sees mst en eae nn
BE tte Sn an a
13. The seo due over plate gen by = À 3 in which she ve nme na
a om at pars nase SRE mn mate
ne
0 de ae RE
ue
GE pd he pil ne
pede pe criada
D pes seen
RSR the an nn nee
a mp nat e y e a ste)
pp Nit moe pi
D Rene ee one in ene
ias
A ee eee
loc weight of stwgea à 28 x A (ans 978 New
19. The pres 0 a guide rat from 60 Mie? 100 Nm and volume decreas by 02 pe cen,
Determine he bal modas a elt. ane 2 10" New
20. Determine he Dal modus uf late oa od which compre in aelinder om a one ot
0009 mu 0 Non pesar lo a volume 0.085 mat 290 Nam pense, [Ans 38% 10 Men]
21, The sara tension a mater contact with ara Ci ven ar 0.0716 Ni, The press fee &
pl of water it be ODLET Now peter than he ous prosas calculo the mse of the
drop of wae {ans 198 mu)
22, Ft aras tenio ina sop babble of 30m amet when te Inside pressure 952 Nh above
mosso Ans, 060738 Nm]
23, The le tension of wate in coat wih as given 00728 Nim The presu outside dept of
vee o dame 02 mm moe (10322), Ce he press win te dr o
wate {an 1.77 New)
Properties of Fluids 38]
7A, Cu ep a in glass ube of 3.0 um diameter wa iene ven ne) wate, and
(0) mercy. Tae sure tios or mercury and ales 00728 Nis and 0.52 N respec a
oma wih ar. Spec ravi or mercy e gen a 136 [Ane.0966 cm, 03275 cm
25, The copay is in he passo se or messing wer el 011 exce OS wm Detemin ts
‘nium se, given ht sao lesen or water a coc io lr = OOMIZ Nm. [Ana 58 in)
28. (SL Unite One ie of eae weighs 0.9 Cause de poi weigh emy ant api gai.
Ans 00 Nam 9798 kg 09786]
27. (SL Units) A piston 796 nm diameter and 200 on works ins cylinder of $0 mm darter ee
anna pce ied with bling oof co 5 <p cen pe), cleus x spool of desc
{tie ito in srl positon. Thc weight os po and anal od are 931. N, tans. 788 mA]
2%, (SL Uni) Find te capllary be of Water ina tube 0.3 m mer. The sue tension of wat
00735 Nm. ane 058 em
29, Calcule We pei weigh density and pile gravity oft res af lui which weigh 5 N
ans 7500 Ni 7643 Kg 7.0681
30. À 150 mun der vera yer oes concentcaly inside ater cer of dancer 151,
Bath he pierre of 280m height. The space Ben the cline ld with a ii of vss
iy 10 ie. Determine he re require vote te ner der a 100 <p. TAs. STN)
2, A stato amer 120 ms tang sie a oul bang of har 122 un aa spe 6 pan
“Te space beter the sat and the Bering il Wich abc Gi visos 6 pole ind oe
over aptos i of Ip best 100 me Fans 11873 WT
32. À at of diameter 100m rotin id à uma eating of diameter 12 rm a a pace of 360
pan. The space hotes the sal and Seating i led wa a rang ol of ici 3 pub. The
length ofthe fering 200 mm Find the power abro in he bing oi (Awe LIL SEW]
33. Asuming tat the fuk mesas of easily of water 207 10° N? At sandal atmosphere
or TL
[Him = 2.07 un
an
crease in pressure (4) K x u
x2 itso neun]
2 mr mn iy
nei
tas
Component of weight slong ts pie W x sin ©
[aol
were sin 8 = BE.
fF Fas,
meer wines wie tts
Now cn bere dey ue LS ma dy = me 110?
Baa,
136615, imo?
5
Pay ¿00007 N 007 pie
or Ant = 00007 23-0892 pot
7A, Cu ep a in glass ube of 3.0 um diameter wa iene ven ne) wate, and
(0) mercy. Tae sure tios or mercury and ales 00728 Nis and 0.52 N respec a
oma wih ar. Spec ravi or mercy e gen a 136 [Ane.0966 cm, 03275 cm
25, The copay is in he passo se or messing wer el 011 exce OS wm Detemin ts
‘nium se, given ht sao lesen or water a coc io lr = OOMIZ Nm. [Ana 58 in)
28. (SL Unite One ie of eae weighs 0.9 Cause de poi weigh emy ant api gai.
Ans 00 Nam 9798 kg 09786]
27. (SL Units) A piston 796 nm diameter and 200 on works ins cylinder of $0 mm darter ee
anna pce ied with bling oof co 5 <p cen pe), cleus x spool of desc
{tie ito in srl positon. Thc weight os po and anal od are 931. N, tans. 788 mA]
2%, (SL Uni) Find te capllary be of Water ina tube 0.3 m mer. The sue tension of wat
00735 Nm. ane 058 em
29, Calcule We pei weigh density and pile gravity oft res af lui which weigh 5 N
ans 7500 Ni 7643 Kg 7.0681
30. À 150 mun der vera yer oes concentcaly inside ater cer of dancer 151,
Bath he pierre of 280m height. The space Ben the cline ld with a ii of vss
iy 10 ie. Determine he re require vote te ner der a 100 <p. TAs. STN)
2, A stato amer 120 ms tang sie a oul bang of har 122 un aa spe 6 pan
“Te space beter the sat and the Bering il Wich abc Gi visos 6 pole ind oe
over aptos i of Ip best 100 me Fans 11873 WT
32. À at of diameter 100m rotin id à uma eating of diameter 12 rm a a pace of 360
pan. The space hotes the sal and Seating i led wa a rang ol of ici 3 pub. The
length ofthe fering 200 mm Find the power abro in he bing oi (Awe LIL SEW]
33. Asuming tat the fuk mesas of easily of water 207 10° N? At sandal atmosphere
or TL
[Him = 2.07 un
an
crease in pressure (4) K x u
x2 itso neun]
2 mr mn iy
nei
tas
Component of weight slong ts pie W x sin ©
[aol
were sin 8 = BE.
fF Fas,
meer wines wie tts
Now cn bere dey ue LS ma dy = me 110?
Baa,
136615, imo?
5
Pay ¿00007 N 007 pie
or Ant = 00007 23-0892 pot
|
PRESSURE AND IS N
MEASUREMENT
»
1 FLUID PRESSURE AT A POINT
(Consider a small aca din lage mass of Mud. I the Hid taonary, the the force exerted by
‘te surround fluid oa the area dA wll always be perpendicular tothe surface dA. Let dis the force
Sing on te na ner in Tun ai LÉ on e nen of
pressure or simply pressure and this rato ls represented by p, Hence mauhematclly the pressure ata
Point à ui a rests
ae
a
ie force (#) sunray distributed over he area (A). en pressure at any points piven by
Force or pressure force,
‘The unis of pressure are: () gl? and kifen © in MKS units, (i) Newton or Nim and
Nm in SI units. Nits nose as Pascal andi represent by Pa. Other commonly used unis of
pressure ae
100 kPa = 10° Nin?
> 22 PASCALS LAW
eats that the pressure ten of pressure a point
in à sai id equal in al dictions. This proved a8
“The id clement is of very small dimensions ke ds, dy
rary id element of wedge shape im a
‘od mass tes as shown in Fig. 2.1. Let he width ofthe best
clement perpendicular tthe plane of paper is ly and pig. 21 Forces on fad lement.
35
PRESSURE AND IS N
MEASUREMENT
»
1 FLUID PRESSURE AT A POINT
(Consider a small aca din lage mass of Mud. I the Hid taonary, the the force exerted by
‘te surround fluid oa the area dA wll always be perpendicular tothe surface dA. Let dis the force
Sing on te na ner in Tun ai LÉ on e nen of
pressure or simply pressure and this rato ls represented by p, Hence mauhematclly the pressure ata
Point à ui a rests
ae
a
ie force (#) sunray distributed over he area (A). en pressure at any points piven by
Force or pressure force,
‘The unis of pressure are: () gl? and kifen © in MKS units, (i) Newton or Nim and
Nm in SI units. Nits nose as Pascal andi represent by Pa. Other commonly used unis of
pressure ae
100 kPa = 10° Nin?
> 22 PASCALS LAW
eats that the pressure ten of pressure a point
in à sai id equal in al dictions. This proved a8
“The id clement is of very small dimensions ke ds, dy
rary id element of wedge shape im a
‘od mass tes as shown in Fig. 2.1. Let he width ofthe best
clement perpendicular tthe plane of paper is ly and pig. 21 Forces on fad lement.
35
36. Fluid Mechanics
and pase pressures or intensity of pressure acing on he face AB. AC and BC respectively. Let
“ABC = 0, Then the forces acting on the lement are:
1. Pressure forces normal tothe surfaces, and
2. Weight of clement in the vertical direction.
‘The forges on he faces ate
Foros on the face AB pox Arc of los AB
ayn
Slory fs one ae AC pp
oa do
Wat of tenet Last eme) x
{Bx AC ox
Ant) xp
ve 9
ling resin action, we have
px D an ED -8)
“ pde cose
Bu ho 4.21, ‘em Dan
5 PP]
« » en
Sins, sai he fos intone bt
yd = dern EE ja xp
« A RA
But ds sin = dr anda clement very mal ná hnos weights epi.
: panies
Her a»
From equons (2.1) nd 22). 0 a
ps as
‘The above equation shows thatthe pressure a an) point in, and directions is equal
Since te choice o Mui element vas completely atirar which means de pressure a any pl
the same i al directions.
à 2.3. PRESSURE VARIATION IN A FLUID AT REST
“The pressure a any point in a lid a rst is sine y the Hydro.
static Law which stats thatthe ae of nrease of pressure na et
aly downward direction must he equal 1 the specific weight othe
aia that point. Tiss prove a:
‘Consider a mall ys clement as shown in Fig. 22
Let A= Cross-setiona tea of element
AZ = Height of tui element
(P= Pressure on face AB
2 = Distance of id clement from fos surfe!
‘The forces ating on the lot element ate
FREE SURFACE OF FLUO
is. 22 Fores om a fluid clement
and pase pressures or intensity of pressure acing on he face AB. AC and BC respectively. Let
“ABC = 0, Then the forces acting on the lement are:
1. Pressure forces normal tothe surfaces, and
2. Weight of clement in the vertical direction.
‘The forges on he faces ate
Foros on the face AB pox Arc of los AB
ayn
Slory fs one ae AC pp
oa do
Wat of tenet Last eme) x
{Bx AC ox
Ant) xp
ve 9
ling resin action, we have
px D an ED -8)
“ pde cose
Bu ho 4.21, ‘em Dan
5 PP]
« » en
Sins, sai he fos intone bt
yd = dern EE ja xp
« A RA
But ds sin = dr anda clement very mal ná hnos weights epi.
: panies
Her a»
From equons (2.1) nd 22). 0 a
ps as
‘The above equation shows thatthe pressure a an) point in, and directions is equal
Since te choice o Mui element vas completely atirar which means de pressure a any pl
the same i al directions.
à 2.3. PRESSURE VARIATION IN A FLUID AT REST
“The pressure a any point in a lid a rst is sine y the Hydro.
static Law which stats thatthe ae of nrease of pressure na et
aly downward direction must he equal 1 the specific weight othe
aia that point. Tiss prove a:
‘Consider a mall ys clement as shown in Fig. 22
Let A= Cross-setiona tea of element
AZ = Height of tui element
(P= Pressure on face AB
2 = Distance of id clement from fos surfe!
‘The forces ating on the lot element ate
FREE SURFACE OF FLUO
is. 22 Fores om a fluid clement
Pressureand its Measurement 37
Pressure force on AB = p AA and ating perpendicular 1 Ice AB in the dowaward direcion
ap
2 mtn ce Sat) xn tg ptt cr re
A. Pressure forces on surfaces BC and AD are equal and opposite. For equilibrium of fd
lement, we have
carpo) aso pe cran ce
A FA
= Baran pupa co
E 2 spiele A ot a
Sous er es
abeto w= Weight density of ui
Equation (24) sates that rate of increase of pressure in a velcal direction ls equal to weight
ems ofthe lid at hat pont. Tiss Hydrostatic a.
‘By imtegating the above equation (28) fr liquid, we gst
Jap Ipadz
or va os
where p isthe pressure above atmosphere pressure and 7 isthe height of he point from fre
Surface,
From equation (2.5), we have 26)
ore Z is called pressure head.
Problem 2:1 hydraulic press has ram of 30 em diometerand a plunger of 4.5 em diometer Find
the weight ited bythe lydruatic press when the force applied ot he plunger i 00 N.
Solution. ¿ven
Dia. of ram,
Dia. of plunger,
Force on plunger
Find weight ited
Ana of rim,
Are af plnger,
Pressure force on AB = p AA and ating perpendicular 1 Ice AB in the dowaward direcion
ap
2 mtn ce Sat) xn tg ptt cr re
A. Pressure forces on surfaces BC and AD are equal and opposite. For equilibrium of fd
lement, we have
carpo) aso pe cran ce
A FA
= Baran pupa co
E 2 spiele A ot a
Sous er es
abeto w= Weight density of ui
Equation (24) sates that rate of increase of pressure in a velcal direction ls equal to weight
ems ofthe lid at hat pont. Tiss Hydrostatic a.
‘By imtegating the above equation (28) fr liquid, we gst
Jap Ipadz
or va os
where p isthe pressure above atmosphere pressure and 7 isthe height of he point from fre
Surface,
From equation (2.5), we have 26)
ore Z is called pressure head.
Problem 2:1 hydraulic press has ram of 30 em diometerand a plunger of 4.5 em diometer Find
the weight ited bythe lydruatic press when the force applied ot he plunger i 00 N.
Solution. ¿ven
Dia. of ram,
Dia. of plunger,
Force on plunger
Find weight ited
Ana of rim,
Are af plnger,
[38 Fluid Mechonics
Pressure intensity due to plunger
Farce on plunger F500 a
tea of punger @ — 00139
De to Pascal law, the intensity of pressure wil be E
sually transmite in all irons Hence the pressure
Snteasity ae am
Du pressure nens at ram
2 Weight 144654 07068 = 22222 N = 22222 KN. Ans.
Problem 2.2. A hydraulic pres has a ram of 20 cm diameter and a plunger of 3 em diameter Its
ted for iting a weight of 30 N. Find the fore required at he plunger
Soon cen
Dasran om
‘Area of ram, C2 = 0.0314
D of rn dim 000m
à ot oat Suan? = 705810 a
Van. w= ON 3410 N
Es
Force _ F
Prssuneimensty developed ducto plunger = POSE = F
[By Pascal Law, this pressure is transite equally Sl scons
nes pressure tansmited athe ram E
resure intensity x Arc of ram
Ego PEO y
a Force acting on ram =
ut fore ating on ram = Weight ited = 30000 N
a 30000 = FRI
106810 ©
30000 x 2068 10"
a 30000 706810" _ 695. N, Ans.
r sae 152N. Ai
Problem 2.3 Calewatethe pressure due to column 0f0. (a) water, (b) an où of sp. gr. and
(eh mercury of sp. gr 13.6. Take density of water, 9 = 1000 kh
Solution. Given
Height of quid column, 2
13 m
Pressure intensity due to plunger
Farce on plunger F500 a
tea of punger @ — 00139
De to Pascal law, the intensity of pressure wil be E
sually transmite in all irons Hence the pressure
Snteasity ae am
Du pressure nens at ram
2 Weight 144654 07068 = 22222 N = 22222 KN. Ans.
Problem 2.2. A hydraulic pres has a ram of 20 cm diameter and a plunger of 3 em diameter Its
ted for iting a weight of 30 N. Find the fore required at he plunger
Soon cen
Dasran om
‘Area of ram, C2 = 0.0314
D of rn dim 000m
à ot oat Suan? = 705810 a
Van. w= ON 3410 N
Es
Force _ F
Prssuneimensty developed ducto plunger = POSE = F
[By Pascal Law, this pressure is transite equally Sl scons
nes pressure tansmited athe ram E
resure intensity x Arc of ram
Ego PEO y
a Force acting on ram =
ut fore ating on ram = Weight ited = 30000 N
a 30000 = FRI
106810 ©
30000 x 2068 10"
a 30000 706810" _ 695. N, Ans.
r sae 152N. Ai
Problem 2.3 Calewatethe pressure due to column 0f0. (a) water, (b) an où of sp. gr. and
(eh mercury of sp. gr 13.6. Take density of water, 9 = 1000 kh
Solution. Given
Height of quid column, 2
13 m
Pressure and ts Measurement 39
“The presure at any point in aqui is piven by equation
2
(0) For water, = 1000 Kom?
= 47, = 1000 9.51 4 03
Shas
2943 Nim?
TE Nic? = 0.2943 Niem?. Ans.
(0) For ol of sp. gr. 08,
From equation (11), we know thatthe density o a Muid is equal to specific gravity of Maid
multiple by density of water.
Density of ih Po= Sp. ge ofl Density ofwater pu Density of oid
£08 xp = 08 x 1000 à 800 gm"
Now pressure, P= DxEXZ,
= 800 «9.81 x03-=23514 À - 2354
ES
N
03554 Nan
de) For mercury, sp. gr 36
From equation (1.14) we know thatthe density of Msi is equal to peci gravity of uk
mali by density of water
Density of mercury.
pect gravity of mercury x Density of water
13.6% 1000 = 13600 kim
rere
13600981 03 «4025 À
002
IAN
Problem 2.4 Jhe pressure intensity at a point in a fluid is given 3.924 Née. Find the correspond-
Ing eg off hen heater nd eof 08,
‘olution ann
Pressure mens pe 3904
eG
“The corresponding height, 7, ofthe Mui give by equation (2.6) as
Fa
ps
(a) For water, = 1000 pm
REC Tu
xe” 1000x987
19
(09 x 1000 = 900 kur
pb. 392410"
DET
sm of water. Ans.
O) Foros.
2 Densliy of it
44m of oil. Ans.
“The presure at any point in aqui is piven by equation
2
(0) For water, = 1000 Kom?
= 47, = 1000 9.51 4 03
Shas
2943 Nim?
TE Nic? = 0.2943 Niem?. Ans.
(0) For ol of sp. gr. 08,
From equation (11), we know thatthe density o a Muid is equal to specific gravity of Maid
multiple by density of water.
Density of ih Po= Sp. ge ofl Density ofwater pu Density of oid
£08 xp = 08 x 1000 à 800 gm"
Now pressure, P= DxEXZ,
= 800 «9.81 x03-=23514 À - 2354
ES
N
03554 Nan
de) For mercury, sp. gr 36
From equation (1.14) we know thatthe density of Msi is equal to peci gravity of uk
mali by density of water
Density of mercury.
pect gravity of mercury x Density of water
13.6% 1000 = 13600 kim
rere
13600981 03 «4025 À
002
IAN
Problem 2.4 Jhe pressure intensity at a point in a fluid is given 3.924 Née. Find the correspond-
Ing eg off hen heater nd eof 08,
‘olution ann
Pressure mens pe 3904
eG
“The corresponding height, 7, ofthe Mui give by equation (2.6) as
Fa
ps
(a) For water, = 1000 pm
REC Tu
xe” 1000x987
19
(09 x 1000 = 900 kur
pb. 392410"
DET
sm of water. Ans.
O) Foros.
2 Densliy of it
44m of oil. Ans.
AO Fluid Mechonics
Problem 2.5 Anoilofap. er 09 is contained ina vessel An a point the high of oi 40m. Find
the corresponding height af water a he po.
Solution. Gien.
Sp. gr of où,
Height of
Density of oi
x 1000 = 900 kg/m?
Intensity of pressure,
+ Corresponding height of w
OH 9 40 = 36 m of water. Ans.
1000 79.81
Problem 2.6 Aopen tak contains water upto a depth of ? m and above it an ol ofp gr 0.9 for
a del of | m. Find the pressure intensity () the nerfce ofthe neo lids, and (6) the bottom
ofthe tank
Solution. Given
Height of water, m
Height of ol, m
Sp. sr ofl 5
Density of water, 000 Kam
Density oral, p.80 il Density of water
9 4 1000 = 900 Keim?
Press itensiy at any point is given hy
Kaxz.
(D Atimertace, ie. at
(Atte ham, Lat
73 + Bi § XZ, = 9009.81 2 10+ 1000 9.81 42.0
ren? = 2.8449 Nem Ans.
Problem 2.7. The diameter of small piston und a large piston e a Iydrauie Jack are 3 em and
10 em ep fre of 80 Ns apie one ml peo. Fin hla ied ye arse
Pisten when
(a) the pistons arr ot the same vel
(9 mal piston i 40 em above the large piston
‘The density ofthe guid in he jack is given as 1000 gyn
Solution. Given
Dia. of smal piston,
Auca of small piston,
Problem 2.5 Anoilofap. er 09 is contained ina vessel An a point the high of oi 40m. Find
the corresponding height af water a he po.
Solution. Gien.
Sp. gr of où,
Height of
Density of oi
x 1000 = 900 kg/m?
Intensity of pressure,
+ Corresponding height of w
OH 9 40 = 36 m of water. Ans.
1000 79.81
Problem 2.6 Aopen tak contains water upto a depth of ? m and above it an ol ofp gr 0.9 for
a del of | m. Find the pressure intensity () the nerfce ofthe neo lids, and (6) the bottom
ofthe tank
Solution. Given
Height of water, m
Height of ol, m
Sp. sr ofl 5
Density of water, 000 Kam
Density oral, p.80 il Density of water
9 4 1000 = 900 Keim?
Press itensiy at any point is given hy
Kaxz.
(D Atimertace, ie. at
(Atte ham, Lat
73 + Bi § XZ, = 9009.81 2 10+ 1000 9.81 42.0
ren? = 2.8449 Nem Ans.
Problem 2.7. The diameter of small piston und a large piston e a Iydrauie Jack are 3 em and
10 em ep fre of 80 Ns apie one ml peo. Fin hla ied ye arse
Pisten when
(a) the pistons arr ot the same vel
(9 mal piston i 40 em above the large piston
‘The density ofthe guid in he jack is given as 1000 gyn
Solution. Given
Dia. of smal piston,
Auca of small piston,
Dia. of age piston, D= Wem
4 Area of argc piston, A2 (107
Force on smal piston,
Lette lod ite
(a) When the pistons are at the same level
Pressure try mal pisa
AS
“mn
‘This transite equally on he age ston
Prue tens on he are pin Tees
so
705
Farc one ge piston Pre x Aa
At, x 7.54 = 88896 Nam.
7068
(6) When the small piston is 40 em above the large piston
Pressure intensity on the smal piston
EN
a 758 me
Pressure intensity at section A-A
= © pressure intensity due to height of 40 em of gui.
‘But pressure inte dus 10 40 em of qui
Ep xgxh= 1000 9.81 X04 Nm?
0009.814P ayy? = 9.3904 Nem?
10
2 Presa mens at section 4-4
so
= 8 sos
7008 *°
1132 = 03924 = 1171 Ni
Pressure intensity transmuted o the large piston = 1171 Men
2 Force on the are piston = Pressure Area ofthe are piston
AL A A = 117147834 = 919.7 N.
» 24 ABSOLUTE, GAUGE, ATMOSPHERIC AND VACUUM PRESSURES
The pressure 0 3 id is measure in two different systems. In one system, ts measured above
‘he absolute zero or complete Vacuum and i is called the absolute pressure and in othe system,
Presse is messured above the atmospheric pressure and Ii called gange pressure. This
1. Absolute pressure is defined asthe pressure wich Is measured wi reference to absolute
‘au pressure.
2. Gauge pressure is defined as the pressure which is measured with the lp ofa pressure mea
‘ring instrument, in which the atmosphere pressure Is taken as datum. The atmosphere pressure an
th sale is marke as ro,
4 Area of argc piston, A2 (107
Force on smal piston,
Lette lod ite
(a) When the pistons are at the same level
Pressure try mal pisa
AS
“mn
‘This transite equally on he age ston
Prue tens on he are pin Tees
so
705
Farc one ge piston Pre x Aa
At, x 7.54 = 88896 Nam.
7068
(6) When the small piston is 40 em above the large piston
Pressure intensity on the smal piston
EN
a 758 me
Pressure intensity at section A-A
= © pressure intensity due to height of 40 em of gui.
‘But pressure inte dus 10 40 em of qui
Ep xgxh= 1000 9.81 X04 Nm?
0009.814P ayy? = 9.3904 Nem?
10
2 Presa mens at section 4-4
so
= 8 sos
7008 *°
1132 = 03924 = 1171 Ni
Pressure intensity transmuted o the large piston = 1171 Men
2 Force on the are piston = Pressure Area ofthe are piston
AL A A = 117147834 = 919.7 N.
» 24 ABSOLUTE, GAUGE, ATMOSPHERIC AND VACUUM PRESSURES
The pressure 0 3 id is measure in two different systems. In one system, ts measured above
‘he absolute zero or complete Vacuum and i is called the absolute pressure and in othe system,
Presse is messured above the atmospheric pressure and Ii called gange pressure. This
1. Absolute pressure is defined asthe pressure wich Is measured wi reference to absolute
‘au pressure.
2. Gauge pressure is defined as the pressure which is measured with the lp ofa pressure mea
‘ring instrument, in which the atmosphere pressure Is taken as datum. The atmosphere pressure an
th sale is marke as ro,
32 Fluid Mechanies
3. Vacuum pressure is defined a the pres y
sure below te atmospheric pressure.
‘gauge pressure and vacuum pressure are shown in RR.
(0 Absolute pressure Pressure.
(i Vacuum pressure
"Atmospheric pressure ~ Absolute pressure
ote (9 The amospere pese a se evel 19 i 10.3 LNA of 1013 Ne
(io The amor preus heu a 760 man of meray or 1088 m ol wa
Problem 28 What are the sauge pressure and ubsoate pressure a a point 3 m below the fre
face ofa liquid having a density of 1.3 % 10" kan’ if the amospheri pressure is equivalent to
750 mm of mercury ? The specific gravity of mercury is 13.6 and density of water = 1000 kei
Solution. Given
Fig. 27 Relationship berseen pressures.
Depth of iguid, zum
Density of qui, Pin 133 « 10° kyla?
[Aumosphsric pressure head, 2 750 mm of Hig
250
250 05 mat ty
1000 ®
mosphere PRESSURE, Pan =P 8X7
whore py = Demy of Hg = Sp. gr. of mercury Density of mate = 136 1000 kgm?
and Zas Poss Read in
Pressure at a point, which i at a depth of 3 m from the fce surface ofthe liquid is given hy,
Po
1.53 1000) x9,
Gage pressure, p= 48028 Nin’. Ans.
[Now absolute pressure ‘Gauge pressure + Atmospheric pressure
= 45028 + 100062 = 145090 Nm”. Ans
» 2.5. MEASUREMENT OF PRESSURE
23 = 45028 Nin?
‘Te pressure of Mui is measured by the following devices
Manometers 2. Mechanical Gauges.
2.5.1. Manometers. Manometers are defined as A devices used for measuring the pressure at
point ina hid by balancing the column of ui by the same or another column o he Mid. They are
late as
(a) Simple Manometen. (6) Diterendat Manometer.
3. Vacuum pressure is defined a the pres y
sure below te atmospheric pressure.
‘gauge pressure and vacuum pressure are shown in RR.
(0 Absolute pressure Pressure.
(i Vacuum pressure
"Atmospheric pressure ~ Absolute pressure
ote (9 The amospere pese a se evel 19 i 10.3 LNA of 1013 Ne
(io The amor preus heu a 760 man of meray or 1088 m ol wa
Problem 28 What are the sauge pressure and ubsoate pressure a a point 3 m below the fre
face ofa liquid having a density of 1.3 % 10" kan’ if the amospheri pressure is equivalent to
750 mm of mercury ? The specific gravity of mercury is 13.6 and density of water = 1000 kei
Solution. Given
Fig. 27 Relationship berseen pressures.
Depth of iguid, zum
Density of qui, Pin 133 « 10° kyla?
[Aumosphsric pressure head, 2 750 mm of Hig
250
250 05 mat ty
1000 ®
mosphere PRESSURE, Pan =P 8X7
whore py = Demy of Hg = Sp. gr. of mercury Density of mate = 136 1000 kgm?
and Zas Poss Read in
Pressure at a point, which i at a depth of 3 m from the fce surface ofthe liquid is given hy,
Po
1.53 1000) x9,
Gage pressure, p= 48028 Nin’. Ans.
[Now absolute pressure ‘Gauge pressure + Atmospheric pressure
= 45028 + 100062 = 145090 Nm”. Ans
» 2.5. MEASUREMENT OF PRESSURE
23 = 45028 Nin?
‘Te pressure of Mui is measured by the following devices
Manometers 2. Mechanical Gauges.
2.5.1. Manometers. Manometers are defined as A devices used for measuring the pressure at
point ina hid by balancing the column of ui by the same or another column o he Mid. They are
late as
(a) Simple Manometen. (6) Diterendat Manometer.
Pressureand ts Measurement 43
2.5.2 Mechanical Gauges, Mechanical gauges ue defined as the devices wed for measuring
‘te pressure by balancing the ld column by the spring or dead weight The commonly used mechan
ca pressure gauges are
(a) Diaphrapo pressure gauge, (0) Bourdon we pressure gauge,
(0) Deadeweight pressure page, and (a) Bellows pressure gauge.
» 2.6 SIMPLE MANOMETERS
A simple manometer consists oa glass tube haviag one of it eds connected 10 pola where
pressure I 1 be measured and othe end remains open 10 umosphere. Common types of simple ma-
1. Piezometer:
2, Une Manometer, and
3. Single Column Manometer
2.6.1 Piezometer. isthe simplest form of manomstr u for
suring gauge pressures, On end ofthis manomster i conncet 0
¡de pint where pressure Ito be measured and other end Is open to the
atmosphere a shown in Fig. 2-8. The rs of guid gives the pressure
head at tat point If ats pot A, the height of gui say water i ia
piezometer tube, hen pressure at A
sexo
Meza Pixma,
2.62 Utube Manometer. It ons of ls tbe beat Up ne nef ch
Send à pn mh plese st be a ud aber tena open to
par u down I ig. 29 True pry cs mar a a Ug vo
‘geile toy b pear cae Scie evo had wos osu be ere
Corregir (Forman pose
Fig. 29. Utube Manometer
(a) For Gauge Pressure. Lot isthe point at which pressure so be measured, whose value fp
"The datum Inc À À
Height of igh liquid above the dasa ine
Height oF heavy liquid above the datum fine
42 Sp. gr of ight liquid
Ps = Density o ligt quid = 1000 x 5,
Sp. ef. of heavy guid
P2= Density of beasy Liquid = 10005;
2.5.2 Mechanical Gauges, Mechanical gauges ue defined as the devices wed for measuring
‘te pressure by balancing the ld column by the spring or dead weight The commonly used mechan
ca pressure gauges are
(a) Diaphrapo pressure gauge, (0) Bourdon we pressure gauge,
(0) Deadeweight pressure page, and (a) Bellows pressure gauge.
» 2.6 SIMPLE MANOMETERS
A simple manometer consists oa glass tube haviag one of it eds connected 10 pola where
pressure I 1 be measured and othe end remains open 10 umosphere. Common types of simple ma-
1. Piezometer:
2, Une Manometer, and
3. Single Column Manometer
2.6.1 Piezometer. isthe simplest form of manomstr u for
suring gauge pressures, On end ofthis manomster i conncet 0
¡de pint where pressure Ito be measured and other end Is open to the
atmosphere a shown in Fig. 2-8. The rs of guid gives the pressure
head at tat point If ats pot A, the height of gui say water i ia
piezometer tube, hen pressure at A
sexo
Meza Pixma,
2.62 Utube Manometer. It ons of ls tbe beat Up ne nef ch
Send à pn mh plese st be a ud aber tena open to
par u down I ig. 29 True pry cs mar a a Ug vo
‘geile toy b pear cae Scie evo had wos osu be ere
Corregir (Forman pose
Fig. 29. Utube Manometer
(a) For Gauge Pressure. Lot isthe point at which pressure so be measured, whose value fp
"The datum Inc À À
Height of igh liquid above the dasa ine
Height oF heavy liquid above the datum fine
42 Sp. gr of ight liquid
Ps = Density o ligt quid = 1000 x 5,
Sp. ef. of heavy guid
P2= Density of beasy Liquid = 10005;
HA Fluid Mechonies
‘As the pressure is te sume forthe horizontal surface. Hence pressure above the horizontal datum
line A-A in the let column and in the sight column of U-tube manometer should be same.
Pressure above A in ie left column ETE
Pressure above A in tc ig cola HT
Hencs equating the to pressures p+ Pig = Push,
P= (ps6 = Dax 8 4h) en
(6) For Vacuum Pressure. For measuring vacuum presu, the Level ofthe avy liquid in the
manometer wil he 38 shown in Tig 2.9 (5). Then
Pressure above AA inthe et column = path + pag +p
Pressure ead in the ight column above À A
Paths à pithy + p= 0
(ahs + push. es
Problem 29 The righ limb of simple Ustube manometer containing mercury is open 10 Me
samosphere white the lf limb is connected toa pipe in which a fluid of sp ar. 09 Is loming. The
‘enue ofthe pipe I 12 em below the evel of mercury In the right mb. Find the pressure of fad in
Ihe pipe if the diference of mercury level ln ih te links I 20 em
Solution. Given
Sp. of id,
Density of Nui,
Sp. ge of mercury,
Density of mercury.
Ditterence of mercury level
Height of uid from Ar
Let p= Pressure of fui in pipe
Tuti the pressure above A.A, we got
P+ path = pat
or 1 à 900 4 9.81 00% = 136% 1000 x 981 x 2
1316 4 1000 x 9.81 X 2 - 900 x 9.81 0.08
16683 706 = 25917 Nim? = 2.597 Nien. Ans.
Problem 2.10. A simple U-tube manometer containing mereur ls connected 10 à pipe in which u
ai ofp. gr. 08 and having vacuum pressure is lowing. The other end ofthe manometer is open to
“amospher. Find the vacuum presure in pipe, i the difference of mercury level inthe two limbs is
40 om and the height o id In the ef from the centre of pipe 1815 cm belo.
Solution. Given
Sp. ge of lid, 5-08
Sp. ar. of mercury, 52136
Density of id,
Density of mercury,
Difference of mercury level = 40 em = 0m, Het of guid nef im. h,
= 15 cm = 0.15 m Let the pressure in pipe = p. Equating pressure above datum
line AA, we pet
ig. 240
oth pur +
‘As the pressure is te sume forthe horizontal surface. Hence pressure above the horizontal datum
line A-A in the let column and in the sight column of U-tube manometer should be same.
Pressure above A in ie left column ETE
Pressure above A in tc ig cola HT
Hencs equating the to pressures p+ Pig = Push,
P= (ps6 = Dax 8 4h) en
(6) For Vacuum Pressure. For measuring vacuum presu, the Level ofthe avy liquid in the
manometer wil he 38 shown in Tig 2.9 (5). Then
Pressure above AA inthe et column = path + pag +p
Pressure ead in the ight column above À A
Paths à pithy + p= 0
(ahs + push. es
Problem 29 The righ limb of simple Ustube manometer containing mercury is open 10 Me
samosphere white the lf limb is connected toa pipe in which a fluid of sp ar. 09 Is loming. The
‘enue ofthe pipe I 12 em below the evel of mercury In the right mb. Find the pressure of fad in
Ihe pipe if the diference of mercury level ln ih te links I 20 em
Solution. Given
Sp. of id,
Density of Nui,
Sp. ge of mercury,
Density of mercury.
Ditterence of mercury level
Height of uid from Ar
Let p= Pressure of fui in pipe
Tuti the pressure above A.A, we got
P+ path = pat
or 1 à 900 4 9.81 00% = 136% 1000 x 981 x 2
1316 4 1000 x 9.81 X 2 - 900 x 9.81 0.08
16683 706 = 25917 Nim? = 2.597 Nien. Ans.
Problem 2.10. A simple U-tube manometer containing mereur ls connected 10 à pipe in which u
ai ofp. gr. 08 and having vacuum pressure is lowing. The other end ofthe manometer is open to
“amospher. Find the vacuum presure in pipe, i the difference of mercury level inthe two limbs is
40 om and the height o id In the ef from the centre of pipe 1815 cm belo.
Solution. Given
Sp. ge of lid, 5-08
Sp. ar. of mercury, 52136
Density of id,
Density of mercury,
Difference of mercury level = 40 em = 0m, Het of guid nef im. h,
= 15 cm = 0.15 m Let the pressure in pipe = p. Equating pressure above datum
line AA, we pet
ig. 240
oth pur +
Pressure and ts Measurement 45
eer
13.6% 1000 «9.81 X04 + 800 «9:81 «0.151
1593664 + 1177.2) =~ 4436 Nim? == 8.454 Nom. Ans
Problem 2.11 À U-Tube manometer is used 1 measure the presure of water ina pipeline, which
sim excess of atmaspheric pressure. The right mb of he manometer conta mercury and Is open 20
‘amosphere. The contact benween water and mercury i in the left mb. Determine the pressure of
Water inthe main tne. Y the diferenceintvel of mercury in the limbs of Use is 10 em and the
free surface of mercury iin level wih the centre of the pipe. If the pressure of mater in pipe Ine à
reduced 10 9810 Nim calculate the new diferene in the level of mercury. Sketch the arrangements
in both cases
Solution. Given
Difference of mercury = 10 em = 0.1 m
‘The arrangement is shown in Fig. 2.11 (0)
Ist Part
Let pa = (pressure of water la pipe line (ea point A)
“The points Band Cleon he same horizontal line, Hence pressure a B should be equal 0 pressure.
at Ce But pressure a D
= Pressure aA + Pressure dus to 10 cm (or 0.1 m)
of water
rn
where 9 = 1000 kg/m and h= 01m
2 a + 1000 x 981 x01
pa + 981 Nim “
Pressure at C = Pressure at D 4 Pressure due to 10 cm of mercury
8% la
where pu fOr mercury = 13.6% 1000 kim?
and ho = Wem = 0.1
Pressure at
401363 1000) 9.81 0.1
mon «in
But pressure a Bis equal to pressure at €. eace equating the equa
tions (and i), we pet
Py 981 = 133416 Fig. 221 Go
a Dam 188416 - 981
= 123606
nd Part
Given, p4 = 9810 Na
Find new difference of mercury level. The amanpement ls shown ia Fig. 211 0). ln ts ease the
pressure at Ai 9810 Nin" which sess han the 12360.6 Nim Hence mercury I let ib wil ise
Theis of mercury i ft mb wil be equal 1 the fll of mercury in Eh ib the ol volume of
mercury remains same.
Let x Rise of mercuy in ft mb in cm
Then fall of mercury in ight mb = à om
‘The points B, Cand D show the iia conditions whercas points 1, C* and D show the
final conditions,
eer
13.6% 1000 «9.81 X04 + 800 «9:81 «0.151
1593664 + 1177.2) =~ 4436 Nim? == 8.454 Nom. Ans
Problem 2.11 À U-Tube manometer is used 1 measure the presure of water ina pipeline, which
sim excess of atmaspheric pressure. The right mb of he manometer conta mercury and Is open 20
‘amosphere. The contact benween water and mercury i in the left mb. Determine the pressure of
Water inthe main tne. Y the diferenceintvel of mercury in the limbs of Use is 10 em and the
free surface of mercury iin level wih the centre of the pipe. If the pressure of mater in pipe Ine à
reduced 10 9810 Nim calculate the new diferene in the level of mercury. Sketch the arrangements
in both cases
Solution. Given
Difference of mercury = 10 em = 0.1 m
‘The arrangement is shown in Fig. 2.11 (0)
Ist Part
Let pa = (pressure of water la pipe line (ea point A)
“The points Band Cleon he same horizontal line, Hence pressure a B should be equal 0 pressure.
at Ce But pressure a D
= Pressure aA + Pressure dus to 10 cm (or 0.1 m)
of water
rn
where 9 = 1000 kg/m and h= 01m
2 a + 1000 x 981 x01
pa + 981 Nim “
Pressure at C = Pressure at D 4 Pressure due to 10 cm of mercury
8% la
where pu fOr mercury = 13.6% 1000 kim?
and ho = Wem = 0.1
Pressure at
401363 1000) 9.81 0.1
mon «in
But pressure a Bis equal to pressure at €. eace equating the equa
tions (and i), we pet
Py 981 = 133416 Fig. 221 Go
a Dam 188416 - 981
= 123606
nd Part
Given, p4 = 9810 Na
Find new difference of mercury level. The amanpement ls shown ia Fig. 211 0). ln ts ease the
pressure at Ai 9810 Nin" which sess han the 12360.6 Nim Hence mercury I let ib wil ise
Theis of mercury i ft mb wil be equal 1 the fll of mercury in Eh ib the ol volume of
mercury remains same.
Let x Rise of mercuy in ft mb in cm
Then fall of mercury in ight mb = à om
‘The points B, Cand D show the iia conditions whercas points 1, C* and D show the
final conditions,
136. Fluid Mechonies
“The pressure at BY = Pressure at C*
or Presure at 4 Pressure due to (10 = x) em of wate
= Pressure at D* + Presse due 10
(10 29 em of mercury
OF et) EX PO + Pix EX
«moon (24)
_
<0. ses) est (Ba)
oe i eae
‘rarer 1
as
260.
0992 cm
Fig. 211 @)
New difference of mercury = 10 - 2x em =10= 2 x 0992
> 8016 cm. Ans.
Problem 2.12 is. 2.12 shows a conical vessel having Is ouler a A t which a Uube manometer
ds connected. The reading ofthe manometer given in the figure shows when the vessel em End the
reading ofthe manometer when the veses completely fled with water.
Solution. Vessel ls empty. Given:
Difference of mercury level hy 4 m
Lane Heo wa re XX i
mm See
Sp. gr of water, S=10 DE}
Deny of ma, pea 186 x un 1
Density of water, Pis 1000 18 y
evan eHow stove date Le EX we ae |
paper "
Pen
DL 27 ma ae
‘Vessel i full water, When vessel ful of water, the Fig. 22
pressure in the right limb will Increase and mercury level in the igh Limb wil go down. Let the
distance though which mercury goes down inthe right limb be, y em as shown in Fi. 2.13. The
mercury wil sei the lft by a distance of y em. Now the datum Ine i 2-2 Equaing the pressure
above the datum ine 222.
Pressure i et mb = Pressure in ht Limb
136 1000 981 « (02 + 29/100)
1000 9.81 x (3+ I, + 3/100)
“The pressure at BY = Pressure at C*
or Presure at 4 Pressure due to (10 = x) em of wate
= Pressure at D* + Presse due 10
(10 29 em of mercury
OF et) EX PO + Pix EX
«moon (24)
_
<0. ses) est (Ba)
oe i eae
‘rarer 1
as
260.
0992 cm
Fig. 211 @)
New difference of mercury = 10 - 2x em =10= 2 x 0992
> 8016 cm. Ans.
Problem 2.12 is. 2.12 shows a conical vessel having Is ouler a A t which a Uube manometer
ds connected. The reading ofthe manometer given in the figure shows when the vessel em End the
reading ofthe manometer when the veses completely fled with water.
Solution. Vessel ls empty. Given:
Difference of mercury level hy 4 m
Lane Heo wa re XX i
mm See
Sp. gr of water, S=10 DE}
Deny of ma, pea 186 x un 1
Density of water, Pis 1000 18 y
evan eHow stove date Le EX we ae |
paper "
Pen
DL 27 ma ae
‘Vessel i full water, When vessel ful of water, the Fig. 22
pressure in the right limb will Increase and mercury level in the igh Limb wil go down. Let the
distance though which mercury goes down inthe right limb be, y em as shown in Fi. 2.13. The
mercury wil sei the lft by a distance of y em. Now the datum Ine i 2-2 Equaing the pressure
above the datum ine 222.
Pressure i et mb = Pressure in ht Limb
136 1000 981 « (02 + 29/100)
1000 9.81 x (3+ I, + 3/100)
Picasa #7]
SOI EEE Te
4 Rem a
. Pre 1
. er T
"The difernce of mercury level in two limbs Ë ”
(20 + 2y) em of mercury. 4
bar
re E
dsd ones utah
Problem 213 A pressure gaute consists of mo eyindrical bulbs Band C each of 1034 em cross.
seconalarca, which are connected by a U-tube with vertical Imbscach of 0.254. cm cross-sectional
area A red liquid of specific gravity 09 i led into C and lear waver ts filled no the surface of
separation being nthe mb anached to C. Find the displacement ofthe surface of separation when the
pressure an the surface in Cis greater than that in by on amount equal to | cu head of water
tea of exch bub Band €,
‘Area ofeach vertical limb,
Sp. ge of red liquid
La XX ta separation vel
= Meigh of red liquid above XX
Ing = Height of water above XX
Pressure ahove CK nthe lf im = 1000 9.81 ip
Pressure above X-X inthe right mb = 900 9.81 he
qui the two pressure, we get,
1000 9.81 hy = 900 9:81 he
hy = 09 he 0
lis density = 900 Kg
When the pressure head over the surface im Cs
inewased hy Y em of water, let the separation level
falls by an amount equal 10 Z Then Y-Y becomes the
‘ina separation evel
"Now fall in surface level of C multiplied by cross
sectional area of bolb € must be equal t the fll in
septation level molipled by cosssetioal area of
tino.
Fallin surface level of € ‘even
Fallin separation evel xa
a
SOI EEE Te
4 Rem a
. Pre 1
. er T
"The difernce of mercury level in two limbs Ë ”
(20 + 2y) em of mercury. 4
bar
re E
dsd ones utah
Problem 213 A pressure gaute consists of mo eyindrical bulbs Band C each of 1034 em cross.
seconalarca, which are connected by a U-tube with vertical Imbscach of 0.254. cm cross-sectional
area A red liquid of specific gravity 09 i led into C and lear waver ts filled no the surface of
separation being nthe mb anached to C. Find the displacement ofthe surface of separation when the
pressure an the surface in Cis greater than that in by on amount equal to | cu head of water
tea of exch bub Band €,
‘Area ofeach vertical limb,
Sp. ge of red liquid
La XX ta separation vel
= Meigh of red liquid above XX
Ing = Height of water above XX
Pressure ahove CK nthe lf im = 1000 9.81 ip
Pressure above X-X inthe right mb = 900 9.81 he
qui the two pressure, we get,
1000 9.81 hy = 900 9:81 he
hy = 09 he 0
lis density = 900 Kg
When the pressure head over the surface im Cs
inewased hy Y em of water, let the separation level
falls by an amount equal 10 Z Then Y-Y becomes the
‘ina separation evel
"Now fall in surface level of C multiplied by cross
sectional area of bolb € must be equal t the fll in
septation level molipled by cosssetioal area of
tino.
