Unit -2b Fluid Dynamics [Compatibility Mode].pdf

Fluid Kinetics

1. Fluid Kinetics

2. Syllabus
Fluiddynamics-equationsofmotion-Euler'sequationalonga
streamline-Bernoulli'sequation–applications-Venturimeter,Orifice
meterandPitottube.Linearmomentumequationanditsapplication.
•Second Half of Unit -2
»Bernoulli Equation(Energy equation)
»Applications:»Applications:
Venturimeter
Orificemeter
Pitot tube
»Momentum Equation
Force on bends
Torque exerted in Lawn Sprinkler

3. Equations Of Motion
In a fluid flow, the following forces are present:
Gravity force, F
g
Pressure force ,F
p
Force due to viscosity,F
v
Force due to turbulence,F
t
Force due to compressibility,F
c

4. Equations Of Motion
According to Newton’s second law of motion ,
the net force F
xacting on a fluid element in the
direction of x is equal to mass m of the fluid
element multiplied by the acceleration a
xin the
x-direction.x-direction.
F
x= m. a
x
Therefore the net force :
F
x= ( F
g )
x+ (F
p)
x+ (F
v)
x+ (F
t)
x+ (F
c)
x

5. Equations Of Motion
F
x= ( F
g)
x+ (F
p)
x+ (F
v)
x+ (F
t)
x+ (F
c)
x
a)If the force due to compressibility, (F
c)is negligible,
the resulting net force
F
x= ( F
g)
x+ (F
p)
x+ (F
v)
x+ (F
t)
x
this equation of motion are called Reynold’s
Equation Of Motion.Equation Of Motion.
b)For flow where (F
t)is negligible, the resulting
equation of motion are known as Navier-Stokes
Equation.
c)If the flow is assumed to be ideal, viscous force
(F
v)is zero and equation of motion are known as
Euler’s Equation Of Motion.

6. Euler’s Equation Of Motion
Based on Law of Conservation of Energy

7. Euler’s Equation Of Motion
Sum of all Forces = Mass x Acceleration
Pr.Force
Body Force

8. Euler’s Equation Of Motion

9. Bernoulli’s Equation From Euler’s Formula
= constant
If flow is incompressible, ρis constant
= constant
Dividing through out by g
 
 vdvgdz
dp

2
2
v
gz
p


Dividing through out by g
= constant -Bernoulli’s equation
Where:
= pressure energy per unit weight of fluid or
pressure head
v
2
/2g = kinetic energy per unit weight or kinetic head
Z = potential energy per unit weight or potential head
g
z
g
p v
2
2


g
p


10. Bernoulli’s Equation
The following are the assumptions made :
The fluid is ideal, i.e viscosity is zero
The flow is steady
The flow is incompressible
The flow is irrotational

11. Example 1
Waterisflowingthroughapipeof5cmdiameterundera
pressureof29.43N/cm
2
(gauge)andwithmeanvelocityof2
m/s.Findthetotalheadortotalenergyperunitweightofthe
wateratacross-section,whichis5mabovethedatumline.
Given:
Diameterofpipe=5cm=0.5mDiameterofpipe=5cm=0.5m
pressure:p=29.43N/cm
2
velocity:v=2m/s
DatumHead:z=5m
Required:totalheadortotalenergyperunitweightofthe
wateratacross-section

12. Example 1
Solution :
Total head = pressure head + kinetic head + datum head
pressure head = = = 30 m
kinetic head = = = 0.204 m
g
p

81.91000
10
4
43.29


v
2
2
2
kinetic head = = = 0.204 m
Total head = =
= 35.204 m
g
v
2
81.92
2

z
gg
pv

2
2

5204.030 

13. Example 2
Thewaterisflowingthroughapipehavingdiameters20cm
and10cmatsections1and2respectively.Therateofflow
throughpipeis35litres/s.Thesection1is6mabovedatum
andsection2is4mabovedatum.Ifthepressureatsection1is
39.24N/cm
2
,findtheintensityofpressureatsection2.
(Neglectthelosses).

14. Example 2
Given:
At section 1 , D
1= 20cm = 0.2m
A
1= = 0.0314 m
2
p
1 = 39.24 N/cm
2
= 39.24 x 10
4
N/m
2
2.0
2
4

= 39.24 x 10 N/m
z
1= 6m
At section 2 , D
1= 10cm = 0.1m
A
1= = 0.0314 m
2
z
2= 4m
1.0
2
4


15. Example 2
Rate of flow : Q = 35 lit/s =
= 0.035 m
3
/s
Required : Intensity of pressure at section 2
Solution :
Q = AV= AV
1000
35
Q = A
1V
1= A
2V
2
= 1.114 m/s
= 4.456 m/s
0314.0
035.0
1
1

A
V
Q
00785.0
035.0
2
2

A
V
Q

16. Example 2
Applying Bernoulli’s equation at section 1 & 2
4
81.92
456.4
2
81.91000
2
6
81.92
114.1
2
81.91000
10
4
24.39








 p
z
g
V
g
p
z
g
V
g
p
2
2
2
22
1
2
2
11


81.9281.9100081.9281.91000 
012.5
9810
2
063.46 
p
051.41
9810
2

p
p
2= 40.27 x10
4
N/m
2
Answer :Intensity of pressure at section 2 = 40.27 N/cm
2

17. Bernoulli’sEquationforrealfluid:
TheBernoulli’sEquationwasderivedon
theassumptionthatfluidisinviscidandthereforefrictionless.
Butalltherealfluidareviscousandhenceofferresistanceto
flow.Thustherearealwayssomelossesinfluidflowsand
hencetheselossesaretakenintoconsideration.hencetheselossesaretakenintoconsideration.
Bernoulli’sEquationfortherealfluidsare:
h
L: loss of energy between two points
hz
Vp
z
Vp
L
gggg

2
2
22
1
2
11
22 

18. Example 3
A pipe of diameter 400 mm carries water at a velocity of 25 m/s.
The pressure at the points A and B are given as 29.43 N/cm
2
respectively while the datum head at A and B are 28 m and 30 m.
Find the loss of head between A and B.

19. Example 3
Given :
Dia. Of pipe,D = 400 mm = 0.4 m
Velocity,V = 25 m/s
Required: loss of head between A and B
Solution:
At point A,p
A= 29.43 N/cm
2
= 29.43 * 10
4
N/m
2
z
A= 28 mz
A= 28 m
v
A= v = 25 m/s
Total energy at A,
= 30 + 31.85 + 28 = 89.85 m
ZA
g
v
A
g
p
A

2
2
EA

28
81.92
25
2
81.91000
10
4
43.29
EA 






20. Example 3
At point B,p
B= 22.563 N/cm
2
= 22.563 * 10
4
N/m
2
z
B= 30 m
v
B= v = v
A = 25 m/s
Total energy at B, ZB
g
v
B
g
p
B

2
2
EB

25
2
10
4
563.22
= 23 + 31.85 + 30 = 84.85 m
Loss of energy:h
L= E
A–E
B= 89.85 –84.85 = 5.0 m
Answer :loss of head between A and B = 5.0 m
30
81.92
25
2
81.91000
10
4
563.22
EB 






21. »24
th
Aug »24Aug

22. Syllabus
Fluiddynamics-equationsofmotion-Euler'sequation
alongastreamline-Bernoulli'sequation–applications-
Venturimeter,OrificemeterandPitottube.Linear
momentumequationanditsapplication.
•Second Half of Unit -2
»Bernoulli Equation(Energy equation)»Bernoulli Equation(Energy equation)
»Applications:
Venturimeter
Orificemeter
Pitot tube
»Momentum Equation

23. Practical applications of bernoulli’s equation
Bernoulli’s equation is applied in all problems of
incompressible fluid flow where energy considerationare
involved. Its application to the following measuring devices
are:
1. Venturimeter. 1. Venturimeter.
2. Orificemeter. To measure discharge(rate of flow)
3. Rotameter
4. Pitot tube. –To measure velocity in a flow of liquid.

24. Pictures
Venturimeter
Rotameter

25. x
20
o
6
o

26. Venturimeter
Itisusedformeasuringtherateofaflowofafluid
flowingthroughthepipe.Thebasicprincipleisthat
byreducingthecross-sectionalareaoftheflow
passage,apressuredifferenceiscreatedandthe
measurementofpressuredifferenceenablesthe
determinationofthedischargethroughpipes.It
consistsofthreeparts.,consistsofthreeparts.,
(i)Ashortconvergingpart.
(ii)Throat.
(iii)Divergingpart.