Fallin surface level of € ‘even
Fallin separation evel xa
a
AB Fluid Mechonics
x0 7028 7
Du “4
Also Eli surface level of ©
Rise in surface love of
The pressure of I em (or AL m) of water = pgh = 1000 x 981 x 0.01= 98,1 Nin
(Consider final separation level YY
Pressure above Y-Y in the lft mb
oxo xoat (20, À)
Pressure shove
Equating the two pressure, we get
sm
ame «9.81 (200392) = (7244-2) ooo «981
( a]
Dividing Wy 9.8, we
um (ze
(zen)
en Jab. ve a zen Zron[zen- À) son
Bu nom equi, y= 09 he
Ze
20900 Ze PE 0920904001
0910 = DE 0920910900
za
x HE Doro
40 40 a
a (1-23)
oa
2.6.3 Single Column Manometer. Single column manometr i a moi form of à Utube
Manometer in which a reservoir, having à age cross scetinal ara (about 100 is) as compared to
bo rc ofthe ths conned one hs ibs (ay If mb) of he mamans a shown m ag 215
Due to larg cross sectional ara ofthe reservar, for ny variation in pressure, he change nthe guid
level nae reservoir wil be very small which may bo mglectt and henc the pressure is given by the
eight of gui nthe other mb. The ber mb may be vertical or nen Thus
of single column manomoor as
1. Ventcal Single Coloma Manometer
2 nel Single Colum Manometer
1. Vertical Single Column Manometer
ig 2.15 shows the vertical single column manometr
nd in the right ib of th manometer, hen i nt connected 1 the pins. When the mano
x0 7028 7
Du “4
Also Eli surface level of ©
Rise in surface love of
The pressure of I em (or AL m) of water = pgh = 1000 x 981 x 0.01= 98,1 Nin
(Consider final separation level YY
Pressure above Y-Y in the lft mb
oxo xoat (20, À)
Pressure shove
Equating the two pressure, we get
sm
ame «9.81 (200392) = (7244-2) ooo «981
( a]
Dividing Wy 9.8, we
um (ze
(zen)
en Jab. ve a zen Zron[zen- À) son
Bu nom equi, y= 09 he
Ze
20900 Ze PE 0920904001
0910 = DE 0920910900
za
x HE Doro
40 40 a
a (1-23)
oa
2.6.3 Single Column Manometer. Single column manometr i a moi form of à Utube
Manometer in which a reservoir, having à age cross scetinal ara (about 100 is) as compared to
bo rc ofthe ths conned one hs ibs (ay If mb) of he mamans a shown m ag 215
Due to larg cross sectional ara ofthe reservar, for ny variation in pressure, he change nthe guid
level nae reservoir wil be very small which may bo mglectt and henc the pressure is given by the
eight of gui nthe other mb. The ber mb may be vertical or nen Thus
of single column manomoor as
1. Ventcal Single Coloma Manometer
2 nel Single Colum Manometer
1. Vertical Single Column Manometer
ig 2.15 shows the vertical single column manometr
nd in the right ib of th manometer, hen i nt connected 1 the pins. When the mano
Praia 8]
connected 10 te pipe, due 10 high pressure at À, he heavy liquid inthe reservoir will be pushed
on and will sein he right Limb,
Let Ah = Fallof heavy liquid in reservo
‘hy Rise of heavy igi ight Limb
In, = Hei of entre of pipe above XX
‘py = Pressure at A, whichis toe menu
‘A= Cross-sectional aca of the reservoir
a = Cross sctiona ars ofthe right lim
5, = Sp. of liquid in pipe
= Sp. gr. hey gui ser and sight in
Pr = Density of guid in pipe ig. 215 Vertical single column
Pa = Density of liquid in reservoir
Fal of beavylgoi in reservoir wil case ars of heavy ui level in be rip mb
Axis ax
x 5
ae o
Now consider the datum line Ys own in Fig. 215. Then pressure inthe ight lim above 1-2,
= ppg (AN +)
Pressure inthe left ib above 7 y cx AA +h) + Pa
quan these pressures, we have
PAINE Rp Xe (AN + + Pa
« a= Ps (A hd pa A
= Abs = el + par he
ash,
‘But rom equation (D. ana do
uation (D. 5
ash,
a mn E lp = pas + hse por 29
‘As te ae A very large as compared o a, Rene ato becomes very small und can be
neglected
Then Pa = ht - Ps 2.10)
‘From equation (2.10), iis clear that as A, is known T
and hence by knowing A, or ie of hey Igo inthe "
Fight limb, the pressure a can be calculated. mM
2. Inclined Single Column Manometer E
Fig 2.16 shows the inclined single clumo manom-
cee This manometer s moe sensitive. Dueto cine
tion be distance moved bythe heavy liquid inthe right Fig 26 Inlined inge column
Tin il be more. manometer
connected 10 te pipe, due 10 high pressure at À, he heavy liquid inthe reservoir will be pushed
on and will sein he right Limb,
Let Ah = Fallof heavy liquid in reservo
‘hy Rise of heavy igi ight Limb
In, = Hei of entre of pipe above XX
‘py = Pressure at A, whichis toe menu
‘A= Cross-sectional aca of the reservoir
a = Cross sctiona ars ofthe right lim
5, = Sp. of liquid in pipe
= Sp. gr. hey gui ser and sight in
Pr = Density of guid in pipe ig. 215 Vertical single column
Pa = Density of liquid in reservoir
Fal of beavylgoi in reservoir wil case ars of heavy ui level in be rip mb
Axis ax
x 5
ae o
Now consider the datum line Ys own in Fig. 215. Then pressure inthe ight lim above 1-2,
= ppg (AN +)
Pressure inthe left ib above 7 y cx AA +h) + Pa
quan these pressures, we have
PAINE Rp Xe (AN + + Pa
« a= Ps (A hd pa A
= Abs = el + par he
ash,
‘But rom equation (D. ana do
uation (D. 5
ash,
a mn E lp = pas + hse por 29
‘As te ae A very large as compared o a, Rene ato becomes very small und can be
neglected
Then Pa = ht - Ps 2.10)
‘From equation (2.10), iis clear that as A, is known T
and hence by knowing A, or ie of hey Igo inthe "
Fight limb, the pressure a can be calculated. mM
2. Inclined Single Column Manometer E
Fig 2.16 shows the inclined single clumo manom-
cee This manometer s moe sensitive. Dueto cine
tion be distance moved bythe heavy liquid inthe right Fig 26 Inlined inge column
Tin il be more. manometer
50 Floid Mechanics
Let Lo Length of heavy liga! moved in ight mb trom XX
= Inclination of right Hib with horizontal
hz Vertical rise of heavy liquid in ight Km rom X= sin 8
From equation (2.10), the pressure at As
Pa pag = pie
Substuting the value of hi, we get
peererrenry ea
Problem 2.14 A single column manometer is connected ta pipe containing a gui of sp. ar. 09
as shown in Fig 2.17 Find the pressure in the pie if the area of the reservo is 100 times the area
Of the tube forthe manometer reading shown in Fig. 2.17. The specific pravty of mercury 8136
Solution. Given
Sp. sof quid in pipe,
2 Densiiy
Sp. xe of hesey qui,
x
Density, x:
4
‘Area of ight tin "a
OS m=20em=
Rise of mercy in right im.
y= dem = 00
La Pa = Pressure in pipe
Using equation (9) we pet
ne Enns pals hos—hoe
105 X OAI3S x 1000 9.81 - 900 > 981] à 04 x LB x 100x981 - 02x00 9.81
os
= 24 tes ea
533.664 + 533664 - 17655 Nin! = 52134 Nin! = 521 Nom, Am
> 2.7 DIFFERENTIAL MANOMETERS
Differential manometers ate the devices used for measuring the difference of pressures between
wo points in a pipe or in two diferent pies. A differential manometer consists ofa Ustube, cota»
ng 3 heavy quid, wok two ends are connected o the points, whose diffrence of pressure so Be
measured. Most commonly types of dtfremial manometer ae
‘Ustube differential manomter and
2. Tnvenod U-tube diferent manometer.
2.1.1 Ustube Differential Manometer. ig. 2.18 shows the diffe
Uuhe ype
Let Lo Length of heavy liga! moved in ight mb trom XX
= Inclination of right Hib with horizontal
hz Vertical rise of heavy liquid in ight Km rom X= sin 8
From equation (2.10), the pressure at As
Pa pag = pie
Substuting the value of hi, we get
peererrenry ea
Problem 2.14 A single column manometer is connected ta pipe containing a gui of sp. ar. 09
as shown in Fig 2.17 Find the pressure in the pie if the area of the reservo is 100 times the area
Of the tube forthe manometer reading shown in Fig. 2.17. The specific pravty of mercury 8136
Solution. Given
Sp. sof quid in pipe,
2 Densiiy
Sp. xe of hesey qui,
x
Density, x:
4
‘Area of ight tin "a
OS m=20em=
Rise of mercy in right im.
y= dem = 00
La Pa = Pressure in pipe
Using equation (9) we pet
ne Enns pals hos—hoe
105 X OAI3S x 1000 9.81 - 900 > 981] à 04 x LB x 100x981 - 02x00 9.81
os
= 24 tes ea
533.664 + 533664 - 17655 Nin! = 52134 Nin! = 521 Nom, Am
> 2.7 DIFFERENTIAL MANOMETERS
Differential manometers ate the devices used for measuring the difference of pressures between
wo points in a pipe or in two diferent pies. A differential manometer consists ofa Ustube, cota»
ng 3 heavy quid, wok two ends are connected o the points, whose diffrence of pressure so Be
measured. Most commonly types of dtfremial manometer ae
‘Ustube differential manomter and
2. Tnvenod U-tube diferent manometer.
2.1.1 Ustube Differential Manometer. ig. 2.18 shows the diffe
Uuhe ype
(a7wo pp état (Aan are au some
Fix 218 Utube diferente! manometer
In Fig. 2.18 a, the two points A and Jaro at different level and alo contains liquid of diffrent
sp. gr Thess points are connected 16 the Utube differential manometer. Let the pressure a À and
a pa and Py.
La = Dieren of mercury esl inthe U-tube
Distance ofthe cence of From the mercury evel nthe right mb.
x= Distance of th centr ofA, fom the mercury evel nthe igh lim.
Pr = Density of Liquid a A.
pa = Density of quid at 3
= Density of heavy liquid or mercury
Taking datum tne at XX.
Pressure above XX in the left Limb = yet 4 2) 4 Pa
here p= pressure a
Pressure above Kin the ight Limb = pyx PX 829 +p
whore pe = Pressure a B.
quit the wo presturs, we have
msthenen,
Pape
Eh + pa + Pe
PX Rh + ET)
% 104-01) pie e
Difference of pressure at A and B= AX (O, PA) + PAR ~ Bute
In Fig, 2.8 (the two points À and are a the same level and contains he same liquid of density
Py. Then
Pressure above Kim ght Hind = PICA 9s X24 Pa
Pressure shove XX in et lib SP + D+ Ry
gig he two pressure
ES 24 Pa
ae Py Eh à page = Pa + D
= ex np, po} 213)
Problem 2.15 A pipe contains ano of sp. gr. 0. A diferent manometer comeced a the v0
‘otis aná B shows difference in ercur eve ox 15 em. Fond the difference of peste athe me
pois
Fix 218 Utube diferente! manometer
In Fig. 2.18 a, the two points A and Jaro at different level and alo contains liquid of diffrent
sp. gr Thess points are connected 16 the Utube differential manometer. Let the pressure a À and
a pa and Py.
La = Dieren of mercury esl inthe U-tube
Distance ofthe cence of From the mercury evel nthe right mb.
x= Distance of th centr ofA, fom the mercury evel nthe igh lim.
Pr = Density of Liquid a A.
pa = Density of quid at 3
= Density of heavy liquid or mercury
Taking datum tne at XX.
Pressure above XX in the left Limb = yet 4 2) 4 Pa
here p= pressure a
Pressure above Kin the ight Limb = pyx PX 829 +p
whore pe = Pressure a B.
quit the wo presturs, we have
msthenen,
Pape
Eh + pa + Pe
PX Rh + ET)
% 104-01) pie e
Difference of pressure at A and B= AX (O, PA) + PAR ~ Bute
In Fig, 2.8 (the two points À and are a the same level and contains he same liquid of density
Py. Then
Pressure above Kim ght Hind = PICA 9s X24 Pa
Pressure shove XX in et lib SP + D+ Ry
gig he two pressure
ES 24 Pa
ae Py Eh à page = Pa + D
= ex np, po} 213)
Problem 2.15 A pipe contains ano of sp. gr. 0. A diferent manometer comeced a the v0
‘otis aná B shows difference in ercur eve ox 15 em. Fond the difference of peste athe me
pois
52 Fluid Mechanics
Solution. Given
Sp. gr. of oi
Difference in mercury eve
Sp. ar. of mercury,
The difference of pressure is given y equation (2.13)
or Pa-Pu = 8% MO, 90
AL 6.15 (13600 - 900) = 18688 Nim?, Ans.
Problem 2.16 A diferenial manometer is connected atthe vo points A and B of vo pipes as
shown in Fig. 2.19. The pipe A contains a liquid of sp. gr. = 13 while pipe B contains a quid of
sp. gr. = 09. Ihe pressures at À and B are À fem and 1.80 kefen? respectively. Find the
Aiference in mercury level in e differential manometer
Solution. Given
Sp. ge of quid at A, 5,
Sp. gr of liquid a,
Pressure at 4 Pi
10% 981 Ni
Presse ar, pp LA kg
8x 10! kg”
2182 100%081 Nm? Cr Tigh 981 *
Density of mereury = 13.6 x 1000 kh"
Taking X-X as datum line
Prosar above XX im the ef Im
13,6 1000 x 981 x h à 150091 x (2 + 3) + pa
= 136 x 1000 x 981 x i+ 7500 9814931 x10°
Pressure above X ln th ight lb = 900 x 9.81 (4 +2) + Pa
= 900 x 9.81% GA + 2)+ 18% 10% 981
Tatin the two pressure, we get
1346 % 1000 x 981A + 7500 x 981 + 9.81 x 10°
900 9.81 (h+ 2) + 18x 108 DT
we get
(+20) «9418
09k + 18+ 182 098 + 198
198-1750 17h + 23
2
Ce Tkgt= 981.8)
Fig. 219
Dividing by 1000 x 9481.
136 +75 + 10
or ht + 175
or 36-09
181 m= 18.1 am. Ans.
127
Problem 2.17. A difereniel manometer i connected at he 190 points A and Bas shown in
Fig. 220, AUB air pressure te 981 Nien? (ab), find the absolute pressure at A
Solution. Given
Ale pressure at 2
or P
81 Nien?
81 10° Nim?
Solution. Given
Sp. gr. of oi
Difference in mercury eve
Sp. ar. of mercury,
The difference of pressure is given y equation (2.13)
or Pa-Pu = 8% MO, 90
AL 6.15 (13600 - 900) = 18688 Nim?, Ans.
Problem 2.16 A diferenial manometer is connected atthe vo points A and B of vo pipes as
shown in Fig. 2.19. The pipe A contains a liquid of sp. gr. = 13 while pipe B contains a quid of
sp. gr. = 09. Ihe pressures at À and B are À fem and 1.80 kefen? respectively. Find the
Aiference in mercury level in e differential manometer
Solution. Given
Sp. ge of quid at A, 5,
Sp. gr of liquid a,
Pressure at 4 Pi
10% 981 Ni
Presse ar, pp LA kg
8x 10! kg”
2182 100%081 Nm? Cr Tigh 981 *
Density of mereury = 13.6 x 1000 kh"
Taking X-X as datum line
Prosar above XX im the ef Im
13,6 1000 x 981 x h à 150091 x (2 + 3) + pa
= 136 x 1000 x 981 x i+ 7500 9814931 x10°
Pressure above X ln th ight lb = 900 x 9.81 (4 +2) + Pa
= 900 x 9.81% GA + 2)+ 18% 10% 981
Tatin the two pressure, we get
1346 % 1000 x 981A + 7500 x 981 + 9.81 x 10°
900 9.81 (h+ 2) + 18x 108 DT
we get
(+20) «9418
09k + 18+ 182 098 + 198
198-1750 17h + 23
2
Ce Tkgt= 981.8)
Fig. 219
Dividing by 1000 x 9481.
136 +75 + 10
or ht + 175
or 36-09
181 m= 18.1 am. Ans.
127
Problem 2.17. A difereniel manometer i connected at he 190 points A and Bas shown in
Fig. 220, AUB air pressure te 981 Nien? (ab), find the absolute pressure at A
Solution. Given
Ale pressure at 2
or P
81 Nien?
81 10° Nim?
Picasa 53]
Demi oft 0.91000 = 90 ka?
Dem of mercury = 134 100 kim
Lette pre A iy
Taking Gar te at KA
se above A nthe ight im
lomo «981 06 + pp
S88 + 38100 toda
Prose above Xin ef m
3.46» 1000» 9.81.0. 490
2981 x02 +,
= HL + ON
Fun the two press heads
Toon 133416 + 176838 +
: a= Wake. SOA 2 ¿res
> 7.888758 Nin? à STERN
10000 cm
Anal presse u A= 887 Ne Ans
2.72 ‘Inverted Utube Diflerential Manometer. lu coms of an ivened Ut
canine «ht qui. The wo ends of the be re connected he plats she eens of
feces be menda nd for cala eens ow rss Pi, 22) shows an.
Ein Ube ral mnomete mec w e tv pa a Bethe pes A
en he rer a Be
La cig quid i et mb below the damn tne XX
eight of qui in righ tind |
À = Difference of light quid
Py = Density of liquidar
Density of lig at
P = Density of tight guid
Pa = Presure at À
= Pressure a B.
“Taking XX as datum line Then presario XX SE >
PAE ig. 221
Pressure in he right ib slow X
OE Ed
Fquating th two pressure
PP X EX = Py 09 X 8 XD, EX
or DR pe 9,8 ers
Problem 2.18 Water is flowing through rue diferent pipes to ich an inverted differential
‘manometer having an où af sp. gr. O38 tr connected. The pressure head i he pipe A In m of water,
{fd the pressure the pipe forthe manometer reading ox shown da Fig. 2.22
Solution. Given
Pressure head at A
= az 9% 72 = 1000 981 x2 = 19620 Na
Fig. 222 shows in arangemont Taking XX s dtu Inc
Pressure below XK in he le lind = PX
2m of war
Demi oft 0.91000 = 90 ka?
Dem of mercury = 134 100 kim
Lette pre A iy
Taking Gar te at KA
se above A nthe ight im
lomo «981 06 + pp
S88 + 38100 toda
Prose above Xin ef m
3.46» 1000» 9.81.0. 490
2981 x02 +,
= HL + ON
Fun the two press heads
Toon 133416 + 176838 +
: a= Wake. SOA 2 ¿res
> 7.888758 Nin? à STERN
10000 cm
Anal presse u A= 887 Ne Ans
2.72 ‘Inverted Utube Diflerential Manometer. lu coms of an ivened Ut
canine «ht qui. The wo ends of the be re connected he plats she eens of
feces be menda nd for cala eens ow rss Pi, 22) shows an.
Ein Ube ral mnomete mec w e tv pa a Bethe pes A
en he rer a Be
La cig quid i et mb below the damn tne XX
eight of qui in righ tind |
À = Difference of light quid
Py = Density of liquidar
Density of lig at
P = Density of tight guid
Pa = Presure at À
= Pressure a B.
“Taking XX as datum line Then presario XX SE >
PAE ig. 221
Pressure in he right ib slow X
OE Ed
Fquating th two pressure
PP X EX = Py 09 X 8 XD, EX
or DR pe 9,8 ers
Problem 2.18 Water is flowing through rue diferent pipes to ich an inverted differential
‘manometer having an où af sp. gr. O38 tr connected. The pressure head i he pipe A In m of water,
{fd the pressure the pipe forthe manometer reading ox shown da Fig. 2.22
Solution. Given
Pressure head at A
= az 9% 72 = 1000 981 x2 = 19620 Na
Fig. 222 shows in arangemont Taking XX s dtu Inc
Pressure below XK in he le lind = PX
2m of war
SA Foi Mechanics
16677 Nim Au
Fa: mr
19620 - 1000 » 9.81 x0,
Pressure below inthe ip in
= Py 1000 4981 x 0.1 - 800x981 x0.12
= 981 ~ 941.76 = py 1922.76
Equating the two pressure, we get
16677 = pe 1922.76
or = 16677 4 1922.76 = 1859.76 Nm?
or ‘Py = 18599 Nic, Ans.
Problem 2.19 In Fig. 223, an inver diferenil manometer is comected 10 wo pipes A and B
whic convey water. The fluid in manometer isa ofp. gr. 03. Forte manometer readings shawn in
Ihe figure find he pressure dference between A ond Re awe
Solution. Cien Sos
Spa of ol =08
Difference ofan the two ins
300 kph
= (80+ 20) 30 =20 em
Taking Jat ine at EX
Prosar helft Limb blow XX
= py - 1000x981 x0 Hip. 228
E20
Pressure in he right i below XX
Pe 1000%9.81 <03 ~ 8009.81 102
Pe 2983 - 15696 = rg 45126
gung the wo pressure pa 2963 = py 45126.
Da P= 4512.6 ~ 2983 = 186946 Nm’. Ans,
Problem 220. Find out the ere rding on eri Ube manometer containing oof
weiße gravi 07 asthe manometric fluid when connected across pipes A and B as shan in Fig. 224
Bet, conveying liquido pei gravis 1.2 and 1.0 end biscbe with manomerie ui. pes A
nd Par rat le so ee an rum Be press nd
Solution. Given
Fig, 224 shows the arrangement, Taking X-X as dat ine
La
Density of quid in pipe A
Density of igi in pipe F
Density of ol
16677 Nim Au
Fa: mr
19620 - 1000 » 9.81 x0,
Pressure below inthe ip in
= Py 1000 4981 x 0.1 - 800x981 x0.12
= 981 ~ 941.76 = py 1922.76
Equating the two pressure, we get
16677 = pe 1922.76
or = 16677 4 1922.76 = 1859.76 Nm?
or ‘Py = 18599 Nic, Ans.
Problem 2.19 In Fig. 223, an inver diferenil manometer is comected 10 wo pipes A and B
whic convey water. The fluid in manometer isa ofp. gr. 03. Forte manometer readings shawn in
Ihe figure find he pressure dference between A ond Re awe
Solution. Cien Sos
Spa of ol =08
Difference ofan the two ins
300 kph
= (80+ 20) 30 =20 em
Taking Jat ine at EX
Prosar helft Limb blow XX
= py - 1000x981 x0 Hip. 228
E20
Pressure in he right i below XX
Pe 1000%9.81 <03 ~ 8009.81 102
Pe 2983 - 15696 = rg 45126
gung the wo pressure pa 2963 = py 45126.
Da P= 4512.6 ~ 2983 = 186946 Nm’. Ans,
Problem 220. Find out the ere rding on eri Ube manometer containing oof
weiße gravi 07 asthe manometric fluid when connected across pipes A and B as shan in Fig. 224
Bet, conveying liquido pei gravis 1.2 and 1.0 end biscbe with manomerie ui. pes A
nd Par rat le so ee an rum Be press nd
Solution. Given
Fig, 224 shows the arrangement, Taking X-X as dat ine
La
Density of quid in pipe A
Density of igi in pipe F
Density of ol
Pressure and ts Measurement 55
Now pressure below XX inthe et im
= 1200 9381 0.3 700 à DA À
Prosar below XX bn ah ight ib
= 1000 + 9381 x (A + 03)
quan the two pressure, we get
a= 1200 9.81 X03 — 700 % 9.81 XA = py = 1000% 9381 (44 03)
Da Pas Gien)
= 1200 9.87 0.3 700 987 k= — 1000 39.81 (14 03)
Dividing by 100004
= 12x03 -078 2 - (h+ 03)
or 03 x 12 + 0.7h= à à 03 or 0.36 - 04
036-030 _ 006 „,
60 0m
% 100 = 20 em. Ans.
Problem 221 An inverted U-sube manometer is connected 10 mo horizonal pipes A and B
‘through which water is lowing. The vertical distance between the aes ofthese pipes s 30 em. When
an oil of specific gravity 0.8 is sed asa gauge Aid, the vertical heights of water columns in he mo
limbs ofthe inverted manomeies (hen measured from the respective centre lines of the pipes) are
{found o be same and equa o 35 om. Determine the diference of pressure between Ihe pipes.
Soon. Gin
Spee gy of using ui = 08 somos
Ti arngenen toma in 224 6)
m Zeno
pa
Sis Eo Do ad
Bec seat CD cal oma a
lement UM
0» 981 (035) wich um
rares
i at (039 - amo xa x03 —b
nn -
100 981 35 Der
‘And pressure at D
Now pressure below XX inthe et im
= 1200 9381 0.3 700 à DA À
Prosar below XX bn ah ight ib
= 1000 + 9381 x (A + 03)
quan the two pressure, we get
a= 1200 9.81 X03 — 700 % 9.81 XA = py = 1000% 9381 (44 03)
Da Pas Gien)
= 1200 9.87 0.3 700 987 k= — 1000 39.81 (14 03)
Dividing by 100004
= 12x03 -078 2 - (h+ 03)
or 03 x 12 + 0.7h= à à 03 or 0.36 - 04
036-030 _ 006 „,
60 0m
% 100 = 20 em. Ans.
Problem 221 An inverted U-sube manometer is connected 10 mo horizonal pipes A and B
‘through which water is lowing. The vertical distance between the aes ofthese pipes s 30 em. When
an oil of specific gravity 0.8 is sed asa gauge Aid, the vertical heights of water columns in he mo
limbs ofthe inverted manomeies (hen measured from the respective centre lines of the pipes) are
{found o be same and equa o 35 om. Determine the diference of pressure between Ihe pipes.
Soon. Gin
Spee gy of using ui = 08 somos
Ti arngenen toma in 224 6)
m Zeno
pa
Sis Eo Do ad
Bec seat CD cal oma a
lement UM
0» 981 (035) wich um
rares
i at (039 - amo xa x03 —b
nn -
100 981 35 Der
‘And pressure at D
56 lid Mechanics
> 2.8 PRESSURE AT A POINT IN COMPRESSIBLE FLUID
For compressible Mulas, density (p) changes with the change of pressure and temperature. Such
problems are encountered in aeronates, oceanography and meteorology where we are concerned
With atmosphere” ait were density. pressure and temperature changes With elevation. Thus or Mus
ith variable density, equation (24) cannot De Integrated, unless the relationship between p and p is.
Know. For pass the equation of states
219)
Now equation (24) is
RS en
tE 015
In equation (24), Z s measured vertical
E
a
‘downward, But i Z is measured vertically up, tien
DE and hence equation (2.16) ecomes
ET am
»
2.8.1 Isothermal Process. Case. temperature Ti constant whichis ru or athermat proc:
Ss. equation (2.17) can he integrated as
ea her Ea
Eee arte re
o og LE 0-78
A
wher ste pressure ere height i If the datum Hoe ten at then Z, = ad Becomes
the pressure at atom ine
mL
»
Lon
m
or pressure a a heim Zi given by p = pg” PT as
2.8.2 Adiabatic Process, If temperature Tis not constant but the process follows adlabatc
law then te relation between pressure and density i piven by
>
Constant = €
=)
© Te rd asp rss, per nd dent eed te STP as Joel ar
rs 101.325 RNA à Tempera = 15°C and Dei = 1225 kp
> 2.8 PRESSURE AT A POINT IN COMPRESSIBLE FLUID
For compressible Mulas, density (p) changes with the change of pressure and temperature. Such
problems are encountered in aeronates, oceanography and meteorology where we are concerned
With atmosphere” ait were density. pressure and temperature changes With elevation. Thus or Mus
ith variable density, equation (24) cannot De Integrated, unless the relationship between p and p is.
Know. For pass the equation of states
219)
Now equation (24) is
RS en
tE 015
In equation (24), Z s measured vertical
E
a
‘downward, But i Z is measured vertically up, tien
DE and hence equation (2.16) ecomes
ET am
»
2.8.1 Isothermal Process. Case. temperature Ti constant whichis ru or athermat proc:
Ss. equation (2.17) can he integrated as
ea her Ea
Eee arte re
o og LE 0-78
A
wher ste pressure ere height i If the datum Hoe ten at then Z, = ad Becomes
the pressure at atom ine
mL
»
Lon
m
or pressure a a heim Zi given by p = pg” PT as
2.8.2 Adiabatic Process, If temperature Tis not constant but the process follows adlabatc
law then te relation between pressure and density i piven by
>
Constant = €
=)
© Te rd asp rss, per nd dent eed te STP as Joel ar
rs 101.325 RNA à Tempera = 15°C and Dei = 1225 kp
Pressure and its Measurement 57
‘where Ks aio of specii constant
ay o
“Thon equation 24) für 7 measured vertically up becomes.
nl
= ante
F
mere ve xt a nat
or | alo
= eral Pe AG conan can ben só]
Dy
* In
But from equation (i), as pe
= Go
Suhsttuting his value of CM above, we get
m
penn
B
« Le 1-7
af 2] He
farm ne tan a whet piesa, temperature and density ae py, Ta en
se | 2412-41
‘where Ks aio of specii constant
ay o
“Thon equation 24) für 7 measured vertically up becomes.
nl
= ante
F
mere ve xt a nat
or | alo
= eral Pe AG conan can ben só]
Dy
* In
But from equation (i), as pe
= Go
Suhsttuting his value of CM above, we get
m
penn
B
« Le 1-7
af 2] He
farm ne tan a whet piesa, temperature and density ae py, Ta en
se | 2412-41
58 Fluid Mechanics
= „cn
ot om en (D.
Subt value of Din quai i, we ge
D
2. 2) [-
a
or Ezy" |
nn
es
equation (219). pa
pressure at ground level where 7 =
Pa = density of air at proud level
A 1
Equation of sae bs 2 Ein
& » am
Susto vals of © 1 gun (219), we gt
n
2.8.3 Temperat
temperature a ny help im al calculated as
tain of sates pound lve and ei tom ground levels wien as
Berta L = aT
m °
Diving des union, we st
(9)
Ro
= „cn
ot om en (D.
Subt value of Din quai i, we ge
D
2. 2) [-
a
or Ezy" |
nn
es
equation (219). pa
pressure at ground level where 7 =
Pa = density of air at proud level
A 1
Equation of sae bs 2 Ein
& » am
Susto vals of © 1 gun (219), we gt
n
2.8.3 Temperat
temperature a ny help im al calculated as
tain of sates pound lve and ei tom ground levels wien as
Berta L = aT
m °
Diving des union, we st
(9)
Ro
Pressure and ts Measurement 39)
Bu from equation (220) i given by
‘so for abate press Le
5
sot ge) ©
A
Temperature Lapse-Rate (L)- It is defined as the rat at which
changes wih lation, To obtain an expression forthe temperature ps rate the tempera given
by equation (2.21) is dtfereatd with respecto Z as
a, (dt
a2 dz\" RE
were eM, and Kat constant
a xx ( )
FAR RP RE
“The temperature pst denoted by and ene
pata em
a" RU
Im equation (2.22).
is constant with big.
(AA > 1. the lapse-ete is negative which means temperature decreases with the increase i
eight
Tn armes, he value o varies wt eight and hence the al of temperature ps us als var
‘rom the sa vel up an levtion of about 11000 m (or 11 m), th temperature of ar decreases
nifomly atthe rate of 1.0065°Cm. fom 11000 m to 32000 m, the temperature remains constant
at 565'C and heno in hi range lap ate er Temperate cs again fe 3200 m m ai
ir
which mans sothermal process, 27. = 0, whieh means temperature
7 ee
Bu from equation (220) i given by
‘so for abate press Le
5
sot ge) ©
A
Temperature Lapse-Rate (L)- It is defined as the rat at which
changes wih lation, To obtain an expression forthe temperature ps rate the tempera given
by equation (2.21) is dtfereatd with respecto Z as
a, (dt
a2 dz\" RE
were eM, and Kat constant
a xx ( )
FAR RP RE
“The temperature pst denoted by and ene
pata em
a" RU
Im equation (2.22).
is constant with big.
(AA > 1. the lapse-ete is negative which means temperature decreases with the increase i
eight
Tn armes, he value o varies wt eight and hence the al of temperature ps us als var
‘rom the sa vel up an levtion of about 11000 m (or 11 m), th temperature of ar decreases
nifomly atthe rate of 1.0065°Cm. fom 11000 m to 32000 m, the temperature remains constant
at 565'C and heno in hi range lap ate er Temperate cs again fe 3200 m m ai
ir
which mans sothermal process, 27. = 0, whieh means temperature
7 ee
(© Fluid Mechonics
Problem 2.22 (SI Unlts) Ifthe amosphere pressure af sea level is 10.148 Nom’, determine the
pressure at a height of 2500 m assuming the pressure variation follows () Hydrosate law, and
{iy othermat Law. The density of ar ts given as 1.208 kam,
Solution. Given
Po st y= 10148 Non = 1048810 Ni?
om. 22280 m
Dayo pee
CO Pr by hydro aw Farhi a, p I med conn ac re
sion 22. =
oy estos Lp
meng, vea Foo =5 -osaz=—os faz
“ ostra
or ao In,
pat o paro
AO LE 20 p= 1208
2 on = 29606 a or TR went
18 Nm, Ans.
(io Pressure by Isothermal Law. Pressure at any help Z by thermal Jaw is given by equation
Ligas
pame om
æ Ty
10.143 x10 À 2 RT and pag =
101d x10 e *
To 1090 tot gf 2802 ar
= 101430 "= 101430 LE = 75743 Nim?
L
ini
SE em? = 2.574 Nom, Ans.
ve
10
Problem 223 The barometric pressure a se levels 760 mm of mercury while that on a mountain
top i5 783 mm. Ifthe density of air in assumed constant at 12 Ref, what is the elevation ofthe
‘mountaintop?
Solution. Given
Pressure” at sen, y= 760 mm oF Hig
101396 Na?
Hee presse bead (2) 5 given a Tan o Hg Hence p= Ta of Hp ee des (por mery
16% 1000 pm Hence presu () wil au io px 4% Ze, 1363 1000 981 2.
= om Hence pesas) wil te cyl op Ze 1863 1000 01 x I,
Problem 2.22 (SI Unlts) Ifthe amosphere pressure af sea level is 10.148 Nom’, determine the
pressure at a height of 2500 m assuming the pressure variation follows () Hydrosate law, and
{iy othermat Law. The density of ar ts given as 1.208 kam,
Solution. Given
Po st y= 10148 Non = 1048810 Ni?
om. 22280 m
Dayo pee
CO Pr by hydro aw Farhi a, p I med conn ac re
sion 22. =
oy estos Lp
meng, vea Foo =5 -osaz=—os faz
“ ostra
or ao In,
pat o paro
AO LE 20 p= 1208
2 on = 29606 a or TR went
18 Nm, Ans.
(io Pressure by Isothermal Law. Pressure at any help Z by thermal Jaw is given by equation
Ligas
pame om
æ Ty
10.143 x10 À 2 RT and pag =
101d x10 e *
To 1090 tot gf 2802 ar
= 101430 "= 101430 LE = 75743 Nim?
L
ini
SE em? = 2.574 Nom, Ans.
ve
10
Problem 223 The barometric pressure a se levels 760 mm of mercury while that on a mountain
top i5 783 mm. Ifthe density of air in assumed constant at 12 Ref, what is the elevation ofthe
‘mountaintop?
Solution. Given
Pressure” at sen, y= 760 mm oF Hig
101396 Na?
Hee presse bead (2) 5 given a Tan o Hg Hence p= Ta of Hp ee des (por mery
16% 1000 pm Hence presu () wil au io px 4% Ze, 1363 1000 981 2.
= om Hence pesas) wil te cyl op Ze 1863 1000 01 x I,
Pressure and ts Measurement 61
Pressure at mountain, p= 735 am of Hp
25 x 136 x 1000 x 981 = 98060 Nim?
1000
Density ofa, pe 12 kya
Let hr Height of the mountain from se level.
We know that as the elevation above the serlvel increases, the atmosphere pressure decreases
Here te density of airs given constant, hence the pressure at any height A above the sea-level
lv by the equation,
pap-pxsxh
= = oz 101396 ~ 98060
px 981
Problem 2:24 Calculate she pressure at a height of 7500 m above sea level te atmospheric
pressure is 10.143 Nem? and temperature is 15°C at the seatvel, assume (air incompressible,
{il pressure variation follows isothermal law, and) pressure variation follows adlaba aw: Take
the density of air atthe sea-level as equal to 1.285 ku. Neglect variation af with alude
Solution. Given
283.03 m. Am
Me an een ¿=1808
Pressure at sea-level, Po = 10.143 Niem? = 10.143 x 10 N/m?
Deny rat enters
DE then al compr
dr
Fée fon à peme-prié- ai
a Papo (= Zen 1
EB 2a
= 101430 - 94543 = 6887 Nim? = 0.688 N;.. Ans.
Fée ar a tas
a ere a
Fame
101830 € M à 101430 5 > Los crane
01430. = 101430 x 39376
939 Nim? or 3393 Neem? Ans.
iy Pressure variation follows adiabatic law: (k= LAL
where y= 1285 kg/m"
Using equation @.19), we have pe |
m
Pressure at mountain, p= 735 am of Hp
25 x 136 x 1000 x 981 = 98060 Nim?
1000
Density ofa, pe 12 kya
Let hr Height of the mountain from se level.
We know that as the elevation above the serlvel increases, the atmosphere pressure decreases
Here te density of airs given constant, hence the pressure at any height A above the sea-level
lv by the equation,
pap-pxsxh
= = oz 101396 ~ 98060
px 981
Problem 2:24 Calculate she pressure at a height of 7500 m above sea level te atmospheric
pressure is 10.143 Nem? and temperature is 15°C at the seatvel, assume (air incompressible,
{il pressure variation follows isothermal law, and) pressure variation follows adlaba aw: Take
the density of air atthe sea-level as equal to 1.285 ku. Neglect variation af with alude
Solution. Given
283.03 m. Am
Me an een ¿=1808
Pressure at sea-level, Po = 10.143 Niem? = 10.143 x 10 N/m?
Deny rat enters
DE then al compr
dr
Fée fon à peme-prié- ai
a Papo (= Zen 1
EB 2a
= 101430 - 94543 = 6887 Nim? = 0.688 N;.. Ans.
Fée ar a tas
a ere a
Fame
101830 € M à 101430 5 > Los crane
01430. = 101430 x 39376
939 Nim? or 3393 Neem? Ans.
iy Pressure variation follows adiabatic law: (k= LAL
where y= 1285 kg/m"
Using equation @.19), we have pe |
m
[62 Fluid Mechanies
pou |
101430 1 26621 #4 191430 «(73379
0 Nin or 3481S Ans
Problem 2.25 Calculate the pressure and density of erat a height of 4000 m from sea-level where
pressure and temperature ofthe air are 10.143 Nem: and 15°C respecte, Ihe temperature lapse
ae à given us 0.0065°C/m. Take density of er ut seadevel equal 1.283 hyn
Solution. ven
Height,
Pressure at sa el, 10.143 10" = 101430
‘Temperate a seatevel,
Tea 1S = 280K
“Temperature lapse 0.0065°K a,
AQ 815
sim
Fa
“This means that te valve of power index À 12
(9 Pressure a 4000 m heights given by equation 2.19) as
22 and py = 1285
4000 1285])22 o
DUT]
1430 +595
Rf
ora
= 101430 11-0918 =
rosıx
= 60350 Nim? = 6035 N. Ans.
pou |
101430 1 26621 #4 191430 «(73379
0 Nin or 3481S Ans
Problem 2.25 Calculate the pressure and density of erat a height of 4000 m from sea-level where
pressure and temperature ofthe air are 10.143 Nem: and 15°C respecte, Ihe temperature lapse
ae à given us 0.0065°C/m. Take density of er ut seadevel equal 1.283 hyn
Solution. ven
Height,
Pressure at sa el, 10.143 10" = 101430
‘Temperate a seatevel,
Tea 1S = 280K
“Temperature lapse 0.0065°K a,
AQ 815
sim
Fa
“This means that te valve of power index À 12
(9 Pressure a 4000 m heights given by equation 2.19) as
22 and py = 1285
4000 1285])22 o
DUT]
1430 +595
Rf
ora
= 101430 11-0918 =
rosıx
= 60350 Nim? = 6035 N. Ans.
Pressure and ts Measurement 63
(0 Density. Using equation of state, we get
Barr
a
here p=Dressue at 4000 m height
= Density at 4009 m heighe
7 = Temperature at 4000 m bight
Now Tis calculated from temperature aprte as
at 4000 = à à A x 4000 = 15 006 000 = 15-26=- IC
78 $= 273 ~ LN = 260°K
D 00 a
Density is given by En = ORS kg/m". Ans.
er Ke” Frags «ae RS ‘ia
Problem 2.26 An weroplane i ling ar an alude of 5000 m. Calealte the pressure around the
aeroplane given he loc. rue in e atmosphere as 0.0065°Kin Neglect variation of with ol.
Take pressure and temperatare at ground level as 10.143 Nem? and 15°C and density of air as
1255 kom.
Solution. Given
Height,
Lapse,
Pressure at round level,
Density,
832 = 2555°K
Firs find the valo of power Inder Las.
a)
From equation (2.22), we have
po 101830
Mn any
T, 1285x288 =
where R
(0 Density. Using equation of state, we get
Barr
a
here p=Dressue at 4000 m height
= Density at 4009 m heighe
7 = Temperature at 4000 m bight
Now Tis calculated from temperature aprte as
at 4000 = à à A x 4000 = 15 006 000 = 15-26=- IC
78 $= 273 ~ LN = 260°K
D 00 a
Density is given by En = ORS kg/m". Ans.
er Ke” Frags «ae RS ‘ia
Problem 2.26 An weroplane i ling ar an alude of 5000 m. Calealte the pressure around the
aeroplane given he loc. rue in e atmosphere as 0.0065°Kin Neglect variation of with ol.
Take pressure and temperatare at ground level as 10.143 Nem? and 15°C and density of air as
1255 kom.
Solution. Given
Height,
Lapse,
Pressure at round level,
Density,
832 = 2555°K
Firs find the valo of power Inder Las.
a)
From equation (2.22), we have
po 101830
Mn any
T, 1285x288 =
where R
A Fluid Mechonies
con oes io
o100[1- (+2
2 win
222 stot)
ora [io OES)
101430 [1 0.11288) = 101430 x 05175 = 52490 Nm"
5249 Nom’. Ans.
HIGHLIGHTS
1 The pressure at any pot in Ba dele a he fore or un te
2 The Pascal's aw stats at est of pres fora al et cgi al tions
3. Prosure variation at pia a gt ss ls by the yds aw which tas hat he ate of
ines of promu in he verily downward econ gl ote apie weight of he i,
Sexe
4 The pressure any point na incompresile Mi (gid equal 0 the product of dens off at
at point, acceleration due dora and vertical eight rm fee surface of fi,
en
Axa rear is the presure In which absolut vacuum preset stan as datum while gage
Pressure ithe pressure in which he atmosphere pressure I ken as datum
Manors isa device wed fer messing pressure at pol in ad.
Manor ar classified as (o) Simple enormes and (0) Din moms
Simple manometes ar usd fo mein proue pon wile ier noms re se or
rein the ferne opresores beten th wo pois na pio, or no dese ips.
A single column manometer (or mcm ls usd fo messing sal! presu hero accus
ote.
10. The pressure at pot ia tae compres is ote by combing wo equations, Le ean
fsa for ps má equation sven by Asa
AL The prose at à height Z in a sae comprewsble Nu (gun) undergoing thermal compression
e)
poe ee
R= Gas constant
T= Absolute temperate
12. The pressure and temperate au eight in à ti compressible fd (gs) undergoing dba
compression (np = cont)
con oes io
o100[1- (+2
2 win
222 stot)
ora [io OES)
101430 [1 0.11288) = 101430 x 05175 = 52490 Nm"
5249 Nom’. Ans.
HIGHLIGHTS
1 The pressure at any pot in Ba dele a he fore or un te
2 The Pascal's aw stats at est of pres fora al et cgi al tions
3. Prosure variation at pia a gt ss ls by the yds aw which tas hat he ate of
ines of promu in he verily downward econ gl ote apie weight of he i,
Sexe
4 The pressure any point na incompresile Mi (gid equal 0 the product of dens off at
at point, acceleration due dora and vertical eight rm fee surface of fi,
en
Axa rear is the presure In which absolut vacuum preset stan as datum while gage
Pressure ithe pressure in which he atmosphere pressure I ken as datum
Manors isa device wed fer messing pressure at pol in ad.
Manor ar classified as (o) Simple enormes and (0) Din moms
Simple manometes ar usd fo mein proue pon wile ier noms re se or
rein the ferne opresores beten th wo pois na pio, or no dese ips.
A single column manometer (or mcm ls usd fo messing sal! presu hero accus
ote.
10. The pressure at pot ia tae compres is ote by combing wo equations, Le ean
fsa for ps má equation sven by Asa
AL The prose at à height Z in a sae comprewsble Nu (gun) undergoing thermal compression
e)
poe ee
R= Gas constant
T= Absolute temperate
12. The pressure and temperate au eight in à ti compressible fd (gs) undergoing dba
compression (np = cont)
Pressure and ts Measurement 65
er
sat nan, Se]
tee py, Ta us prs and tempers se evel = 1 or st
18. The rate which te teperature changes Wits coat shows as Temperature Lapse Ra I
Bene
a)
it k= temperate e200
(69 À temperature decreases wih the cree of eight
EXERCISE
(A) THEORETICAL PROBLEMS
Dee presu Oben a expression fr the press Inter aa pin Mi
Waa do you uti by Hydrate Lan 7
Pier between () Absolute ad gage proue, i) Simple manemeier an diferent manom
Sen and (i) ere and presse engen
Wha do yon mea by vacuum peu 2
7. Wat do yoo mea by gl cou manos? Mow are ey wed forthe menureren of neun?
e Whit ie the deren between Udo differenti monometere and Inverted Une área!
ramo ? Were we hey und?
9 Distinguish between manometars and mechanics gages, WRAL ar he dilo pes of mechanical
peste gauges?
10. Devivean expression or ie pressure at igh rm seu eve fra sti ar when the compresion of
‘eas assed othe. Te presse ad temperature Seen af pan Y, opaco.
11, rove he pretend tpt for an daha process bg Zen e or fora sae it
bee py nd Ty ae the presse and temperuue at xa eve
22, Wi you til by e cr ones Lap Ra Oda an opos or e
13, What is hydrous gresare distin? Give one example here presas shuren is
ron hydro
14, Esplin bly the working precip of Bourdon Pressure Gauge wit a nat ice
INTL, Hipieraod. 5 2002)
(8) NUMERICAL PROBLEMS
1. A dante pres has asar of 30 ameter nd aplunge ofS cm ameter Find the iaa ie by
the ae pros when ne ore applied the large ik ON. Tams 44 ENT
2. A slo press has a ram of 20 em date and à linge of 4 cs dite i sd or ing &
Weight of 20 AN, Find the fore egies at he lungen ans. 9005)
er
sat nan, Se]
tee py, Ta us prs and tempers se evel = 1 or st
18. The rate which te teperature changes Wits coat shows as Temperature Lapse Ra I
Bene
a)
it k= temperate e200
(69 À temperature decreases wih the cree of eight
EXERCISE
(A) THEORETICAL PROBLEMS
Dee presu Oben a expression fr the press Inter aa pin Mi
Waa do you uti by Hydrate Lan 7
Pier between () Absolute ad gage proue, i) Simple manemeier an diferent manom
Sen and (i) ere and presse engen
Wha do yon mea by vacuum peu 2
7. Wat do yoo mea by gl cou manos? Mow are ey wed forthe menureren of neun?
e Whit ie the deren between Udo differenti monometere and Inverted Une área!
ramo ? Were we hey und?
9 Distinguish between manometars and mechanics gages, WRAL ar he dilo pes of mechanical
peste gauges?
10. Devivean expression or ie pressure at igh rm seu eve fra sti ar when the compresion of
‘eas assed othe. Te presse ad temperature Seen af pan Y, opaco.
11, rove he pretend tpt for an daha process bg Zen e or fora sae it
bee py nd Ty ae the presse and temperuue at xa eve
22, Wi you til by e cr ones Lap Ra Oda an opos or e
13, What is hydrous gresare distin? Give one example here presas shuren is
ron hydro
14, Esplin bly the working precip of Bourdon Pressure Gauge wit a nat ice
INTL, Hipieraod. 5 2002)
(8) NUMERICAL PROBLEMS
1. A dante pres has asar of 30 ameter nd aplunge ofS cm ameter Find the iaa ie by
the ae pros when ne ore applied the large ik ON. Tams 44 ENT
2. A slo press has a ram of 20 em date and à linge of 4 cs dite i sd or ing &
Weight of 20 AN, Find the fore egies at he lungen ans. 9005)
Ti ai
on po Im A DOT
Ine dey oe = ENE, Am 0390 Ne) 0389 No 39 New
‘Hep ey apna ern NP cren ht hl
ern rer
BE On a pe ae ea
EE Re Ten
a open nk conato alert a devo LS mandabove an it. ge 08 for deu of? Fad
fhe pese em (a einer oe wo gids, and a Bo othe ak
ms ST Ne 504 Nem
“Th dacs of so ison md a ups po o Rr re 2 cn 10 cepo, À
[ers 9 upload MB byt ag icon hen =) son are
Ale am eve (6) sal pon 20e above e arg pan, Te dense Bald nh ck
¿iento ‘hm DO sn)
Dew an ine pp ls 20 sim ri we, Ts
ES thn tm Ne
A pie marc se 6 meme the pressure o Gp. gr. = O8) wing in apipe ie I eh
imo open te amanphere and et inh e connected e pie. Te ero teppei cm bo
Ih level of mercury op. 136) the ig me. he ilernee fm level ah ob 15
(Om deemin he sole press ft ee pio La Ne (Ams 12088 Ncw]
‘single marometr (Ute eomaring merry L comece pipe in which an al fs. O8
Pind the acu, pren in pipe. the lees of trey lvl ne to ibe 20m an high
‘fo the kin rn he ent ofthe pipe 9 em ben. Tans. 2186 New|
A Ange sun vers ramamter (Le, ret) 1 connected o pipe containing i Fs...
e ve of me reser eH rh rc of te mana bo. Te reser orale merry oF
rene of merry esla inthe reset and igh in i O em. Fi the prete in I is
TAs 74 Nn]
A pipe cos an op. 8:08. A liven manner competed a the two pis À and Bol be
: ane 25113680)
A Uitte dierenal manometer contes wo presre pipes A and M; Pipe A coins can
99. 3. under a ru of LTR Nk. The pipe A he. 25 m above pipe Fina the ference of
rue mney mery as ad ling Uwe Tans 3136 of mre
IR 1848 Nl (he), nd the abet pesar A (ane. 697 New
on po Im A DOT
Ine dey oe = ENE, Am 0390 Ne) 0389 No 39 New
‘Hep ey apna ern NP cren ht hl
ern rer
BE On a pe ae ea
EE Re Ten
a open nk conato alert a devo LS mandabove an it. ge 08 for deu of? Fad
fhe pese em (a einer oe wo gids, and a Bo othe ak
ms ST Ne 504 Nem
“Th dacs of so ison md a ups po o Rr re 2 cn 10 cepo, À
[ers 9 upload MB byt ag icon hen =) son are
Ale am eve (6) sal pon 20e above e arg pan, Te dense Bald nh ck
¿iento ‘hm DO sn)
Dew an ine pp ls 20 sim ri we, Ts
ES thn tm Ne
A pie marc se 6 meme the pressure o Gp. gr. = O8) wing in apipe ie I eh
imo open te amanphere and et inh e connected e pie. Te ero teppei cm bo
Ih level of mercury op. 136) the ig me. he ilernee fm level ah ob 15
(Om deemin he sole press ft ee pio La Ne (Ams 12088 Ncw]
‘single marometr (Ute eomaring merry L comece pipe in which an al fs. O8
Pind the acu, pren in pipe. the lees of trey lvl ne to ibe 20m an high
‘fo the kin rn he ent ofthe pipe 9 em ben. Tans. 2186 New|
A Ange sun vers ramamter (Le, ret) 1 connected o pipe containing i Fs...
e ve of me reser eH rh rc of te mana bo. Te reser orale merry oF
rene of merry esla inthe reset and igh in i O em. Fi the prete in I is
TAs 74 Nn]
A pipe cos an op. 8:08. A liven manner competed a the two pis À and Bol be
: ane 25113680)
A Uitte dierenal manometer contes wo presre pipes A and M; Pipe A coins can
99. 3. under a ru of LTR Nk. The pipe A he. 25 m above pipe Fina the ference of
rue mney mery as ad ling Uwe Tans 3136 of mre
IR 1848 Nl (he), nd the abet pesar A (ane. 697 New
Pressureand its Measurement 67
15. An invented len masonic canin an ol op. 8-0 ls comet od he fase of
rares al vo point al pipe containing water Ub manors rang iO em. fin te drone
pres Ans 3924 Nia]
16, In abv Fig. 226 sows u ise deena manner ont tot pipes an erating
alr Te im marr ic Cap. 08. Fo man einge om in de gus i the
een of peste ead een ah Va. 0.26 of water}
17, Ihe snopes proue a reel 10.43 Men”, demise the pressure a ight of 2000 m
suring thal ie prepare veran fl () Hyon aw, and (4) ea Te density ot
Ari gen as 1208 ms 277 Ne 803 NA]
1%, Cal the proue a a ol 00 m above se eve te simon pres 113 An?
and tempera à LC the sx lve sung () ar soca res 49 pre arto allows
able a, and) pesao varon follows em aw, Take te deny ofa these evel as
a0 1285 Ag. Nglc vain of g wih alte
(Ams (6078 Ni, i) 313304 (i) 3748 NP
19. Calculate he pose and deal of aa high 300 m above se vel whee posse and ten
ero har ar 10188 Ne and 15°C oeste, The tempera lp ra lie 065°
Kn Tato easly ofa a ve opal o 1285 ka [Ans 6 896 Nit 0937 hn)
20. An sepas sing an lide of 4000 m. Calculate the pressure aroun te aeopa, give
lune ral in the simospbere as 0.069". Neglc arian of g wil able Take prose and
lemprtize a ground leves 10.149 NA ad 15°C respectively. The ess ofa a ground reis
ven a 1285 kp : (Ans 88 Sen
peste 000 1 above save, sumi (ar Incomprosii, them varian of pressure
And dens, and (i) aaa variation of presse und desi, Asme den ol ar at seve as
28 kg. Neg van lg wih lade : “
(Ans. (95013 Ni, 6) 3745 NA, i LS AN]
22. An io. à under pra of 1972 KN”
(Wha she posar he expressed ete of ae
Ai) Whats te presse ad expres in me of? (Ans) 14m Gi) 175)
23, Te aonb rent tthe sc leel is 013 KN nd oper 13°C. Cleat the reste
00 m above sieve, suming (bots vara of pres aad desk. an (1) alba
‘aration of presse ad ds. Assure denso sr al satel 1.285 An lc aston of
Derve Le (onal at you may we ans (937454 LSA
24. What ae te gauge presse a sso resize a pol à m blow the ie surface Fa li of
spent gray 1.9. mos pesao equivale o 750 mol mere .
(As. 60037 Nh nd 16000 Na
25. Find he gage pressure und bolt pres in Nia poi Sm below tee stc gui of
spo 12e spero presi segue! o 250 am of mere .
Ans $7088 Nh 7150 Mal
hic sm teow tee sta ofthe gud. Te asp peso Mea seit o 769m
orme ns 17092 Nr à 18696 Na
15. An invented len masonic canin an ol op. 8-0 ls comet od he fase of
rares al vo point al pipe containing water Ub manors rang iO em. fin te drone
pres Ans 3924 Nia]
16, In abv Fig. 226 sows u ise deena manner ont tot pipes an erating
alr Te im marr ic Cap. 08. Fo man einge om in de gus i the
een of peste ead een ah Va. 0.26 of water}
17, Ihe snopes proue a reel 10.43 Men”, demise the pressure a ight of 2000 m
suring thal ie prepare veran fl () Hyon aw, and (4) ea Te density ot
Ari gen as 1208 ms 277 Ne 803 NA]
1%, Cal the proue a a ol 00 m above se eve te simon pres 113 An?
and tempera à LC the sx lve sung () ar soca res 49 pre arto allows
able a, and) pesao varon follows em aw, Take te deny ofa these evel as
a0 1285 Ag. Nglc vain of g wih alte
(Ams (6078 Ni, i) 313304 (i) 3748 NP
19. Calculate he pose and deal of aa high 300 m above se vel whee posse and ten
ero har ar 10188 Ne and 15°C oeste, The tempera lp ra lie 065°
Kn Tato easly ofa a ve opal o 1285 ka [Ans 6 896 Nit 0937 hn)
20. An sepas sing an lide of 4000 m. Calculate the pressure aroun te aeopa, give
lune ral in the simospbere as 0.069". Neglc arian of g wil able Take prose and
lemprtize a ground leves 10.149 NA ad 15°C respectively. The ess ofa a ground reis
ven a 1285 kp : (Ans 88 Sen
peste 000 1 above save, sumi (ar Incomprosii, them varian of pressure
And dens, and (i) aaa variation of presse und desi, Asme den ol ar at seve as
28 kg. Neg van lg wih lade : “
(Ans. (95013 Ni, 6) 3745 NA, i LS AN]
22. An io. à under pra of 1972 KN”
(Wha she posar he expressed ete of ae
Ai) Whats te presse ad expres in me of? (Ans) 14m Gi) 175)
23, Te aonb rent tthe sc leel is 013 KN nd oper 13°C. Cleat the reste
00 m above sieve, suming (bots vara of pres aad desk. an (1) alba
‘aration of presse ad ds. Assure denso sr al satel 1.285 An lc aston of
Derve Le (onal at you may we ans (937454 LSA
24. What ae te gauge presse a sso resize a pol à m blow the ie surface Fa li of
spent gray 1.9. mos pesao equivale o 750 mol mere .
(As. 60037 Nh nd 16000 Na
25. Find he gage pressure und bolt pres in Nia poi Sm below tee stc gui of
spo 12e spero presi segue! o 250 am of mere .
Ans $7088 Nh 7150 Mal
hic sm teow tee sta ofthe gud. Te asp peso Mea seit o 769m
orme ns 17092 Nr à 18696 Na
if
HYDROSTATIC FORCES!
ONFSURFACES
> 3.1 INTRODUCTION
‘This chapter deals with the Mids (Le, liquids and gases) at rest. This means that there wi e no
relative motion between adjacent or neighbouring Maid ayers. The velocity gradient which sequal to
Be ye ey el e es on,
‘te
stl zen or À <0, The shear stress which is aqua ot À will aloe zur. Then the Force ct
u aut .
on the Mui particles wil be
1. due to pressure of ud normal 0 Ih surface,
2. due o gravity (or self avia of uid packs).
> 3.2 TOTAL PRESSURE AND CENTRE OF PRESSURE
‘Total pressure is defined a the force exerted by a tie aon a surface lier plane or curved
‘wh the Md comes im contact with the surface This force always dts normal othe Surface
Centre of pressure i defined asthe point of aplication ofthe toral pressure on he surface There
ar our eases of submerged surfaces on which he total pressure Force and centre of Pressures Be
determined. The submerged surfaces may bo
1. Verial plano surface,
2. Horizontal plane surface,
3. Inclined plane surface. und
4. Curvo surco
> 33. VERTICAL PLANE SURFACE SUBMERGED IN LIQUID
Considera plane vertical surface of scary shape immerse in a liquid as sos in Fp, 3.1
Let A= Total ard of the surface
Fy = Distance of C.. ofthe asa from fee surface of Hii
6 = Comte of gravity of plane surface
P = Contr of pressure
= Distance o centre of pressure om ce surface of hg
69
HYDROSTATIC FORCES!
ONFSURFACES
> 3.1 INTRODUCTION
‘This chapter deals with the Mids (Le, liquids and gases) at rest. This means that there wi e no
relative motion between adjacent or neighbouring Maid ayers. The velocity gradient which sequal to
Be ye ey el e es on,
‘te
stl zen or À <0, The shear stress which is aqua ot À will aloe zur. Then the Force ct
u aut .
on the Mui particles wil be
1. due to pressure of ud normal 0 Ih surface,
2. due o gravity (or self avia of uid packs).
> 3.2 TOTAL PRESSURE AND CENTRE OF PRESSURE
‘Total pressure is defined a the force exerted by a tie aon a surface lier plane or curved
‘wh the Md comes im contact with the surface This force always dts normal othe Surface
Centre of pressure i defined asthe point of aplication ofthe toral pressure on he surface There
ar our eases of submerged surfaces on which he total pressure Force and centre of Pressures Be
determined. The submerged surfaces may bo
1. Verial plano surface,
2. Horizontal plane surface,
3. Inclined plane surface. und
4. Curvo surco
> 33. VERTICAL PLANE SURFACE SUBMERGED IN LIQUID
Considera plane vertical surface of scary shape immerse in a liquid as sos in Fp, 3.1
Let A= Total ard of the surface
Fy = Distance of C.. ofthe asa from fee surface of Hii
6 = Comte of gravity of plane surface
P = Contr of pressure
= Distance o centre of pressure om ce surface of hg
69
70 Fluid Mechanics
(a) Total Pressure (F). The total pressure on the surface FREE SURFACE OF LOUD
may be determined by dividing teenie surface nt a aumber
of smal parle tips. The force on small stip is then calcu-
at ad the total pressure force onthe whole aa i called.
by Integrating the force on sal stip.
Considera tip of thickness ah and wid 0 at depth of
fom fie surface of quid as shown in Fig. 3.1
Pressure intensity onthe sip. = pat
(Ses equation 29)
Ares of he strip, anda
Total pressure force on suip, dF =p X Area
= pan x xd
ae ee art
far fosse -prfecicat
sa Joxaxan= jou
= Moment of surface arca about the fee surface of guid
‘Area of surface x Distance of CG. from tee surface
xh
a Fa pea Gay
For water the value of p = 1000 kg and 4 = 9.81 mis. The force vil bein Newton,
(6) Centre of Pressure he). Centre of pressure is calculated by using the “Principle of Moment
hich sites that the moment ofthe resultat force about an ais is equal othe sum of moments OF the
components about he same axis
The resultant force F is acing at. ata distance from tee surface ofthe liquid as shown
Fig. 3.1 Hence moment ofthe force F about re surco ofthe lid = F 32)
Moment of force dF, acting on a sip about foo surface of liquid
a aF xh Lo d= pgh xb xa)
= pgh xb ah
Sum of moments of al such forces about fee surface of iui
José ane fox ira
y foe ps fax
bw Joan fun
foment of Inertia ofthe surface about fee surface of Hig
Sum of moments about fee surface
Pal ry
(a) Total Pressure (F). The total pressure on the surface FREE SURFACE OF LOUD
may be determined by dividing teenie surface nt a aumber
of smal parle tips. The force on small stip is then calcu-
at ad the total pressure force onthe whole aa i called.
by Integrating the force on sal stip.
Considera tip of thickness ah and wid 0 at depth of
fom fie surface of quid as shown in Fig. 3.1
Pressure intensity onthe sip. = pat
(Ses equation 29)
Ares of he strip, anda
Total pressure force on suip, dF =p X Area
= pan x xd
ae ee art
far fosse -prfecicat
sa Joxaxan= jou
= Moment of surface arca about the fee surface of guid
‘Area of surface x Distance of CG. from tee surface
xh
a Fa pea Gay
For water the value of p = 1000 kg and 4 = 9.81 mis. The force vil bein Newton,
(6) Centre of Pressure he). Centre of pressure is calculated by using the “Principle of Moment
hich sites that the moment ofthe resultat force about an ais is equal othe sum of moments OF the
components about he same axis
The resultant force F is acing at. ata distance from tee surface ofthe liquid as shown
Fig. 3.1 Hence moment ofthe force F about re surco ofthe lid = F 32)
Moment of force dF, acting on a sip about foo surface of liquid
a aF xh Lo d= pgh xb xa)
= pgh xb ah
Sum of moments of al such forces about fee surface of iui
José ane fox ira
y foe ps fax
bw Joan fun
foment of Inertia ofthe surface about fee surface of Hig
Sum of moments about fee surface
Pal ry
Hydrostatic Forces on Surfaces 71
Equating G2) and (33) we get
Fx I= ply
Da F=pgañ
PRA xi = palo
a me Ph do 84)
Dean an
y the ore of parallel axis, we bare
eier AXE
whore 1 = Moment of Inert of teu about an axis passing trou the €. ofthe area and pale
Wo the fee surface of Il,
Substittng A im equation (3), we got
wre ñ as
ak an
An equation (3.9. isthe distance of CG. ofthe area ofthe vertical surface fom fre surface of
‘te ld. Hence rom equation (3.5), itis clear that
(0) Centre of pressure (he. 19) les below te cetro of gravity ofthe verda! surface.
i) The distance of cent of pressure rom fee surface of Il independent ofthe density ofthe
liquid
Tube The moments o net and other geamelee proper of some important plane surfaces
omen era | Moment
Plane face CG sm | Ara | “trough amd | has (hu)
ase orale as (do)
oe E ce
a 3
We.
=
ela
1. Rectangle
Contd
Equating G2) and (33) we get
Fx I= ply
Da F=pgañ
PRA xi = palo
a me Ph do 84)
Dean an
y the ore of parallel axis, we bare
eier AXE
whore 1 = Moment of Inert of teu about an axis passing trou the €. ofthe area and pale
Wo the fee surface of Il,
Substittng A im equation (3), we got
wre ñ as
ak an
An equation (3.9. isthe distance of CG. ofthe area ofthe vertical surface fom fre surface of
‘te ld. Hence rom equation (3.5), itis clear that
(0) Centre of pressure (he. 19) les below te cetro of gravity ofthe verda! surface.
i) The distance of cent of pressure rom fee surface of Il independent ofthe density ofthe
liquid
Tube The moments o net and other geamelee proper of some important plane surfaces
omen era | Moment
Plane face CG sm | Ara | “trough amd | has (hu)
ase orale as (do)
oe E ce
a 3
We.
=
ela
1. Rectangle
Contd
72. Fluid Mechanics
Women of wera | Moment of
Plane face €6.jrom me | Area | aboatam ats passing | inert abou
base through CO and | base te)
paral boe o)
3 One
IL bes eto ap) -
Problem 31 A rectangulor plane surface ie 2 m wide and 3 m deep. I les in vertical plane in
water, Determine th oral pressure and postion of centre of pressure on the plane surface chen its
"per edge is horizontal aná (a) coincides with water surface, (9) 2S m below the free water surface.
‘Solution. Gives,
Width of plane surface, 6 =2m
Depth of plane suce, d= 3m
(a) Upper edge coinchles with water surface (Pg. 3.2). Total pressure s given by equation (3.1)
ae REE WATER SURFACE
where sim ms
@-15m 5
1000 «981 x 6 x 18 | ot | om
8820 No Ans Le
Depth ot cet of pressures given y equation (3.5) as
igs |
me ler
sí amd
whore fy Pout CG of the are of surface a
Women of wera | Moment of
Plane face €6.jrom me | Area | aboatam ats passing | inert abou
base through CO and | base te)
paral boe o)
3 One
IL bes eto ap) -
Problem 31 A rectangulor plane surface ie 2 m wide and 3 m deep. I les in vertical plane in
water, Determine th oral pressure and postion of centre of pressure on the plane surface chen its
"per edge is horizontal aná (a) coincides with water surface, (9) 2S m below the free water surface.
‘Solution. Gives,
Width of plane surface, 6 =2m
Depth of plane suce, d= 3m
(a) Upper edge coinchles with water surface (Pg. 3.2). Total pressure s given by equation (3.1)
ae REE WATER SURFACE
where sim ms
@-15m 5
1000 «981 x 6 x 18 | ot | om
8820 No Ans Le
Depth ot cet of pressures given y equation (3.5) as
igs |
me ler
sí amd
whore fy Pout CG of the are of surface a
Hydrostatic Forces on Surfaces 73)
ma AS 415205 +15 «20m. Ans
Bas
(0) Upper edge is 25 m below water surface (Fig, 3). Total pressure () is given by (3.1)
waren suarace
Distance of CS. from fee surface of water
3 25m
s+i |
a F ill
Centre of pressure ls given by I y 2e Y
where p= 45,4 = 6.0. À 2 40
* ham
we Fig 33
E +10
60x40
1875 à 40 = 4.1875 = 41878 m. Ans.
Problem 3.2 Determine the nal pressure on a circular plate of diameter 1.5 m which i placed
verrcalls in water im such a way thatthe centre ofthe plate is 3m below the ree surface of water ind
‘he poston of centre of presure ar FRE SURFACE
Solution. Given: Dis. of pate, d = LS m
kan |
2 Ama,
“Total pressure ls piven by equation (3.1)
= peat
LOU x 981» 1767 3.0 N
2 52002581 N, Ans.
Position of centre of pressure (h*) I given by equation (3.5. ee.
etek Fig 34
rat must A
where las EEE > 02485 n°
oF ot
$ yo BS | 39 2 00468 + 30
6730
= 2.0468 m. Ans
ma AS 415205 +15 «20m. Ans
Bas
(0) Upper edge is 25 m below water surface (Fig, 3). Total pressure () is given by (3.1)
waren suarace
Distance of CS. from fee surface of water
3 25m
s+i |
a F ill
Centre of pressure ls given by I y 2e Y
where p= 45,4 = 6.0. À 2 40
* ham
we Fig 33
E +10
60x40
1875 à 40 = 4.1875 = 41878 m. Ans.
Problem 3.2 Determine the nal pressure on a circular plate of diameter 1.5 m which i placed
verrcalls in water im such a way thatthe centre ofthe plate is 3m below the ree surface of water ind
‘he poston of centre of presure ar FRE SURFACE
Solution. Given: Dis. of pate, d = LS m
kan |
2 Ama,
“Total pressure ls piven by equation (3.1)
= peat
LOU x 981» 1767 3.0 N
2 52002581 N, Ans.
Position of centre of pressure (h*) I given by equation (3.5. ee.
etek Fig 34
rat must A
where las EEE > 02485 n°
oF ot
$ yo BS | 39 2 00468 + 30
6730
= 2.0468 m. Ans
74 Pid Mean
Problem 3:3 à rca ie gar sad one eel wal ofa Te ert sie
fine sce meta eng and dh cer te areas Sl he wee face
Prove ese pen en (2) Sue
A e
Sotuton. inca r
Dep serial xe 1
Lette wilh tee
‘Area, i
Fig. 35
(free)
Problem 3.4 À circular opening, 3 m diameter, ina vertical side of à tank i lose by a is of
3 m diameter which con rate about a horizontal diameter. Colealte
(i) the force on the disc, and
(i) me torque required to maintain the disc in equilibrium i he vertical position when the head
of water above the horzonal diameter is 4m.
Soon Gen
D fing am
aves Lace 70688 0?
Dear ©. “in
(Force on te die I given by equation (3.1) as
PEAR = 1000 x 9.81 x 7.0685 x 40
277368 N = 277.368 KN. Ans.
4 To fi the torque required 10 malntin the die in equilibrium, fest ealelat the point of
pplication of force acting onthe dise, Le. centre of pressure ofthe force F The dept of centre of
Pressure (i) i piven by equation (3.5) as
_ En
Du PER 40
Fax
Fo
Loro E ca
TOO
Problem 3:3 à rca ie gar sad one eel wal ofa Te ert sie
fine sce meta eng and dh cer te areas Sl he wee face
Prove ese pen en (2) Sue
A e
Sotuton. inca r
Dep serial xe 1
Lette wilh tee
‘Area, i
Fig. 35
(free)
Problem 3.4 À circular opening, 3 m diameter, ina vertical side of à tank i lose by a is of
3 m diameter which con rate about a horizontal diameter. Colealte
(i) the force on the disc, and
(i) me torque required to maintain the disc in equilibrium i he vertical position when the head
of water above the horzonal diameter is 4m.
Soon Gen
D fing am
aves Lace 70688 0?
Dear ©. “in
(Force on te die I given by equation (3.1) as
PEAR = 1000 x 9.81 x 7.0685 x 40
277368 N = 277.368 KN. Ans.
4 To fi the torque required 10 malntin the die in equilibrium, fest ealelat the point of
pplication of force acting onthe dise, Le. centre of pressure ofthe force F The dept of centre of
Pressure (i) i piven by equation (3.5) as
_ En
Du PER 40
Fax
Fo
Loro E ca
TOO
Hydrostati Forces on Surfaces 75 |
Fig 36
‘The force F is acting ata distance of 4:13 m from fee surface, Moment of tis force about
horizontal diameter XX
= Fx (=F) = 277368 4.14 40) = 36831 Nm. Ans
ones à orque of 38831 Nm mus be applied the disc in the clocks dicton.
Problem 3.5 A pipe line which is min diameter contain a gate valve. The pressure at he centre
ofthe pipe is 196 Nien? If the pipe is led with ol of sp. gr 0.07, ind the force exerted hy the oil
"pon the gane and position of centre of pressure.
Solution. Given
Dia. of pine,
Sp. 36 ofall
Density of ol
‘Weight density of oi,
Pressure a the centre of pipe, p = 19.6 Niem? = 19.6% 10° Nin?
rss head at he cons a Bm 125210" m
Pete eta PRIT ns
“The eight of equivalent fre ol surface fom te centre of pipe = 2.988 m.
‘The depth of CG. ofthe gate valve from fs il surface f= 22.988 m.
{D Now the force exened by the oil onthe gute is piven by
Papua
whore p = dy of ol = 870 g/m
F-81039 81% 4m 22988 = 2465800 N= 2.465 MN. Ans
(6) Peston of conte of pressure (4) vem by (35) a6
Fig 36
‘The force F is acting ata distance of 4:13 m from fee surface, Moment of tis force about
horizontal diameter XX
= Fx (=F) = 277368 4.14 40) = 36831 Nm. Ans
ones à orque of 38831 Nm mus be applied the disc in the clocks dicton.
Problem 3.5 A pipe line which is min diameter contain a gate valve. The pressure at he centre
ofthe pipe is 196 Nien? If the pipe is led with ol of sp. gr 0.07, ind the force exerted hy the oil
"pon the gane and position of centre of pressure.
Solution. Given
Dia. of pine,
Sp. 36 ofall
Density of ol
‘Weight density of oi,
Pressure a the centre of pipe, p = 19.6 Niem? = 19.6% 10° Nin?
rss head at he cons a Bm 125210" m
Pete eta PRIT ns
“The eight of equivalent fre ol surface fom te centre of pipe = 2.988 m.
‘The depth of CG. ofthe gate valve from fs il surface f= 22.988 m.
{D Now the force exened by the oil onthe gute is piven by
Papua
whore p = dy of ol = 870 g/m
F-81039 81% 4m 22988 = 2465800 N= 2.465 MN. Ans
(6) Peston of conte of pressure (4) vem by (35) a6
76. Fluid Mechanics
rat
a m 4
made ño
Da Fra are
4
= 0043 4 22988 = 28.081 m. Ans.
‘Or cen of presu is below the centre of the pipe by a distance of 0.013 m. Ans
Problem 3.6 | Determine he total pressure and cenre of pressure onan soscels triangular plate
FS base 4 mand altitude 4 m when itis immersed vertical i. ol of sp. gr. O39. The base ofthe plate
coincides withthe fee surface of ot FREE OL 4m
Solution. ven surge AM
Base of plate. beam
eight of pte, heim
Dich 4x4 gm
aces, An eh som
Sp. ge of oi 5-09 E
Density of oi p= 900 gin
“The distance of C.G. from fee surface of ei,
Total pressure (Pi given by P= pat
= 9009.81 8.0% 1.33 N = 9597.6.N. Ans
‘Cente of pressure (4) fom ce surface fol is given by
ty
weak
ah
where fg = MO. of wangula section abou its CG,
LR oat
EE
ara 741.33 2 06667 + 1.33 = 1.99 m. Ans.
CES
Problem 3.7. A serial slice gute is usd 1 cover an opening in adam. The opening i 2 mide
and 12 m high On the stream ofthe gate, the liquid ef sp. gr. LAS, les upto a height of 1.5 m
“above the top ofthe gate, where on the downstream side the water is available upto a height touch.
ling the top of the got. Find the resultant force acting onthe gate and position of centre of pressure
Find also the force acting horizontally atthe tp ofthe gate which i capable of opening Assume
‘ha he gate is hinged at he bottom.
Solution. Given
Wii of gate beam
Depth at 4
2 Area, Amon
Sp. ge oF iid
rat
a m 4
made ño
Da Fra are
4
= 0043 4 22988 = 28.081 m. Ans.
‘Or cen of presu is below the centre of the pipe by a distance of 0.013 m. Ans
Problem 3.6 | Determine he total pressure and cenre of pressure onan soscels triangular plate
FS base 4 mand altitude 4 m when itis immersed vertical i. ol of sp. gr. O39. The base ofthe plate
coincides withthe fee surface of ot FREE OL 4m
Solution. ven surge AM
Base of plate. beam
eight of pte, heim
Dich 4x4 gm
aces, An eh som
Sp. ge of oi 5-09 E
Density of oi p= 900 gin
“The distance of C.G. from fee surface of ei,
Total pressure (Pi given by P= pat
= 9009.81 8.0% 1.33 N = 9597.6.N. Ans
‘Cente of pressure (4) fom ce surface fol is given by
ty
weak
ah
where fg = MO. of wangula section abou its CG,
LR oat
EE
ara 741.33 2 06667 + 1.33 = 1.99 m. Ans.
CES
Problem 3.7. A serial slice gute is usd 1 cover an opening in adam. The opening i 2 mide
and 12 m high On the stream ofthe gate, the liquid ef sp. gr. LAS, les upto a height of 1.5 m
“above the top ofthe gate, where on the downstream side the water is available upto a height touch.
ling the top of the got. Find the resultant force acting onthe gate and position of centre of pressure
Find also the force acting horizontally atthe tp ofthe gate which i capable of opening Assume
‘ha he gate is hinged at he bottom.
Solution. Given
Wii of gate beam
Depth at 4
2 Area, Amon
Sp. ge oF iid
Hydrostatic Forces on Surfaces 77]
a= LAS 1000 = 1450 Kg"
Fi = Fore exerted by the Mid of sp. gr LAS on gate
FE = Force exerted by water on th gate
2 Density of gui,
La
The foros Ph given by. Fy = pug XA fh
where py = LAS x 1000 = 1450 kg/m?
Fi = Depth of CG. of pate tom fe surface of aid
12
52
2
24m
ree suneace oF uaup AN
Figo,
1430 x 9.81 6242.1
pas e
Similar,
where pa 1000 kg/m?
Tis = Dept of CG. of gate from fee surface of water
4x12=060
y= 1000 #931 x 24 x06 = 14126 N
(Resultant force on the gate = F — F = 71691 ~ 14126 = 57565 N. Ans.
(i) Position of centre of pressure of resaltan force, The fore: F, will be acting u depth of
In? from fe surface of liquid, given by the elation
te i,
lem
ay
bd? par ñ
see fg En PE 0288 0
288
+ ne +21 200871 +21 = 2.1571 m
age les s
Distance of F fom binge
(154 1.2) yt 2 27-2157 = 05829 m
“Te force wil be acing at depa of," from fos surface of war and Is given by
a= LAS 1000 = 1450 Kg"
Fi = Fore exerted by the Mid of sp. gr LAS on gate
FE = Force exerted by water on th gate
2 Density of gui,
La
The foros Ph given by. Fy = pug XA fh
where py = LAS x 1000 = 1450 kg/m?
Fi = Depth of CG. of pate tom fe surface of aid
12
52
2
24m
ree suneace oF uaup AN
Figo,
1430 x 9.81 6242.1
pas e
Similar,
where pa 1000 kg/m?
Tis = Dept of CG. of gate from fee surface of water
4x12=060
y= 1000 #931 x 24 x06 = 14126 N
(Resultant force on the gate = F — F = 71691 ~ 14126 = 57565 N. Ans.
(i) Position of centre of pressure of resaltan force, The fore: F, will be acting u depth of
In? from fe surface of liquid, given by the elation
te i,
lem
ay
bd? par ñ
see fg En PE 0288 0
288
+ ne +21 200871 +21 = 2.1571 m
age les s
Distance of F fom binge
(154 1.2) yt 2 27-2157 = 05829 m
“Te force wil be acing at depa of," from fos surface of war and Is given by
78. Fluid Mechanics
were (57024800. F,=06m.4=24 0
PE
24x06
Distance of From hinge =12-08=04m
‘The sut fore $7565 N wl eating a dinos given by
_ 1691 x-5129 — 1412604
51565
MESA
57505
= 0578 m above
(i or at the top of gate which capable of opening the gale. Let Fs the force required
onthe wp ofthe gate o ape as shown in ig 59. Thing the moments FFF nd bau the
hinge, we pet
Feiner xode Fx 509
Fx SDF x04
12
TOS 5129 OA | 38921-50514
E 12 2
= 2785 Nam.
Problem3.8 Acaistonforciosng he entrance ay dock of rapecidal form 16m wide at the
top and 10 m wide ar he boton and 6 m dep. Fin the ta presa aná centre of pressure on the
son e mater o e cute jut eve wih the op and dacs emp
Solution. Given e”
hy +06=02+06=08m
or Fe
wa stop ion =
Yt tom un \
Depth, d=6m 7
RE apa ae, BES
an BEA) eg L
a nn
1018 reat
WEN rg me
DC oad ae ABCD om ao wae
non IN ox xs
i
7
150.36
= M atom mr
0 cay su
(0 Total Pressure (E). Total pressure, Fis given by
were (57024800. F,=06m.4=24 0
PE
24x06
Distance of From hinge =12-08=04m
‘The sut fore $7565 N wl eating a dinos given by
_ 1691 x-5129 — 1412604
51565
MESA
57505
= 0578 m above
(i or at the top of gate which capable of opening the gale. Let Fs the force required
onthe wp ofthe gate o ape as shown in ig 59. Thing the moments FFF nd bau the
hinge, we pet
Feiner xode Fx 509
Fx SDF x04
12
TOS 5129 OA | 38921-50514
E 12 2
= 2785 Nam.
Problem3.8 Acaistonforciosng he entrance ay dock of rapecidal form 16m wide at the
top and 10 m wide ar he boton and 6 m dep. Fin the ta presa aná centre of pressure on the
son e mater o e cute jut eve wih the op and dacs emp
Solution. Given e”
hy +06=02+06=08m
or Fe
wa stop ion =
Yt tom un \
Depth, d=6m 7
RE apa ae, BES
an BEA) eg L
a nn
1018 reat
WEN rg me
DC oad ae ABCD om ao wae
non IN ox xs
i
7
150.36
= M atom mr
0 cay su
(0 Total Pressure (E). Total pressure, Fis given by
Hydrostatic Forces on Surfaces 79
= PEAR = 1000 x 9.81 x 78 x 2769 N
2118783 N = 2.118783 MN. Ans.
D Centre of Pressure (he). Centre of pressure is given by equation (3.5) as
dr
ak
where fg = M.04. of trapezoidal ABCD about is C6,
La La = MOU, of rectangle FBCE about ts CG.
Le, = MOLL of to As ABF and ECD about its C6.
ig PRET
NC
La, isthe MO. of the rectangle about the avi pasing though 6,
180m"
3 MOL. of the rectangle about the aus passing though the C.G. ofthe wapezokdal 4, + Ares ol
rectangle x
‘wheres distance between the CG. of rectangle and
80-2709 = 0231 m
MLOLL ol FBCE passing trough CG. of trapezoidal
= 180 10% 6% (023) = 180 + 320 = 18320 mt
ei
‘of tapezodal
Now
MO: of AABD in Fp. 311 about G>
(6-10)x6?
%
"The distance between the CG. of tangle and C.G. of taperoial
(0109-20) =0.769
MOL ofthe two As about an avis passing through CG. of wapceoidl
Lo, + Area of angles x (769°
3604 928 cre !
36.04 10.64 = 46.64 4.
2 lo 7 MO operada tout CC
MOL age on ue ei om
1 MLO arg une CO orde meri
MAS
x wee ah se
ai ie an
vine Armin
ms
EM 2109210682270) «S488. An
78X2769 nul
Alternate Method
‘he distance of the CG, of the trapezoidal channel trom surface AD is given by (er to Table 3.1
on page 71)
= PEAR = 1000 x 9.81 x 78 x 2769 N
2118783 N = 2.118783 MN. Ans.
D Centre of Pressure (he). Centre of pressure is given by equation (3.5) as
dr
ak
where fg = M.04. of trapezoidal ABCD about is C6,
La La = MOU, of rectangle FBCE about ts CG.
Le, = MOLL of to As ABF and ECD about its C6.
ig PRET
NC
La, isthe MO. of the rectangle about the avi pasing though 6,
180m"
3 MOL. of the rectangle about the aus passing though the C.G. ofthe wapezokdal 4, + Ares ol
rectangle x
‘wheres distance between the CG. of rectangle and
80-2709 = 0231 m
MLOLL ol FBCE passing trough CG. of trapezoidal
= 180 10% 6% (023) = 180 + 320 = 18320 mt
ei
‘of tapezodal
Now
MO: of AABD in Fp. 311 about G>
(6-10)x6?
%
"The distance between the CG. of tangle and C.G. of taperoial
(0109-20) =0.769
MOL ofthe two As about an avis passing through CG. of wapceoidl
Lo, + Area of angles x (769°
3604 928 cre !
36.04 10.64 = 46.64 4.
2 lo 7 MO operada tout CC
MOL age on ue ei om
1 MLO arg une CO orde meri
MAS
x wee ah se
ai ie an
vine Armin
ms
EM 2109210682270) «S488. An
78X2769 nul
Alternate Method
‘he distance of the CG, of the trapezoidal channel trom surface AD is given by (er to Table 3.1
on page 71)
‘80 Fluid Mechanics
FW
Gens
Leona, .
mr) 3 7
%
"Baron
2 =
toman, Fou ham
‘at
ak
Cr) y fe rémto6 16),
la 36(a+h) u 3610 +16)
100220, a
sten
2 a
ee ap Anan’
782769 N ?
mans
Prob 29 A race 3 8 he bam an nd ha espe
Dons
1 de tecno on
D Socal oo hee ch se chet hen al of ae
Sehen. Gren
ai
eo ==
een Gee >
rei, 2 r
Kemer. 2 E
ot on ma E, EN
a
Arca of wapevoidal ABCD,
FW
Gens
Leona, .
mr) 3 7
%
"Baron
2 =
toman, Fou ham
‘at
ak
Cr) y fe rémto6 16),
la 36(a+h) u 3610 +16)
100220, a
sten
2 a
ee ap Anan’
782769 N ?
mans
Prob 29 A race 3 8 he bam an nd ha espe
Dons
1 de tecno on
D Socal oo hee ch se chet hen al of ae
Sehen. Gren
ai
eo ==
een Gee >
rei, 2 r
Kemer. 2 E
ot on ma E, EN
a
Arca of wapevoidal ABCD,
Hydrostatic Forces on Surfaces 81
Depth of CG. of two tñangles ABF and ECD trom water surface,
Depth of C6. of trapezoidal ABCD fom fre suce of water
ja MX Aja 208 41038883
(ra) er)
(D Total Pressure (E) Total pressure Fi given by
F= pea
1000 4931 4 3.0% 0.48444 = 1307.9 N. Ans.
MO. of rectangle FBCE about is Ch,
PIE FR
ae
MO..OFFBCE about an axis passing through the CG of trapezoidal
or To = o, + Aux Distance between C.G. of rectangle and C6.
of tpeoldal?
En
MOL ofthe wo wiangles about the CG. of wapezoldl,
Tag = Fo, + Asx (Distance between CG. of tangles a
cc.
of taped?
= CID = 00555 + LED?
0558 + 101234 = 0.06789 m“
MO. ofthe rpezoida about ts CG
Los la 94 le, 92 ATIT + 06789
24059 m“
“Then conte of pressure (4) onthe yetcal trapezoidal,
lo gj = 028059
an a
= 0.625 m. Aas.
ye
+ 4444018046 + 4444 = 0.6248
Depth of CG. of two tñangles ABF and ECD trom water surface,
Depth of C6. of trapezoidal ABCD fom fre suce of water
ja MX Aja 208 41038883
(ra) er)
(D Total Pressure (E) Total pressure Fi given by
F= pea
1000 4931 4 3.0% 0.48444 = 1307.9 N. Ans.
MO. of rectangle FBCE about is Ch,
PIE FR
ae
MO..OFFBCE about an axis passing through the CG of trapezoidal
or To = o, + Aux Distance between C.G. of rectangle and C6.
of tpeoldal?
En
MOL ofthe wo wiangles about the CG. of wapezoldl,
Tag = Fo, + Asx (Distance between CG. of tangles a
cc.
of taped?
= CID = 00555 + LED?
0558 + 101234 = 0.06789 m“
MO. ofthe rpezoida about ts CG
Los la 94 le, 92 ATIT + 06789
24059 m“
“Then conte of pressure (4) onthe yetcal trapezoidal,
lo gj = 028059
an a
= 0.625 m. Aas.
ye
+ 4444018046 + 4444 = 0.6248
CG. ofthe trapezoidal channel from sunface AD i given by (eer o Table 3.1
Ga+b),h_@x2+4 ds
(a+b) 37 Bea)
oaum
h=x=0444m
‘Total pressure, F2 pgAÑ = 1000 x 9.81 x 3.0% 84 & 4230
= 13079 N. Ans.
ese
mare of presse, Wt GR
Cente of pr 4
where fg fom Table 3.1 is given by
ea], heated) 3 a
RS uen a Fxg ON
aan
7 + AAA = 0425 m. Ans.
E
Problem 9:10 A square aperture inthe vertical side of a tank has one diagonal vertical and is
completely covered by a plane plate hinged along one ofthe upper sides of he aperture. The diagonals
(ofthe aperture are 2 m long and the tank contains a liquid of specific gravity 1.5. The centre of
apertre 1s 1.5 m Below the free surface. Caleulae the thrust exerted an the pate by he liquid and
Position of ts centre of pressure.
‘Solution. Given: Diagonal of aperture, AC = BD = 2m
2 fea of square aperture, A = Area of AACB + Atea of AACD
Sp. ae of iui Lis
Density of Nguid, p= 1.15% 1000.
Dept of centre of aperture from fee surface,
B= 15m.
Ga+b),h_@x2+4 ds
(a+b) 37 Bea)
oaum
h=x=0444m
‘Total pressure, F2 pgAÑ = 1000 x 9.81 x 3.0% 84 & 4230
= 13079 N. Ans.
ese
mare of presse, Wt GR
Cente of pr 4
where fg fom Table 3.1 is given by
ea], heated) 3 a
RS uen a Fxg ON
aan
7 + AAA = 0425 m. Ans.
E
Problem 9:10 A square aperture inthe vertical side of a tank has one diagonal vertical and is
completely covered by a plane plate hinged along one ofthe upper sides of he aperture. The diagonals
(ofthe aperture are 2 m long and the tank contains a liquid of specific gravity 1.5. The centre of
apertre 1s 1.5 m Below the free surface. Caleulae the thrust exerted an the pate by he liquid and
Position of ts centre of pressure.
‘Solution. Given: Diagonal of aperture, AC = BD = 2m
2 fea of square aperture, A = Area of AACB + Atea of AACD
Sp. ae of iui Lis
Density of Nguid, p= 1.15% 1000.
Dept of centre of aperture from fee surface,
B= 15m.
Hydrostatic Forces on Surfaces 83]
(9 The thrust on te plate i given by
F2 puAñ = 115091 x2 x 1.5= 338448, Ans.
D) Centre of pressure (h) is given by
wee gk
‘Ale
MOL. of ABCD about diagonal AC
= MO. of tangle ABC about AC + MO. of angle ACD about AC
ACx OP, ACxoD"
cry e
[e Motors
zur
[2
Ly
ans
Problem 3.11 A rank contains water up a height of 0.5 m above the base. An immiscible liquid of
sp. 87.0. filed on he top of water upto m height. Caleta
(i) tout pressure on ome side ofthe tak,
(i) the position of centre of pressure for one side ofthe tank, which i 2 m wide.
ses
maxis th
S611 m. Ans.
Solution. Given
Depth wate sm
Depts ru m
Sp sr of guid 5
Deny of gu $1000 = 800 Kom?
Dena of wate, 0 kgm
Wid of tank A
(0 Total premureom ome side scaled y drawing prs gran, wich shown in Fig 34,
Inensiy of rear on fp. P
Intensity of pressure on D (or DD), F
Intensity of pressure on base (or BO).
Now free fi
(9 The thrust on te plate i given by
F2 puAñ = 115091 x2 x 1.5= 338448, Ans.
D) Centre of pressure (h) is given by
wee gk
‘Ale
MOL. of ABCD about diagonal AC
= MO. of tangle ABC about AC + MO. of angle ACD about AC
ACx OP, ACxoD"
cry e
[e Motors
zur
[2
Ly
ans
Problem 3.11 A rank contains water up a height of 0.5 m above the base. An immiscible liquid of
sp. 87.0. filed on he top of water upto m height. Caleta
(i) tout pressure on ome side ofthe tak,
(i) the position of centre of pressure for one side ofthe tank, which i 2 m wide.
ses
maxis th
S611 m. Ans.
Solution. Given
Depth wate sm
Depts ru m
Sp sr of guid 5
Deny of gu $1000 = 800 Kom?
Dena of wate, 0 kgm
Wid of tank A
(0 Total premureom ome side scaled y drawing prs gran, wich shown in Fig 34,
Inensiy of rear on fp. P
Intensity of pressure on D (or DD), F
Intensity of pressure on base (or BO).
Now free fi
34. Fluid Mechanics
Force Fy = Area of rectangle DBFE x Wil of tak
05 x 7818.2 TN
Fa = Ares of AERC x Width of tank
POS x 4905 «20 = 2452.5 N
1
Lxerxrex
+ x20
Total pressure RATA
= 7848 4 7848 à 24925 = 181485 N. Ans
(65) Centre of Pressure (4). Taking the moments ofall for about A, we get
Pain aner ane Lane nn. 2 aD)
2
snes man 07 (1925) à ss (1062
= 5232 4 96104 3270 = 18312
Problem 8.12 A cubic tank has sides of 1S m. I contains water for the lower 06 m depth. The
upper remaining partis fled with eof specie gravity 0.9. Callate fr one vertical side of he tank:
a) total pressure, amd
10) postion of centre of pressure
Solution. Gives
Cuba tank of ides 1.5 m means the dimensions ofthe tank ao 15 m x LS mx 15m.
Depth of water 206m
Depth of quid =15-06=09m
Sp. ne ol iguid 09
Density of quid, 09 1000 = 900 kg/m!
Density of water, 1000 kg/m
pressure on one venical ie is calculated by drawing pressure diagram, whic is shown
Force Fy = Area of rectangle DBFE x Wil of tak
05 x 7818.2 TN
Fa = Ares of AERC x Width of tank
POS x 4905 «20 = 2452.5 N
1
Lxerxrex
+ x20
Total pressure RATA
= 7848 4 7848 à 24925 = 181485 N. Ans
(65) Centre of Pressure (4). Taking the moments ofall for about A, we get
Pain aner ane Lane nn. 2 aD)
2
snes man 07 (1925) à ss (1062
= 5232 4 96104 3270 = 18312
Problem 8.12 A cubic tank has sides of 1S m. I contains water for the lower 06 m depth. The
upper remaining partis fled with eof specie gravity 0.9. Callate fr one vertical side of he tank:
a) total pressure, amd
10) postion of centre of pressure
Solution. Gives
Cuba tank of ides 1.5 m means the dimensions ofthe tank ao 15 m x LS mx 15m.
Depth of water 206m
Depth of quid =15-06=09m
Sp. ne ol iguid 09
Density of quid, 09 1000 = 900 kg/m!
Density of water, 1000 kg/m
pressure on one venical ie is calculated by drawing pressure diagram, whic is shown
Hydrostatic Forces on Surfaces 85)
‘of pressure at A, 7, =O
of pressure at Dp = pur 4h = 900 x 9.81 X 09 = 7946.1 Nm?
‘of pressure a B,pg= Dich, + path = 900 À OAI X LD + 1000 x 9.81 4 6
= 7946.1 4 5886 = 13832.1 Nin?
Hence in pressure diagram
DE = 7946.1 Nim", BC = 13832.1 Nim’, FC = 5886 Nin?
‘The pressure digram is sl nto tangle ADE, rectangle BDEF and tangle EFC. The toa pes
sur free const of te following components
(For Fi = Arca of angle ADE x Width of tank
= (x ADx DE) x15
wast = LS m)
= (509% 79461) x LS N
62.6 N
“Tis force wil be ating atthe CG, ofthe tangle ADE, eat adm of 3 09=06 m Blow A
(9 Force Fy = Area of rectangle BDEF x Wi of tank
(BD x DE) LS = (06 7946.1) x LS = 71515
12m
Tis Force will he acting atthe C.G. of the rectangle BDEF ive ata distance of 09 + ES =
below A
(i) Force Fa Area of angle BFC With of tank
Lx Ex FC) x 15 = (806% 5000) x 1.5 = 20187 N
"This force willbe ating athe CG. ofthe
gk EC te, adhance 0008, 2408
ii à
2" Fa eurem ne ee af en,
PF 4 ht hy
ss 268 ENGEN
(Aut tito pres
Le fe Fey ht oe rc ri om
"ing omen fest Ae
Pao Ra a
x06+ 6 x1245 x13
nr
S36R6 6 + TISLS 124 26487913,
> 3.4 HORIZONTAL PLANE SURFACE SUBMERGED IN LIQUID
Considera pln horizontal surface immerscd in a static ud, As every point of the surface sa the
same depth fem the fee surface ofthe iui, the pressure intensity wil equal on the entire surface
and equa 0 p = pl where hs dep of surface
‘of pressure at A, 7, =O
of pressure at Dp = pur 4h = 900 x 9.81 X 09 = 7946.1 Nm?
‘of pressure a B,pg= Dich, + path = 900 À OAI X LD + 1000 x 9.81 4 6
= 7946.1 4 5886 = 13832.1 Nin?
Hence in pressure diagram
DE = 7946.1 Nim", BC = 13832.1 Nim’, FC = 5886 Nin?
‘The pressure digram is sl nto tangle ADE, rectangle BDEF and tangle EFC. The toa pes
sur free const of te following components
(For Fi = Arca of angle ADE x Width of tank
= (x ADx DE) x15
wast = LS m)
= (509% 79461) x LS N
62.6 N
“Tis force wil be ating atthe CG, ofthe tangle ADE, eat adm of 3 09=06 m Blow A
(9 Force Fy = Area of rectangle BDEF x Wi of tank
(BD x DE) LS = (06 7946.1) x LS = 71515
12m
Tis Force will he acting atthe C.G. of the rectangle BDEF ive ata distance of 09 + ES =
below A
(i) Force Fa Area of angle BFC With of tank
Lx Ex FC) x 15 = (806% 5000) x 1.5 = 20187 N
"This force willbe ating athe CG. ofthe
gk EC te, adhance 0008, 2408
ii à
2" Fa eurem ne ee af en,
PF 4 ht hy
ss 268 ENGEN
(Aut tito pres
Le fe Fey ht oe rc ri om
"ing omen fest Ae
Pao Ra a
x06+ 6 x1245 x13
nr
S36R6 6 + TISLS 124 26487913,
> 3.4 HORIZONTAL PLANE SURFACE SUBMERGED IN LIQUID
Considera pln horizontal surface immerscd in a static ud, As every point of the surface sa the
same depth fem the fee surface ofthe iui, the pressure intensity wil equal on the entire surface
and equa 0 p = pl where hs dep of surface
86. Fluid Mechanics
La A= Total ares of surace
‘Then total free, F. on the surface
= nx Area = pg xh xA= ped
where = Depth of CG. tom fie surface of guid = h
ao =Depthof conte of pressure from feo surface = A
Problem 3.13 Fig. 3.17 shows a tank fl of water. Find +
(O) Total pressure om he bottom of tank ig. 6
HD. Weight of water inthe tank.
(ii) Hydrostatic paradox been the results of and (i), Width of tank 1 2 m.
Solution. Given gum
Depth water on bottom of tank
3406-360
With of tank
Length or ak at boom
‘Ate tthe boina,
1000 «9.81 x8 x 36
282528 N. Ans.
ax Volume of tank
1000 49.81 x 13 X 0.4 42 + 44.621
1000 À 981 124 + 4.8} = 70632 N. Ans.
(di) From the resulto (and Gi, ts observed thatthe total weight of water inthe tanks much
Jess than the total pressure at the bottom ofthe Lak, This i known a Hydrostatic parado.
> 355. INCLINED PLANE SURFACE SUBMERGED IN LIQUID
Consider a plane surface of ri shape immersed in a gui in such away dat the plane ofthe
surface makes an angle 0 with the fre surface ofthe liq as shown in Fig. 3.18.
Fig. 38 Inclined immersed surface.
Let A = Total area of inclined surface
À = Depth of CG. of incined area from fee surface
1° = Distance of eens of pressure fom fre surface of quid
9 = Angle made by the plane of the surface with re gu surface.
La A= Total ares of surace
‘Then total free, F. on the surface
= nx Area = pg xh xA= ped
where = Depth of CG. tom fie surface of guid = h
ao =Depthof conte of pressure from feo surface = A
Problem 3.13 Fig. 3.17 shows a tank fl of water. Find +
(O) Total pressure om he bottom of tank ig. 6
HD. Weight of water inthe tank.
(ii) Hydrostatic paradox been the results of and (i), Width of tank 1 2 m.
Solution. Given gum
Depth water on bottom of tank
3406-360
With of tank
Length or ak at boom
‘Ate tthe boina,
1000 «9.81 x8 x 36
282528 N. Ans.
ax Volume of tank
1000 49.81 x 13 X 0.4 42 + 44.621
1000 À 981 124 + 4.8} = 70632 N. Ans.
(di) From the resulto (and Gi, ts observed thatthe total weight of water inthe tanks much
Jess than the total pressure at the bottom ofthe Lak, This i known a Hydrostatic parado.
> 355. INCLINED PLANE SURFACE SUBMERGED IN LIQUID
Consider a plane surface of ri shape immersed in a gui in such away dat the plane ofthe
surface makes an angle 0 with the fre surface ofthe liq as shown in Fig. 3.18.
Fig. 38 Inclined immersed surface.
Let A = Total area of inclined surface
À = Depth of CG. of incined area from fee surface
1° = Distance of eens of pressure fom fre surface of quid
9 = Angle made by the plane of the surface with re gu surface.
Hydrostatic Forces on Surfaces 87]
Let the plane ofthe surface, if produced met the tree liquid surface at 0, Then 0-0 ¥s the axis
‘perpendicular othe plane of the surface,
La $ = distance ofthe CG. ofthe inclined surface from 0-0
= distance ofthe cen of pressure ftom 0-0.
depth A rom fr seas ad st a distance y rom the axis
Considera smal rip of anc dA a
(0-0 a6 shown in Fig. 318
Pres Inensy on the sip, poh
Pressure foros, don the stip, A =p x Arca of stip = ph x JA
Ta promu fc one wholes. = fu ph
as
[RE
‘Buc from Pig. 3.18,
Jou eas
Disane 0.6. rom ais 0-0
F ape sin 3a
esi €
Fin 0) 486)
Centre of Pressure (4)
Presse foros on the A = ph
= pay an 844 Weysne)
Moment ofthe fore, A, shout axis 0.0 .
"= dP y = pay sin 8 dA xy = pg in BUA
PR anne
+ Jovsiney"at=prsine fat
MOI of the surface about 0-0 2 fy
a fra
Sumo? momen of a cs ou 0-0» gún, an
om ot ae Pa hu phen
ae as
vie = Dito cnt rr tom 0.0
oy de aos ¿nen un) an)
Wo pas
META
men 09
and bythe theorem of parallel axis = / + AFA
Let the plane ofthe surface, if produced met the tree liquid surface at 0, Then 0-0 ¥s the axis
‘perpendicular othe plane of the surface,
La $ = distance ofthe CG. ofthe inclined surface from 0-0
= distance ofthe cen of pressure ftom 0-0.
depth A rom fr seas ad st a distance y rom the axis
Considera smal rip of anc dA a
(0-0 a6 shown in Fig. 318
Pres Inensy on the sip, poh
Pressure foros, don the stip, A =p x Arca of stip = ph x JA
Ta promu fc one wholes. = fu ph
as
[RE
‘Buc from Pig. 3.18,
Jou eas
Disane 0.6. rom ais 0-0
F ape sin 3a
esi €
Fin 0) 486)
Centre of Pressure (4)
Presse foros on the A = ph
= pay an 844 Weysne)
Moment ofthe fore, A, shout axis 0.0 .
"= dP y = pay sin 8 dA xy = pg in BUA
PR anne
+ Jovsiney"at=prsine fat
MOI of the surface about 0-0 2 fy
a fra
Sumo? momen of a cs ou 0-0» gún, an
om ot ae Pa hu phen
ae as
vie = Dito cnt rr tom 0.0
oy de aos ¿nen un) an)
Wo pas
META
men 09
and bythe theorem of parallel axis = / + AFA
88. Fluid Mechanics
Substiting these valves in equation (3.9, we get
HT BESO,
aro peak Le,
PRE
Da Emo o
510
10 = 90, equation (3.10) Becomes same as equation (3.5) which Is applicable to vercally plane
submerged surfaces.
In equation (3.10) f= M.OL finie surfaces about an ai asin trough Gand parallel 0 0-0
Problem 3.14 (9) A rectangular plane surface 2 m wie and 3m deep isin water in such a way
‘hat its plane makes an angle of 30° with the fre surface of water. Determine the total pressure and
position of centre of pressure when the upper edge is 1.5 m below the fee water surface
Solution. Gien FREE WATER SUREACE
Wiath of plane surface, b= 2m
Depth, m
Angle, 0
Distance o upper edge from re wate surface = 1.5 8
(Total pressure oros ls given y equation (36) 36
capes
where p = 1000 kgs?
+ EB = 18 + BC in 0° = LS + LS in 30°)
2150154 =225m
E F
D Centre of pressure (19)
‘Using equation (3.10), we have
000 9.81 6 x 2.25 = 132436 N. Ans.
10833 4 2.25 = 2.3933 m- Ans
Substiting these valves in equation (3.9, we get
HT BESO,
aro peak Le,
PRE
Da Emo o
510
10 = 90, equation (3.10) Becomes same as equation (3.5) which Is applicable to vercally plane
submerged surfaces.
In equation (3.10) f= M.OL finie surfaces about an ai asin trough Gand parallel 0 0-0
Problem 3.14 (9) A rectangular plane surface 2 m wie and 3m deep isin water in such a way
‘hat its plane makes an angle of 30° with the fre surface of water. Determine the total pressure and
position of centre of pressure when the upper edge is 1.5 m below the fee water surface
Solution. Gien FREE WATER SUREACE
Wiath of plane surface, b= 2m
Depth, m
Angle, 0
Distance o upper edge from re wate surface = 1.5 8
(Total pressure oros ls given y equation (36) 36
capes
where p = 1000 kgs?
+ EB = 18 + BC in 0° = LS + LS in 30°)
2150154 =225m
E F
D Centre of pressure (19)
‘Using equation (3.10), we have
000 9.81 6 x 2.25 = 132436 N. Ans.
10833 4 2.25 = 2.3933 m- Ans
Hydrostatic Forces on Surfaces _ 89
Problem 3.14 (6) A rectangular plane surface 3m wide and 4m deep ies in water in such a way
‘that i plano makes an angle of 30° with he fee surface of water Determine the total pressure force
and postion of centre of pressure, when the upper edge is 2 m below the free surface
Solution. Gien free aca tar
b= 3m, d=4m,0= 30"
Distance of upper edge fom fre surface of wate
{D Total pressure foree is given by equation (3.6) as
seat
A=bxd=3x
and = Deptnof CG. of pate from
nos water surface
224 BE=260C sno
isla
2242030 Fig. 329 a)
53.167 KN. Ans.
a 000 9.81 12
Ki Centre of pressure (1)
167
Using equation @.10, we have ne = LENS 47
wba ge EEE teat
na
1
16%
po 16300 mes
CEE ese sant mem.
Problem 3.16 (8) A circular plate 3.0 m diameter is immersed in water in such a way hat its
greatest and leas! dep below ihe fre suface are 4 m and 1S m respective}. Determine the total
Pressure on one face fthe plate and position of te centre of pressure,
Solution. Gives nv EEE
Dia. of plate, >
aces,
Distance DC = 15m. BE =
Distance of C4, rom fre surface
40-DC_40-15
But
m
Rense 15% AIS 154120 22.709 m
Fig. 320
Problem 3.14 (6) A rectangular plane surface 3m wide and 4m deep ies in water in such a way
‘that i plano makes an angle of 30° with he fee surface of water Determine the total pressure force
and postion of centre of pressure, when the upper edge is 2 m below the free surface
Solution. Gien free aca tar
b= 3m, d=4m,0= 30"
Distance of upper edge fom fre surface of wate
{D Total pressure foree is given by equation (3.6) as
seat
A=bxd=3x
and = Deptnof CG. of pate from
nos water surface
224 BE=260C sno
isla
2242030 Fig. 329 a)
53.167 KN. Ans.
a 000 9.81 12
Ki Centre of pressure (1)
167
Using equation @.10, we have ne = LENS 47
wba ge EEE teat
na
1
16%
po 16300 mes
CEE ese sant mem.
Problem 3.16 (8) A circular plate 3.0 m diameter is immersed in water in such a way hat its
greatest and leas! dep below ihe fre suface are 4 m and 1S m respective}. Determine the total
Pressure on one face fthe plate and position of te centre of pressure,
Solution. Gives nv EEE
Dia. of plate, >
aces,
Distance DC = 15m. BE =
Distance of C4, rom fre surface
40-DC_40-15
But
m
Rense 15% AIS 154120 22.709 m
Fig. 320
9 Fluid Mechanics
(Total pressure (E)
P= peat
1000 x 9.81 x 7.0685 X 2.749 = 190621 N. Am.
(D Centre of pressure (49)
Using equation 3.10), we nave an = JASCO gg
she Le Ea E yt = 3976 m'
916 x (5338) 8333,
“685 2749
me #2289 = 0.1400 + 2.789
991m. Ans.
Problem 3.15 (b) Ifin the above problem, the given circular plate shaving a concentric circular
hole of diameter 1.5 , then callate the toral pressure and positon of the centre of pressure on ane
face of the plate E
Solution. Given Refer Fig. 320 (a)
Dia. of plat, a
Area of solid plate
Dia. of ole loe plate, dy = 1.50
2 Ares of hole
de> Easy
Easy
2 Area of the piven plate, A = Area of soli plate Area of hole
10685 1.671 = 53014 m?
Distance CD= 15, BE=
Distance of CG. rom the fee surac
Fig. 320 (2)
000 9.81 x S014 2.75
LOIS N = 143.018 KN. Ans.
(Total pressure (E)
P= peat
1000 x 9.81 x 7.0685 X 2.749 = 190621 N. Am.
(D Centre of pressure (49)
Using equation 3.10), we nave an = JASCO gg
she Le Ea E yt = 3976 m'
916 x (5338) 8333,
“685 2749
me #2289 = 0.1400 + 2.789
991m. Ans.
Problem 3.15 (b) Ifin the above problem, the given circular plate shaving a concentric circular
hole of diameter 1.5 , then callate the toral pressure and positon of the centre of pressure on ane
face of the plate E
Solution. Given Refer Fig. 320 (a)
Dia. of plat, a
Area of solid plate
Dia. of ole loe plate, dy = 1.50
2 Ares of hole
de> Easy
Easy
2 Area of the piven plate, A = Area of soli plate Area of hole
10685 1.671 = 53014 m?
Distance CD= 15, BE=
Distance of CG. rom the fee surac
Fig. 320 (2)
000 9.81 x S014 2.75
LOIS N = 143.018 KN. Ans.
Hydrostati Fores on Surfaces 91
(6) Position of centre of pressure (1%)
Using equation (310) we have
a2) as
psp
due ja y
Solution. Ge var
Diane.
An 1005
Dane. im
AD DEAR BE-DC_20-10_1 u
nac, sng = 48
oie Be” BC 30 Re
‘Te core o gavi ofthe plates at te mid of BC, at stance 1.5 m fom €.
‘The distance of ent of gravity om the fs arc ofthe water ion y
Feb 4 CGsin0~ 1041544 ©
= 15m
(0 Total pesar the oat Face the plates ven by
P= peak
2 100 9.81 10685 x 15 = 104013 N. Ans
(Let the distan of the ents of pres from hs ro surface of he aer e AT. Then using
gun G10), we hve
nen es Og
a
(6) Position of centre of pressure (1%)
Using equation (310) we have
a2) as
psp
due ja y
Solution. Ge var
Diane.
An 1005
Dane. im
AD DEAR BE-DC_20-10_1 u
nac, sng = 48
oie Be” BC 30 Re
‘Te core o gavi ofthe plates at te mid of BC, at stance 1.5 m fom €.
‘The distance of ent of gravity om the fs arc ofthe water ion y
Feb 4 CGsin0~ 1041544 ©
= 15m
(0 Total pesar the oat Face the plates ven by
P= peak
2 100 9.81 10685 x 15 = 104013 N. Ans
(Let the distan of the ents of pres from hs ro surface of he aer e AT. Then using
gun G10), we hve
nen es Og
a
Substutng the values, we gst
Ex
&
415.0864
E S416 m. Ans.
Texoxis
Problem 3.17 A rectangular gate Sm x2 m is hinged atts hase and inclined a 60°10 the horizon
alas shown in Fig. 3.22. To keep the gate in stable position, a counterweight of 000 Ag attached
at the upper end ofthe gate as shown in gure Find he depth of mater at which the gate begins 1 fall
Neglect the weigh of the gate and front he hinge and pale
Solution. Gives
Length of gate
Width of gate
Wett,
= 5000 9.81 N
49050 N CE Lkgf = 9.81 ND
Aste pully icones, the foros acting at B= 49050 N. Fst
find the eral force F ating On Ihe gate AB fora given depth of
fom i, e
COS
a
cn gate mens water, A= AD Wie DE 22 Fw
»
so depa ofthe C.G. ofthe inmersed area
sh
h 19820
1000 x9.81 x eN
$e oe
“The centre of pressure ofthe immersed surface, ls ven by
2 Total force Fis given by F2 pgAñ
ie tam gg
a
wi 192 MOI of emma
AD „2 (a Ÿ a
h nl bez
16h an
DARAS
Ex
&
415.0864
E S416 m. Ans.
Texoxis
Problem 3.17 A rectangular gate Sm x2 m is hinged atts hase and inclined a 60°10 the horizon
alas shown in Fig. 3.22. To keep the gate in stable position, a counterweight of 000 Ag attached
at the upper end ofthe gate as shown in gure Find he depth of mater at which the gate begins 1 fall
Neglect the weigh of the gate and front he hinge and pale
Solution. Gives
Length of gate
Width of gate
Wett,
= 5000 9.81 N
49050 N CE Lkgf = 9.81 ND
Aste pully icones, the foros acting at B= 49050 N. Fst
find the eral force F ating On Ihe gate AB fora given depth of
fom i, e
COS
a
cn gate mens water, A= AD Wie DE 22 Fw
»
so depa ofthe C.G. ofthe inmersed area
sh
h 19820
1000 x9.81 x eN
$e oe
“The centre of pressure ofthe immersed surface, ls ven by
2 Total force Fis given by F2 pgAñ
ie tam gg
a
wi 192 MOI of emma
AD „2 (a Ÿ a
h nl bez
16h an
DARAS
Hydrostatic Forces on Surfaces 93 |
UN
ta
Now in the ACHD,
ee mn
ES
Taking the moments about inge, we ast
ac.
49080 x 50
or
Bes
or 243250)
9245250
39240
a 6.25)"
Problem 3.18 An inclined rectangular slice gate AB, 1.2 m by 3 m sie as shown in Fig. 3.23 is
installed ip control he discharge of water. The end À Ming. Determine the force normal to the gate
applied at to open it
‘Solution. Given
A 2 Area of gate = 12502 60 m
Dept of CG. of ie gate fon fee surface a the water =
1G = BC BE FREEWATERSURFACE © © 0
2502 BG sin 45°
1
06% + =4576m
0-06 % Jz = 4576
“The tora pressure farce (£) tng onthe gate,
F= pra
00 «9.81 X60 x 4.76
269343 N
“This force is acting at, where the deal of from Fig. 823
free surface I given by
UN
ta
Now in the ACHD,
ee mn
ES
Taking the moments about inge, we ast
ac.
49080 x 50
or
Bes
or 243250)
9245250
39240
a 6.25)"
Problem 3.18 An inclined rectangular slice gate AB, 1.2 m by 3 m sie as shown in Fig. 3.23 is
installed ip control he discharge of water. The end À Ming. Determine the force normal to the gate
applied at to open it
‘Solution. Given
A 2 Area of gate = 12502 60 m
Dept of CG. of ie gate fon fee surface a the water =
1G = BC BE FREEWATERSURFACE © © 0
2502 BG sin 45°
1
06% + =4576m
0-06 % Jz = 4576
“The tora pressure farce (£) tng onthe gate,
F= pra
00 «9.81 X60 x 4.76
269343 N
“This force is acting at, where the deal of from Fig. 823
free surface I given by
94 Fluid Mechanics
whore Jy = MOLL of gate
2 Dep of centre of pressure ht =
‘But rom Fig. 323 (a
Distance,
Distance, Bo. se
Distance, oH 197 - 6:89 = 0.582 m
Distance AH = AB BH = 12-0582 = 0618 m
Taking the moments about the hinge A
PRA = FA AM)
‘where is the Force normal 1 the gate app at 2
Px 12 = 269348 x 0618,
pe ROMERO ar am
Problem 3.18 ote aporto wes shown ni 24 Find he hf mt hat
Pen chu hinge Toe he lane et ety
Sion: Ge tri
Distance, act
4
where = Dept of water
‘The gate will tt tipping about hinge A the resultant pressure force ct a IF the resent
pressure force pases tough a pont which is ying from B wo C anywhere on the gate the pate will ip
‘ver the hinge Heace limiting cas is when the resultant fore passes through, tt resultat force
passes tnouph the centre of pressure. Heace for the given postion, pont B becomes the centre of
pressure Hence dept of cae of pressure.
CEE FREE WATER SURFACE
fos
But isal given by
a a Ice
Taking width of gate ua. Then
eN
Arc, acacia
ES 2
net Fig 326
sn mac a), mw, am
BR TS
whore Jy = MOLL of gate
2 Dep of centre of pressure ht =
‘But rom Fig. 323 (a
Distance,
Distance, Bo. se
Distance, oH 197 - 6:89 = 0.582 m
Distance AH = AB BH = 12-0582 = 0618 m
Taking the moments about the hinge A
PRA = FA AM)
‘where is the Force normal 1 the gate app at 2
Px 12 = 269348 x 0618,
pe ROMERO ar am
Problem 3.18 ote aporto wes shown ni 24 Find he hf mt hat
Pen chu hinge Toe he lane et ety
Sion: Ge tri
Distance, act
4
where = Dept of water
‘The gate will tt tipping about hinge A the resultant pressure force ct a IF the resent
pressure force pases tough a pont which is ying from B wo C anywhere on the gate the pate will ip
‘ver the hinge Heace limiting cas is when the resultant fore passes through, tt resultat force
passes tnouph the centre of pressure. Heace for the given postion, pont B becomes the centre of
pressure Hence dept of cae of pressure.
CEE FREE WATER SURFACE
fos
But isal given by
a a Ice
Taking width of gate ua. Then
eN
Arc, acacia
ES 2
net Fig 326
sn mac a), mw, am
BR TS
Hydrostatic Forces on Surfaces 95)
a
3
Equation the wo values of,
7 ñ
mala op Les
3 3
4
2 Helpht of water foc ping the gate = 9m. Ans.
Problem 3.20 A rectangular sluice gute AB, 2 m wide and 3m long is hinged at A us shown in
Fig. 3.25. skp closed by a weight ed 1 the gate. The tal weight of he gate and weight ied to
the gots 343350 N. Find he height ofthe water which wil js cause the gate to open The centre
of gravity ofthe weight and gate iat G.
x329m
Solution. Given
id of gate. b=2m: Length of gate L= 3 m
2 men Ar2x3=6m
Weight of gate and W=343350N
Angle of inclination, 82 45%
Let vis te require height of water.
Depth of C4. of te gate and weight = À
From Fig. 325 (0.
—ED=h-(AD-AE)
-ano-56un0 {ru
= sin 45° 0.6 tn 45°)
= @121 ~06)= (r= 1.521) m
“Te total presu Fores Fi given by
F = pg = 1000 9.81 x6 x 1521)
= 58860 (b= 1.520 N.
The total force Fis ating atthe centre o pressure as shown in Fi, 325 (6) at A. The dep of H
‘nom fic surface is given by AE which i gunl o
Hase
pt er 107
mt er
Assis
ana ASI sen
6x (A= L521) : 2
a
3
Equation the wo values of,
7 ñ
mala op Les
3 3
4
2 Helpht of water foc ping the gate = 9m. Ans.
Problem 3.20 A rectangular sluice gute AB, 2 m wide and 3m long is hinged at A us shown in
Fig. 3.25. skp closed by a weight ed 1 the gate. The tal weight of he gate and weight ied to
the gots 343350 N. Find he height ofthe water which wil js cause the gate to open The centre
of gravity ofthe weight and gate iat G.
x329m
Solution. Given
id of gate. b=2m: Length of gate L= 3 m
2 men Ar2x3=6m
Weight of gate and W=343350N
Angle of inclination, 82 45%
Let vis te require height of water.
Depth of C4. of te gate and weight = À
From Fig. 325 (0.
—ED=h-(AD-AE)
-ano-56un0 {ru
= sin 45° 0.6 tn 45°)
= @121 ~06)= (r= 1.521) m
“Te total presu Fores Fi given by
F = pg = 1000 9.81 x6 x 1521)
= 58860 (b= 1.520 N.
The total force Fis ating atthe centre o pressure as shown in Fi, 325 (6) at A. The dep of H
‘nom fic surface is given by AE which i gunl o
Hase
pt er 107
mt er
Assis
ana ASI sen
6x (A= L521) : 2
96 Fluid Mechanics
Fee WATER SURFACED
o
Fig 3.25
Now taking moments about hinge A, we got
343350 x EG = Fx AM
aK
ans
or asso x 06 =F x
[rm as sara anis can]
SSR AK
Er
343350 x06 xin AS", 035357
aK
‘38860(h—1521) "Ins “
E
su Anm a0 um 1.500 - À Go
[ar De 2
Da AC = CD AD =~ AB sin 45° = = 3 Xs 48° = h- 2121
Saban this valu i i, we get
315
40192) -(0-221)
CIÓN Las
as ys
a = Buue wi
risa sai *°% un
using he to vals of AR from (and Ci)
Fee WATER SURFACED
o
Fig 3.25
Now taking moments about hinge A, we got
343350 x EG = Fx AM
aK
ans
or asso x 06 =F x
[rm as sara anis can]
SSR AK
Er
343350 x06 xin AS", 035357
aK
‘38860(h—1521) "Ins “
E
su Anm a0 um 1.500 - À Go
[ar De 2
Da AC = CD AD =~ AB sin 45° = = 3 Xs 48° = h- 2121
Saban this valu i i, we get
315
40192) -(0-221)
CIÓN Las
as ys
a = Buue wi
risa sai *°% un
using he to vals of AR from (and Ci)
Hydrostatic Forces on Surfaces _ 97]
03535%7 _ ass
a +06
CIS
or 03535 «7 =0375 + 06 (h~ 1521) = 03784 06-06% 1.321
or 6h = 2.748 - 3754 06% 1321 = 20998 + 09126= 20121
0 na OT Ep m. Ans.
06
Problem 3.21 Find he tl pressure und positon of centre of pressure on a triangular plate of
ae 2m and eight 3 wc és immerse in water in uch a way hat he plane ofthe plate makes an
ange of 60° with the fre surface of the water. The Das ofthe plate is paralelo water surfe aná ot
dep of 2.5 m from water surface FREE WATER SURFACE
Solution, Given
Base of pat.
eit of pte,
2 As
Inclination,
Fig. 326
So high of ante}
(0 Total pressure force (E)
= pgA = 10009 81 4 3% 3.366 = 9061.38 Ans
(0 Centre of pressure (h*). Depth of cote of pressure {rom fee surface of waters given by
erie
wn ese
CRIA
ES
yt = LO 5.366 = 0.111 4 3.366 = 3.477 m. Ans.
“pase
> 3.6 CURVED SURFACE SUB-MERGED IN LIQUID
Consider a curved surface AB, sub-morged in a stati ud a shown In Fig. 327. Lot dA isthe area
‘ofa small sip at dept of A rom water Surface.
“Thon pressure nens om the aa dA is = pgh
and pressure fore AP Ep ree
"This foros aci normal to the surface
ones total pressure for om th curved surface should he
F=f paran a
gh cds om
03535%7 _ ass
a +06
CIS
or 03535 «7 =0375 + 06 (h~ 1521) = 03784 06-06% 1.321
or 6h = 2.748 - 3754 06% 1321 = 20998 + 09126= 20121
0 na OT Ep m. Ans.
06
Problem 3.21 Find he tl pressure und positon of centre of pressure on a triangular plate of
ae 2m and eight 3 wc és immerse in water in uch a way hat he plane ofthe plate makes an
ange of 60° with the fre surface of the water. The Das ofthe plate is paralelo water surfe aná ot
dep of 2.5 m from water surface FREE WATER SURFACE
Solution, Given
Base of pat.
eit of pte,
2 As
Inclination,
Fig. 326
So high of ante}
(0 Total pressure force (E)
= pgA = 10009 81 4 3% 3.366 = 9061.38 Ans
(0 Centre of pressure (h*). Depth of cote of pressure {rom fee surface of waters given by
erie
wn ese
CRIA
ES
yt = LO 5.366 = 0.111 4 3.366 = 3.477 m. Ans.
“pase
> 3.6 CURVED SURFACE SUB-MERGED IN LIQUID
Consider a curved surface AB, sub-morged in a stati ud a shown In Fig. 327. Lot dA isthe area
‘ofa small sip at dept of A rom water Surface.
“Thon pressure nens om the aa dA is = pgh
and pressure fore AP Ep ree
"This foros aci normal to the surface
ones total pressure for om th curved surface should he
F=f paran a
gh cds om
98. Fluid Mechanics
D w
©
Fig. 327
‘But here as he direction ofthe forces 0 the sl aens ar not In the same direction, but vaies
{rom point 0 point Hence integration of equation (3.11) fr curved surface is impossible. The problem
an. however, be solved by resolving the force dF in two components dF, and df, in the 1 ad y
direction respectively. The ttl force in the x and y directions, Le. F, and F, ae obtained by
Integrating dE, and dF, The total force 0 the curved surface is
ary
dicton of sata with horizontal is tan 9 = E as
oF a
Resolving the force dF give by equation (3.11) ax an y directions
ou fre exar y iio
Fea ar, [ot sn =e J ran sn cas
Pr 6,7] ué,= panas coso= pe | acoso 0
Fig. 327 0) shows he enlarge arca dA From this igure, Le, AEF,
“Ths in eqution (3.15), sin = FG:
Yor pin hr nes onen
a nO orense ns o e pjs ate he cad fe on he
vertical plane. Thus
E, Total pressure force am th projected arc of the curve surface on vertical plane. 3.17)
Also dA cos 8 = EG = horizontal pojcetion of dA and hones hdA cos 9 is the volume of he liquid
cov o semen rn dA pie sre of th iu Tins JA cs 8 ie wa
o url srs exited is mas
Mence pf os 01 he leg supported he ave ice. Th
x J nth os 0
= wol of iad sappy th cud ra u re ut of Ng aus
5,
D w
©
Fig. 327
‘But here as he direction ofthe forces 0 the sl aens ar not In the same direction, but vaies
{rom point 0 point Hence integration of equation (3.11) fr curved surface is impossible. The problem
an. however, be solved by resolving the force dF in two components dF, and df, in the 1 ad y
direction respectively. The ttl force in the x and y directions, Le. F, and F, ae obtained by
Integrating dE, and dF, The total force 0 the curved surface is
ary
dicton of sata with horizontal is tan 9 = E as
oF a
Resolving the force dF give by equation (3.11) ax an y directions
ou fre exar y iio
Fea ar, [ot sn =e J ran sn cas
Pr 6,7] ué,= panas coso= pe | acoso 0
Fig. 327 0) shows he enlarge arca dA From this igure, Le, AEF,
“Ths in eqution (3.15), sin = FG:
Yor pin hr nes onen
a nO orense ns o e pjs ate he cad fe on he
vertical plane. Thus
E, Total pressure force am th projected arc of the curve surface on vertical plane. 3.17)
Also dA cos 8 = EG = horizontal pojcetion of dA and hones hdA cos 9 is the volume of he liquid
cov o semen rn dA pie sre of th iu Tins JA cs 8 ie wa
o url srs exited is mas
Mence pf os 01 he leg supported he ave ice. Th
x J nth os 0
= wol of iad sappy th cud ra u re ut of Ng aus
5,
Hydrostatic Forces on Surfaces 99 |
In Fig. 3:28, th curved srfae AB is not supporting any ol, Im
such caes, Fs equal othe weight ofthe imaginary qui supported
y AB up fie surface of liquid. The direction of F, willbe takeo in
upward direcion.
Problem 3.22 Compute Me horizontal and vertical components
ofthe total force ating on a curved surface AB, which is In the form
Ja quadrant o cree ofradi 2 mas shown in Fi. 329, Take the
wid ofthe gate us uni
Solution. Given
Width of gate =10m
Radios ofthe gate = 20m
= Distance A0=08B=2m
A lo FREE SURFACE OF WATER
Horizontal fore, F, exerted by water on gate ds given by
Sol Et
ation G17 3
E22 Ton promu force on the projet are of carted
use AB on vere plane
= Total pressure force on 08. 4
ji are of euved sri on sera plane = 08 1)
= pai
1000 984 x2 crx (is
Eo Ava of 0B = A= 20x12
À = Depth of CG. of OB from fre surface = 15+ #1
= 9.81 x 2000 x 2. = 49080 N. Ans.
‘The point of application of Fi given by tm LE
1333 à 25= 2.633 m from fee surface,
Verica fore, E exerted by waters piven by equation (3.18)
= Weight of portion DABOC
= Weight of DAOC 4 Weight of water AOD
= pg [Volume of DAOC + Volume of 408]
10981 [4040x148 40) xt
In Fig. 3:28, th curved srfae AB is not supporting any ol, Im
such caes, Fs equal othe weight ofthe imaginary qui supported
y AB up fie surface of liquid. The direction of F, willbe takeo in
upward direcion.
Problem 3.22 Compute Me horizontal and vertical components
ofthe total force ating on a curved surface AB, which is In the form
Ja quadrant o cree ofradi 2 mas shown in Fi. 329, Take the
wid ofthe gate us uni
Solution. Given
Width of gate =10m
Radios ofthe gate = 20m
= Distance A0=08B=2m
A lo FREE SURFACE OF WATER
Horizontal fore, F, exerted by water on gate ds given by
Sol Et
ation G17 3
E22 Ton promu force on the projet are of carted
use AB on vere plane
= Total pressure force on 08. 4
ji are of euved sri on sera plane = 08 1)
= pai
1000 984 x2 crx (is
Eo Ava of 0B = A= 20x12
À = Depth of CG. of OB from fre surface = 15+ #1
= 9.81 x 2000 x 2. = 49080 N. Ans.
‘The point of application of Fi given by tm LE
1333 à 25= 2.633 m from fee surface,
Verica fore, E exerted by waters piven by equation (3.18)
= Weight of portion DABOC
= Weight of DAOC 4 Weight of water AOD
= pg [Volume of DAOC + Volume of 408]
10981 [4040x148 40) xt
100 Fluid Mechanics
1000 x 981 [30 + RIN = 60249: N. Ans.
Problem 3.23 Fig 3.30 shows a gate having a quadran shape of radius 2 m. Find the resultont
Jorce due 1 water per metre length ofthe gate Find also the angle a which the oa force wil ac
Solution. Gives WATERSURFACE Ao
Radius of gate m = À lu
wt ee “a CE
3
Horizontal Force
A
ore on te projected are ofthe
‘curved surface on vertical plane
= Fees on BO = pat Fig. 330
where A = Ares of 80 =2%
r
ia do 2 nom erie
heb xte tm
mie his:
1000 9.81 42% 1 = 19620 N
Vertical Force, F,
= 1000 9.61 À À (E o L0= 30819 N
AR 4x20
“This wilt at tun of SF 2
2 Resuhanı force, Fis given by
Pf
RO SOS - SON HT
265344 N Ans
‘The angle mad y th result A natal ive by
3089
E, 19620
: Ol 17082 ST A Ans,
Problem 3.24 Find the magnitude and direcion of the resultant force due 10 water ating on a
roller gate of eyndrcal form of 40 m diameter, when he gat placed on he dam in sch awa hat
ater Just going 0 spl Take the length of he gate as Em.
‘Solution
Dia. of gate
Radios.
Length of pte,
= 0.848 m from OB.
tan 0. 5708
1000 x 981 [30 + RIN = 60249: N. Ans.
Problem 3.23 Fig 3.30 shows a gate having a quadran shape of radius 2 m. Find the resultont
Jorce due 1 water per metre length ofthe gate Find also the angle a which the oa force wil ac
Solution. Gives WATERSURFACE Ao
Radius of gate m = À lu
wt ee “a CE
3
Horizontal Force
A
ore on te projected are ofthe
‘curved surface on vertical plane
= Fees on BO = pat Fig. 330
where A = Ares of 80 =2%
r
ia do 2 nom erie
heb xte tm
mie his:
1000 9.81 42% 1 = 19620 N
Vertical Force, F,
= 1000 9.61 À À (E o L0= 30819 N
AR 4x20
“This wilt at tun of SF 2
2 Resuhanı force, Fis given by
Pf
RO SOS - SON HT
265344 N Ans
‘The angle mad y th result A natal ive by
3089
E, 19620
: Ol 17082 ST A Ans,
Problem 3.24 Find the magnitude and direcion of the resultant force due 10 water ating on a
roller gate of eyndrcal form of 40 m diameter, when he gat placed on he dam in sch awa hat
ater Just going 0 spl Take the length of he gate as Em.
‘Solution
Dia. of gate
Radios.
Length of pte,
= 0.848 m from OB.
tan 0. 5708
Hydrostatic Fores on Surfaces 101
Horizontal force, F, sting on the gate is ware
Fa PRAÍ = Force on projected area of curved surface
ACE on vera pane
= Force on vertical area AB
where A= Area of AOB = 4.0% 80 = 320m?
= Depth of CG. 0f AOB = AR = 2.0m
E = 1000x981 320% 2.0
= e270, Fig. 331
Vertical force, F, à piven by
F, = Weight of water enclosed or support (actually or imaginary) hy
te curved surface ACB
x Volume af portion ACB
= pg x Area of ACB XE
10009. ee) 80 omo oyo 99106
In will acting inthe upward direction.
Result force, = + BS = GRO ION = 798328 N. Ans.
CRT
Dico o resultat fore is give by tan 20285
RTE, o
0= 31° 8 Am.
Problem 225. Find e hoot nd vera component of ater pres acing en fic fa
Ito gu 790° sector of ras ms shown it Pate hf te oly.
Soliton, Gea pl
Radi of pate, R-im ua,
ron compone! offre acting on hep i
ra occa reo ue
roto vn plane
Face on aes ADB
= pet
wee AAD wie pu
on care ENE
2x4 x sin 45° = 8% 707 = 5.656 m’ 1 AD =4 sin 45°)
AS
2s a
E = 1000 X 9.81 5.656 2828 N = 156911 N. Ams
Vertical component
E, = Weight of water supported or cnclosd hy the curved surface
= Weight of water in portion ACBDA
= py x Arca of ACBDA x With of gate
= 1000 x 981 x [Area of sector ACBOA — Arca of AABO] x 1
Horizontal force, F, sting on the gate is ware
Fa PRAÍ = Force on projected area of curved surface
ACE on vera pane
= Force on vertical area AB
where A= Area of AOB = 4.0% 80 = 320m?
= Depth of CG. 0f AOB = AR = 2.0m
E = 1000x981 320% 2.0
= e270, Fig. 331
Vertical force, F, à piven by
F, = Weight of water enclosed or support (actually or imaginary) hy
te curved surface ACB
x Volume af portion ACB
= pg x Area of ACB XE
10009. ee) 80 omo oyo 99106
In will acting inthe upward direction.
Result force, = + BS = GRO ION = 798328 N. Ans.
CRT
Dico o resultat fore is give by tan 20285
RTE, o
0= 31° 8 Am.
Problem 225. Find e hoot nd vera component of ater pres acing en fic fa
Ito gu 790° sector of ras ms shown it Pate hf te oly.
Soliton, Gea pl
Radi of pate, R-im ua,
ron compone! offre acting on hep i
ra occa reo ue
roto vn plane
Face on aes ADB
= pet
wee AAD wie pu
on care ENE
2x4 x sin 45° = 8% 707 = 5.656 m’ 1 AD =4 sin 45°)
AS
2s a
E = 1000 X 9.81 5.656 2828 N = 156911 N. Ams
Vertical component
E, = Weight of water supported or cnclosd hy the curved surface
= Weight of water in portion ACBDA
= py x Arca of ACBDA x With of gate
= 1000 x 981 x [Area of sector ACBOA — Arca of AABO] x 1
102 Fluid Mechanics
a nom ar angst
ES 29] 48796 N. Ans
“7
Problem 3.26. Calculate the horizontal and vertical components ofthe water pressure eered on a
Jaimes goteo raus 8m as shown in Fig, 3.33. Take wid of got unir.
Solution. The horizontal component of water pressure Is given by
E peli = Foros oo te are projected on vertical plane
Fores on he vera arca of BD
BD x Width of gue= 40 = 4.0
1000 x 9.81 44,0 x 2.0 = BON. Ans Fe das
Venical component ofthe water prssue is given hy
F, = Weight of water supported cnelosd (imaginary) hy curved
surface CB
Weight of water in the portion CRDC
= pe (Area of portion CRDCI x Width of gate
ng x (Area of sector CHO ~ Area of the tingle BOD] 1
E |
360 2
_ 40x88 ons. 30°
?
De,
=98t0x[ Les?
il igo
DO = BO con 30" = 8 x<08 30°]
= 9810 {16.755 - 13356] = 28839 N. Ans.
Problem 3.27 A crtindrical gate of 4 m diameter 2 m long has mater om ts oth is at shown in
Fig. 5.24, Determine the magnitude, location and direction of the resultant force exerted by the water
on e gate. Find aso Ihe leas weight ofthe cinder so tha it may not be led ax fromthe lor.
Solution. Gives ‘WATER SURFACE A
Dia. of gate =4m =
Radios 2m
{D The forces acting on the left ie ofthe cylinder are
‘The horizontal componen,
where F,, = Fore of water on bra projected on vertical
plane
= Force on arca AOC
pen where A
1000 49.81 x 842.
= 156960 N
a nom ar angst
ES 29] 48796 N. Ans
“7
Problem 3.26. Calculate the horizontal and vertical components ofthe water pressure eered on a
Jaimes goteo raus 8m as shown in Fig, 3.33. Take wid of got unir.
Solution. The horizontal component of water pressure Is given by
E peli = Foros oo te are projected on vertical plane
Fores on he vera arca of BD
BD x Width of gue= 40 = 4.0
1000 x 9.81 44,0 x 2.0 = BON. Ans Fe das
Venical component ofthe water prssue is given hy
F, = Weight of water supported cnelosd (imaginary) hy curved
surface CB
Weight of water in the portion CRDC
= pe (Area of portion CRDCI x Width of gate
ng x (Area of sector CHO ~ Area of the tingle BOD] 1
E |
360 2
_ 40x88 ons. 30°
?
De,
=98t0x[ Les?
il igo
DO = BO con 30" = 8 x<08 30°]
= 9810 {16.755 - 13356] = 28839 N. Ans.
Problem 3.27 A crtindrical gate of 4 m diameter 2 m long has mater om ts oth is at shown in
Fig. 5.24, Determine the magnitude, location and direction of the resultant force exerted by the water
on e gate. Find aso Ihe leas weight ofthe cinder so tha it may not be led ax fromthe lor.
Solution. Gives ‘WATER SURFACE A
Dia. of gate =4m =
Radios 2m
{D The forces acting on the left ie ofthe cylinder are
‘The horizontal componen,
where F,, = Fore of water on bra projected on vertical
plane
= Force on arca AOC
pen where A
1000 49.81 x 842.
= 156960 N
Hydrostatic Forces on Surfaces 108]
‘weight of water enclosed by ABCOA
vow ost [Em] 20 «9810 8 22°20 = 176,
ight Side of the Cylinder
Fra
2x2x2f
1000x981 x242x 2 | A
Force on vertical ares CO
coxı-axı
=39240 N
E, = Weight of water enclosed hy DOCD
=n [fe «witht gate
= 1000 49.81 x 2 x2= 61638 N
Result fore la the diecia of x,
EEE = 156960 ~ 39240 = 117720 N
Resultant free in the dico Of
F =F, + E, 123276 à 61638 = 194914 N
(9 Resultant Force, Fi given as
Fz {RR = (uray aaa)”
(6) Direction of resultant force i given by
uno À IS
NTE
a 028731 Ans.
(Gu) Location ofthe resultat force
219206 N. Ans
= 15707
sat dans 0 23624 1.3 mom o sa ns ic of ns sta
foe fin th cima vla distance from e Room as
CRETE
a 1177209 = 156960 x LAS 39240 67 = 2087565 - 20090 = 182406
132866
1.55 m from the boom,
Try ee
Face a es on 40€ a ns 22 a rom AOC oval
was
ia aoc
Ao ae 0348800 A oma AOC Te eae
Fil act at distance + fom AOC whieh is given by
‘weight of water enclosed by ABCOA
vow ost [Em] 20 «9810 8 22°20 = 176,
ight Side of the Cylinder
Fra
2x2x2f
1000x981 x242x 2 | A
Force on vertical ares CO
coxı-axı
=39240 N
E, = Weight of water enclosed hy DOCD
=n [fe «witht gate
= 1000 49.81 x 2 x2= 61638 N
Result fore la the diecia of x,
EEE = 156960 ~ 39240 = 117720 N
Resultant free in the dico Of
F =F, + E, 123276 à 61638 = 194914 N
(9 Resultant Force, Fi given as
Fz {RR = (uray aaa)”
(6) Direction of resultant force i given by
uno À IS
NTE
a 028731 Ans.
(Gu) Location ofthe resultat force
219206 N. Ans
= 15707
sat dans 0 23624 1.3 mom o sa ns ic of ns sta
foe fin th cima vla distance from e Room as
CRETE
a 1177209 = 156960 x LAS 39240 67 = 2087565 - 20090 = 182406
132866
1.55 m from the boom,
Try ee
Face a es on 40€ a ns 22 a rom AOC oval
was
ia aoc
Ao ae 0348800 A oma AOC Te eae
Fil act at distance + fom AOC whieh is given by
104 Fluid Mechanics
EAS E, BASS — F,, x BAB
or 184914 x x= 123276 8488 61638 2 ABS = 8488 |
Sos
HMS
ler. The resultat fore in he upwar direction is
E, = ADI N
“Thus the weight of cylinder should not e les han the upwand force. Hence leas weight of
inde should heat east
3276-61638] = 323184
02829 m rom AOC.
(69) Least weight of 6
184914N. Ans.
Problem 3.28 Fig. 3.35 shows the cross section fa tank fll f water under pressure. The length
ofthe tank ie? m. An empty cinder les along the length ofthe tank om one of te corner as show
Find the horizontal and verical components ofthe force acting on the curved surface ABC ofthe
inde
‘Solution. Radius, 02 katie?
gt a,
femur, 25.981 Ven
1362 Nem = 962 10 Nm?
Pr tent, ha Pn O Lg
be 1000x9381 ©
roe surface of water will e ata hsight of 2 m from
he top ofthe tank
Fig. 336 shows the equivalent fre surface of wate,
(0 Horizontal Component of Force
F2 pra
where À Atea projected on vertical plane
9
215
F,= 1000 9.81 x302.5,
809325 N. Ans.
(Gd Vertical Component of Force
= Weight of water enclosed or sported
actual or imaginary by curved surface ABC
Weight of water in he portion CODE ABC
Weight of waterin CODFBC Weight of water in AEF
‘But weight of water in CODFBC
Weight of w
Fix 336
me Bu
= + 80x00) x: Bras |
oe ry | 4 |
648585
Weight of water in AEP = pg [Area of AEF) x 2.0
EAS E, BASS — F,, x BAB
or 184914 x x= 123276 8488 61638 2 ABS = 8488 |
Sos
HMS
ler. The resultat fore in he upwar direction is
E, = ADI N
“Thus the weight of cylinder should not e les han the upwand force. Hence leas weight of
inde should heat east
3276-61638] = 323184
02829 m rom AOC.
(69) Least weight of 6
184914N. Ans.
Problem 3.28 Fig. 3.35 shows the cross section fa tank fll f water under pressure. The length
ofthe tank ie? m. An empty cinder les along the length ofthe tank om one of te corner as show
Find the horizontal and verical components ofthe force acting on the curved surface ABC ofthe
inde
‘Solution. Radius, 02 katie?
gt a,
femur, 25.981 Ven
1362 Nem = 962 10 Nm?
Pr tent, ha Pn O Lg
be 1000x9381 ©
roe surface of water will e ata hsight of 2 m from
he top ofthe tank
Fig. 336 shows the equivalent fre surface of wate,
(0 Horizontal Component of Force
F2 pra
where À Atea projected on vertical plane
9
215
F,= 1000 9.81 x302.5,
809325 N. Ans.
(Gd Vertical Component of Force
= Weight of water enclosed or sported
actual or imaginary by curved surface ABC
Weight of water in he portion CODE ABC
Weight of waterin CODFBC Weight of water in AEF
‘But weight of water in CODFBC
Weight of w
Fix 336
me Bu
= + 80x00) x: Bras |
oe ry | 4 |
648585
Weight of water in AEP = pg [Area of AEF) x 2.0
Hydrostatic Forces on Surfaces _ 105]
2 100 2 81 (eso WEF y AGH - A 20
an 05
740710
BI 2 B00 = 10-0 cos 8 10-1 60630" à 0134
An DI = cc 80 doen A
EN
360
ln anno, sio
s 2 02
WHO _ aR? 05%.866
2 ma
= 0.0458
Weight of water in AEFB
9810 X IAE x AG + AG X AH - 004531 «20
9810 X (20 «134 4 138 X 5 — 0853] x 2.0
9810 x (268 + 067 - 0453] X 2.0 = 5684 N
A E, = 614585 ~ 5658 = 587745 N. Ans.
Problem 3.29 Find he magnitde and direction of the resultant water pressure acting on a curved
Jace fa dam whichis shaped rconding te relation = as shown in Fi. 3.37. The high fe
water retained bythe dam is 10 m. Consider the width ofthe dam as unity.
‘Solution. Equation of curve AB is
3 ud
Las
5 2
Height of water, ke 10m
width, balm 7
"The horizontal component, is given y Tig. 337
Fi = Pressure due 0 water on the curved rca pros on vertical plane
Pressure on aca BC
peat
where A=BCXx1 210x108,
F,= 1000 «9.81 x 10%5 = 490500 N
‘Vertical component, F, à iven by
E, = Weight of water supported by the curve AB
= Weight of wate in he portion ABC
Area of ABC x With of dam
oe [feo] x10 [Arcor
bet
+ cas fa)
somos x [a as zu)
oe ann
2300 | =2900% 3 7
= 19620 x 31.622 = 620439 N
2 100 2 81 (eso WEF y AGH - A 20
an 05
740710
BI 2 B00 = 10-0 cos 8 10-1 60630" à 0134
An DI = cc 80 doen A
EN
360
ln anno, sio
s 2 02
WHO _ aR? 05%.866
2 ma
= 0.0458
Weight of water in AEFB
9810 X IAE x AG + AG X AH - 004531 «20
9810 X (20 «134 4 138 X 5 — 0853] x 2.0
9810 x (268 + 067 - 0453] X 2.0 = 5684 N
A E, = 614585 ~ 5658 = 587745 N. Ans.
Problem 3.29 Find he magnitde and direction of the resultant water pressure acting on a curved
Jace fa dam whichis shaped rconding te relation = as shown in Fi. 3.37. The high fe
water retained bythe dam is 10 m. Consider the width ofthe dam as unity.
‘Solution. Equation of curve AB is
3 ud
Las
5 2
Height of water, ke 10m
width, balm 7
"The horizontal component, is given y Tig. 337
Fi = Pressure due 0 water on the curved rca pros on vertical plane
Pressure on aca BC
peat
where A=BCXx1 210x108,
F,= 1000 «9.81 x 10%5 = 490500 N
‘Vertical component, F, à iven by
E, = Weight of water supported by the curve AB
= Weight of wate in he portion ABC
Area of ABC x With of dam
oe [feo] x10 [Arcor
bet
+ cas fa)
somos x [a as zu)
oe ann
2300 | =2900% 3 7
= 19620 x 31.622 = 620439 N
106 Fv Mechanics
‘Resultant water pressure on dam
P= fe +8 fons) +2089"
{NUT N = 790907 AN. Ans.
Dirsction ofthe reatant ssc hy
E, 02089 neg
490500
SU 40% Ans.
Problem 3.30 _ À dam has praho shape y 6] as shown In Fi, 338 Below having = 6m
and yy = 9m. The id is water with density = 1000 kg/m’. Compute the horizontal, vertical and the
esta trast exerted by water per metre length ofthe dam.
Solution, cio a
Equation of be cuve OA is
ia of dam, Im.
(9 Horizontal thrust exerted by water
= Force exer by water on veicalsorface
. (ré
(OB, ie he surface oincd hy projcting ORGN ALO
‘ths curved surface on vertical plano
= pati Fig, 338
1000 «9.81 «(9 1)
(6) Vertical thrust exerted by water
= Weight of water supported y curved surface OA upto free surface of
Weight of water the portion ABO.
pu Area oF OAB Width of dam
saox0a'x[ xa] ie
2%)
sosa [992.0] 10
9620 x{ | = 19620x 2 19%)
won [22] md
19620 x 3 #27 = 353160 N. Am.
‘Resultant water pressure on dam
P= fe +8 fons) +2089"
{NUT N = 790907 AN. Ans.
Dirsction ofthe reatant ssc hy
E, 02089 neg
490500
SU 40% Ans.
Problem 3.30 _ À dam has praho shape y 6] as shown In Fi, 338 Below having = 6m
and yy = 9m. The id is water with density = 1000 kg/m’. Compute the horizontal, vertical and the
esta trast exerted by water per metre length ofthe dam.
Solution, cio a
Equation of be cuve OA is
ia of dam, Im.
(9 Horizontal thrust exerted by water
= Force exer by water on veicalsorface
. (ré
(OB, ie he surface oincd hy projcting ORGN ALO
‘ths curved surface on vertical plano
= pati Fig, 338
1000 «9.81 «(9 1)
(6) Vertical thrust exerted by water
= Weight of water supported y curved surface OA upto free surface of
Weight of water the portion ABO.
pu Area oF OAB Width of dam
saox0a'x[ xa] ie
2%)
sosa [992.0] 10
9620 x{ | = 19620x 2 19%)
won [22] md
19620 x 3 #27 = 353160 N. Am.
Hydrostatic Forces on Surfaces 107]
Problem 3.31 A cylinder 3m in diameter and 4m long retains water on one side. The elinder is
supported as shown in Fig. 3.39. Determine the horizontal reaction at and the vertical reaction at B.
The slider weighs 196.2 EN. Ignore icon. WATER SURFACE _C
Solution. Give E
Dia. o eylinder
Length of elinder
Weight of cylinder,
Horizontal force exerted by war
F, = Force on voit rea BOC
bag Fig. 339
ods.
2
196200 N
where A=BOCxI=3%4= 127,
sm
= 1000 «9.81 12 1
Te verti force exerted by water
F, = Weight of water enclosed in BDCOB
176580 8
2)
=00x (Ea?) xr
Force E, ls ating in the vpvar dicto
For the equilibrium of eylinder
orion action a À = F,= 176580 N
Vertical waction at B= Weight of ind,
196000 = 138688 = 57516 N- Ans.
000 x 9,81 x Ex x 4 = 138684 N
» 3.7. TOTAL PRESSURE AND CENTRE OF PRESSURE ON LOCK GATES
Lock gates ae the devices used for changing the water lve in a canal oF ve for navigation
ig, 340 shows plan and clevation ofa par of lock gates. Lot AB and JC bo the two lock gates. Each
at is supported on two hinges fixed on thelr op and bottom atthe ends À and €. Inthe closed
positon, the gates meet at.
Let” F = Resultant force dueto water on the gate AB or BC acting ae sight angles tothe gate
R =Reaction a the lower and upper binge
P = Reaction a the common contact surface ofthe wo gates and setngpeepeadielarto te
contact sure
Lette force P and F meet at 0, Then the rection R most pass through 0 as the gate AB sl the
slim under the aston of ie forces. Let Os the inclination ofthe Tock gate with he normal to
Ue side ofthe lock,
Problem 3.31 A cylinder 3m in diameter and 4m long retains water on one side. The elinder is
supported as shown in Fig. 3.39. Determine the horizontal reaction at and the vertical reaction at B.
The slider weighs 196.2 EN. Ignore icon. WATER SURFACE _C
Solution. Give E
Dia. o eylinder
Length of elinder
Weight of cylinder,
Horizontal force exerted by war
F, = Force on voit rea BOC
bag Fig. 339
ods.
2
196200 N
where A=BOCxI=3%4= 127,
sm
= 1000 «9.81 12 1
Te verti force exerted by water
F, = Weight of water enclosed in BDCOB
176580 8
2)
=00x (Ea?) xr
Force E, ls ating in the vpvar dicto
For the equilibrium of eylinder
orion action a À = F,= 176580 N
Vertical waction at B= Weight of ind,
196000 = 138688 = 57516 N- Ans.
000 x 9,81 x Ex x 4 = 138684 N
» 3.7. TOTAL PRESSURE AND CENTRE OF PRESSURE ON LOCK GATES
Lock gates ae the devices used for changing the water lve in a canal oF ve for navigation
ig, 340 shows plan and clevation ofa par of lock gates. Lot AB and JC bo the two lock gates. Each
at is supported on two hinges fixed on thelr op and bottom atthe ends À and €. Inthe closed
positon, the gates meet at.
Let” F = Resultant force dueto water on the gate AB or BC acting ae sight angles tothe gate
R =Reaction a the lower and upper binge
P = Reaction a the common contact surface ofthe wo gates and setngpeepeadielarto te
contact sure
Lette force P and F meet at 0, Then the rection R most pass through 0 as the gate AB sl the
slim under the aston of ie forces. Let Os the inclination ofthe Tock gate with he normal to
Ue side ofthe lock,
108 Fluid Mechanics
lo ZABO, — ZOAB=ZAB0
Resolving al force long the gate AB and puting equal to zero, we get
Roos P cos = Oor R= P
wwarensuarace [7 atenos
a
Fig. 340
Resolving forces normal 1 ih gate AB
Ron 0+ Pain F=0
or F=Rsin + Pain 0 2 sa 0
Pe
Pano
To calcule P and R
aus)
22
620
In equation (320) can be calculated it F and O re known. The vale of Qs calculated from the
ange between the lock gates, The angle between the to lock pate I equal to 180” ~ 20. Hence 0 ean
be clculated, The value of Fis cuculatd as
La 1, = igh of water on the upercam sido
1, = Height of water on the downstcam side
cr pressure om the gate on upsrcam sido
7 wid of gate
Xow Cave
m
ooo
Pr fs etx
, E P
Stat, root ora Me
2 Resultat tt? _ pall
lore FFF,
Ir pressure om th gate on dowastcam side of the gate
a
2
2. à
Substutag the value of and Fin equation (320), te value of P aod R canbe calculated
Resctlons atthe top and bottom hinges
La = Reaction of te top binge
lo ZABO, — ZOAB=ZAB0
Resolving al force long the gate AB and puting equal to zero, we get
Roos P cos = Oor R= P
wwarensuarace [7 atenos
a
Fig. 340
Resolving forces normal 1 ih gate AB
Ron 0+ Pain F=0
or F=Rsin + Pain 0 2 sa 0
Pe
Pano
To calcule P and R
aus)
22
620
In equation (320) can be calculated it F and O re known. The vale of Qs calculated from the
ange between the lock gates, The angle between the to lock pate I equal to 180” ~ 20. Hence 0 ean
be clculated, The value of Fis cuculatd as
La 1, = igh of water on the upercam sido
1, = Height of water on the downstcam side
cr pressure om the gate on upsrcam sido
7 wid of gate
Xow Cave
m
ooo
Pr fs etx
, E P
Stat, root ora Me
2 Resultat tt? _ pall
lore FFF,
Ir pressure om th gate on dowastcam side of the gate
a
2
2. à
Substutag the value of and Fin equation (320), te value of P aod R canbe calculated
Resctlons atthe top and bottom hinges
La = Reaction of te top binge
Hydrostatic Forces on Surfaces 109]
Reaction ofthe botom hinge
Then BR + Ra
‘The resultant water pressure F acts normal to the gate Mal o the value of Fis resisted by the
hinges of one lek gates and oer halt Wil be resisted bythe hinges of oer Lock gate. Also F, a a
m #
aisance of “fom bot vil ts at distance of À om bou
3
“Taking moments about the lower hinge
Ah
Rosina Aine o
5 ©
where H = Distance Betwsen two inges
Resolving faces harisomally
CNA
Amor an 0 AB m
rom equations () and (id, we can find R, and Ry
Problem 3,32 Each gate of a luck is 6 m high and is supported by two hinges placed on the op
(and bottom of the gate. Wien the gates are closed, they make an angle of 120° The wid of lock i
‘Sm. fthe water levels are4 mand 2 mon the upstream and downstreom sides respectively determine
the magnitude ofthe forces on the hinges due to water pressure.
Solution. Given me A
eight of lok om 7
io lock =5m %
With of each lock pate = AB 8
AD _ 25 im =
> 9 25 me |:
a aa NER
22887 m
Angle between gates 1207
5 ME
Su: or
Hei of water on upstream side
oe FA AE
20 Fig. 34
“Total water pressure on upstream side
E, =pgA hi where Ay = Hy x1 4042887 m0
and
covers flan)
= 226571 N .
fe vii ast = = 3 mn
Sint rm nn de
1000 x 981 x 2x 2.887 x 1.0 A
Reaction ofthe botom hinge
Then BR + Ra
‘The resultant water pressure F acts normal to the gate Mal o the value of Fis resisted by the
hinges of one lek gates and oer halt Wil be resisted bythe hinges of oer Lock gate. Also F, a a
m #
aisance of “fom bot vil ts at distance of À om bou
3
“Taking moments about the lower hinge
Ah
Rosina Aine o
5 ©
where H = Distance Betwsen two inges
Resolving faces harisomally
CNA
Amor an 0 AB m
rom equations () and (id, we can find R, and Ry
Problem 3,32 Each gate of a luck is 6 m high and is supported by two hinges placed on the op
(and bottom of the gate. Wien the gates are closed, they make an angle of 120° The wid of lock i
‘Sm. fthe water levels are4 mand 2 mon the upstream and downstreom sides respectively determine
the magnitude ofthe forces on the hinges due to water pressure.
Solution. Given me A
eight of lok om 7
io lock =5m %
With of each lock pate = AB 8
AD _ 25 im =
> 9 25 me |:
a aa NER
22887 m
Angle between gates 1207
5 ME
Su: or
Hei of water on upstream side
oe FA AE
20 Fig. 34
“Total water pressure on upstream side
E, =pgA hi where Ay = Hy x1 4042887 m0
and
covers flan)
= 226571 N .
fe vii ast = = 3 mn
Sint rm nn de
1000 x 981 x 2x 2.887 x 1.0 A
LO Fluid Mechanics
= 56613 N
4,
Fz wil act ata distance or #2 = 2 = 0,67 m from botom,
. D
Resultant water pressure on each gate
= Fy = 226871 = 56643 = 169928 N,
Lots sight of fom the honor, then aking moments of E, and about the bot, we have
Fy x13 F,x067
= 226571 x 1.38 = 56643 x 0.67
226871 x 133 $6683% 0.67 _ 301339 37950
From equation (3.19),
AR, and Ry ar the actions at hs top and otto hinges, en
“Taking movement of hinge scans Ry, A and about the hatom hinges, we have
Rrx60 + Ry X0=RX1SS
169928 «1.5
60
R 43998.
A Ry =R— R= 169928 — 43898 = 126030 N. Ans.
Problem 8.38. The end gates ABC of a lock are 9 m high and when closed include on angle of
120°, The with of the lck 1 m. Each gate ts supported by mo hinges located À m und 6 m
above the bottom ofthe lock The dep of water on the two sides are Em and 4m respectively. Find
(i) Resultant water force on each gute,
(i) Reaction benveen the gates AB and BC, and
(i) Force on each hing, considering the reaction ofthe gate acting inthe same horizontal plane
a resultant water pressure
Solution. Given
eight of gate
Inclination of gate
9m
m man 09 ELEVATION
Fig 3.42
= 56613 N
4,
Fz wil act ata distance or #2 = 2 = 0,67 m from botom,
. D
Resultant water pressure on each gate
= Fy = 226871 = 56643 = 169928 N,
Lots sight of fom the honor, then aking moments of E, and about the bot, we have
Fy x13 F,x067
= 226571 x 1.38 = 56643 x 0.67
226871 x 133 $6683% 0.67 _ 301339 37950
From equation (3.19),
AR, and Ry ar the actions at hs top and otto hinges, en
“Taking movement of hinge scans Ry, A and about the hatom hinges, we have
Rrx60 + Ry X0=RX1SS
169928 «1.5
60
R 43998.
A Ry =R— R= 169928 — 43898 = 126030 N. Ans.
Problem 8.38. The end gates ABC of a lock are 9 m high and when closed include on angle of
120°, The with of the lck 1 m. Each gate ts supported by mo hinges located À m und 6 m
above the bottom ofthe lock The dep of water on the two sides are Em and 4m respectively. Find
(i) Resultant water force on each gute,
(i) Reaction benveen the gates AB and BC, and
(i) Force on each hing, considering the reaction ofthe gate acting inthe same horizontal plane
a resultant water pressure
Solution. Given
eight of gate
Inclination of gate
9m
m man 09 ELEVATION
Fig 3.42
Hydrostatic Fores on Surfaces 111
‘With of tock = 10m
5
With of exch lock person
Dept of water on upsueam side, = 8m
Dept of water on downstream sie, Hy = 4m
Water pressure an upstream side
y= pra
she Ave Hh 281738 246188 mF = Mio ann
2
= 1000 9.81 46.184 4.0 = 1812260 N = 1812.26 kN
Water pressure on downstream sie,
Fe pete
4
where Ay = ix = 5.773 4 = 28.092 ms = 20
Fy = 1000 x 9.81 x 23092 x 20
Resalta water pressure
2 Fi ~ Fan 1812.26 ~ 493068 = 1359.195 kN
(iD Reaction between the gates AB and BC. The reaction (P) between the gates AB and BC is
given by equation 3.20) as
153065 N
CESTO
Bain" 2sin 30°
(di) Force on cach hinge. If Ry and Ry ar the reactions a te top and bottom hinges then
R= Ry=R
‘But from equation (3.19), R = P = 1359.195
= Ry = Ry= 1359.195
Ms
1359.195 kN. Ans.
“Te force Fasting at
2.67 m from bottom and Fy
33 m fom boto,
373
‘The resultant free £ will ata a distance from boton ls gica by
ES
E 1. ORO FXIS8_18122602.67~ 4530657133
F 139195
AS38731 — 602576
= SESS 5603
Hence Rs also acting a a distance 3.11 m from bottom.
“Taking moments off and aout he boom ings
Rx (60-10) = RG LO)
EXI) 138819522211
Re 50 so
= Ris R—R, = 1359.95 - 57358
RSS KN. Ans.
2573588
‘With of tock = 10m
5
With of exch lock person
Dept of water on upsueam side, = 8m
Dept of water on downstream sie, Hy = 4m
Water pressure an upstream side
y= pra
she Ave Hh 281738 246188 mF = Mio ann
2
= 1000 9.81 46.184 4.0 = 1812260 N = 1812.26 kN
Water pressure on downstream sie,
Fe pete
4
where Ay = ix = 5.773 4 = 28.092 ms = 20
Fy = 1000 x 9.81 x 23092 x 20
Resalta water pressure
2 Fi ~ Fan 1812.26 ~ 493068 = 1359.195 kN
(iD Reaction between the gates AB and BC. The reaction (P) between the gates AB and BC is
given by equation 3.20) as
153065 N
CESTO
Bain" 2sin 30°
(di) Force on cach hinge. If Ry and Ry ar the reactions a te top and bottom hinges then
R= Ry=R
‘But from equation (3.19), R = P = 1359.195
= Ry = Ry= 1359.195
Ms
1359.195 kN. Ans.
“Te force Fasting at
2.67 m from bottom and Fy
33 m fom boto,
373
‘The resultant free £ will ata a distance from boton ls gica by
ES
E 1. ORO FXIS8_18122602.67~ 4530657133
F 139195
AS38731 — 602576
= SESS 5603
Hence Rs also acting a a distance 3.11 m from bottom.
“Taking moments off and aout he boom ings
Rx (60-10) = RG LO)
EXI) 138819522211
Re 50 so
= Ris R—R, = 1359.95 - 57358
RSS KN. Ans.
2573588
112 Fluid Mechanics
> 3.8 PRESSURE DISTRIBUTION IN A LIQUID SUBJECTED TO CONSTANT
HORIZONTAL/VERTICAL ACCELERATION
In chapters 2 and 3, the containers which contains liquida, are assumed to beat rest Hence the
liquids re alo at rest They are in sate eqiibrivm wih respect 0 containers. But if the container
containing a god mode to move with constant acceleration, the Liquid partite niall will move
relative o eachother and after some time, there wil ot be any relative motion Between the goid
particles and boundaries ofthe container. Te liquid wil tke up new position under the etc of
acceleration imparted to its container. The liquid will ome ont in his new positon relative to the
ontaince. The entire Mak mass moves asa single unit Sine the qui afer Mining a new position
58 insti condition relative w he container, laws of hydro canbe appli o determine the
Tui pressure. As thor sn ative motion hc the liquid parcs, Renee the shear rasen and
shar forces between quid parcs wl esr, Te presse wil o normal tthe surface in contact
withthe guid
"The following are th important cases under consideracion
(Liquid containers subject constant horizontal accleraton
(6) Liquid containers subject to constant vericl acceleration.
3.8.1 Liquid Containers Subject to Constant Horizontal Acceleration. ig. 3.43 (u)
‘Shows tank containing a liquid upo à certain depth. The tank is ttlonary and e surface liquid
Fs horiomal Let his tank is moving with constan ale a nth rirontal dicton towards
Fight as shown in Fig, 343 (6), The ntl re surco of quid which was horizontal, now takes the
shape as shown in Fig. 343 (9). Now AB represents the new fre suce o the gui. Thus the fee
Surface of qu ue to horizontal acceleration will become a downward sloping inclined plane, with
te quid ii atthe back end. the Liquid falling atthe ron end. The equation forthe fee liquid
surface can be derived by considering the equilibrium of a ld element C Tying on the re surface
“Te tores acing on the element Car:
a ee
Mrs
="
E
o
Fig 3.43
(the pressure force P exerted by the surouoding fluid oa the element C. This force is normal to
the fee surco
(6) the weight ofthe Mui element Le. m x acting verally downward
D accclruing force de, m a seing in horizontal direcion.
> 3.8 PRESSURE DISTRIBUTION IN A LIQUID SUBJECTED TO CONSTANT
HORIZONTAL/VERTICAL ACCELERATION
In chapters 2 and 3, the containers which contains liquida, are assumed to beat rest Hence the
liquids re alo at rest They are in sate eqiibrivm wih respect 0 containers. But if the container
containing a god mode to move with constant acceleration, the Liquid partite niall will move
relative o eachother and after some time, there wil ot be any relative motion Between the goid
particles and boundaries ofthe container. Te liquid wil tke up new position under the etc of
acceleration imparted to its container. The liquid will ome ont in his new positon relative to the
ontaince. The entire Mak mass moves asa single unit Sine the qui afer Mining a new position
58 insti condition relative w he container, laws of hydro canbe appli o determine the
Tui pressure. As thor sn ative motion hc the liquid parcs, Renee the shear rasen and
shar forces between quid parcs wl esr, Te presse wil o normal tthe surface in contact
withthe guid
"The following are th important cases under consideracion
(Liquid containers subject constant horizontal accleraton
(6) Liquid containers subject to constant vericl acceleration.
3.8.1 Liquid Containers Subject to Constant Horizontal Acceleration. ig. 3.43 (u)
‘Shows tank containing a liquid upo à certain depth. The tank is ttlonary and e surface liquid
Fs horiomal Let his tank is moving with constan ale a nth rirontal dicton towards
Fight as shown in Fig, 343 (6), The ntl re surco of quid which was horizontal, now takes the
shape as shown in Fig. 343 (9). Now AB represents the new fre suce o the gui. Thus the fee
Surface of qu ue to horizontal acceleration will become a downward sloping inclined plane, with
te quid ii atthe back end. the Liquid falling atthe ron end. The equation forthe fee liquid
surface can be derived by considering the equilibrium of a ld element C Tying on the re surface
“Te tores acing on the element Car:
a ee
Mrs
="
E
o
Fig 3.43
(the pressure force P exerted by the surouoding fluid oa the element C. This force is normal to
the fee surco
(6) the weight ofthe Mui element Le. m x acting verally downward
D accclruing force de, m a seing in horizontal direcion.
Hydrostatic Forces on Surfaces 118]
Resolving the frees horizontally. we pet
Prints mx
“ Prin =~ ma 0
Resten the foes seen, we $
Peon Ome
a“ Peo Dem k 0)
Diving (by D. we ae
mot (m names) G20)
8 Le
‘The above equation. gives the slope ofthe fre surface ofthe liquid which is contained in à tank
‘whlch Is subjected horizontal constat acceleration, The tera (a/) constant and hence tan O will
constant. The ve sgn shows that he fee surface of liquid stoping downwards Hones he fee
Surface fea straight plane inclined down at an angle along he disco of acceleration
‘Now tus find the expresion forthe pressure at any point D in the quid mass subject to
horizontal acceleration Let the point D is at a depth of from the fre surface. Consider an
slemetary prism DE of height “A and crose-sctona arca dA as shown in Fig. 344.
Der
Consier the equilibrium ofthe elementary prism DE.
“The Frees acting on this prism DE in the vera direction ae
(0) the amaspheric pressure force (py 4A) atthe ap end of the prism acting downwards,
(Gi) ine weight ofthe element (px 4 2 x dA) at he CG. ofthe element acting In the downward
cc, and
(i the pressure force (pA athe boto end ofthe prise ating upwards
Since thet is mo vercal acceleration given to the tank, hence ne force ating verily should he
DXAA ~ pa AA = pgh dd =O
or P=p-pgh=0 oF pp Ros
or P= Po= ah
gauge pressure at point Dis given by
= ah
fo pressure bead at point D. 2
be
mo
Resolving the frees horizontally. we pet
Prints mx
“ Prin =~ ma 0
Resten the foes seen, we $
Peon Ome
a“ Peo Dem k 0)
Diving (by D. we ae
mot (m names) G20)
8 Le
‘The above equation. gives the slope ofthe fre surface ofthe liquid which is contained in à tank
‘whlch Is subjected horizontal constat acceleration, The tera (a/) constant and hence tan O will
constant. The ve sgn shows that he fee surface of liquid stoping downwards Hones he fee
Surface fea straight plane inclined down at an angle along he disco of acceleration
‘Now tus find the expresion forthe pressure at any point D in the quid mass subject to
horizontal acceleration Let the point D is at a depth of from the fre surface. Consider an
slemetary prism DE of height “A and crose-sctona arca dA as shown in Fig. 344.
Der
Consier the equilibrium ofthe elementary prism DE.
“The Frees acting on this prism DE in the vera direction ae
(0) the amaspheric pressure force (py 4A) atthe ap end of the prism acting downwards,
(Gi) ine weight ofthe element (px 4 2 x dA) at he CG. ofthe element acting In the downward
cc, and
(i the pressure force (pA athe boto end ofthe prise ating upwards
Since thet is mo vercal acceleration given to the tank, hence ne force ating verily should he
DXAA ~ pa AA = pgh dd =O
or P=p-pgh=0 oF pp Ros
or P= Po= ah
gauge pressure at point Dis given by
= ah
fo pressure bead at point D. 2
be
mo
114 Fluid Mechanics
From the above equation, is clear that pressure ead a any point in liquid subjected to à
constant horizontal acceleration is equal tothe height ofthe guid column above that point. Therefore
te pressure disuibuton in a liquid subjected 10 a constant horizontal acceleration is same a5
yarostatc pressure dstibution The planes of constant pressure are therefore, parle othe inclined
surface as shown io ip. 344. This igure also shows te vario of pressure of the fea and foot
end ofthe an.
If hy = Dept of quid at he rer end of the tank
Depa of igi tthe ron nd of the tank
Total pressure fore exerted by liquid on te rar sie of the tak
Total pressure force euere y quid on the front side ofthe tam
(Are of angle AML) x Wid
then
= IMAM xB) = pag x=
and P= (Area of lange BNO) x Wid
Pe
“whore b = Width of tank perpendicular to the plane of the paper.
‘The values of F and ca also be obtained as
{eter Fi 344 (a)
BN NO) $y x px
y= 9x9 A x, where Ay = hy Xb ad
2
hol a
prenne ted pg. bh?
PreK xx ET
where A
1b and By
am ESP XERAL
h
o
Fig. 3.448)
cel, hen same ofthe id wl si Oa ar 1S ni and New re tse whe pe gr BY
= will be developed.
een rere rere neue
of he go Bathe equation tn 8
ego a
Eon casa
Go) Tee example fr nk wi Ik subjected à cosas anal cleus, ul ak owas
plane during tae on
From the above equation, is clear that pressure ead a any point in liquid subjected to à
constant horizontal acceleration is equal tothe height ofthe guid column above that point. Therefore
te pressure disuibuton in a liquid subjected 10 a constant horizontal acceleration is same a5
yarostatc pressure dstibution The planes of constant pressure are therefore, parle othe inclined
surface as shown io ip. 344. This igure also shows te vario of pressure of the fea and foot
end ofthe an.
If hy = Dept of quid at he rer end of the tank
Depa of igi tthe ron nd of the tank
Total pressure fore exerted by liquid on te rar sie of the tak
Total pressure force euere y quid on the front side ofthe tam
(Are of angle AML) x Wid
then
= IMAM xB) = pag x=
and P= (Area of lange BNO) x Wid
Pe
“whore b = Width of tank perpendicular to the plane of the paper.
‘The values of F and ca also be obtained as
{eter Fi 344 (a)
BN NO) $y x px
y= 9x9 A x, where Ay = hy Xb ad
2
hol a
prenne ted pg. bh?
PreK xx ET
where A
1b and By
am ESP XERAL
h
o
Fig. 3.448)
cel, hen same ofthe id wl si Oa ar 1S ni and New re tse whe pe gr BY
= will be developed.
een rere rere neue
of he go Bathe equation tn 8
ego a
Eon casa
Go) Tee example fr nk wi Ik subjected à cosas anal cleus, ul ak owas
plane during tae on
Hydrostatic Forces on Surfaces 118]
Problem 8.84 A rectangular tank is moving horizonaly in the direction of ts eng with a
constant acceleration of 24 m/e. The teng width and depth ofthe tank are 6 m. 25 m and 2 m
respectively. Ifthe del of water in the tank is 4 m and tank is pen ot the op then caleulte
(D ie angle ofthe water surface to the horizontal,
(i) the maximun and minimum pressure intensities at the bottom,
(di) the oat force due to water ating on each end ofthe tank
Solution. Given
Constan accckrtion, a= 24 mé
Length = 6m ; Width = 25 m and depth = 2m. 20
Dept of water in nk, = 1m
(The angle ofthe water surface to Ihe
horizontal
Let @ =the angle of water surface to the horizontal
Using equation 3.20), we get
2.24 Fig. 345
ajar ~~ OHS
(ine -ve sign shows thatthe fre surface of waters sloping downward as showa in Fig. 345)
tan 8 = 02446 (slope downward)
= 9 tan! 02116 = 13.7446" oF 13° 44.6 Ans.
D The masimum and minimum pressure intensities at the bottom of the tank
From the Fig. 34,
Dept of water atthe front end,
hye 13am 0=
Depth of water tthe car end,
‘hy 4 Stam Ba 1 4.302446 = 1.2338 m
The pressure intensity wil be maximum a he tom, where dept of water is maxim
Now the maximum pressure intensity athe bottom wil beat pot À and it is given 3,
Pus PARA
1000 9.81 x 1.2338 Nin? = 170085 Nm”. Ans.
"The minimum pressure intensity atthe Boom wil he a point B and itis given by
Pasa = BX 8% hy
1000 x 9.81 0.2662 = 26114 Nim. Ans.
(di) The total force due to water acting on each end of the tank
La Fi = total foros actag onthe front side (Le, on face BD)
tanos
2 3 0.2446 = 02662 m
= tual force acting on the war side (Le, on face AC)
Then Fe pti, where A, = BD x width of tank = hy x28 = 02662 425
Ge BD _ 02662
mi Fe Dh 13310
1000 x 981 (02662 x 25) x 013
868.95 N. Ans.
Problem 8.84 A rectangular tank is moving horizonaly in the direction of ts eng with a
constant acceleration of 24 m/e. The teng width and depth ofthe tank are 6 m. 25 m and 2 m
respectively. Ifthe del of water in the tank is 4 m and tank is pen ot the op then caleulte
(D ie angle ofthe water surface to the horizontal,
(i) the maximun and minimum pressure intensities at the bottom,
(di) the oat force due to water ating on each end ofthe tank
Solution. Given
Constan accckrtion, a= 24 mé
Length = 6m ; Width = 25 m and depth = 2m. 20
Dept of water in nk, = 1m
(The angle ofthe water surface to Ihe
horizontal
Let @ =the angle of water surface to the horizontal
Using equation 3.20), we get
2.24 Fig. 345
ajar ~~ OHS
(ine -ve sign shows thatthe fre surface of waters sloping downward as showa in Fig. 345)
tan 8 = 02446 (slope downward)
= 9 tan! 02116 = 13.7446" oF 13° 44.6 Ans.
D The masimum and minimum pressure intensities at the bottom of the tank
From the Fig. 34,
Dept of water atthe front end,
hye 13am 0=
Depth of water tthe car end,
‘hy 4 Stam Ba 1 4.302446 = 1.2338 m
The pressure intensity wil be maximum a he tom, where dept of water is maxim
Now the maximum pressure intensity athe bottom wil beat pot À and it is given 3,
Pus PARA
1000 9.81 x 1.2338 Nin? = 170085 Nm”. Ans.
"The minimum pressure intensity atthe Boom wil he a point B and itis given by
Pasa = BX 8% hy
1000 x 9.81 0.2662 = 26114 Nim. Ans.
(di) The total force due to water acting on each end of the tank
La Fi = total foros actag onthe front side (Le, on face BD)
tanos
2 3 0.2446 = 02662 m
= tual force acting on the war side (Le, on face AC)
Then Fe pti, where A, = BD x width of tank = hy x28 = 02662 425
Ge BD _ 02662
mi Fe Dh 13310
1000 x 981 (02662 x 25) x 013
868.95 N. Ans.
116 Fluid Mechanics
Pasha, whee Ay = AB widh of tank = hy x25 = 17338 x25
Am 18
E
1000 x 9.81 (1.7338 x 2.8) x 0.8660
36861: N. Ans.
2 Resultant force =F ~ Fy
6861.8 N — 86595
23500288 N
of te nk equ to te fore nen o accio Ie
Tid mas TS eu be proved as sven ben
same fe ACDRA a shown in Fig, 3.46, The rt foro ating on
‘he cool volume e ral detente ese
‘he proue of mass he ld coo volume und cc A a
ion ofthe ig A Á 0 —
FM Fig. 346
(Bol of onto volume) x a
= x Ars of ABDCE win) «24
[oc ans nu
0.669 m
¡PRE
wo
sora
EN
Eo AC= hy 1.7338, BD =A, = 02662 and A= 6, vd 2,5)
‘Tus above fre sweaty te soe the rene of he ares ating onthe two eof e ak, (Le
an = 9000),
Problom 3.35. The rectangular tank ofthe above problem contain water 1 a depth of 1.5 m. Find
e horkontal acceleration which may be imparted 1 he tak in the direction of length so that
(6) Ihe spilling of wate from the tank is us on the verge of taking place,
(i the front bottom corner of the tank sust exposed,
Gi) the bottom ofthe tank is exposed upto ts mid pain
‘Also calculate the total forces exerted hy th water on each end of the tank in each case. Also prove
‘that the ference Benen these force is equal othe force necessary to cceleate the mass of water
unk
Solution. Gien
Dimensions of the tank from previous problem,
= 6.m, wid (0) = 2.5m and depth = 2 m
Pasha, whee Ay = AB widh of tank = hy x25 = 17338 x25
Am 18
E
1000 x 9.81 (1.7338 x 2.8) x 0.8660
36861: N. Ans.
2 Resultant force =F ~ Fy
6861.8 N — 86595
23500288 N
of te nk equ to te fore nen o accio Ie
Tid mas TS eu be proved as sven ben
same fe ACDRA a shown in Fig, 3.46, The rt foro ating on
‘he cool volume e ral detente ese
‘he proue of mass he ld coo volume und cc A a
ion ofthe ig A Á 0 —
FM Fig. 346
(Bol of onto volume) x a
= x Ars of ABDCE win) «24
[oc ans nu
0.669 m
¡PRE
wo
sora
EN
Eo AC= hy 1.7338, BD =A, = 02662 and A= 6, vd 2,5)
‘Tus above fre sweaty te soe the rene of he ares ating onthe two eof e ak, (Le
an = 9000),
Problom 3.35. The rectangular tank ofthe above problem contain water 1 a depth of 1.5 m. Find
e horkontal acceleration which may be imparted 1 he tak in the direction of length so that
(6) Ihe spilling of wate from the tank is us on the verge of taking place,
(i the front bottom corner of the tank sust exposed,
Gi) the bottom ofthe tank is exposed upto ts mid pain
‘Also calculate the total forces exerted hy th water on each end of the tank in each case. Also prove
‘that the ference Benen these force is equal othe force necessary to cceleate the mass of water
unk
Solution. Gien
Dimensions of the tank from previous problem,
= 6.m, wid (0) = 2.5m and depth = 2 m
Hydrostatic Forces on Surfaces 117]
Dept of water in nk,
Horizontal acceleration imparted to the tank.
© @) When the siling of woter fom the tank is
just onthe verge of taking place
Let a= required horizontal acceleration
sm
ox
Wien the spiling of water ftom the tank is just on e |
verge of taking place, the water would ie upto the rear
top comer ofthe tank a shown in Fi 347 (a)
ac 0-15) os an —
ae 95 20.1067
407 TO pig 347 (a) Spilling of water is jst on
the serge of taking place.
But fom equation (320) an 0 = À (Numerical)
= tan 0 981 x0.1667 = 1.638 mA, Ans.
(9) Tota forces exerted by water on each eno th tak
The force exerted by water on the end CE of the tank is
E = pra, where A = CE x width ofthe tak = 2% 2.5
3e 2
n-E-3-1m
= 1000 «9.81 x 2 42.) «1
49050 N. Ans.
“Tae force exerted by water on the end FD ofthe tank is
Fa mp4 xT where,
CsAC=BD=05m, 2 FD
NA
= 122625 N. Ans
(6) Difference ofthe forces is equal a the force necessary 1 accelerate he mass of water in the tank
Difference ofthe forecs = E
(9050 - 12262.5= 6787.5 N
Volume of water inthe tank before acceleration is imparted 1
228X15= 25m.
Te foros necessary to acer the mass of water in de tank
dep of water
= Mass of wate in tank X Acceleration
= (px volume of water) x 1.635 Ce a= 1.635 mis)
1000 22:5 1.635 [There is no spiling of water and volume of
2.5 m1
3678758
Dept of water in nk,
Horizontal acceleration imparted to the tank.
© @) When the siling of woter fom the tank is
just onthe verge of taking place
Let a= required horizontal acceleration
sm
ox
Wien the spiling of water ftom the tank is just on e |
verge of taking place, the water would ie upto the rear
top comer ofthe tank a shown in Fi 347 (a)
ac 0-15) os an —
ae 95 20.1067
407 TO pig 347 (a) Spilling of water is jst on
the serge of taking place.
But fom equation (320) an 0 = À (Numerical)
= tan 0 981 x0.1667 = 1.638 mA, Ans.
(9) Tota forces exerted by water on each eno th tak
The force exerted by water on the end CE of the tank is
E = pra, where A = CE x width ofthe tak = 2% 2.5
3e 2
n-E-3-1m
= 1000 «9.81 x 2 42.) «1
49050 N. Ans.
“Tae force exerted by water on the end FD ofthe tank is
Fa mp4 xT where,
CsAC=BD=05m, 2 FD
NA
= 122625 N. Ans
(6) Difference ofthe forces is equal a the force necessary 1 accelerate he mass of water in the tank
Difference ofthe forecs = E
(9050 - 12262.5= 6787.5 N
Volume of water inthe tank before acceleration is imparted 1
228X15= 25m.
Te foros necessary to acer the mass of water in de tank
dep of water
= Mass of wate in tank X Acceleration
= (px volume of water) x 1.635 Ce a= 1.635 mis)
1000 22:5 1.635 [There is no spiling of water and volume of
2.5 m1
3678758
118 Fluid Mechanics
Henge the difference between the forces on the two ends ofthe tank is equal 10 the force Necessary
tw accelerate the mass of water he tak,
Vote rie uaa abba vom = (SEF) wie ert
Fig. 347 (al un
sn rc
(UD (a) Horizontal acceiration when the front
otto corner ofthe tank i just exposed
Refer ta Tig. 37 (9). ln this cate the fre sur
‘ac of water nthe tank wil be along CD.
Let a= required horizontal acceleration, 5
Mise, e
ES
‘But rom equation (3.17), E
Fig. 347 4)
cute Orga
A
america)
An 02981 x 327 ml, Ans,
(©) Tort forces exerted by water om each end af the tank
‘The force exerted by water on the end CE of the tank is
Espa kA có
49050 N. Ans.
"The force exerts hy war on the end BD of hs tank s ro as ther iso water aginst ace BD
Difference of forces = 49080 D 49050 N
(6) Digerence of forces is equal nthe force necessary to accelerate he mass of water in the tank
Volume of water in the tank = Area of CED x Wi of tank.
ee
7
Jas (With ok 225
TE
:
Bo man ee i lv a rk
on ve nu ar
= 1000 LS «3.27 = 49050 N
Ditterence of two forces is also = 49050 N
Henge the difference between the forces on the two ends ofthe tank is equal 10 the force Necessary
tw accelerate the mass of water he tak,
Vote rie uaa abba vom = (SEF) wie ert
Fig. 347 (al un
sn rc
(UD (a) Horizontal acceiration when the front
otto corner ofthe tank i just exposed
Refer ta Tig. 37 (9). ln this cate the fre sur
‘ac of water nthe tank wil be along CD.
Let a= required horizontal acceleration, 5
Mise, e
ES
‘But rom equation (3.17), E
Fig. 347 4)
cute Orga
A
america)
An 02981 x 327 ml, Ans,
(©) Tort forces exerted by water om each end af the tank
‘The force exerted by water on the end CE of the tank is
Espa kA có
49050 N. Ans.
"The force exerts hy war on the end BD of hs tank s ro as ther iso water aginst ace BD
Difference of forces = 49080 D 49050 N
(6) Digerence of forces is equal nthe force necessary to accelerate he mass of water in the tank
Volume of water in the tank = Area of CED x Wi of tank.
ee
7
Jas (With ok 225
TE
:
Bo man ee i lv a rk
on ve nu ar
= 1000 LS «3.27 = 49050 N
Ditterence of two forces is also = 49050 N
Hydrostatic Forces on Surfaces 119]
‘Henge difference between the forces on the two ends ee oe
ofthe tank is equal to the force necessary 10 accelerate
‘he mas of wate inthe tan,
(dl) (a) Horizont acceleration when the boom of
the tank exposed upto ts midpoint
Refer to Fig. 347 () In this ease the fee surface of 2
‘wate im he tank will be along CD, where DY the
mid point of ED.
Let a = require horizontal aceceraton from
Fig, 347 (its lear that
|
ur
mo 2 — BE
CE no
‘But from equation (3.20) numerically
uno:
n
a amg
981 x = 654 mé. Am.
(0) Tata forces exerted by water on each end of te tank
“The fore exerted hy water on the end CE of the tank is
Esp xg xa xa
whore Ay = CE With =2 4252 5m
PR
ye Eheim
1000 «9.81 x5 x 1
49050 N. Ans.
“Tne foros exerted y watson the end BD iso a there is no war against he face BD.
à F,20
[Difference ofthe Forces = Fy — Fy = 49050 - 0 49050 N
(6) Diference of the two forces ts equal 1o the force necessary to accelerate the mass of water
‘remaining in the tank
‘Volume of water the
ok = Arca CED® x Width of nk
CEXED? „2x3 ;
= Et vase D «252 75m
Force necessary to accelerate he mass af water he tan
Mass of water x Aceckraon
= p x Volume af water 6.54 €
= 1000 47.8 x 6.84
49050 N
is the same ore a the diflrence of te to forges on Ihe two ends he tank.
Problem 3.96 A rectangular tank fleng 6 m, with 25m and high 2m is completly lle with
water when a res. The tank is open a he to. The tanh i subjected 1 à horizontal constant near
acceleration of 24 mds in e direction of ts length Find the volume of water spilled fom he tank
‘Henge difference between the forces on the two ends ee oe
ofthe tank is equal to the force necessary 10 accelerate
‘he mas of wate inthe tan,
(dl) (a) Horizont acceleration when the boom of
the tank exposed upto ts midpoint
Refer to Fig. 347 () In this ease the fee surface of 2
‘wate im he tank will be along CD, where DY the
mid point of ED.
Let a = require horizontal aceceraton from
Fig, 347 (its lear that
|
ur
mo 2 — BE
CE no
‘But from equation (3.20) numerically
uno:
n
a amg
981 x = 654 mé. Am.
(0) Tata forces exerted by water on each end of te tank
“The fore exerted hy water on the end CE of the tank is
Esp xg xa xa
whore Ay = CE With =2 4252 5m
PR
ye Eheim
1000 «9.81 x5 x 1
49050 N. Ans.
“Tne foros exerted y watson the end BD iso a there is no war against he face BD.
à F,20
[Difference ofthe Forces = Fy — Fy = 49050 - 0 49050 N
(6) Diference of the two forces ts equal 1o the force necessary to accelerate the mass of water
‘remaining in the tank
‘Volume of water the
ok = Arca CED® x Width of nk
CEXED? „2x3 ;
= Et vase D «252 75m
Force necessary to accelerate he mass af water he tan
Mass of water x Aceckraon
= p x Volume af water 6.54 €
= 1000 47.8 x 6.84
49050 N
is the same ore a the diflrence of te to forges on Ihe two ends he tank.
Problem 3.96 A rectangular tank fleng 6 m, with 25m and high 2m is completly lle with
water when a res. The tank is open a he to. The tanh i subjected 1 à horizontal constant near
acceleration of 24 mds in e direction of ts length Find the volume of water spilled fom he tank
120 Fluid Mechanics
Solution, Given
sb = 2.5 mand height, = 2m
Hocizonialageleruion. a= 24 mA
‘The slope ofthe fre surface of water afer the tank is subjected to linear constant acceleration is
given by equation 3.20) as douce
(Sumercaly)
el
From Fig. 3.48,
Volume of water spilled = Area of ABC x Wid of tank
XABX BC) «25 id = 2.5 m)
$6 «14676 425 LE BC= 14676 m)
1.007 mi. Ans.
3.8.2 Liquid Container Subjected to Constant Vertical Acceleration. Fig. 349 shows
tank containing aliquid andthe tank is moving vertically upward witha constant aceertion, The
Liquid nthe tank wil he subjected to the same vertical acceleration, To obtain the expression for the
pressure at any point nth liquid mass subjecid to vertical upward aceciraion, consider a vertical
‘lena pra of ligukd CDFE.
mr men
| cle ;
T I
vey
BZ:
Kamen
Dre
Der
Let dA= Cross-sectional area of pis
= Height ot prism
o = Atmospheric pressure acting on the face CE
[p= Pressure ata dep h acting on the ace DE
Solution, Given
sb = 2.5 mand height, = 2m
Hocizonialageleruion. a= 24 mA
‘The slope ofthe fre surface of water afer the tank is subjected to linear constant acceleration is
given by equation 3.20) as douce
(Sumercaly)
el
From Fig. 3.48,
Volume of water spilled = Area of ABC x Wid of tank
XABX BC) «25 id = 2.5 m)
$6 «14676 425 LE BC= 14676 m)
1.007 mi. Ans.
3.8.2 Liquid Container Subjected to Constant Vertical Acceleration. Fig. 349 shows
tank containing aliquid andthe tank is moving vertically upward witha constant aceertion, The
Liquid nthe tank wil he subjected to the same vertical acceleration, To obtain the expression for the
pressure at any point nth liquid mass subjecid to vertical upward aceciraion, consider a vertical
‘lena pra of ligukd CDFE.
mr men
| cle ;
T I
vey
BZ:
Kamen
Dre
Der
Let dA= Cross-sectional area of pis
= Height ot prism
o = Atmospheric pressure acting on the face CE
[p= Pressure ata dep h acting on the ace DE
Hydrostatic Fores on Surfaces 121
“The forces acting onthe elementary prism are =
(0) Pressure force equal o pax dA acting on the face CE vertical downward
(a) Pressure force equal 10 pA acting a the face DF vereally upward
(Ud Weight ofthe pis equal to p x 4 X dA x h acting though CG ofthe element vertically
downward.
Acconling o Newton's second aw of motion. the net
mas multiplied by acceleration inthe same direction.
‘Net fore in vertical upward iretion = Mas x acceleration
ce ating on the element mus be equal to
ro (2 Mase pxach
a oia ¿Caca at mb)
id peer
sales] om
+
But py) is he gauge pressure: Hence gauge pressure at any point in the Higukd mass subjected o
constant eral upward acceleration sen by
A (32%
= pan + pha 6224)
whore py =P ~Po = save pressure
In equation (322) 9. and ae constant. Hence var of puago pressure is linear Alo when
h = 0.p, = 0. This means - pp = Var p= Ps Hence when h =O, the pressure I qual to amaspheric
Pressure Hence re surface of lguldsobscted 1 constat vertical acceleration will be horizontal
From equation (3.224) is also clear tat the reste a any pont inthe liquid mas greater Man
‘he hydrostatic pressure (hydrosaic pressure = pg) hy an amount of X a
ig. 3:49 shows the variation of pressure fo he liquid mass subjected ta constant vertical upward
sccoeration,
ft tank containing Liquid is moving vertically downward with a constant acceleration, the the
auge pressure at any pol inthe guid at a depth of A rom the fee surface will be piven by
cromos [1-2] ES
ère cue is ie pas y
point alma testo he ma pes
ba amsn of pla 930 ust vien vs
€ presas forthe ul mass stbjected 0 a constant we
Cereal ward acc
Tre wacom Bq movia ova wi
+ alien egw gay Mae = puss
monito a or p= Tn ese
ee dt
noes pre, Te wil ben fe on te Pe
Selena te beet nk,
hi ant em sie jc a man stn ine cl, e
TS be ha eon ne in The a of ee eet
e per tnd stress pe
“The forces acting onthe elementary prism are =
(0) Pressure force equal o pax dA acting on the face CE vertical downward
(a) Pressure force equal 10 pA acting a the face DF vereally upward
(Ud Weight ofthe pis equal to p x 4 X dA x h acting though CG ofthe element vertically
downward.
Acconling o Newton's second aw of motion. the net
mas multiplied by acceleration inthe same direction.
‘Net fore in vertical upward iretion = Mas x acceleration
ce ating on the element mus be equal to
ro (2 Mase pxach
a oia ¿Caca at mb)
id peer
sales] om
+
But py) is he gauge pressure: Hence gauge pressure at any point in the Higukd mass subjected o
constant eral upward acceleration sen by
A (32%
= pan + pha 6224)
whore py =P ~Po = save pressure
In equation (322) 9. and ae constant. Hence var of puago pressure is linear Alo when
h = 0.p, = 0. This means - pp = Var p= Ps Hence when h =O, the pressure I qual to amaspheric
Pressure Hence re surface of lguldsobscted 1 constat vertical acceleration will be horizontal
From equation (3.224) is also clear tat the reste a any pont inthe liquid mas greater Man
‘he hydrostatic pressure (hydrosaic pressure = pg) hy an amount of X a
ig. 3:49 shows the variation of pressure fo he liquid mass subjected ta constant vertical upward
sccoeration,
ft tank containing Liquid is moving vertically downward with a constant acceleration, the the
auge pressure at any pol inthe guid at a depth of A rom the fee surface will be piven by
cromos [1-2] ES
ère cue is ie pas y
point alma testo he ma pes
ba amsn of pla 930 ust vien vs
€ presas forthe ul mass stbjected 0 a constant we
Cereal ward acc
Tre wacom Bq movia ova wi
+ alien egw gay Mae = puss
monito a or p= Tn ese
ee dt
noes pre, Te wil ben fe on te Pe
Selena te beet nk,
hi ant em sie jc a man stn ine cl, e
TS be ha eon ne in The a of ee eet
e per tnd stress pe
122 Floid Mechanics
Problem 3.87 A tank contining water upto à depth of 500 mm is moving vertically upward with
a constant acceleration of 245 mie. Find the force exerted by water on the side ofthe tank. Also
calelate the force on the side ofthe tank when the width of tank is 2 m and
1) tank is moving vericaly downward witha constant acceleration nf 245 mi and
(i) the tank no moving tal,
Solution. Given
Depth of ware, 00 mm = 05 m oan
Venica cestas, 45 mi?
Width of nk, m
To find the free exer by water on the sido
Fine tank when moving vercally upward, Lt us
fist And the pressure a he boto of the tank. CB A
"The gauge pressure au he bottom (i, point) PIPA
for this cas I ven by equations Hot à
m
al )
n
245
res (1248 as
“This pressure i represented hy In BC
[Now the fore onthe sde AB = Area of tangle ABC = Width of tank
Ge ABx-BC) x»
GxOSXG1SL2S) x2
30656 Na Ans.
(0 Force on the side of the tank, when tank Is moving vericlly downward.
‘The pressure variation is shown in Fig. 3.82. Fortis ease, the pressure a the bottom ofthe tak
(eat pont B) is iven by equation (323) as
moi) JE
ë 7
245
ons 051-28)
367825 Nin? a
“This pressure present by Ine BC. s
Now th force on th side AB = Arcaof mingle ABCXWidth fe gg AY
= (J ABC) x pan
= (x0s3367875) x2 (= BC= 367875,6=2)
183937 N. Ans.
Problem 3.87 A tank contining water upto à depth of 500 mm is moving vertically upward with
a constant acceleration of 245 mie. Find the force exerted by water on the side ofthe tank. Also
calelate the force on the side ofthe tank when the width of tank is 2 m and
1) tank is moving vericaly downward witha constant acceleration nf 245 mi and
(i) the tank no moving tal,
Solution. Given
Depth of ware, 00 mm = 05 m oan
Venica cestas, 45 mi?
Width of nk, m
To find the free exer by water on the sido
Fine tank when moving vercally upward, Lt us
fist And the pressure a he boto of the tank. CB A
"The gauge pressure au he bottom (i, point) PIPA
for this cas I ven by equations Hot à
m
al )
n
245
res (1248 as
“This pressure i represented hy In BC
[Now the fore onthe sde AB = Area of tangle ABC = Width of tank
Ge ABx-BC) x»
GxOSXG1SL2S) x2
30656 Na Ans.
(0 Force on the side of the tank, when tank Is moving vericlly downward.
‘The pressure variation is shown in Fig. 3.82. Fortis ease, the pressure a the bottom ofthe tak
(eat pont B) is iven by equation (323) as
moi) JE
ë 7
245
ons 051-28)
367825 Nin? a
“This pressure present by Ine BC. s
Now th force on th side AB = Arcaof mingle ABCXWidth fe gg AY
= (J ABC) x pan
= (x0s3367875) x2 (= BC= 367875,6=2)
183937 N. Ans.
Hydrostatic Forces on Surfaces 123]
(69 Force on the side of the tank, when
The pressure a point Bis given by. -
Pa Oh = 1000 x 9.81 x 0.5 = 4905 Nin’
“Tis pressure i represented hy ine BD in Fig. 3.52
Force onthe sde AB = Area of tingle ABD Width
= (dx anx an) x»
05 4908) «2
nk Is stationary.
24525 No Ans.
For this case, the foros om AB can aso be ohtincd as
Fan PA
where A= AB x Width = 05 42 = 1m?
Ge ABS 025 m= 1009.81 x 1025
= 24525 N. Ams.
Problem 3.38 A onk contains water upo a depth of 1S m. The length and width ofthe tank are
4m and 2 m respectively. The tank is moving up an incline plane wih a constant acceleration of
4 mé. The inchnaton ofthe plane withthe horizontal is 30° as shown in Fig. 3.53. Find,
(i) the angle made by he fee surface of water with the horizontal.
(iy ie pressure at the bottom ofthe tank a the front and rear ends
Solution. Gives
Deptt wat, Sm la brand F
Wa b= 2 ie
Constant acceleration along th nena plan, ite
A i
mis
Inclination of plane, = 30°
Let @= Angle made hy the fee surface of water
te the acccleraton is imparted 1 th nk
a = Pressure atthe bottom of he tank at he frontend
Sd pp = Pressure at he bottom ofthe tank athe rae
ond. PEC]
‘Tis problem can he dons by resolving the given acceleration along the horizomal direction and
vertical discon, Thon cach ofthese cass may be separately anayssl according othe st procdure
Horizontal and vertical components f the sccleraton are
m A 08 30° = 364 m
4, = ain & 2 4 sin 30° 2 2 me
‘When te tan is staonary onthe incline plane, fee surface of liquid wil be along EF as shown
An Fig 3.53. Bu when the tank Is moving upwar along the inlined plane the fee surface of guid
Al be along BC. When the tank containing ago is moving up an inclined plane with a constant
scceletation, the angle made bythe fee surface ofthe Liquid withthe horizontal is piven by
(69 Force on the side of the tank, when
The pressure a point Bis given by. -
Pa Oh = 1000 x 9.81 x 0.5 = 4905 Nin’
“Tis pressure i represented hy ine BD in Fig. 3.52
Force onthe sde AB = Area of tingle ABD Width
= (dx anx an) x»
05 4908) «2
nk Is stationary.
24525 No Ans.
For this case, the foros om AB can aso be ohtincd as
Fan PA
where A= AB x Width = 05 42 = 1m?
Ge ABS 025 m= 1009.81 x 1025
= 24525 N. Ams.
Problem 3.38 A onk contains water upo a depth of 1S m. The length and width ofthe tank are
4m and 2 m respectively. The tank is moving up an incline plane wih a constant acceleration of
4 mé. The inchnaton ofthe plane withthe horizontal is 30° as shown in Fig. 3.53. Find,
(i) the angle made by he fee surface of water with the horizontal.
(iy ie pressure at the bottom ofthe tank a the front and rear ends
Solution. Gives
Deptt wat, Sm la brand F
Wa b= 2 ie
Constant acceleration along th nena plan, ite
A i
mis
Inclination of plane, = 30°
Let @= Angle made hy the fee surface of water
te the acccleraton is imparted 1 th nk
a = Pressure atthe bottom of he tank at he frontend
Sd pp = Pressure at he bottom ofthe tank athe rae
ond. PEC]
‘Tis problem can he dons by resolving the given acceleration along the horizomal direction and
vertical discon, Thon cach ofthese cass may be separately anayssl according othe st procdure
Horizontal and vertical components f the sccleraton are
m A 08 30° = 364 m
4, = ain & 2 4 sin 30° 2 2 me
‘When te tan is staonary onthe incline plane, fee surface of liquid wil be along EF as shown
An Fig 3.53. Bu when the tank Is moving upwar along the inlined plane the fee surface of guid
Al be along BC. When the tank containing ago is moving up an inclined plane with a constant
scceletation, the angle made bythe fee surface ofthe Liquid withthe horizontal is piven by
ata 24981
= tan! 02933 = 16.846" oF 16° 208", Ams.
Now let fit find the dep of lig a the front and ear nd of te nk
‘Depth liquid at frontend = hy = AB
‘Dept of liquid a rar end = hy = CD
= 0258
ce
oF. ne = CE
CE= £0 an 8 = 202933, (£0 = 2m, tan 0 = 02933)
= 0.5866 m
2 CD= y= ED + CE = 1,5 + 05866 = 20866 m
Sim y= AB = AF BF
= 150.5866
= 09134 m
‘The pressure atthe httom of tank atthe ear ed given by,
mm
‘From Fig. 3.53, in a
A= 15, BP = CE = 05866)
2
on ona (10,2) ar mt am.
sx
‘The re at Rom of nk a Gon end en y
mes)
5
210x081 co (1453, = 107872 Nat Am
= tan! 02933 = 16.846" oF 16° 208", Ams.
Now let fit find the dep of lig a the front and ear nd of te nk
‘Depth liquid at frontend = hy = AB
‘Dept of liquid a rar end = hy = CD
= 0258
ce
oF. ne = CE
CE= £0 an 8 = 202933, (£0 = 2m, tan 0 = 02933)
= 0.5866 m
2 CD= y= ED + CE = 1,5 + 05866 = 20866 m
Sim y= AB = AF BF
= 150.5866
= 09134 m
‘The pressure atthe httom of tank atthe ear ed given by,
mm
‘From Fig. 3.53, in a
A= 15, BP = CE = 05866)
2
on ona (10,2) ar mt am.
sx
‘The re at Rom of nk a Gon end en y
mes)
5
210x081 co (1453, = 107872 Nat Am
Hydrostatic Forces on Surfaces 125]
8. The conte of press for a plano veal surface esa à depth of wohin the Regt of the
immer ses
6. The toa ore on a curved rice i given by Pa EPS
Wher f= Horizontal os on cured surface td is qual 0 Il pesao freon the projet rca
‘ofthe cursed surface on the vet plans,
= pet
and E Veta fre on submerged curved surface and qua othe weight of gud actully
ot imaginary support bythe curved uae
2.
E
“The inclinado ofthe esi ron cure src vi aot a 0 = 2
The restant force on ice gto, E F,— Fy
where = retire fre onthe upsream sid of the sic gate and
Fa Pressure fore othe downs sid ofthe ie at.
For ac athe ration betwee te two gute I qual ote ton tthe hing,
Pind
whee = Rosal wae promu on he lock gato = 3
and 0 Totton of the ate with te norma othe eof he lock,
to the est between he wo gas,
(A) THEORETICAL PROBLEMS
1. Wu do you undersund by "Total Preso and Cente of Press”?
2 Derive an expen for he force exer on à mb mer venal plano fae y the stale ils
and locate the poston of sente of pressure
Prove it he cee of pesar of à comp sub meros plane sac i lays blow the cease
‘of gly of the aubeteged surface or ton clade with the on of gravy when tbe plo
Suri i horizon,
Prove tas the itl pressure exe hy a sui liquid on an icin plane sb-merges sac se
‘Sime ae he fre exes on a ra plane utes lamp a the dep of e ete fav oF Te
5. Dee an expres or he dep of ene of pressure fom fe sae of au ofan nine pane
surface su megas in we Hous
6) Bam void yo te hors and veia copos fh ein pres cn a
med eared suce?
(0) Bla proceso finding hydrostatic forces on cures faces
(Delhi Universi, Dee 2002)
2. Explain how you would nd the rest peer om a saved race immersing
‘Wy te resin pressure ona carved sumerge face Is dsemines by fist fining horizontal
and vere! forces om the syed suce Way ete same med ot adopt or à line end
suce sb-merged in à quid ?
8. The conte of press for a plano veal surface esa à depth of wohin the Regt of the
immer ses
6. The toa ore on a curved rice i given by Pa EPS
Wher f= Horizontal os on cured surface td is qual 0 Il pesao freon the projet rca
‘ofthe cursed surface on the vet plans,
= pet
and E Veta fre on submerged curved surface and qua othe weight of gud actully
ot imaginary support bythe curved uae
2.
E
“The inclinado ofthe esi ron cure src vi aot a 0 = 2
The restant force on ice gto, E F,— Fy
where = retire fre onthe upsream sid of the sic gate and
Fa Pressure fore othe downs sid ofthe ie at.
For ac athe ration betwee te two gute I qual ote ton tthe hing,
Pind
whee = Rosal wae promu on he lock gato = 3
and 0 Totton of the ate with te norma othe eof he lock,
to the est between he wo gas,
(A) THEORETICAL PROBLEMS
1. Wu do you undersund by "Total Preso and Cente of Press”?
2 Derive an expen for he force exer on à mb mer venal plano fae y the stale ils
and locate the poston of sente of pressure
Prove it he cee of pesar of à comp sub meros plane sac i lays blow the cease
‘of gly of the aubeteged surface or ton clade with the on of gravy when tbe plo
Suri i horizon,
Prove tas the itl pressure exe hy a sui liquid on an icin plane sb-merges sac se
‘Sime ae he fre exes on a ra plane utes lamp a the dep of e ete fav oF Te
5. Dee an expres or he dep of ene of pressure fom fe sae of au ofan nine pane
surface su megas in we Hous
6) Bam void yo te hors and veia copos fh ein pres cn a
med eared suce?
(0) Bla proceso finding hydrostatic forces on cures faces
(Delhi Universi, Dee 2002)
2. Explain how you would nd the rest peer om a saved race immersing
‘Wy te resin pressure ona carved sumerge face Is dsemines by fist fining horizontal
and vere! forces om the syed suce Way ete same med ot adopt or à line end
suce sb-merged in à quid ?
126 Fi Mechanics
8, Describe bly with sehe the various methods usd for menting press excited by id
10. Prove ht the vera componca of the tesulan pressure on a submerged cui ste seal o
the meg o held supported y the cured sac.
I. What sth ferne between sic ue and lock gue?
122 Prove that he ection ern he pts o Jocs equal he ecto at he hinge
E
hr P Resalta va esr noc gi nain aw normal the ie ck,
14, When wil ene of pressure and conte fav an imss plane surface cli 7
AS. Find a expression for be ore exerted and eet o pesas fra comp sb merged ined plane
soca. Can te same moto be apd fring the salu re on a ered eae Immer in.
Le quid? hy?
16. Wade po understand by he hydra quo ? Wil he help of Mis equation drive ie expesons
{ore tll ns on sheng plas ara adhe bet fre ating ono sb-mergd od,
13, Devine a exc forthe recon einen he gts as 9
(8) NUMERICAL PROBLEMS
Det the al preste und dept of ene of pressure on le rectangular sue of ndo
and 3m dsp wben is oper ees hora a) eines wi sae surface 4) 2 m belo the
Tre water sce. has) 4845, 20m. (9) 109005837142]
Detemie the oa pressure clear ple of damier 1.8 m which is la vean war a
suc a way tat om of plate 5 2 m Below te Hee surf of wate Fide poston of eae of
presi abo [As 3460858. 2 07m
A ocunglesllc gts uned ote vera Wl fa lock. The vic ofthe sde 6
in kn and ep of cnt of wen is 8 m blow the war surface. Provo tthe de of cent of
‘omar ave by Bam.
44 Accu opening. 3 lee, in a vena ie of tank closed by a di fm dtr mich en
Toate aout a hol ameter. Calculo >) free om ie ie. ad (ti equ ru o
ii ie ase eli i he vera psi when te bead of water above the bore
arte ism Tans DS OSAN, 4) 908 Na
and led with ue vale. Find te force exec by th ci om the ul ad psi ol cen of
rosa Ans. 208 MN, 016m blow ent pipe
Deine the oa pressure and conte of pressure on u seco angular plato of base $ mund
aloe $m mes Le ple Is meted vera in an oo sp. gf 08, The ase of he pt À
Below te fee surface of water. (Ams. 261927 NO
7. The open dar is m wide und 2m high A vera sue gt ls sc o cover Le opens. On
Le up of D gle, i gu ot pr. LS o po à Siga 2.0 m ae he wp ofthe ae
cas on he unse side, the wars avaible up à Bight of be lp of the gale in te
relat fre acting om Le gal und poston of ene 0 press. Asse te gai er a
the totem. ans. 206010 0968 m above einge
8, Describe bly with sehe the various methods usd for menting press excited by id
10. Prove ht the vera componca of the tesulan pressure on a submerged cui ste seal o
the meg o held supported y the cured sac.
I. What sth ferne between sic ue and lock gue?
122 Prove that he ection ern he pts o Jocs equal he ecto at he hinge
E
hr P Resalta va esr noc gi nain aw normal the ie ck,
14, When wil ene of pressure and conte fav an imss plane surface cli 7
AS. Find a expression for be ore exerted and eet o pesas fra comp sb merged ined plane
soca. Can te same moto be apd fring the salu re on a ered eae Immer in.
Le quid? hy?
16. Wade po understand by he hydra quo ? Wil he help of Mis equation drive ie expesons
{ore tll ns on sheng plas ara adhe bet fre ating ono sb-mergd od,
13, Devine a exc forthe recon einen he gts as 9
(8) NUMERICAL PROBLEMS
Det the al preste und dept of ene of pressure on le rectangular sue of ndo
and 3m dsp wben is oper ees hora a) eines wi sae surface 4) 2 m belo the
Tre water sce. has) 4845, 20m. (9) 109005837142]
Detemie the oa pressure clear ple of damier 1.8 m which is la vean war a
suc a way tat om of plate 5 2 m Below te Hee surf of wate Fide poston of eae of
presi abo [As 3460858. 2 07m
A ocunglesllc gts uned ote vera Wl fa lock. The vic ofthe sde 6
in kn and ep of cnt of wen is 8 m blow the war surface. Provo tthe de of cent of
‘omar ave by Bam.
44 Accu opening. 3 lee, in a vena ie of tank closed by a di fm dtr mich en
Toate aout a hol ameter. Calculo >) free om ie ie. ad (ti equ ru o
ii ie ase eli i he vera psi when te bead of water above the bore
arte ism Tans DS OSAN, 4) 908 Na
and led with ue vale. Find te force exec by th ci om the ul ad psi ol cen of
rosa Ans. 208 MN, 016m blow ent pipe
Deine the oa pressure and conte of pressure on u seco angular plato of base $ mund
aloe $m mes Le ple Is meted vera in an oo sp. gf 08, The ase of he pt À
Below te fee surface of water. (Ams. 261927 NO
7. The open dar is m wide und 2m high A vera sue gt ls sc o cover Le opens. On
Le up of D gle, i gu ot pr. LS o po à Siga 2.0 m ae he wp ofthe ae
cas on he unse side, the wars avaible up à Bight of be lp of the gale in te
relat fre acting om Le gal und poston of ene 0 press. Asse te gai er a
the totem. ans. 206010 0968 m above einge
Hydrostatic Forces on Surfaces 127]
8. A cabsom fr chia he ern 1 ey och of pool frm 16. wie al be ip and 12m
wide a th tots and 8 m dep. Fad the oil presse and cono of presse on he caisson I the
a an the cul 1 m Blow he tp level We eon und deck s epi.
[Ans 3168 MN 4.56 m Blow aer suce
A Sin gate 2 m wide an 15m igh een ver plane and sa fen: of fon of 2
einen sc ud godes. If he ga weighs one tons in the era re rele ab the a
supper ge st dep of om se sure of wae. [Ame ISSN)
10, A nk contas water ul a eight fm abs ih base, An misc igus ofp gs en
tie top of wer apo 3m high Calle (teal presas am ane oth tank. be positon o
cee reste or Oe sie ofthe nk, Ml 3m wide. [ANS 5S1BN, 1.685 fom wp]
A A rocangolt an 4m ong LS m wide contains wat uo a gh 2 m. Cll the fee dueto
ater presu on the hase of he ak. Find ls the dept cnt of presi on fe sce
Ana TON, 2 om fe ee]
12. rectangular loe cic we ad m dep sin water in such a may at plane mates an ape
the une peo the plate 12 flow he fre wae sae (Ane 0325 N, 2318]
A hr plate 30 meter ls mmersd in water In such a way at he plano ofthe pie maker an
ange "wit he re sac of water, Determine ie pressure and po o cee o presse
‘nen the upper edge ofthe pe 2m below the fee water sae
{Ans 22.0 AN, 3.27 fom ee sce]
14, A rectangular gt 6m 2m ing a as and nie ado horn as show in Fi. 34
"o esp the gate ina sable pose, a cum weigh of 29430 ide upper end of au
Find the dep of war a wich the pate gis o al Nel the weight of he gat an lo fon
the hinge an ple. (Ans. 348m)
ware suprace
ig 354 ig 385
15. An cine rectangle gl of wilh Sm and dept 1.5 m sal cote charge of war a
‘own in Fg. 385. The end A sig. Determine he ore normal he gi pi a o pen
ans. 974358)
16, A gue supporting water i shown in Fp. 3.55. Find he Height
Take te wid of te gle as wy. ame 3 AF a
17. Find the tata preste and dep of cet of pressure on à
agar lt fase 3 man hip m ch mesa in
tater in sah a ay Ma plane ofthe ple matt an angle of 6° Fig. 356
dh fe sus, The as oft pi parallel Wo wale suce anda doi 0 2 m fom wer
suce. Ans 120 524.200}
8. A cabsom fr chia he ern 1 ey och of pool frm 16. wie al be ip and 12m
wide a th tots and 8 m dep. Fad the oil presse and cono of presse on he caisson I the
a an the cul 1 m Blow he tp level We eon und deck s epi.
[Ans 3168 MN 4.56 m Blow aer suce
A Sin gate 2 m wide an 15m igh een ver plane and sa fen: of fon of 2
einen sc ud godes. If he ga weighs one tons in the era re rele ab the a
supper ge st dep of om se sure of wae. [Ame ISSN)
10, A nk contas water ul a eight fm abs ih base, An misc igus ofp gs en
tie top of wer apo 3m high Calle (teal presas am ane oth tank. be positon o
cee reste or Oe sie ofthe nk, Ml 3m wide. [ANS 5S1BN, 1.685 fom wp]
A A rocangolt an 4m ong LS m wide contains wat uo a gh 2 m. Cll the fee dueto
ater presu on the hase of he ak. Find ls the dept cnt of presi on fe sce
Ana TON, 2 om fe ee]
12. rectangular loe cic we ad m dep sin water in such a may at plane mates an ape
the une peo the plate 12 flow he fre wae sae (Ane 0325 N, 2318]
A hr plate 30 meter ls mmersd in water In such a way at he plano ofthe pie maker an
ange "wit he re sac of water, Determine ie pressure and po o cee o presse
‘nen the upper edge ofthe pe 2m below the fee water sae
{Ans 22.0 AN, 3.27 fom ee sce]
14, A rectangular gt 6m 2m ing a as and nie ado horn as show in Fi. 34
"o esp the gate ina sable pose, a cum weigh of 29430 ide upper end of au
Find the dep of war a wich the pate gis o al Nel the weight of he gat an lo fon
the hinge an ple. (Ans. 348m)
ware suprace
ig 354 ig 385
15. An cine rectangle gl of wilh Sm and dept 1.5 m sal cote charge of war a
‘own in Fg. 385. The end A sig. Determine he ore normal he gi pi a o pen
ans. 974358)
16, A gue supporting water i shown in Fp. 3.55. Find he Height
Take te wid of te gle as wy. ame 3 AF a
17. Find the tata preste and dep of cet of pressure on à
agar lt fase 3 man hip m ch mesa in
tater in sah a ay Ma plane ofthe ple matt an angle of 6° Fig. 356
dh fe sus, The as oft pi parallel Wo wale suce anda doi 0 2 m fom wer
suce. Ans 120 524.200}
128 Fluid Mechanics
18, Fd he ral and eral components ofthe Lil ore ang oa à cul sue AB, whi s
te form of «quran fa cio o as 2 an shown in Fig 37 ake Ue véd of he gue 2,
(Aas f= ITZA, = MEN
waren SURFACE WATERSURFACE Once
Fig 387 Fig 3.58
19. Fi. 358 toms a gate having a quasar shape of ras of 3 m. Find te est forse ae to wae per
tree let ft at, ld a he ange a wih e al fre wil ac, Pans $2201 AN, B= 87:31
20, A oler gis shown ia Fi. 389. Isla fo of 60 darter Tis places on he dam. Find
‘the magro and nein fe estan force ue o var ating om the fate when he Mat just
going oil. The Inch gt piven 10 (Ane 2265 MN 0= 34°81
be,
Hace
ig 359 Fig 3.60
21, Find he horizontal and ental components of he water pressure exci ea ter ae ais
“Amst stown in Fig. 360. Consider wilt ofthe gat uni [AF = 19.524N, f= 71024)
Pose acing on a curved face of dm wie spo
accent te ri y= = a shown in Fig. 34. The
igh of water eased by he dam 5 12 m. Take he wit
da as ny Ams OTTERS, 0 43°19
23. ach gate ola lock e Sm high and is suport dy Io Fig. 361
inge paola the Lp und Dtm ofthe ga: Wien the
‘es ae cool ey make an angle of 120" The wit fhe ck me dps of aleron he wo
‘Ses ol Ie gs are 4m and 3m spectively dein: he gate of estat peste on cach
ges and (i) anid ofthe Hinge reactions. [Ans 79279 i) ym 27924 KN. RSI SSA
24, The eat gates ABC of «lock re 8 m Hh ad when closed ake an ange of 120 Tae wd of Look
Ie 10m Each ue pp by to ops ca I m and above the fom fe ack. The
ep of water on he upucc and dora sen ofthe lok re 6m and 4m respectively. Find
("Resta wer fore on exch tle,
18, Fd he ral and eral components ofthe Lil ore ang oa à cul sue AB, whi s
te form of «quran fa cio o as 2 an shown in Fig 37 ake Ue véd of he gue 2,
(Aas f= ITZA, = MEN
waren SURFACE WATERSURFACE Once
Fig 387 Fig 3.58
19. Fi. 358 toms a gate having a quasar shape of ras of 3 m. Find te est forse ae to wae per
tree let ft at, ld a he ange a wih e al fre wil ac, Pans $2201 AN, B= 87:31
20, A oler gis shown ia Fi. 389. Isla fo of 60 darter Tis places on he dam. Find
‘the magro and nein fe estan force ue o var ating om the fate when he Mat just
going oil. The Inch gt piven 10 (Ane 2265 MN 0= 34°81
be,
Hace
ig 359 Fig 3.60
21, Find he horizontal and ental components of he water pressure exci ea ter ae ais
“Amst stown in Fig. 360. Consider wilt ofthe gat uni [AF = 19.524N, f= 71024)
Pose acing on a curved face of dm wie spo
accent te ri y= = a shown in Fig. 34. The
igh of water eased by he dam 5 12 m. Take he wit
da as ny Ams OTTERS, 0 43°19
23. ach gate ola lock e Sm high and is suport dy Io Fig. 361
inge paola the Lp und Dtm ofthe ga: Wien the
‘es ae cool ey make an angle of 120" The wit fhe ck me dps of aleron he wo
‘Ses ol Ie gs are 4m and 3m spectively dein: he gate of estat peste on cach
ges and (i) anid ofthe Hinge reactions. [Ans 79279 i) ym 27924 KN. RSI SSA
24, The eat gates ABC of «lock re 8 m Hh ad when closed ake an ange of 120 Tae wd of Look
Ie 10m Each ue pp by to ops ca I m and above the fom fe ack. The
ep of water on he upucc and dora sen ofthe lok re 6m and 4m respectively. Find
("Resta wer fore on exch tle,
Hydrostatic Forces on Surfaces 129]
Gi) Reacion berween the gates AB and BC and
(ai) vce on och bing, comida ren of the gale ating in he same hon! plane at
relat water presi [Ans S653 N, () 566.35 EN aná (Ry = 173.64, By 392. UN]
25. A ol cul plat ol Zem and 1 m see amis mer verily in war uch at
A cone pi xp on war steed the il rears en ep of come of rose
(ans 02 S08RN, £07835)
26. A retar oping 2 id an 1 m dsp in sra ie oa nk sc y lu a of sae
A. Te gc a lot oral ens a Demi: e al oie se Be
‘ne eng on ese pas Than of war ave he app ee oe ps1
Ta) ER A SN
27, Demin te wa fre an eso of cet of presen one fc Of Fee SURFACE 0 U
the pl shown in Fi. 342 mena a ei ei avi 09. PAGE SNAEAGE.OF OUD.
Ame SEEN OU
A la opening, m mts, nthe vera sd of ater tn coed
by aie of 3m diner wich can ote by a art meer 7
Cats: (te freon he di, a D he gu og à mata
te übe a uti ln Be vera poston when i Real valer
sve ie ll dance fi (Anse ZN and) MAN 0]
fae dancer lo a a a valve which conti Be dicts png
aeg it can ote wou a ol dee. I bao water
toe scene 520 m Bd Wa ee ag on ed an he
dog eg o mint i a et poston.
430. A cour sum 18 eater nd 12m eight submerged wih sa vera ns pp en at
dep of An beau wate ie. Determine +
(0 wa ren or top, htm and cares surfaces ofthe dam,
(i roe pressure om he whole rta, u
i dep of see of press on cerned surta.
3. Acer pate Flame 3 merce water In uch a way at eas and tet dent fn
Be re sae of al re 1s and D eget Por ie Men de fe la fi kal fe
‘xed y wate and 1) the poston of eae of pressure (Ame 386448 10)2 128m)
32. Atk coms trop «eg ut 10m One lo is of ie Lk inc, The angle been
fre sae of water cime sie The wii fhe km. Find: (efor exe PY
water om icine side and (potion of cere of presare (An. () 28 901 XN G67
3, Alar pat of 3 m dint sur water whi plas making an angle 0 30" wah ie wa
Surface Ih tp edge of he pie. m blow the wa ra, ft fore on on ic o pate
Andi oan NT, Hydro 5 2060) res
[iit = 3m, 30 ig pe LÀ Lt
Poor mom x (Ans) cris nnasan.
testo gal?)
A pssomasataim)
Mas
Ders
Gi) Reacion berween the gates AB and BC and
(ai) vce on och bing, comida ren of the gale ating in he same hon! plane at
relat water presi [Ans S653 N, () 566.35 EN aná (Ry = 173.64, By 392. UN]
25. A ol cul plat ol Zem and 1 m see amis mer verily in war uch at
A cone pi xp on war steed the il rears en ep of come of rose
(ans 02 S08RN, £07835)
26. A retar oping 2 id an 1 m dsp in sra ie oa nk sc y lu a of sae
A. Te gc a lot oral ens a Demi: e al oie se Be
‘ne eng on ese pas Than of war ave he app ee oe ps1
Ta) ER A SN
27, Demin te wa fre an eso of cet of presen one fc Of Fee SURFACE 0 U
the pl shown in Fi. 342 mena a ei ei avi 09. PAGE SNAEAGE.OF OUD.
Ame SEEN OU
A la opening, m mts, nthe vera sd of ater tn coed
by aie of 3m diner wich can ote by a art meer 7
Cats: (te freon he di, a D he gu og à mata
te übe a uti ln Be vera poston when i Real valer
sve ie ll dance fi (Anse ZN and) MAN 0]
fae dancer lo a a a valve which conti Be dicts png
aeg it can ote wou a ol dee. I bao water
toe scene 520 m Bd Wa ee ag on ed an he
dog eg o mint i a et poston.
430. A cour sum 18 eater nd 12m eight submerged wih sa vera ns pp en at
dep of An beau wate ie. Determine +
(0 wa ren or top, htm and cares surfaces ofthe dam,
(i roe pressure om he whole rta, u
i dep of see of press on cerned surta.
3. Acer pate Flame 3 merce water In uch a way at eas and tet dent fn
Be re sae of al re 1s and D eget Por ie Men de fe la fi kal fe
‘xed y wate and 1) the poston of eae of pressure (Ame 386448 10)2 128m)
32. Atk coms trop «eg ut 10m One lo is of ie Lk inc, The angle been
fre sae of water cime sie The wii fhe km. Find: (efor exe PY
water om icine side and (potion of cere of presare (An. () 28 901 XN G67
3, Alar pat of 3 m dint sur water whi plas making an angle 0 30" wah ie wa
Surface Ih tp edge of he pie. m blow the wa ra, ft fore on on ic o pate
Andi oan NT, Hydro 5 2060) res
[iit = 3m, 30 ig pe LÀ Lt
Poor mom x (Ans) cris nnasan.
testo gal?)
A pssomasataim)
Mas
Ders
|
BUOYANCY AND
EIOATATION
» 4.1 INTRODUCTION
In bis chapter, the equim ofthe Most and submerged bodes wil be considered. Thus the
chapter will include ; L Buoyancy, 2. Centre of buoyancy. 3. Metacenne, 4. Metaceic height
5, Analytical metho! for determining metacen height 6. Conditions of equilibrium of a foating
and sub-merged body, and 7. Experimental method for metcenri eight
» 4.2 BUOYANCY
Wien a body Is immersed in Mud, an upward force is exeted by the Moi on the body. This
‘upward force is equal o the weight ofthe fluid displaced by the body and is called the force of
‘buoyancy or simply buoyancy,
» 4.3 CENTRE OF BUOYANCY
is defined as the point, tough which the force of buoyancy Is supposed to act, As the for of
buoyancy ls eral free and is equal to the weit ofthe id displaced by the body, the cent of
‘buoyancy vil e centre Of gravity ofthe Nui displaced.
Problem 4.1. Find the volune of the water displaced and postion of centre of buoyancy for a
wooden back of wid 25 m and of depth 15 m, when i floats horizontals in water The dent of
‘wooden block is 650 Kg" and length 6.0m.
‘Solution. Given
vn -25n
De isn a
teh Lan +
Volume of the block 2.5% 1.5 x 6.0= 22.50 m? 5, mn
Deny wo. pois .
‘Wagar ck 2 pgs Volume ==
98125086 N Feat
131
BUOYANCY AND
EIOATATION
» 4.1 INTRODUCTION
In bis chapter, the equim ofthe Most and submerged bodes wil be considered. Thus the
chapter will include ; L Buoyancy, 2. Centre of buoyancy. 3. Metacenne, 4. Metaceic height
5, Analytical metho! for determining metacen height 6. Conditions of equilibrium of a foating
and sub-merged body, and 7. Experimental method for metcenri eight
» 4.2 BUOYANCY
Wien a body Is immersed in Mud, an upward force is exeted by the Moi on the body. This
‘upward force is equal o the weight ofthe fluid displaced by the body and is called the force of
‘buoyancy or simply buoyancy,
» 4.3 CENTRE OF BUOYANCY
is defined as the point, tough which the force of buoyancy Is supposed to act, As the for of
buoyancy ls eral free and is equal to the weit ofthe id displaced by the body, the cent of
‘buoyancy vil e centre Of gravity ofthe Nui displaced.
Problem 4.1. Find the volune of the water displaced and postion of centre of buoyancy for a
wooden back of wid 25 m and of depth 15 m, when i floats horizontals in water The dent of
‘wooden block is 650 Kg" and length 6.0m.
‘Solution. Given
vn -25n
De isn a
teh Lan +
Volume of the block 2.5% 1.5 x 6.0= 22.50 m? 5, mn
Deny wo. pois .
‘Wagar ck 2 pgs Volume ==
98125086 N Feat
131
132 Fluid Mechanics
For equilibrium the weight of water displaced = Weight of wooden block
= MTN
Volume of water displaced
__ eight of water displaced 143471
‘Weight density of water” 1000x981
(Weight deny of wate
Position of Centre of Buoyancy. Volume of wooden block in water
Volume of water diplacod
or 25h 60 = 14625 m where is dep of wooden Block in water
14625
25x60
de Centre of Buoyancy = oe = 0.4875 m from base.
14625 m‘. Ans.
1000 «9.81 Nim)
975
Problem 4.2 A wooden log of 6 m diameter and 5 m length ls floating ln river water. Find the
dein ofthe wooden login water when the sp. gravity af the og 507.
Solution. Gives 2
Dia. of lop 05m
Length L=5m
sr 5207
a Density of log = 0.71000 = 700 kg/m
2 Weight density of og, w= p xe
= 100 49381 Nin?
Find depth of immersion or
Weight of wooden log = Weight density x Volume of log
= 700 «9:81 x (Dy? xt
oy
= 1009.81 x À (6) XS N= 9896%981 N
For equiibium,
Weight of wooden log = Weight of water displaced
= Weight donsty of water x Volume of water displaced
Volume of water displaced
(Weight density of water = 1000 9.81 Nin!)
[Levins te depth of immersion
Volume of log inside water = Arca of ADCA x Length
= Arca of ADCA 450
Bot volume of lo inside water = Volume of water diplaced = 0.9896 m’
For equilibrium the weight of water displaced = Weight of wooden block
= MTN
Volume of water displaced
__ eight of water displaced 143471
‘Weight density of water” 1000x981
(Weight deny of wate
Position of Centre of Buoyancy. Volume of wooden block in water
Volume of water diplacod
or 25h 60 = 14625 m where is dep of wooden Block in water
14625
25x60
de Centre of Buoyancy = oe = 0.4875 m from base.
14625 m‘. Ans.
1000 «9.81 Nim)
975
Problem 4.2 A wooden log of 6 m diameter and 5 m length ls floating ln river water. Find the
dein ofthe wooden login water when the sp. gravity af the og 507.
Solution. Gives 2
Dia. of lop 05m
Length L=5m
sr 5207
a Density of log = 0.71000 = 700 kg/m
2 Weight density of og, w= p xe
= 100 49381 Nin?
Find depth of immersion or
Weight of wooden log = Weight density x Volume of log
= 700 «9:81 x (Dy? xt
oy
= 1009.81 x À (6) XS N= 9896%981 N
For equiibium,
Weight of wooden log = Weight of water displaced
= Weight donsty of water x Volume of water displaced
Volume of water displaced
(Weight density of water = 1000 9.81 Nin!)
[Levins te depth of immersion
Volume of log inside water = Arca of ADCA x Length
= Arca of ADCA 450
Bot volume of lo inside water = Volume of water diplaced = 0.9896 m’
‘Buoyancy and Floatation 133]
0.9896 = Area of ADCA 5.0
2 Area of ADCA 99896 0,1979 n°
30
Dar ares of ABCA
ea of curved surface ADCOA + Arca of AOC
=e arn
lernen
Eje orenoss
0.1979 = 2827 - 00157 8+ 09 cos 0 in @
oF 0187 8.09 cos sin 8 = 2827 = 1070
0-57.32 608 @ sin 9= S801
7.32 0080 in 0 SOL
, (60 57.32 x05 x 866 - 5401
1-57.32 x 342 x 09396 - 5601
2 257.32 4 309 4 951 ~ $401 = 72~ 1684 - 54.01
71 257.32 x 325 x 9485 - S401 7.61 580
LS = S732 3173 X 948 S401 = TS 1724 — SOI = à 248
CAES
103 + 0.3 03178 = 0.98 m. Ans.
Problem 4.8 A stone weighs 392.4 N in air and 196.2 N in water. Compute the volume of sone
‘and ts specific gravity.
Solution. Gives
Weight of stone in air man
Weight of stone in water = 1962.
For equilibrium,
‘Weight i ac = Weight of tone in water = Weight of water spaced
or 3924 - 1962 = 196.2 = 1000 9,81 x Volume of water displaced
Volume of water displaced
CE
TETE TES
x 10e = 2x 10! em! Ans.
= Volume of sone
Volume of stone = 2310 em Ans
0.9896 = Area of ADCA 5.0
2 Area of ADCA 99896 0,1979 n°
30
Dar ares of ABCA
ea of curved surface ADCOA + Arca of AOC
=e arn
lernen
Eje orenoss
0.1979 = 2827 - 00157 8+ 09 cos 0 in @
oF 0187 8.09 cos sin 8 = 2827 = 1070
0-57.32 608 @ sin 9= S801
7.32 0080 in 0 SOL
, (60 57.32 x05 x 866 - 5401
1-57.32 x 342 x 09396 - 5601
2 257.32 4 309 4 951 ~ $401 = 72~ 1684 - 54.01
71 257.32 x 325 x 9485 - S401 7.61 580
LS = S732 3173 X 948 S401 = TS 1724 — SOI = à 248
CAES
103 + 0.3 03178 = 0.98 m. Ans.
Problem 4.8 A stone weighs 392.4 N in air and 196.2 N in water. Compute the volume of sone
‘and ts specific gravity.
Solution. Gives
Weight of stone in air man
Weight of stone in water = 1962.
For equilibrium,
‘Weight i ac = Weight of tone in water = Weight of water spaced
or 3924 - 1962 = 196.2 = 1000 9,81 x Volume of water displaced
Volume of water displaced
CE
TETE TES
x 10e = 2x 10! em! Ans.
= Volume of sone
Volume of stone = 2310 em Ans
134 Fluid Mechanics
Specine Gravity of Stone
Mass of stone
Density of stone
„ Density ofstone 2000
Density of water © 1000
Problem 4.4 A body of dimensions 4.5 m X LO m X2 m, weighs 1962 N in water Find is weight
in air. What wil be its specific gravity ?
Solution. Given
Volume of body.
Weight of ody in water 1962 N
Volume of te water displaced = Volume ofthe body = 3.0 m°
2. Weight of water displaced = 1000 x 9.1 x30 = 29430 N
For the equilibrium of the body.
‘Weight of body in ae — Weight of water spaced = Weight in water
» Wa 29430 = 1962
Wa, = 29430 + 1962 = 31392 N
2 Sp. ge of sone = 20 Ans.
om
S0x10x20=
os gi AD gag
sa
Deny tty 2 2 2 ner
‘oa
Sp. gravity ofthe boty = 100697 à 1.067. Am
Problem 4.5 Find ine density of a metallic body which floats at the interface of mercury of
a re 136 and water such that 40% 0 Us volume i sub-merged in mercury and 60% in water,
Solution. Let the volume ofthe body = Vii?
“Then volume of body submerge in mercury
Volume of body sub-merged in water
60 A
Dv 06 vm"
Fig. 43
For the equilibrium of the body
Total buoyant force (upward force) = Weight ofthe body
‘But worl buoyant force = Force of buoyancy due 10 water + Force of buoyancy due 10 mercury
‘Weight of water displaced by body
= Density of water g X Volume of water space
1000 xx Volume of ody in wate
Force of buoyancy due to water
Specine Gravity of Stone
Mass of stone
Density of stone
„ Density ofstone 2000
Density of water © 1000
Problem 4.4 A body of dimensions 4.5 m X LO m X2 m, weighs 1962 N in water Find is weight
in air. What wil be its specific gravity ?
Solution. Given
Volume of body.
Weight of ody in water 1962 N
Volume of te water displaced = Volume ofthe body = 3.0 m°
2. Weight of water displaced = 1000 x 9.1 x30 = 29430 N
For the equilibrium of the body.
‘Weight of body in ae — Weight of water spaced = Weight in water
» Wa 29430 = 1962
Wa, = 29430 + 1962 = 31392 N
2 Sp. ge of sone = 20 Ans.
om
S0x10x20=
os gi AD gag
sa
Deny tty 2 2 2 ner
‘oa
Sp. gravity ofthe boty = 100697 à 1.067. Am
Problem 4.5 Find ine density of a metallic body which floats at the interface of mercury of
a re 136 and water such that 40% 0 Us volume i sub-merged in mercury and 60% in water,
Solution. Let the volume ofthe body = Vii?
“Then volume of body submerge in mercury
Volume of body sub-merged in water
60 A
Dv 06 vm"
Fig. 43
For the equilibrium of the body
Total buoyant force (upward force) = Weight ofthe body
‘But worl buoyant force = Force of buoyancy due 10 water + Force of buoyancy due 10 mercury
‘Weight of water displaced by body
= Density of water g X Volume of water space
1000 xx Volume of ody in wate
Force of buoyancy due to water
Buoyancy and Floatation 135]
1000 x 8 x06 x VN
and Force of Duoyaney due fo mercuty = Weight of mercury displaced by body
4 Density of mercury x Volume of mercury displaced
= à 4 136 1000 Volume of body in mercury
£2136 % 100004 YN
Weight ofthe body = Density ¢ Volume of hay = px. V
where pis the dena of the bay
2 For equliium, we
Tora buoyant free = Weight of the body
1000 x 206 XV + 13.6 x 1000 x g x À V= PAR XV
or 1 =600 à 13600 x 4 = 600 + 51400 = 6040.00 Kg
13 Density of the body = 6040.00 kg/m’ Ans.
Problem 4.8 A loa valve regulates the flow ofoi of sp. gr. 0.8 into a cistern. The spherical float
ig 15 cm in diameter, AOD sa weighless lnk carrying the loa a one end, and a valve atthe other
end which closes the pipe though which il los nt he cistern. The ink is mounted in a fetonless
hinge at O and the angle AOB ts 135". The length of OA is 201, and he distance between the centre
(ofthe float ad the hing ix 50 em. When the low is stopped AO will be vertical. The valve is I be
pressed on tothe seat wilh force of 9.81 N lo completely op the lw of ol into the ister. was
‘serve that the flow of ol stopped woken the free surface of i inthe cistern i 35 cm below the
Inge: Determine the weight ofthe flat.
Solution. Given
Sp. gt of ol bra En
Density of oi, a >
Dia. of os. 3
Fores,
08 = cn
Find the weight ofthe flat. Let is equal to W.
‘When the flow of ei is topped the cent of flats shown in Fig. 4.4
‘Tae lve o ils also shown, The cnt of float Blow the en of bya dpa
0D OC+CD_ 84h
rom ABOD, sings’ = OP . OC+CD_ 4h
Hmm ane, oo oF a:
a 50 sin 5° 2 384 4
1
or 250% 1-95 = 35355-35= 0.355 cm = 00355 m.
Æ
‘The weigh of float is ating tough 8 but Ut upward buoyant force js acting through he centre
of weight of ol displaced
wen
Volume of oi displaced
1000 x 8 x06 x VN
and Force of Duoyaney due fo mercuty = Weight of mercury displaced by body
4 Density of mercury x Volume of mercury displaced
= à 4 136 1000 Volume of body in mercury
£2136 % 100004 YN
Weight ofthe body = Density ¢ Volume of hay = px. V
where pis the dena of the bay
2 For equliium, we
Tora buoyant free = Weight of the body
1000 x 206 XV + 13.6 x 1000 x g x À V= PAR XV
or 1 =600 à 13600 x 4 = 600 + 51400 = 6040.00 Kg
13 Density of the body = 6040.00 kg/m’ Ans.
Problem 4.8 A loa valve regulates the flow ofoi of sp. gr. 0.8 into a cistern. The spherical float
ig 15 cm in diameter, AOD sa weighless lnk carrying the loa a one end, and a valve atthe other
end which closes the pipe though which il los nt he cistern. The ink is mounted in a fetonless
hinge at O and the angle AOB ts 135". The length of OA is 201, and he distance between the centre
(ofthe float ad the hing ix 50 em. When the low is stopped AO will be vertical. The valve is I be
pressed on tothe seat wilh force of 9.81 N lo completely op the lw of ol into the ister. was
‘serve that the flow of ol stopped woken the free surface of i inthe cistern i 35 cm below the
Inge: Determine the weight ofthe flat.
Solution. Given
Sp. gt of ol bra En
Density of oi, a >
Dia. of os. 3
Fores,
08 = cn
Find the weight ofthe flat. Let is equal to W.
‘When the flow of ei is topped the cent of flats shown in Fig. 4.4
‘Tae lve o ils also shown, The cnt of float Blow the en of bya dpa
0D OC+CD_ 84h
rom ABOD, sings’ = OP . OC+CD_ 4h
Hmm ane, oo oF a:
a 50 sin 5° 2 384 4
1
or 250% 1-95 = 35355-35= 0.355 cm = 00355 m.
Æ
‘The weigh of float is ating tough 8 but Ut upward buoyant force js acting through he centre
of weight of ol displaced
wen
Volume of oi displaced
136 Fluid Mechanics
2 à x mx (OTS) + 00858 x (075) = 0.000945 m°
cht of ol displaced
4 Volume of ei
00 981 x 000945 = 7416 N
“The buoyant fore and weight ofthe float pass trough the same vr
[Lethe weight of lati W. Then net vertical Fores on Moat
yaa force - Weight of float = 7.416 W)
“aking moments about the hinge O, we get
2 Booyant force
Hie, passing trough B.
X20 = (416 - W) x BD = (7416 — W) x 50x08 48°
or 9.81 x20= 0.416 - W) 435.355
a we ate DSL 7416-5552 1866 N. Ans.
35355
> 4.4. META-CENTRE
is defined asthe point about which a body starts oscilating when the body is lid by a small
angle. The meta-cetre may also be defined asthe point at whieh the lie of ation of the force of
buoyancy will meet the normal aus ofthe body when the Body is vena small angular displacement
Considera body Hating in a Jquid as shown in Fig. 45 (a). Lethe body sn equilibrium and Cis
te centre of gravity and B he cenue of buoyancy For equlbchum, both te pints leon the normal
and, whieh is vera
Noma, wes un
Fig. 45 Metacenre
Lette body is given a small angular displacement in the clockwise direction as shown in Fi 45 (0,
‘Tne centre of buoyaney, which i the cent of pravity ofthe displaced god or cen of gravity o the
portion ofthe Body sub-merged in liquid, will now be sifted towards right rom the normal ais. Let
list By as sbown in Fig, 4.5 (). The line of ation ofthe force o buoyaney in this new position, will
interet the normal ais of the body at some point say M. This point Ai alld Meta-centre.
> 4.5 META-CENTRIC HEIGHT
‘The distance MG, ¿e the distance between the meta-entre af a floating body and the centre of
gravity ofthe body Is called meta-enuic height
2 à x mx (OTS) + 00858 x (075) = 0.000945 m°
cht of ol displaced
4 Volume of ei
00 981 x 000945 = 7416 N
“The buoyant fore and weight ofthe float pass trough the same vr
[Lethe weight of lati W. Then net vertical Fores on Moat
yaa force - Weight of float = 7.416 W)
“aking moments about the hinge O, we get
2 Booyant force
Hie, passing trough B.
X20 = (416 - W) x BD = (7416 — W) x 50x08 48°
or 9.81 x20= 0.416 - W) 435.355
a we ate DSL 7416-5552 1866 N. Ans.
35355
> 4.4. META-CENTRE
is defined asthe point about which a body starts oscilating when the body is lid by a small
angle. The meta-cetre may also be defined asthe point at whieh the lie of ation of the force of
buoyancy will meet the normal aus ofthe body when the Body is vena small angular displacement
Considera body Hating in a Jquid as shown in Fig. 45 (a). Lethe body sn equilibrium and Cis
te centre of gravity and B he cenue of buoyancy For equlbchum, both te pints leon the normal
and, whieh is vera
Noma, wes un
Fig. 45 Metacenre
Lette body is given a small angular displacement in the clockwise direction as shown in Fi 45 (0,
‘Tne centre of buoyaney, which i the cent of pravity ofthe displaced god or cen of gravity o the
portion ofthe Body sub-merged in liquid, will now be sifted towards right rom the normal ais. Let
list By as sbown in Fig, 4.5 (). The line of ation ofthe force o buoyaney in this new position, will
interet the normal ais of the body at some point say M. This point Ai alld Meta-centre.
> 4.5 META-CENTRIC HEIGHT
‘The distance MG, ¿e the distance between the meta-entre af a floating body and the centre of
gravity ofthe body Is called meta-enuic height
‘Buoyancy and Floatation 137]
> 4.6 ANALYTICAL METHOD FOR META-CENTRE HEIGHT
Fig 4.6 (a shows the postion of foating body in equilibrium. The location of centre of praviy
and cenue of buoyaney la this positon ls at G and B. The floating body is given a small soplar
displacement inthe clockwise direction. Ths shown in Fi. 16 (). The new cee of buoyancy is
At The verücl ine tzouph By cuts he normal axis a 1. Hence M the metacen and GM is
metre heh.
| | (PLAN OF BODY AT WATERLINE
ig. 46. Aecacentre bight of floating bod.
‘The angular displacement of the body in the clockwise tection causes the wedgo-sbued prism
BOB on ie ight of the aso go inside the water while the Mental wedge-shaped prism represented
by AOA" emerges out ofthe water on te let ofthe ans, These wedges representa gala ln Buoyant
force 0n the right side and a corresponding loss of buoyant force onthe left ide. The gain is
represented by a veical force dFy acting tough the C.G. ofthe pris BOB" while the loss is
represented by an equal and opposite force dF ating vertically downward though the centro of
AOX. The couple due 1 these Buoyant forces tends to rotate the ship in the counterclockwise
“recio. Aso the moment caused by the displacement ofthe centre of buoyancy from Bo is also
in the counerlockwise direction, Ths hese two coupes must be qual
Couple Due to Wedges. Consider towards the sit ofthe ari a small sip of thickness de at à
sistance «from 0 as own in Fig. 4.5 (0). The height of suip ZBOB? =.
Le 2808 = ZAON = am. = 0)
Aca of sip = Height x Thickness
IL isthe length of he oaing body, then
Volume ofstip = Araxt.
xOx Lx dr
RTE
3 Weight of stp = py x Volume = pex 0L de
Similarly, if small strip of thickness dí aa distance» from O towards he Ice of the ais is
considered the weight of stip will Re pgaß Ld. The to welghts are acting in the oppose direcion
And hence constitu a couple
> 4.6 ANALYTICAL METHOD FOR META-CENTRE HEIGHT
Fig 4.6 (a shows the postion of foating body in equilibrium. The location of centre of praviy
and cenue of buoyaney la this positon ls at G and B. The floating body is given a small soplar
displacement inthe clockwise direction. Ths shown in Fi. 16 (). The new cee of buoyancy is
At The verücl ine tzouph By cuts he normal axis a 1. Hence M the metacen and GM is
metre heh.
| | (PLAN OF BODY AT WATERLINE
ig. 46. Aecacentre bight of floating bod.
‘The angular displacement of the body in the clockwise tection causes the wedgo-sbued prism
BOB on ie ight of the aso go inside the water while the Mental wedge-shaped prism represented
by AOA" emerges out ofthe water on te let ofthe ans, These wedges representa gala ln Buoyant
force 0n the right side and a corresponding loss of buoyant force onthe left ide. The gain is
represented by a veical force dFy acting tough the C.G. ofthe pris BOB" while the loss is
represented by an equal and opposite force dF ating vertically downward though the centro of
AOX. The couple due 1 these Buoyant forces tends to rotate the ship in the counterclockwise
“recio. Aso the moment caused by the displacement ofthe centre of buoyancy from Bo is also
in the counerlockwise direction, Ths hese two coupes must be qual
Couple Due to Wedges. Consider towards the sit ofthe ari a small sip of thickness de at à
sistance «from 0 as own in Fig. 4.5 (0). The height of suip ZBOB? =.
Le 2808 = ZAON = am. = 0)
Aca of sip = Height x Thickness
IL isthe length of he oaing body, then
Volume ofstip = Araxt.
xOx Lx dr
RTE
3 Weight of stp = py x Volume = pex 0L de
Similarly, if small strip of thickness dí aa distance» from O towards he Ice of the ais is
considered the weight of stip will Re pgaß Ld. The to welghts are acting in the oppose direcion
And hence constitu a couple
138 Fluid Mechanics
‘Moment ofthis couple = Weight of cc sip x Distance between tess two weights
= par BL 1 + al
= pa OL dex 2e= 2pe BL de
Moment ofthe couple for the whole wege
Lips eL ax an
Moment of coupe du to shifting of centre Of buoyaney from B 40 By
Fy «BB,
= Fax EM x 0 "BB, = BM X it 0s very mal
“wann TES
But dese two couples are the sume, Hence equating equations (41) and (42), we get
Wx BM x 0 | 2997 8 Las
We BM = 2940 J Las
Wx BM = 2pg | x'Lax
Now Las = Elemental arca om the water ne shown in Tig. 46 (c and = dA
5 Wx BM = 298 | ua
Bu rom Fig. 4.5 (cits clear that 2° dA isthe second moment of arca of the plan ofthe body
at vor surface about the axis Y. Therefore
Wc BM = pal where t= 2 1.2 44)
pa = PL
w
Da W= Weight of the body
Weight of the uid displaced by the body
"= pg x Volume ofthe ui diplac bythe body.
= pa Volume ofthe body sub-merged in water
= pax
pex
a mm BXL 43)
pexv
om= am -n6= 1-86
1
2 Mewcentie beigt = GM = À - 86. a
in + aa)
Problem 4.7 A rectangular pontoon ls 5 m long, 3 m wide and 1.20 m high. The depth of
immersion of he pontoon is 0.80 m in sea water the centre of gravity i 0.6 m above the beton of
Ihe pontoon, determine the meta centric height. The density for seu water = 1025 ky
Solution. Given
Dimension of pontoon Smx3mxi20m
Depth of immersion osm
‘Moment ofthis couple = Weight of cc sip x Distance between tess two weights
= par BL 1 + al
= pa OL dex 2e= 2pe BL de
Moment ofthe couple for the whole wege
Lips eL ax an
Moment of coupe du to shifting of centre Of buoyaney from B 40 By
Fy «BB,
= Fax EM x 0 "BB, = BM X it 0s very mal
“wann TES
But dese two couples are the sume, Hence equating equations (41) and (42), we get
Wx BM x 0 | 2997 8 Las
We BM = 2940 J Las
Wx BM = 2pg | x'Lax
Now Las = Elemental arca om the water ne shown in Tig. 46 (c and = dA
5 Wx BM = 298 | ua
Bu rom Fig. 4.5 (cits clear that 2° dA isthe second moment of arca of the plan ofthe body
at vor surface about the axis Y. Therefore
Wc BM = pal where t= 2 1.2 44)
pa = PL
w
Da W= Weight of the body
Weight of the uid displaced by the body
"= pg x Volume ofthe ui diplac bythe body.
= pa Volume ofthe body sub-merged in water
= pax
pex
a mm BXL 43)
pexv
om= am -n6= 1-86
1
2 Mewcentie beigt = GM = À - 86. a
in + aa)
Problem 4.7 A rectangular pontoon ls 5 m long, 3 m wide and 1.20 m high. The depth of
immersion of he pontoon is 0.80 m in sea water the centre of gravity i 0.6 m above the beton of
Ihe pontoon, determine the meta centric height. The density for seu water = 1025 ky
Solution. Given
Dimension of pontoon Smx3mxi20m
Depth of immersion osm
Buoyancy and Floatation 139)
Diane Ao=dom an
Disa A0 4° Dep ot mension PTE
Zixaron rs
Density forsea water = 1025 km FLE mn
nee Bi GM, ny en 1) 5 E
ouh
au L 06 FIT
were f= M. el fe pl fe poten ab a a ee
dessine Sat !
Rn “a
= Vat of te body ses water
3x08x50= 120m
ros
Bo = AG A= 06-04 =02 m bape
Bet gg ess
. ue 5 «1-02-18 02-0077 -02-a7018m. Ans
ce Sc 15-02 8S -02=0905-02=070%8 m à
Problem 48 à ane hay of ie 3m ong 2 m wide 1m dep as in water Whats he
Trig tab ae econ 087 Deere e men on hea
Sohuton. one Sem
Dimension of body =3x2x1 a)
Dentotinmenion —04m
Find (Weight of body, W
i) Mesa-cemri hight, GM
(9) Weight of Body, W
Weight of water displaced
= pe x Volume of water displaced
2 1000 0.1 Volume of body water Eh,
2100098132 «08 8 [
27088 N. Ans reno
(D Meet Meg, GM Des
Using uo) wege
er
ou 1-06
whats MO shot Fai the ln ofthe dy
Lande D soot
wants 22 on
12 R e
Y Volume of body In water
23X24082 48m
BG =AG~ AB.
20
jm = 29 01 2 04167 - 01 = 03167 m. Ans
ou 22 01 = 0161-0,
Diane Ao=dom an
Disa A0 4° Dep ot mension PTE
Zixaron rs
Density forsea water = 1025 km FLE mn
nee Bi GM, ny en 1) 5 E
ouh
au L 06 FIT
were f= M. el fe pl fe poten ab a a ee
dessine Sat !
Rn “a
= Vat of te body ses water
3x08x50= 120m
ros
Bo = AG A= 06-04 =02 m bape
Bet gg ess
. ue 5 «1-02-18 02-0077 -02-a7018m. Ans
ce Sc 15-02 8S -02=0905-02=070%8 m à
Problem 48 à ane hay of ie 3m ong 2 m wide 1m dep as in water Whats he
Trig tab ae econ 087 Deere e men on hea
Sohuton. one Sem
Dimension of body =3x2x1 a)
Dentotinmenion —04m
Find (Weight of body, W
i) Mesa-cemri hight, GM
(9) Weight of Body, W
Weight of water displaced
= pe x Volume of water displaced
2 1000 0.1 Volume of body water Eh,
2100098132 «08 8 [
27088 N. Ans reno
(D Meet Meg, GM Des
Using uo) wege
er
ou 1-06
whats MO shot Fai the ln ofthe dy
Lande D soot
wants 22 on
12 R e
Y Volume of body In water
23X24082 48m
BG =AG~ AB.
20
jm = 29 01 2 04167 - 01 = 03167 m. Ans
ou 22 01 = 0161-0,
140 Fluid Mechanics
Problem 4.9 A biock of wou of specfe gravity 0.7 floats in water. Determine the mera-cenrc
Sohn. ine
Dm #2x1x08 1
Leveepofinmenioa Sm Un
Sp. gr. of wood =07 |
Weight of wooden piece = Weight density of wood* x Volume za Y :
Forma LA XI #08
Weigh of water dspacd = Weight dent of wate et H
A Volume of the wood submergein watt LL z
= 1000 x 9.81 X2 X NN h— 10+
Witt of wooden piece = Weight of water space
1009.81 421 X08 = 1000%9.81 2 TX
— WOX981x2 1X08
1000 9.8121
Distance of centre of Buoyancy from bottom, e.
a » -07%03=036m
and
a 1O=AG-AB=04-026=012m
‘The ete hs ven by eqn 44) oF
7
mel
aueh.
1
where te bat
R a 6
Volume of wood in water
Bm 2x Lx SO
012=0.1488-012= 00268 m. Ans
Problem 4.10 À soli cyúnder of diameter 4.0 m has à height of 3 metres. Find the mer-cenri
eight of me einer when I is floating in water with ls aus vertical. The sp r. of the cylinder
= a6.
Solution. Given
Dia. of eyinder, D=30m
High of cier, he 30m
= Weigh de of wood =p, wise p= dera af wand
2071000 = "00 Epa! Hence for wo = 7009.81 Nm
Problem 4.9 A biock of wou of specfe gravity 0.7 floats in water. Determine the mera-cenrc
Sohn. ine
Dm #2x1x08 1
Leveepofinmenioa Sm Un
Sp. gr. of wood =07 |
Weight of wooden piece = Weight density of wood* x Volume za Y :
Forma LA XI #08
Weigh of water dspacd = Weight dent of wate et H
A Volume of the wood submergein watt LL z
= 1000 x 9.81 X2 X NN h— 10+
Witt of wooden piece = Weight of water space
1009.81 421 X08 = 1000%9.81 2 TX
— WOX981x2 1X08
1000 9.8121
Distance of centre of Buoyancy from bottom, e.
a » -07%03=036m
and
a 1O=AG-AB=04-026=012m
‘The ete hs ven by eqn 44) oF
7
mel
aueh.
1
where te bat
R a 6
Volume of wood in water
Bm 2x Lx SO
012=0.1488-012= 00268 m. Ans
Problem 4.10 À soli cyúnder of diameter 4.0 m has à height of 3 metres. Find the mer-cenri
eight of me einer when I is floating in water with ls aus vertical. The sp r. of the cylinder
= a6.
Solution. Given
Dia. of eyinder, D=30m
High of cier, he 30m
= Weigh de of wood =p, wise p= dera af wand
2071000 = "00 Epa! Hence for wo = 7009.81 Nm
‘Buoyancy and Floatation 141
Sp. gr of eylinger 06 ann
Dept of immersion of cylinder x
06x30
MOL. about YY aus ofthe plan ofthe hody
Da
o
o Eo
má Y= Volume of yn in water
3 0° pam ot nmin
=F ap xism’
ay
ne ae
ETA
e a
i6 “is is
= ve sign means that met-cene, (M) I below the enue of gravity (6).
Problem 4.11 A body hash cylindrical upper portion of 3m diameter and 1.8m deep. The lower
Portion sa curved one, which displaces a volume of 0.6m! of water. The centre of buryancy ofthe
Curved portion i af distance of 1.93 m below the top of the cylinder The centre of gravity ofthe
whole Body is 120m belo the top ofthe cylinder, The total displacement of water is 39 tonnes. Find
‘the meta-cenric height of the body
Solution. Gien
Dia. of body
Depth body
Volume displaced by curved potion
06 = 055-06 =~ 008 m. Ans.
Let B sth conte of huoyancy ofthe curved surface and G ls the conte of gravity ofthe whole
body
Sp. gr of eylinger 06 ann
Dept of immersion of cylinder x
06x30
MOL. about YY aus ofthe plan ofthe hody
Da
o
o Eo
má Y= Volume of yn in water
3 0° pam ot nmin
=F ap xism’
ay
ne ae
ETA
e a
i6 “is is
= ve sign means that met-cene, (M) I below the enue of gravity (6).
Problem 4.11 A body hash cylindrical upper portion of 3m diameter and 1.8m deep. The lower
Portion sa curved one, which displaces a volume of 0.6m! of water. The centre of buryancy ofthe
Curved portion i af distance of 1.93 m below the top of the cylinder The centre of gravity ofthe
whole Body is 120m belo the top ofthe cylinder, The total displacement of water is 39 tonnes. Find
‘the meta-cenric height of the body
Solution. Gien
Dia. of body
Depth body
Volume displaced by curved potion
06 = 055-06 =~ 008 m. Ans.
Let B sth conte of huoyancy ofthe curved surface and G ls the conte of gravity ofthe whole
body
= 3900 49.81 N = 38259 N
Find meta-centrie height of the body
ter the Rig o e fay above te wa srs m Ta
ci ot warded y bo
ch deny twa Ve of wae pil
0002931 Volume fe hd nwa}
= 9810 [Volume of cylindrical part in water + Volume seven
‘of curved portion] “
«10 [Fo Depot ein pat war
+ Vote shy cond prion
“ as em EE xas-03000]
Zoxus-0 406
Exeo
“ 18
218-000 120m
Lt Be xe boyy fell par ne cee of span e mle
von,
“hen dep fina pain wate = LA 0467 m
Ener 212 à = 150050.
“The distance of the centre of huoyaney of he whole ody from the top of the eylindrcal part is
sven as
(CH = (Volume of curved portion X CB + Volume of cylndricl par in water CB)
+ (Total volume of water displaced)
06195 + 3315635 _ M17 +5159
26m.
(06 +33) 35
Then BG = CB-CG= 1623-1202 423m,
Meta ent height, GM. is given by
i
om=L-06
Find meta-centrie height of the body
ter the Rig o e fay above te wa srs m Ta
ci ot warded y bo
ch deny twa Ve of wae pil
0002931 Volume fe hd nwa}
= 9810 [Volume of cylindrical part in water + Volume seven
‘of curved portion] “
«10 [Fo Depot ein pat war
+ Vote shy cond prion
“ as em EE xas-03000]
Zoxus-0 406
Exeo
“ 18
218-000 120m
Lt Be xe boyy fell par ne cee of span e mle
von,
“hen dep fina pain wate = LA 0467 m
Ener 212 à = 150050.
“The distance of the centre of huoyaney of he whole ody from the top of the eylindrcal part is
sven as
(CH = (Volume of curved portion X CB + Volume of cylndricl par in water CB)
+ (Total volume of water displaced)
06195 + 3315635 _ M17 +5159
26m.
(06 +33) 35
Then BG = CB-CG= 1623-1202 423m,
Meta ent height, GM. is given by
i
om=L-06
Buoyancy and Floatation 143]
where 1= MJ. of the plan ofthe body at water surface about Jo axis
= Geils Sst
= Volume ofthe body in water = 3.9.0?
i= Ex À 4230 1019 425 = 0596. Ans.
CT
» 4.7. CONDITIONS OF rau
BODIES
RIUM OF A FLOATING AND SUB-MERGED
A susmerged ora floating ody is ak o o stale if comes ack tos original position afer a
stigh disturance. The reine postion of the contr of gravity (G) and centre of buoyancy (Dy) ola
body determines the sabi of Sub-merged boy
4.4. Stability of a Sub-merged Body. The positon of ente of gravity and centre of buoy
ane in case of a completly sob-merged body are fixed. Considera balon, whieh is completely sub
mergsd im ar. Lor the lower portion of the balloon contains heavier materi so tha is entre of
{ait lower han is conte of buoyancy as shown Im Tig. 412 a) Lot the weight af the balloon is
The weight Wis acting throuph G, vertical ia the dowaward direction, while the buoyant force Fr
ls acting vertical up, through B. For th equlbium of te balloon W =F, Ihe ballon is given an
angular displacement In the clockwise direction as shown in Fig. 4.12 (a, then W and F const a
couple acting Inthe anti-clockwise direction and tags the balloon inthe oiinal poston. Thus the
balloon nthe poston, showa by Fig. 4.12 (a) si stable ego.
73 ee) ©)
a o ©
STAPLEEQUILGRIUM UNSTARLEFAULIBRUM NEUTRAL EQUILIBRIUM.
ig 42 Stabiltesofsubmerged bodies.
(a) Stable Equilibrium. When W = Fy and point B is above Gs the body is sid 10 be in stable
quia
(6) Unstable Equilibrium, 1 W = F but the cen of buoyancy (B) is below cen of gravity (6),
¡ds body Is in unstable eqllibrium as shown in Fig 4.12 (8). A slight displacement othe body, nthe
clockwise ditcetion, gives the couple due o Wand Fy als 3 the clockwise direction, Thos the body
‘does not retum 0 original positon nd hence the Mody isin Unstable elie.
(©) Neutral Equilibrium. If Fy = Wand Band Gare atthe same point aho im Fig. 4.12) the
body id 10 in neural ogiihrium.
4.1.2. Stability of Floating Body. The stability ofa Mostin body determined from the post
tio of Meta-entre (M) In ane of floating body the Weight ofthe body is equal othe wei of Id
displaced,
where 1= MJ. of the plan ofthe body at water surface about Jo axis
= Geils Sst
= Volume ofthe body in water = 3.9.0?
i= Ex À 4230 1019 425 = 0596. Ans.
CT
» 4.7. CONDITIONS OF rau
BODIES
RIUM OF A FLOATING AND SUB-MERGED
A susmerged ora floating ody is ak o o stale if comes ack tos original position afer a
stigh disturance. The reine postion of the contr of gravity (G) and centre of buoyancy (Dy) ola
body determines the sabi of Sub-merged boy
4.4. Stability of a Sub-merged Body. The positon of ente of gravity and centre of buoy
ane in case of a completly sob-merged body are fixed. Considera balon, whieh is completely sub
mergsd im ar. Lor the lower portion of the balloon contains heavier materi so tha is entre of
{ait lower han is conte of buoyancy as shown Im Tig. 412 a) Lot the weight af the balloon is
The weight Wis acting throuph G, vertical ia the dowaward direction, while the buoyant force Fr
ls acting vertical up, through B. For th equlbium of te balloon W =F, Ihe ballon is given an
angular displacement In the clockwise direction as shown in Fig. 4.12 (a, then W and F const a
couple acting Inthe anti-clockwise direction and tags the balloon inthe oiinal poston. Thus the
balloon nthe poston, showa by Fig. 4.12 (a) si stable ego.
73 ee) ©)
a o ©
STAPLEEQUILGRIUM UNSTARLEFAULIBRUM NEUTRAL EQUILIBRIUM.
ig 42 Stabiltesofsubmerged bodies.
(a) Stable Equilibrium. When W = Fy and point B is above Gs the body is sid 10 be in stable
quia
(6) Unstable Equilibrium, 1 W = F but the cen of buoyancy (B) is below cen of gravity (6),
¡ds body Is in unstable eqllibrium as shown in Fig 4.12 (8). A slight displacement othe body, nthe
clockwise ditcetion, gives the couple due o Wand Fy als 3 the clockwise direction, Thos the body
‘does not retum 0 original positon nd hence the Mody isin Unstable elie.
(©) Neutral Equilibrium. If Fy = Wand Band Gare atthe same point aho im Fig. 4.12) the
body id 10 in neural ogiihrium.
4.1.2. Stability of Floating Body. The stability ofa Mostin body determined from the post
tio of Meta-entre (M) In ane of floating body the Weight ofthe body is equal othe wei of Id
displaced,
144 Fluid Mechanics
(a) Stable Equilibrium. th pont fi shove G, te Monti Body wil be in stable equilibrium as
sown in Fig. 413 (a). If slight angular displacement is given othe Heating body in the clockwise
Action, the centre of buoyancy shits from B to By such that the verca ie through 3, cus at
‘Thea the buoyant free Fy tough B, and weight through G const couple aca Inthe anti
clockwise direction and thus rinpi the long body inthe original poston,
N x Source wy
1
y @ r
a
(0 Sale lib Ais above G (0) Unie equim M bow 6
Fig 415 Stability of ating bodies
(0) Unstable Equilibrium. the point Ms helow G, th lating body will hin unstable equ
sium a shown in Fig. 413 (The disturbing couples acting in the clockwise direction. The couple
due to buoyant force Fy and W is also acting in the clockwise direction and thus overtuming the
ostia bod.
(6) Neutral Equilibrium. If the pont A the cen of paviy ofthe body, the Hosting body will
be in neural equilibrium,
Problem 4.12. A solid cylinder of diameter 40m has a eight 0.0 m. Find the meta centric height
ofthe epinde ifthe specific gravity ofthe materia of elder = 06 and itis floating in water with its
taxis vertical State whether the equilibrium is stable or unstable
Solution, Given D=4m ge a
Heit heim
Spat =06 N
Dept o clinder in water 2 Sp. ge 3
06 440 = 24m
25 Distance of cent of huoyancy (B) from A
24
2
or an 24 1m
Distance of centre of gravity (6) trom À
a
or 46
B6=AG-AB=20-12=08m
[Now the meta-centi height GM is given by
1
CAE
(a) Stable Equilibrium. th pont fi shove G, te Monti Body wil be in stable equilibrium as
sown in Fig. 413 (a). If slight angular displacement is given othe Heating body in the clockwise
Action, the centre of buoyancy shits from B to By such that the verca ie through 3, cus at
‘Thea the buoyant free Fy tough B, and weight through G const couple aca Inthe anti
clockwise direction and thus rinpi the long body inthe original poston,
N x Source wy
1
y @ r
a
(0 Sale lib Ais above G (0) Unie equim M bow 6
Fig 415 Stability of ating bodies
(0) Unstable Equilibrium. the point Ms helow G, th lating body will hin unstable equ
sium a shown in Fig. 413 (The disturbing couples acting in the clockwise direction. The couple
due to buoyant force Fy and W is also acting in the clockwise direction and thus overtuming the
ostia bod.
(6) Neutral Equilibrium. If the pont A the cen of paviy ofthe body, the Hosting body will
be in neural equilibrium,
Problem 4.12. A solid cylinder of diameter 40m has a eight 0.0 m. Find the meta centric height
ofthe epinde ifthe specific gravity ofthe materia of elder = 06 and itis floating in water with its
taxis vertical State whether the equilibrium is stable or unstable
Solution, Given D=4m ge a
Heit heim
Spat =06 N
Dept o clinder in water 2 Sp. ge 3
06 440 = 24m
25 Distance of cent of huoyancy (B) from A
24
2
or an 24 1m
Distance of centre of gravity (6) trom À
a
or 46
B6=AG-AB=20-12=08m
[Now the meta-centi height GM is given by
1
CAE
Buoyancy and Floatation 145
where (= MOL. ofthe plan ofthe body about YY axis
os
Y = Volume of lindern water
Exa
q o
A, xD x Dep of elder wae.
aras 1624724
Peur
1
2 22-06
om
4167-03
A167
03633 m. Ans.
ve sig means thatthe met conte (M is helw the centre of gravity (O)- Thus the cylinder sin
usable equllihrlum. Ans.
Problem 4.13 À solid cylinder of 10 cm diameter and 40 cm Long, consists of to parts made of
¿ler materials. The frst arts the has is Le long and of pee
nity
2. The other part
of te esinder is made ofthe material having specific gravity 06, State, fl can flat vertical in
Solution. Give D=10em
Length, L=40cm
Length of Ist pat, y= 10m
Spats 51260
Density of Ist par. = 6 x 1000 = 6000 kph?
Length of 2nd pan 1,240-10=39.00n
ES 5,206
Density of 2nd pat, py = 0.6 1000 = 600 pin?
‘The cylinder wil oat vertically in water Wits meta conri Height GM is
poshive. T find meta mie height, find the location of centre of gravity
(G) and centre of buoyancy (8) of the combine solid cier. The distance
‘ofthe centre of gravity ofthe solid cylinder from Ais iven as
AG = Weight of Ist par x Distance of C.G, of Ist par from A)
"(Weight of 2nd par of eyinder
‘Distance of CG, of Za par from AD
[Weight o Is par + weight of 2ad pat}
soc
uw
+
HS
(50° x1ox60x03) «(07 x390x06x(10%392)
(Fo xt0xe0+ pt x0x06)
a i )
10X60x05+ 390% 6x(205)
10x60 + 90x06
3044797 | 4827
nthe Numerator and Denon = SI» A827
= 1642
where (= MOL. ofthe plan ofthe body about YY axis
os
Y = Volume of lindern water
Exa
q o
A, xD x Dep of elder wae.
aras 1624724
Peur
1
2 22-06
om
4167-03
A167
03633 m. Ans.
ve sig means thatthe met conte (M is helw the centre of gravity (O)- Thus the cylinder sin
usable equllihrlum. Ans.
Problem 4.13 À solid cylinder of 10 cm diameter and 40 cm Long, consists of to parts made of
¿ler materials. The frst arts the has is Le long and of pee
nity
2. The other part
of te esinder is made ofthe material having specific gravity 06, State, fl can flat vertical in
Solution. Give D=10em
Length, L=40cm
Length of Ist pat, y= 10m
Spats 51260
Density of Ist par. = 6 x 1000 = 6000 kph?
Length of 2nd pan 1,240-10=39.00n
ES 5,206
Density of 2nd pat, py = 0.6 1000 = 600 pin?
‘The cylinder wil oat vertically in water Wits meta conri Height GM is
poshive. T find meta mie height, find the location of centre of gravity
(G) and centre of buoyancy (8) of the combine solid cier. The distance
‘ofthe centre of gravity ofthe solid cylinder from Ais iven as
AG = Weight of Ist par x Distance of C.G, of Ist par from A)
"(Weight of 2nd par of eyinder
‘Distance of CG, of Za par from AD
[Weight o Is par + weight of 2ad pat}
soc
uw
+
HS
(50° x1ox60x03) «(07 x390x06x(10%392)
(Fo xt0xe0+ pt x0x06)
a i )
10X60x05+ 390% 6x(205)
10x60 + 90x06
3044797 | 4827
nthe Numerator and Denon = SI» A827
= 1642
146 Fluid Mechanics
To find the centre of buoyancy ofthe combined two pars or ofthe cide, determine the depth
‘of immersion of the eylnde. Let the depth of immersion ofthe cllader ish. Theo
‘Weight ofthe eylinder= Weight of water displaced
Fcc? 22 6009.81 Ec O x 6000981 = ©
APR 60009381 à © (ay? A x00 981 = €
7 x 1000 x9.81
Fe hisioemt
1000 9.81
PEL uroughout we gt
390x064 10x60=h of he 2344 60= 294
+: The distance ofthe centre ofthe buoyaney Bo the cylinder from A is
EN
2
cx ont Ei
AB= Hn ur
BG = AG - AB = 1642 - 1470.
Meta senti height G is given by
1
= La
ous £16
Where 1= M.OJ.of plan ofthe Body about YY ais
yt
4
Y Volume of oylinder in water
0) em*
EE opa
z & 0? «2.
Eee
er
a GM= 0212 - 172 =~ 1.508 em
‘As GM i ve, it means thatthe Meta-centee M is below the centre of gravity (G). Thus 0
«lider in unstable eqiibiu and sot cant hat vertically in wate. Ans.
Problem 4.14 A rectangular pontoon 10.9 long, 7 m broad und 2.5 m deep weighs 686.7 AN. I
aries on lis upper deck um empty bole of 50m diameter weighing 588.6 AN. The centre of gravity
ofthe boier andthe pontoon are at thelr respective centre along a vertca tne. Find the metacentie
eight Weight density of tea water 110.108 Nm
100
PAPAS a a
oan
Solution. iver: Diana tn = 07x25 5
Welt of pom M = 67 WY AÑ
Din of i, D=som le
Worms M Se
ones 100 vin of
e ki :
a =
poate Lt yan ec pen on a Dors
ol respectively. Then
To find the centre of buoyancy ofthe combined two pars or ofthe cide, determine the depth
‘of immersion of the eylnde. Let the depth of immersion ofthe cllader ish. Theo
‘Weight ofthe eylinder= Weight of water displaced
Fcc? 22 6009.81 Ec O x 6000981 = ©
APR 60009381 à © (ay? A x00 981 = €
7 x 1000 x9.81
Fe hisioemt
1000 9.81
PEL uroughout we gt
390x064 10x60=h of he 2344 60= 294
+: The distance ofthe centre ofthe buoyaney Bo the cylinder from A is
EN
2
cx ont Ei
AB= Hn ur
BG = AG - AB = 1642 - 1470.
Meta senti height G is given by
1
= La
ous £16
Where 1= M.OJ.of plan ofthe Body about YY ais
yt
4
Y Volume of oylinder in water
0) em*
EE opa
z & 0? «2.
Eee
er
a GM= 0212 - 172 =~ 1.508 em
‘As GM i ve, it means thatthe Meta-centee M is below the centre of gravity (G). Thus 0
«lider in unstable eqiibiu and sot cant hat vertically in wate. Ans.
Problem 4.14 A rectangular pontoon 10.9 long, 7 m broad und 2.5 m deep weighs 686.7 AN. I
aries on lis upper deck um empty bole of 50m diameter weighing 588.6 AN. The centre of gravity
ofthe boier andthe pontoon are at thelr respective centre along a vertca tne. Find the metacentie
eight Weight density of tea water 110.108 Nm
100
PAPAS a a
oan
Solution. iver: Diana tn = 07x25 5
Welt of pom M = 67 WY AÑ
Din of i, D=som le
Worms M Se
ones 100 vin of
e ki :
a =
poate Lt yan ec pen on a Dors
ol respectively. Then
Buoyancy and Floatation 147]
‘he sane of common con of gravity fom Ais given as
MAG. + WA
age MAG AG. |
al we
TIAS 586250 _ 54
(67380
Lan riot dp € imc. To
“Total weight of pontoon and oler = Weigh of sa water displaced mA
or (686.7 + 884) = x Volume of ne pot m mate PE ads
= 10.104 x L x 0 Depth of immersion bte
# 12783 = 10104 x 107 x
12753.
23 1208
ET
The distance of the common cents of buoyancy B rom As
PEA
0785 m1» 2.078 m
E
Met centre high is given by G = LG
igh is gion by GM =
where 1= MO of the plan ofthe body a the water level along Y-Y
1
a 10497 as
x 10.0 79 A yy
poa 2
Y = Volume of the body in water
SL Xb = 1007 «1857
1 Wax a9
YT Tao tox 71897" 1241857
28m
A GM E RG = 2.198 - 2078 = 0.12.
Met centre height ofboth the pontoon and hole = 0.12 m. An
Problem 4.15. A wooden cylinder ofp. gr = 06 and circular in cross section require to float
in oil (op. 47. = 090) Find the L/D rato forthe esinder to lat with is longitudinal ans vertical
in ol, where Lis the height of eylinder and D is its diameter.
Solution. Given
Dia. o eylinder =p
eight of eylinder L
Sp. ge of cylinder,
‘he sane of common con of gravity fom Ais given as
MAG. + WA
age MAG AG. |
al we
TIAS 586250 _ 54
(67380
Lan riot dp € imc. To
“Total weight of pontoon and oler = Weigh of sa water displaced mA
or (686.7 + 884) = x Volume of ne pot m mate PE ads
= 10.104 x L x 0 Depth of immersion bte
# 12783 = 10104 x 107 x
12753.
23 1208
ET
The distance of the common cents of buoyancy B rom As
PEA
0785 m1» 2.078 m
E
Met centre high is given by G = LG
igh is gion by GM =
where 1= MO of the plan ofthe body a the water level along Y-Y
1
a 10497 as
x 10.0 79 A yy
poa 2
Y = Volume of the body in water
SL Xb = 1007 «1857
1 Wax a9
YT Tao tox 71897" 1241857
28m
A GM E RG = 2.198 - 2078 = 0.12.
Met centre height ofboth the pontoon and hole = 0.12 m. An
Problem 4.15. A wooden cylinder ofp. gr = 06 and circular in cross section require to float
in oil (op. 47. = 090) Find the L/D rato forthe esinder to lat with is longitudinal ans vertical
in ol, where Lis the height of eylinder and D is its diameter.
Solution. Given
Dia. o eylinder =p
eight of eylinder L
Sp. ge of cylinder,
148 Fluid Mechanics
Sp. ge or ail 5209
Let in dep of cylinder immersed in oil = À
Forth principle of buoyaney
Weight of esinder = wt of ol displaced
E
Eph x06 x om AI À D 09 4 100 1 a
osx „2 a
w 3" —
1 Den
‘The dance ocn of gravy 6 tom A. AG = À
“The dance of cms of buoyancy Brom A,
nn
21/27
a
4 c= ac ane L
a
‘The mt em sigh GM e hen y
owe! no
y
where = © D* and Y = Volume of eylinder in oil = E D? x à
me ind 5 ox cylinder 77 h
A E 2
PRÉC TOP 12 ae En
leer ba
wee
. PREM
1 €
For sable cquibrium, GM shouldbe +ve or
" LD < 54, Ans.
Problem 4.16. Show sha a cylindrical buoy of I m diameter and 2.0 m height weighing 7.848 AN
vil not float vertically in sea water of density 1030 Kg. Find the force necessary in a vertical
‘chain attached atthe centre af base ofthe buoy thot wil keep i vertical
Sp. ge or ail 5209
Let in dep of cylinder immersed in oil = À
Forth principle of buoyaney
Weight of esinder = wt of ol displaced
E
Eph x06 x om AI À D 09 4 100 1 a
osx „2 a
w 3" —
1 Den
‘The dance ocn of gravy 6 tom A. AG = À
“The dance of cms of buoyancy Brom A,
nn
21/27
a
4 c= ac ane L
a
‘The mt em sigh GM e hen y
owe! no
y
where = © D* and Y = Volume of eylinder in oil = E D? x à
me ind 5 ox cylinder 77 h
A E 2
PRÉC TOP 12 ae En
leer ba
wee
. PREM
1 €
For sable cquibrium, GM shouldbe +ve or
" LD < 54, Ans.
Problem 4.16. Show sha a cylindrical buoy of I m diameter and 2.0 m height weighing 7.848 AN
vil not float vertically in sea water of density 1030 Kg. Find the force necessary in a vertical
‘chain attached atthe centre af base ofthe buoy thot wil keep i vertical
Buoyancy and Floatation 149)
Solution. Given; Dia, of buoy, D = Im
cie H=20m
Weight, W= 7.818 kN i”
= 7.848 1000 = 7848 N
Density = 1030 kp?
(© Show the cylinder will ot oat vertically.
(i) Find he for in he ean.
Part . The cplade wil at float if meta-centic height is ve. ei
Let th dep of immersion be A la
‘Then for equilibeum, Weight of eyinder
Weight of water displaced
=Demiy Valımeofeplderin war [0
2 7818 1080x9.81x% Dh Fig. 409
1010437 Wyk
110432
42508
a = LTS ago.
To1043 xx
The distance of entre of buoyancy B fom A
» 0989
an 2 098 „uam
Ant me fa of tam AG 2 10m
2 NG =AG-AB
0-494 = 506m,
Now monte ge Gen by cite
whee t= pe cyt
and Y= Volume of eylinderin water =
Ext Ext
1.6" 0
YO Epix o
4 4
o
Lx 0068
16 9 * Toxo
a GM 063 = 506 = = 0.449 m. Ans.
[AS the meta.cenrc height is ae, the point M les blow G and Hones the cylinder wil ho im
‘unstable equilibrium and hence eyinder will ot leat vertically.
Solution. Given; Dia, of buoy, D = Im
cie H=20m
Weight, W= 7.818 kN i”
= 7.848 1000 = 7848 N
Density = 1030 kp?
(© Show the cylinder will ot oat vertically.
(i) Find he for in he ean.
Part . The cplade wil at float if meta-centic height is ve. ei
Let th dep of immersion be A la
‘Then for equilibeum, Weight of eyinder
Weight of water displaced
=Demiy Valımeofeplderin war [0
2 7818 1080x9.81x% Dh Fig. 409
1010437 Wyk
110432
42508
a = LTS ago.
To1043 xx
The distance of entre of buoyancy B fom A
» 0989
an 2 098 „uam
Ant me fa of tam AG 2 10m
2 NG =AG-AB
0-494 = 506m,
Now monte ge Gen by cite
whee t= pe cyt
and Y= Volume of eylinderin water =
Ext Ext
1.6" 0
YO Epix o
4 4
o
Lx 0068
16 9 * Toxo
a GM 063 = 506 = = 0.449 m. Ans.
[AS the meta.cenrc height is ae, the point M les blow G and Hones the cylinder wil ho im
‘unstable equilibrium and hence eyinder will ot leat vertically.
150 Fluid Mechanics
Part Let the force applied in a verte chain attacked te entre
‘of the base ofthe buoy is 7 0 keep the buoy vertical.
[Now find the combined positon of centre of prvi (6) and conte ol
buoyancy (8). For the combined centre of buoyancy. et
{= depth of immersion when te force T ls applied. Then of Y
“Total dowanand force = Weight of water displaced 5
ax CA x= Deny of water» Ve of einer in water |
= 1080 x 981 x À OPH | where = depth of tuner | ry
„ser 784867 MT, Fig 4.20
ma 70
4 a
A ihe
70330 Lise
‘The combined conte of gravity (6 due 10 weight of eylinder and due 1 tension 7 in the chain
rom Ais
AG’ = [Wt of cylinder x Distance oF CG oF cylinder from A
+ Tx Distance of CG. of 7 from AJ + [Weight of eylinde + TI
u.
mar
1 (184821)
(r8t8x<3 +70) us sn
e sen) 158718
eme ns en = =e
ee = Ext Exit Am
Rune Ex OD
NR va
au. 2359 | 78 (78487)
DUT ara
For table equilibrium GM should be postive
on 6m> 0
29359, Re Lea
doser) [men 158718 |?
Part Let the force applied in a verte chain attacked te entre
‘of the base ofthe buoy is 7 0 keep the buoy vertical.
[Now find the combined positon of centre of prvi (6) and conte ol
buoyancy (8). For the combined centre of buoyancy. et
{= depth of immersion when te force T ls applied. Then of Y
“Total dowanand force = Weight of water displaced 5
ax CA x= Deny of water» Ve of einer in water |
= 1080 x 981 x À OPH | where = depth of tuner | ry
„ser 784867 MT, Fig 4.20
ma 70
4 a
A ihe
70330 Lise
‘The combined conte of gravity (6 due 10 weight of eylinder and due 1 tension 7 in the chain
rom Ais
AG’ = [Wt of cylinder x Distance oF CG oF cylinder from A
+ Tx Distance of CG. of 7 from AJ + [Weight of eylinde + TI
u.
mar
1 (184821)
(r8t8x<3 +70) us sn
e sen) 158718
eme ns en = =e
ee = Ext Exit Am
Rune Ex OD
NR va
au. 2359 | 78 (78487)
DUT ara
For table equilibrium GM should be postive
on 6m> 0
29359, Re Lea
doser) [men 158718 |?
‘Buoyancy and Floatation 151
EEE
io 7) m)" son
1935916. 708 | (7818 7)
ion STS
sn OST),
ion)" sss
388-7)
158718
ausm >
« [re
>1
> 0302.3)"
1848 > 108023
T> 108023 - 7848
> 29543 N. Ans.
“The once in the chain must at lest 2984.3 N So iba he eylindeca buoy can be Kept in
vertical position, Ans
Problem 4.17 A solid cone lots in water with its upex donnsards, Determine the last pex angle
of cone for sabe equim. The specific gravity ofthe material of the cone is given 08.
Solution. Given
Sp. 36 of cone A
Density of cono, p= 0.84 1000 = 800 kp“
La Dia, of te one
= Dia. o one at water level
20 Apex ange of cone
Height of one
‘he Depth of cone in wa
Centre of gravity ofthe come
= Centre of buoyancy ofthe cone
For the cone the distance o cet of gravity fom the pex Als
“sight of cone = 24
alo 2 depth of come in water = 3
Volume of water displaced = À RP x 4
Volume of com Pe
+ Weight of cono 800 x x a 2m CHT
Now ftom MEF, tan
Simian,
EEE
io 7) m)" son
1935916. 708 | (7818 7)
ion STS
sn OST),
ion)" sss
388-7)
158718
ausm >
« [re
>1
> 0302.3)"
1848 > 108023
T> 108023 - 7848
> 29543 N. Ans.
“The once in the chain must at lest 2984.3 N So iba he eylindeca buoy can be Kept in
vertical position, Ans
Problem 4.17 A solid cone lots in water with its upex donnsards, Determine the last pex angle
of cone for sabe equim. The specific gravity ofthe material of the cone is given 08.
Solution. Given
Sp. 36 of cone A
Density of cono, p= 0.84 1000 = 800 kp“
La Dia, of te one
= Dia. o one at water level
20 Apex ange of cone
Height of one
‘he Depth of cone in wa
Centre of gravity ofthe come
= Centre of buoyancy ofthe cone
For the cone the distance o cet of gravity fom the pex Als
“sight of cone = 24
alo 2 depth of come in water = 3
Volume of water displaced = À RP x 4
Volume of com Pe
+ Weight of cono 800 x x a 2m CHT
Now ftom MEF, tan
Simian,
152 Fluid Mechanics
2 Weight of cone 2 800 4 rx (an BH
Weight of water dspace
000 8 À por
100 x gx x tan 0
30
1000 2x § an 8) x
For uilibiam
Weight of one Weight of water displaced
800 x x RH wo" xn? xt?
For stable equilibrium, Metacenric height GM should be positive, But GM is given by
1
owe tag
y
where P= Má
cone at watsetine Ed‘
“
bs ranune)
aod
GM = fun (42m)
For stable equilibrium GM should be positive or
=U) >0 or ean? (= >0
or Ha 0> (1 or AD
hn’
= DR
sa 198)” um
+a
wos ado o
ot or ones = os
cos 0> 0.9635
01530 or 20238
Apex angle (20) shouldbe atleast 31° Ans.
2 Weight of cone 2 800 4 rx (an BH
Weight of water dspace
000 8 À por
100 x gx x tan 0
30
1000 2x § an 8) x
For uilibiam
Weight of one Weight of water displaced
800 x x RH wo" xn? xt?
For stable equilibrium, Metacenric height GM should be positive, But GM is given by
1
owe tag
y
where P= Má
cone at watsetine Ed‘
“
bs ranune)
aod
GM = fun (42m)
For stable equilibrium GM should be positive or
=U) >0 or ean? (= >0
or Ha 0> (1 or AD
hn’
= DR
sa 198)” um
+a
wos ado o
ot or ones = os
cos 0> 0.9635
01530 or 20238
Apex angle (20) shouldbe atleast 31° Ans.
Buoyancy and Floatation 153]
Problem 4.18 A cone of specie gravity S, is lating in water wit ts apex downwards thas a
ont and et ih Show or sb trio othe sone < IP |
Bolton, Gen aa
tes nue
Let G= Centre of gravity of cone
= Centre of buoyaney
28 Apor angle
Apex of the cone
Depth of immersion
Dia. of come at water surface
3
Tea 10-4
hz +.
Also weight of cone =Weight of water displace
Ph
100 Sex RE 100 o SE
Han 8,72 hand
SX Etam x
(rian OF
SP tan? OH SH
Dar
(HN = SH 0
Distance, DG=AG-AB
Es a
Fost 5
TN
Sms
Jus o
AO. nern of plan of body a water surface
en
E
Aso
ram LE a pa
Volume of cone in water
Problem 4.18 A cone of specie gravity S, is lating in water wit ts apex downwards thas a
ont and et ih Show or sb trio othe sone < IP |
Bolton, Gen aa
tes nue
Let G= Centre of gravity of cone
= Centre of buoyaney
28 Apor angle
Apex of the cone
Depth of immersion
Dia. of come at water surface
3
Tea 10-4
hz +.
Also weight of cone =Weight of water displace
Ph
100 Sex RE 100 o SE
Han 8,72 hand
SX Etam x
(rian OF
SP tan? OH SH
Dar
(HN = SH 0
Distance, DG=AG-AB
Es a
Fost 5
TN
Sms
Jus o
AO. nern of plan of body a water surface
en
E
Aso
ram LE a pa
Volume of cone in water
154 Fluid Mechanics
Lo a wt
veo TA
Un
TT
Now Mec gt Git i gen
sod 16.H.5
Lui ve o sl quo or > 0
ae
= ater a 5 >0
CO »
ans 6
Tens © 4 ( ” a
‘ko we know Re un Ganar han 6
EE
Du „D gon
‘Swbsttting the valve of d'in equation (3) we pet
ETES
EXPERIMENTAL METHOD OF DETERMINATION OF META-CENTRIC
HEIGHT
‘The meta-centic height o a Noaing vessel can be determined. provided we know the centre of
gravity ofthe floating vs. Let wi known weight placed over the centre of the veses shown
Tig. 423 (a) and the yes is oating
(a) at boty a Tad boar
ig. 423 Metscentic height
Lo a wt
veo TA
Un
TT
Now Mec gt Git i gen
sod 16.H.5
Lui ve o sl quo or > 0
ae
= ater a 5 >0
CO »
ans 6
Tens © 4 ( ” a
‘ko we know Re un Ganar han 6
EE
Du „D gon
‘Swbsttting the valve of d'in equation (3) we pet
ETES
EXPERIMENTAL METHOD OF DETERMINATION OF META-CENTRIC
HEIGHT
‘The meta-centic height o a Noaing vessel can be determined. provided we know the centre of
gravity ofthe floating vs. Let wi known weight placed over the centre of the veses shown
Tig. 423 (a) and the yes is oating
(a) at boty a Tad boar
ig. 423 Metscentic height
Buoyancy and Floatation 155
La
‘Weight of vessel including #4
= Centre of gravity ofthe vessel
B = Centr of buoyancy of the vessel
“The weight wis moved actos he vee towards ight trough a distance xs show In Fig. 4.23)
‘Tne ve wll he td. The angle of hes} @ Is measured by means of a plumbin and a preeactor
tached on the vesel The new centre of gravity ofthe vessel wil shit to G, asthe weight hasbeen
‘moved towards the right. Also the conte of buoyancy wil ehange to B, asthe vessel has ted. Under
‘uli, the moment caused bythe movement ofthe load y Hough distance x must o equal 10
‘he moment caused bythe shit o the conte of gravity fom Gto Gy. Thus
‘The moment due to change of G = GG, x W= Wx Gi tan ©
‘The moment das to movement OF m mi
5 x= WGM an 0
Wind
Problem 4:19 A ship 70 m long and 10 m broad has a displacement of 19620 AN. A weight of
343.5 AN is moved across the deck trough a distance of 6 m. The ship ts ted through 6°. The
‘moment of inertia of the ship at waterline about is fore and aft ass is 75% of M.O.L ofthe
Cicumscribing rectangle. The centre of buoyancy i 2.28 m below woterstin. Find the metacentric
‘eight and postion of centre of gravity of ship, Specific weight of seawater is 10104 N/m
Solution. Given
Hence om: as
Length of sip, m
Brcadth of sip, 10m
Displacement, W= 19620 kN
Angle of el one
MOL. of ship at water = 156 of M.O11 of ireumerbing rectangle
we for sa-water 10104 Nin? = 10.108 EN
Movable welt, ey = B35 kN
Distance moved by Wy,
‘Cente of buoyancy =225 m below water surface
Find (@) Meta comi height, GM
i) Position of comas of gravity, G.
(0 Meta-cenei height, GM is glen by equation (45)
me, MABSKN X60.
Al om=
Wan "19620 kN an 6°
ESE A
19620 KN x Jos
(6 Position of Centre of Gravity, G
1
owe tag
y
where T= MOL, of the sip at water in about 97
La
‘Weight of vessel including #4
= Centre of gravity ofthe vessel
B = Centr of buoyancy of the vessel
“The weight wis moved actos he vee towards ight trough a distance xs show In Fig. 4.23)
‘Tne ve wll he td. The angle of hes} @ Is measured by means of a plumbin and a preeactor
tached on the vesel The new centre of gravity ofthe vessel wil shit to G, asthe weight hasbeen
‘moved towards the right. Also the conte of buoyancy wil ehange to B, asthe vessel has ted. Under
‘uli, the moment caused bythe movement ofthe load y Hough distance x must o equal 10
‘he moment caused bythe shit o the conte of gravity fom Gto Gy. Thus
‘The moment due to change of G = GG, x W= Wx Gi tan ©
‘The moment das to movement OF m mi
5 x= WGM an 0
Wind
Problem 4:19 A ship 70 m long and 10 m broad has a displacement of 19620 AN. A weight of
343.5 AN is moved across the deck trough a distance of 6 m. The ship ts ted through 6°. The
‘moment of inertia of the ship at waterline about is fore and aft ass is 75% of M.O.L ofthe
Cicumscribing rectangle. The centre of buoyancy i 2.28 m below woterstin. Find the metacentric
‘eight and postion of centre of gravity of ship, Specific weight of seawater is 10104 N/m
Solution. Given
Hence om: as
Length of sip, m
Brcadth of sip, 10m
Displacement, W= 19620 kN
Angle of el one
MOL. of ship at water = 156 of M.O11 of ireumerbing rectangle
we for sa-water 10104 Nin? = 10.108 EN
Movable welt, ey = B35 kN
Distance moved by Wy,
‘Cente of buoyancy =225 m below water surface
Find (@) Meta comi height, GM
i) Position of comas of gravity, G.
(0 Meta-cenei height, GM is glen by equation (45)
me, MABSKN X60.
Al om=
Wan "19620 kN an 6°
ESE A
19620 KN x Jos
(6 Position of Centre of Gravity, G
1
owe tag
y
where T= MOL, of the sip at water in about 97
Dee ig 425
se of 1 x 70 x 10°
2
sx L x70 410 = 4375 00
2
Weiphtofsiip 19620
and Y= Volume of spin water = —_etehtof ship _ 1941.74 m?
‘Weight density of water 10104
Le sn
Y" paa
GM = 2253 - BG or 999 = 2.288 - BG
BG = 2.253 - 999 = 1.288 m
From Fig, 425, iis clear thatthe distance of G from tee surface ofthe water = distance of B
fom water surface - HC
25 - 1.254 = 0996 m. Ans.
Problem 4.20. A pontoon of 13696 AN displacement i floating a water. A weight of 24525 KN is
moved through a distance of m across the deck of pontoon, which úl he pontoon through an angle
1 Find mer.cenri height of the pontoon
Solution. Given
Weight of pontoon = Displacement
or 5696 N
Movable weight, y= 28525 4N
Distance moved by weight wy,
Angle of hc,
‘The meter height, GA is piven by equation (4.5
E mx 24525 kN x8
m >
15656 400699 © !*
> 49 OSCILLATION (ROLLING) OF A FLOATING BODY
Considera Mating body, whichis ited through an angle by an ovetuniny couple as shown in
ig. 4.26, Lette overturning couple is suddeny removed. The body will star oselating, Thus. the
se of 1 x 70 x 10°
2
sx L x70 410 = 4375 00
2
Weiphtofsiip 19620
and Y= Volume of spin water = —_etehtof ship _ 1941.74 m?
‘Weight density of water 10104
Le sn
Y" paa
GM = 2253 - BG or 999 = 2.288 - BG
BG = 2.253 - 999 = 1.288 m
From Fig, 425, iis clear thatthe distance of G from tee surface ofthe water = distance of B
fom water surface - HC
25 - 1.254 = 0996 m. Ans.
Problem 4.20. A pontoon of 13696 AN displacement i floating a water. A weight of 24525 KN is
moved through a distance of m across the deck of pontoon, which úl he pontoon through an angle
1 Find mer.cenri height of the pontoon
Solution. Given
Weight of pontoon = Displacement
or 5696 N
Movable weight, y= 28525 4N
Distance moved by weight wy,
Angle of hc,
‘The meter height, GA is piven by equation (4.5
E mx 24525 kN x8
m >
15656 400699 © !*
> 49 OSCILLATION (ROLLING) OF A FLOATING BODY
Considera Mating body, whichis ited through an angle by an ovetuniny couple as shown in
ig. 4.26, Lette overturning couple is suddeny removed. The body will star oselating, Thus. the
‘Buoyancy and Floatation 157]
‘ody wil be ina state of oscillation asf suspended atthe mea-cente Mf Tiss similar to te case of
pendulum. The ony force acting on te body Is due 1 the restoring couple duet the weight W of,
te body force of booyaney Fi
Fig 426
Restoring couple = Wx Distance GA
We GM sin 0 0
“This coupe rie 1 decease the angle
do
a
ve sip as been introduce as the restoring couple ts to decrease the angle 0
Torque d to inertia foment of nera about ¥-¥ Angola acer
nl)
Ca
Angular accleation of th Body, u ==
7.
i ie
sen 2 We td; Ra pd tom
in ba ze
Cal re
equating (D and (i, we get
W x Git sin
For smal ange 0,
Dividing by “> we et x
"The above equation is à diem equation f cond degree, The solution s
= an TE us jen ET a
‘ody wil be ina state of oscillation asf suspended atthe mea-cente Mf Tiss similar to te case of
pendulum. The ony force acting on te body Is due 1 the restoring couple duet the weight W of,
te body force of booyaney Fi
Fig 426
Restoring couple = Wx Distance GA
We GM sin 0 0
“This coupe rie 1 decease the angle
do
a
ve sip as been introduce as the restoring couple ts to decrease the angle 0
Torque d to inertia foment of nera about ¥-¥ Angola acer
nl)
Ca
Angular accleation of th Body, u ==
7.
i ie
sen 2 We td; Ra pd tom
in ba ze
Cal re
equating (D and (i, we get
W x Git sin
For smal ange 0,
Dividing by “> we et x
"The above equation is à diem equation f cond degree, The solution s
= an TE us jen ET a
158 Fluid Mechanics
where C, and Cy as constants of integration,
“The values of Cy and C; are obtained from boundary conditions which are
oar
CE
‘where isthe ime period of one complete scillaion.
Substuing the Ist boundary condition a (i), we get
26x04 Cox 1 sin 82 0,608 = 1}
G
Substituting And boundary conditions in (i, we pet
(comes
sin (HE 7
‘But €; cannot equal o zero and so the other alerte is
Dasn (sin x 2 0]
2 LE “as
ag
Time period o oscillation is given by equation (1.6)
Problom 4,21. Tne east radius of aration of a ship 18m and meta-cenie height 70 em. Cale
{ate ie me period of osllaton ofthe ship
Solution. Gien
Least radis of gyration, Km
Meta.cent eight, GM = 70 em = 0.20 m
‘The tine period of cin s given by equation (4.6.
Ey, [5
M CT CETTE
Problem 4.22 The time period of rolling of à ship of weight 29430 AN in sea water is 10 seconds
The centre of bunyaney ofthe ship 1 15 m below the centre of gravity. Find the radia of gyration of
‘he ship if the momen! of inertia ofthe hip at he water line about fore and aft ats is 1000 m. Take
Specie weigh of sen water as = 10100 Nin
Solution. Given
Time period, T= ose
Distance between cente of buoyancy and cent of gravity, BG = 1.5 m
Moment of laces, 0000 mt
weint, W= 29430 kN
Lette radius of gyration = K
Firs calculate the meta-<entie eight GM, whichis given as
Te
19.18 ses. Ans
29430 x 1000 N
= BM
where C, and Cy as constants of integration,
“The values of Cy and C; are obtained from boundary conditions which are
oar
CE
‘where isthe ime period of one complete scillaion.
Substuing the Ist boundary condition a (i), we get
26x04 Cox 1 sin 82 0,608 = 1}
G
Substituting And boundary conditions in (i, we pet
(comes
sin (HE 7
‘But €; cannot equal o zero and so the other alerte is
Dasn (sin x 2 0]
2 LE “as
ag
Time period o oscillation is given by equation (1.6)
Problom 4,21. Tne east radius of aration of a ship 18m and meta-cenie height 70 em. Cale
{ate ie me period of osllaton ofthe ship
Solution. Gien
Least radis of gyration, Km
Meta.cent eight, GM = 70 em = 0.20 m
‘The tine period of cin s given by equation (4.6.
Ey, [5
M CT CETTE
Problem 4.22 The time period of rolling of à ship of weight 29430 AN in sea water is 10 seconds
The centre of bunyaney ofthe ship 1 15 m below the centre of gravity. Find the radia of gyration of
‘he ship if the momen! of inertia ofthe hip at he water line about fore and aft ats is 1000 m. Take
Specie weigh of sen water as = 10100 Nin
Solution. Given
Time period, T= ose
Distance between cente of buoyancy and cent of gravity, BG = 1.5 m
Moment of laces, 0000 mt
weint, W= 29430 kN
Lette radius of gyration = K
Firs calculate the meta-<entie eight GM, whichis given as
Te
19.18 ses. Ans
29430 x 1000 N
= BM
Buoyancy and Floatation 159]
‘where 1= MO. Inertia
and Volume of water dspace
Weight of ship _ 29430 «1000
29126 m
Sp wei of sea water
m= WM _ 1,5. 3433 -1.5= 1999 m.
Using equation (4.6), we get
HIGHLIGHTS
1. The upward fre exeted by lid on a body when the body ese in the gu own as
baoyany or free of buoyancy
2. The pint trough which force of buoyancy capped tac al centre of bone
3. The point about which hoy are sling when the Body es known as meinte
A The distance between the meta cee and centre of ravi 6 known as mebrcenrke Big
£ The metan eight (GM) I given by GM
whee = Memento lets fe sing body pla) at water ui about he as Y
Y = Volume ofthe body abend in water
RG = Distance Between centre of gray and conte of anya.
6. Condos of eum ofa ating and sbmerped Day are
Erin] Flowing Body | Submerged Body
TO Sue ion ri above @ HITS
i Unsable Equirion Ari deln Riedel G
i Nesta Evie Mand G ence Band Gere
7. The vale of mei ce eeh GM, experimental ren a = GE
wees wy = Movable welt
= Distance though which m i moved
Wea Weight of se ship or Hating body ini
0 Are a Be por Fig By ie he we one
4. Tn aio an cig soni nay uy pu E
hse Rady, i= eer ih
= Time of one complet silo.
‘where 1= MO. Inertia
and Volume of water dspace
Weight of ship _ 29430 «1000
29126 m
Sp wei of sea water
m= WM _ 1,5. 3433 -1.5= 1999 m.
Using equation (4.6), we get
HIGHLIGHTS
1. The upward fre exeted by lid on a body when the body ese in the gu own as
baoyany or free of buoyancy
2. The pint trough which force of buoyancy capped tac al centre of bone
3. The point about which hoy are sling when the Body es known as meinte
A The distance between the meta cee and centre of ravi 6 known as mebrcenrke Big
£ The metan eight (GM) I given by GM
whee = Memento lets fe sing body pla) at water ui about he as Y
Y = Volume ofthe body abend in water
RG = Distance Between centre of gray and conte of anya.
6. Condos of eum ofa ating and sbmerped Day are
Erin] Flowing Body | Submerged Body
TO Sue ion ri above @ HITS
i Unsable Equirion Ari deln Riedel G
i Nesta Evie Mand G ence Band Gere
7. The vale of mei ce eeh GM, experimental ren a = GE
wees wy = Movable welt
= Distance though which m i moved
Wea Weight of se ship or Hating body ini
0 Are a Be por Fig By ie he we one
4. Tn aio an cig soni nay uy pu E
hse Rady, i= eer ih
= Time of one complet silo.
1160 Fluid Mechanics
EXERCISE
(A) THEORETICAL PROBLEMS
1. Dane the tems “suyaney” and “cate of buoys!
2. Explain the terme "meta ene” and meta conti Big
3. Desve an enpresin forthe meten height ola fon body.
4 how ta ie dance Between he ets and on gay I en y a = E
shee = Moment of inert f the plan te losing body a we surface sot longa! ait
Y = Volume of the Body sa merge gu
5. What ae the condi of equim of a fling Body and a amer body ?
ow wil you dtemine te mea on height of a ating body experimental? Explain wit neat
7. Seles he cost tement
(a) The Boyan ove fr a Noting bey pases tg he
@ cent a gray fe body (cena of volume Ub body
(iy meten of he bey o) sens of gay o submerged pt of he By
do) senti te places volume
(©) body submerged ql seul va
(9 mes cent above the cone of gra
sms ene I above he en of boyy
(i ls cent gravity above te cent af uoyaney
(i) Hs cea Muoyaey is above the cen fav
(© noe of es ams ao.
Dassen expression fre ne prod of teen oa Mon body cms of ras grato
and mia cet Hg of he ang bey
9. Define the tens: meta cote, centre of banyan. mew ees sigh ug pressure amd soluto
10. Wa do you understand y e rai gui ? With bl of is quan drive expression
I. Wit sat sett, exp te condi 0 elit for sg ad ab meros bodies
(9 Dynamic cosy nd Kim viscosty Absolute and gage presse (i) Simple and
steal manne (M) Cese of gv and ie of buno)
Dein Unveri Dec. 2002)
(8) NUMERICAL PROBLEMS
A wooden lok of wit 2 m, dept 15m a ent 4m fot orion in wate, Find te wom
‘of water placo nd postin of cece af Boyancy The msc ravi of he wonden ck 107,
[Ans 80052 m rom he bse
EXERCISE
(A) THEORETICAL PROBLEMS
1. Dane the tems “suyaney” and “cate of buoys!
2. Explain the terme "meta ene” and meta conti Big
3. Desve an enpresin forthe meten height ola fon body.
4 how ta ie dance Between he ets and on gay I en y a = E
shee = Moment of inert f the plan te losing body a we surface sot longa! ait
Y = Volume of the Body sa merge gu
5. What ae the condi of equim of a fling Body and a amer body ?
ow wil you dtemine te mea on height of a ating body experimental? Explain wit neat
7. Seles he cost tement
(a) The Boyan ove fr a Noting bey pases tg he
@ cent a gray fe body (cena of volume Ub body
(iy meten of he bey o) sens of gay o submerged pt of he By
do) senti te places volume
(©) body submerged ql seul va
(9 mes cent above the cone of gra
sms ene I above he en of boyy
(i ls cent gravity above te cent af uoyaney
(i) Hs cea Muoyaey is above the cen fav
(© noe of es ams ao.
Dassen expression fre ne prod of teen oa Mon body cms of ras grato
and mia cet Hg of he ang bey
9. Define the tens: meta cote, centre of banyan. mew ees sigh ug pressure amd soluto
10. Wa do you understand y e rai gui ? With bl of is quan drive expression
I. Wit sat sett, exp te condi 0 elit for sg ad ab meros bodies
(9 Dynamic cosy nd Kim viscosty Absolute and gage presse (i) Simple and
steal manne (M) Cese of gv and ie of buno)
Dein Unveri Dec. 2002)
(8) NUMERICAL PROBLEMS
A wooden lok of wit 2 m, dept 15m a ent 4m fot orion in wate, Find te wom
‘of water placo nd postin of cece af Boyancy The msc ravi of he wonden ck 107,
[Ans 80052 m rom he bse
‘Buoyancy and Floatation 161
2
A den lg 10% a ameter ad © m engl sang la ver wate. Fi te dpt of wooden
login waist ws ie sp. gr. ofthe wooden gs 0 1205.03 0
N one weighs 4903 N in lr and 1962 N in wale, Dori the vole of sone and ls specific
a. Ans. 8m oF 3 0° 167)
‘boy of dimensions 20 mx LO x 31 weighs 3924 N in weer. ind is Weigh i ait What Wil be
ls mei ray ? Ham. 2784 N. 10067)
A metal Do ns he race of mercury ofp ar. 13.6 and wate in sch away thal 30% Fis
Sola su meras in mercury nd 70% in war, id te deny othe ml Do.
[am 780 ker
A Ba of diem 05. 0.8m x Om and of pf. 4 I immer wer. tein he est
Forget fe the body Tans 4908 NT
‘rectangular pomo id m lng 3 m wie and 140 igh. The depth immersion of the orton it
1.0m in rw. te cent fra 0.70 m bone the Dtm 0 the pontoon deeming the et
ceri eg. Take the density of ca mer a 1030 Ep. (Ane. 05 mt
¿epi of mern ien6m 7Detemine te meto cemic hipo. [Ane ATISEN03S5 ml
In oc of won of pei pray DR a in ate, Determine te met senti ht of he lok If
fee km mxt Ans. mt
A soli older o meter 30 m has eh of 2 m Find eme cen eight of he eine when
Icing In water wi te ani vena, he sp. ofthe ler 0.7. Ans 0.017 ml
A Body hs ya pp parton 4 m dat and 2 ep The lower prin i cred on,
ic dise volume of 09 ol wae. Te ente f buoyancy ofthe cred poro st tas
12.10 m teow he up of ti ein. The en oF gravy of he whole beds 130 m blow th top
‘ofthe spline, Te total dsplemen of water 4 tonnes, Find e mel cee ght ofthe by
Lans 97m
A slender f mete SO m as à bg of 5.0 m id the mu nt Night ol Older I
tee soci gr fe tal a lides 0 nd I tng in wae sad veel. Si
tice the egitim sabe or uns Ams, 0306 m, Unsahie Equ]
A soi vine 13 em diameter and 0 em Hong consis of puto made of diferent mei, The
fin pat atthe as is 1.20 em long and of pei gr = 5. Thee eso the yin sade
‘ofthe mail having specie ri US. Se an ont very in wale
Ans. GM = 526. UnsublEquiltrion]
rectangular pontoon 80m og, 7 rsd and 30m dep gs SEE AN. I cute on Upper desk
“an pay bo of 401m ae weighing 3924 KN, The cen o grviy oh ler and th pont
Se at he spe ones aan verte ln, Find bemol eii belt. Weg dena of sn
eaters 10108 Ni (ans 0328)
[Roden cylinder os. and car in conection ruiz oat in ip. 08) Find
tee LID rat forthe einer t Host with mp eral whee Le the Height of
inde and D dete Tan (um < 060
bn es motera den 1030 pm. Fhe oe ncesy Ina era in tact he entr the
ae of he oy th wil kcp Yee (Ane 106005
A ai omelets in water itsapex Command, Determine the leas apex angle of one or sae gu
Fg. The sect pri ofthe mateo he one I ine 07. ans. 39°71
Ani rm og and 12 broad has placement of 1062 EN. A weight of 2943 RN is moved aros
tee ack Hough aise of 68m. The sip se tough 5°. The moment of inertia o the shpat
2
A den lg 10% a ameter ad © m engl sang la ver wate. Fi te dpt of wooden
login waist ws ie sp. gr. ofthe wooden gs 0 1205.03 0
N one weighs 4903 N in lr and 1962 N in wale, Dori the vole of sone and ls specific
a. Ans. 8m oF 3 0° 167)
‘boy of dimensions 20 mx LO x 31 weighs 3924 N in weer. ind is Weigh i ait What Wil be
ls mei ray ? Ham. 2784 N. 10067)
A metal Do ns he race of mercury ofp ar. 13.6 and wate in sch away thal 30% Fis
Sola su meras in mercury nd 70% in war, id te deny othe ml Do.
[am 780 ker
A Ba of diem 05. 0.8m x Om and of pf. 4 I immer wer. tein he est
Forget fe the body Tans 4908 NT
‘rectangular pomo id m lng 3 m wie and 140 igh. The depth immersion of the orton it
1.0m in rw. te cent fra 0.70 m bone the Dtm 0 the pontoon deeming the et
ceri eg. Take the density of ca mer a 1030 Ep. (Ane. 05 mt
¿epi of mern ien6m 7Detemine te meto cemic hipo. [Ane ATISEN03S5 ml
In oc of won of pei pray DR a in ate, Determine te met senti ht of he lok If
fee km mxt Ans. mt
A soli older o meter 30 m has eh of 2 m Find eme cen eight of he eine when
Icing In water wi te ani vena, he sp. ofthe ler 0.7. Ans 0.017 ml
A Body hs ya pp parton 4 m dat and 2 ep The lower prin i cred on,
ic dise volume of 09 ol wae. Te ente f buoyancy ofthe cred poro st tas
12.10 m teow he up of ti ein. The en oF gravy of he whole beds 130 m blow th top
‘ofthe spline, Te total dsplemen of water 4 tonnes, Find e mel cee ght ofthe by
Lans 97m
A slender f mete SO m as à bg of 5.0 m id the mu nt Night ol Older I
tee soci gr fe tal a lides 0 nd I tng in wae sad veel. Si
tice the egitim sabe or uns Ams, 0306 m, Unsahie Equ]
A soi vine 13 em diameter and 0 em Hong consis of puto made of diferent mei, The
fin pat atthe as is 1.20 em long and of pei gr = 5. Thee eso the yin sade
‘ofthe mail having specie ri US. Se an ont very in wale
Ans. GM = 526. UnsublEquiltrion]
rectangular pontoon 80m og, 7 rsd and 30m dep gs SEE AN. I cute on Upper desk
“an pay bo of 401m ae weighing 3924 KN, The cen o grviy oh ler and th pont
Se at he spe ones aan verte ln, Find bemol eii belt. Weg dena of sn
eaters 10108 Ni (ans 0328)
[Roden cylinder os. and car in conection ruiz oat in ip. 08) Find
tee LID rat forthe einer t Host with mp eral whee Le the Height of
inde and D dete Tan (um < 060
bn es motera den 1030 pm. Fhe oe ncesy Ina era in tact he entr the
ae of he oy th wil kcp Yee (Ane 106005
A ai omelets in water itsapex Command, Determine the leas apex angle of one or sae gu
Fg. The sect pri ofthe mateo he one I ine 07. ans. 39°71
Ani rm og and 12 broad has placement of 1062 EN. A weight of 2943 RN is moved aros
tee ack Hough aise of 68m. The sip se tough 5°. The moment of inertia o the shpat
162 Fluid Mechanics
ser ne ov is ne and als 17% of momen oferta he housing sg. Te en
buoyancy 1 2.75 m blow wir is. Find mec cg nd poso of cen of pany of
‘Sp. Take specie weight ofsc wer = 10108 N [Ams 1.114303 m Blow war src
19. A poston of 1500 tonces placements feu In wat. A eig of 20 ones is moved dough
à since of 6m across the deck of poro, which ls he pontoon uh an angle of 5°. Find
eu conce heit of e porto Va 0945
20, Find le me period loli o sols llar ler ol aus 2. m and SO ng. Te pei
rai of the ler $509 and ing in water wits sais vera Tate 038 2]
ser ne ov is ne and als 17% of momen oferta he housing sg. Te en
buoyancy 1 2.75 m blow wir is. Find mec cg nd poso of cen of pany of
‘Sp. Take specie weight ofsc wer = 10108 N [Ams 1.114303 m Blow war src
19. A poston of 1500 tonces placements feu In wat. A eig of 20 ones is moved dough
à since of 6m across the deck of poro, which ls he pontoon uh an angle of 5°. Find
eu conce heit of e porto Va 0945
20, Find le me period loli o sols llar ler ol aus 2. m and SO ng. Te pei
rai of the ler $509 and ing in water wits sais vera Tate 038 2]
KINEMATICS OF FLOWS
ANDIDEAL FLOW
A. KINEMATICS OF FLOW
» 5.1 INTRODUCTION
Kinemaes ls defined as that branch of science whlch deals with motion of particles without
considering the forces causing the mouion. The velochy at any pont ia flow Held at any tne is
‘tied in this brane of fluid mochanics. Once the velocity is known thon the pressure distribution
“nd hones force ating on he id can he determined. In this Chapter, the methods of determining
velocity and acceleration ano discussed.
> 5.2. METHODS OF DESCRIBING FLUID MOTION
‘The Mid motion is der by two methods. They are —() Lagrangian Method, and i) Eulerian
Method. In the Lagrangian method, a single Muid partici is followed daring is motion and its
velocity, acceleration, density te ate described. In ease of Eulerian method, the velocity, sccelea»
tion, pressure, density et, are described ata point inflow fel. The Eulerian method is commonly
used in ok! means.
» 5.3 TYPES OF FLUID FLOW
‘The ad lw is cassie as +
(0) Steady and unstcady flows :
(6) Unterm and non-uniform flows:
(di) Laminar and urban Mows
(is) Compressible and incompressible hows
(0) Rotational and oration lows: and
(i) One, two and teeesinensional How.
5.3.1. Steady and Unsteady Flows. Steady row isdefined as hat type of low la whieh the ld
characterises lke velocity, pressure, density, ec, a a pot do not change with tne. Thus for
steady flow, mathematically, we have
163
ANDIDEAL FLOW
A. KINEMATICS OF FLOW
» 5.1 INTRODUCTION
Kinemaes ls defined as that branch of science whlch deals with motion of particles without
considering the forces causing the mouion. The velochy at any pont ia flow Held at any tne is
‘tied in this brane of fluid mochanics. Once the velocity is known thon the pressure distribution
“nd hones force ating on he id can he determined. In this Chapter, the methods of determining
velocity and acceleration ano discussed.
> 5.2. METHODS OF DESCRIBING FLUID MOTION
‘The Mid motion is der by two methods. They are —() Lagrangian Method, and i) Eulerian
Method. In the Lagrangian method, a single Muid partici is followed daring is motion and its
velocity, acceleration, density te ate described. In ease of Eulerian method, the velocity, sccelea»
tion, pressure, density et, are described ata point inflow fel. The Eulerian method is commonly
used in ok! means.
» 5.3 TYPES OF FLUID FLOW
‘The ad lw is cassie as +
(0) Steady and unstcady flows :
(6) Unterm and non-uniform flows:
(di) Laminar and urban Mows
(is) Compressible and incompressible hows
(0) Rotational and oration lows: and
(i) One, two and teeesinensional How.
5.3.1. Steady and Unsteady Flows. Steady row isdefined as hat type of low la whieh the ld
characterises lke velocity, pressure, density, ec, a a pot do not change with tne. Thus for
steady flow, mathematically, we have
163
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