27. Venturimeter
Expression for rate of flow through venturimeter
Let d
1= diameter at inlet or at section (1)
p
1= pressure at section (1)
v
1= velocity of fluid at section at section (1)
a
1= area at section (1) =
And d
1, p
1, v
1, a
1= corresponding values at section (2)
d
2
1
4


28. Venturimeter
Applying Bernoulli’s equation at section (1) & (2)
As pipe is horizontal, hence z
1= z
2
z
g
V
g
p
z
g
V
g
p
2
2
2
22
1
2
2
11


VpVp
2
22
2
11

-(i)
Now applying continuity equation at section (1) & (2)
g
V
g
p
g
V
g
p
2
22
2
11


g
V
g
V
g
pp
2
2
1
2
2
221



h
g
pp



21
g
a
va
g
V
h
2
1
22
2
2
2
2

1
22
12211
a
orQ
va
vvava


29. Venturimeter

30. Venturimeter
Actual discharge will be less than theoretical discharge
Where C
d= co-efficient of venturimeter and its value is less than 1
Value of h is given by differential u –tube manometer
gh
aa
aa
Cd
Q
act
2
2
2
2
1
21



Value of h is given by differential u –tube manometer
Case i: Let the differential manometercontains a liquid which is
heavier than the liquid flowing through the pipe.
Let F
ℎ=ui.dynamcs-ecℎqℎqnamqytmofmE
F
-=ui.dynamcs-etmofmEet-rm'dcℎqimiq
g=Emeeqyq'Bq-ecℎqℎqnamqytmofmEB-tf–'m'f−cfpq

31. To find piezometric head h
Pressure
Manometers
Pressure at two
points
Pressure head at
two points
Difference in manometer
(Specific gravity of manometeric liquid
is greater thanspecific gravity of liquid
flowing in pipe)
Difference in manometer
(Specific gravity of manometeric
liquid is less thanspecific gravity of
liquid flowing in pipe)
Possible datas given in problems

32. Venturimeter
Case i: This case related to the inclined venturimeterhaving differential u-
tube manometer. Let the differential manometer contains heavier
liquid, then ‘h’ is given by
1
2
2
1
1



















So
Sh
x
gg
h z
p
z
p

Case ii: This case related to the inclined venturimeterhaving differential u-



























S
S
o
l
gg
h xz
p
z
p
1
2
2
1
1

Case ii: This case related to the inclined venturimeterhaving differential u-
tube manometer. Let the differential manometer contains lighter
liquid, then ‘h’ is given by
z
g
V
g
p
z
g
V
g
p
2
2
2
22
1
2
2
11


Case iii: Incase velocity given in problem, substitute in Bernoulli equation and
find h

33. Example 1
Ahorizontalventurimeterwithinletandthroatdiameters30cmand
15cmrespectivelyisusedtomeasuretheflowofwater.The
readingofdifferentialmanometerconnectedtotheinletandthe
throatis20cmofmercury.Determinetherateofflow.TakeC
d=
0.98.
Given:
diaatinlet:d
1=30cmdiaatinlet:d
1=30cm
areaatinlet:a
1= = =0.0706m
2
diaatthroat:d
2=15cm
areaatthroat:a
2= =0.017m
2
C
d=0.98
readingofdifferentialmanometer=x=20cmofmercury
required:rateofflow
Solution:
d
2
1
4

30
2
.0
4

15
2
.0
4


34. Example 1
Difference of pressure head :
Where:
S
h= sp. gr of mercury = 13.6
S
0= sp. gr of water = 1
1
So
Sh
xh

= 2.52 m of water
The discharge through venturimeter :
= 0.12496m
3
/s = 125 lit/s




 1
1
6.13
20.0h
gh
aa
aa
Cd
Q
act
2
2
2
2
1
21



52.281.92
2
0176.0
2
0706.0
0176.00706.0
98.0 





35. Example 2
Ahorizontalventurimeterwithinletdiameter20cmandthroatdiameter10cmis
usedtomeasuretheflowofwater.Thepressureatinletis17.658N/cm
2
andthe
vacuumpressureatthethroatis30cmofmercury.Findthedischargeofwater
throughventurimeter.TakeC
d=0.98.
Given:diaatinlet:d
1=20cm
areaatinlet:a
1= =0.031416cm
2
diaatthroat:d
2=10cm
areaatthroat:a= =0.007854cm
2
20
2
.0
4

10
2
.0

areaatthroat:a
2= =0.007854cm
2
C
d=0.98
p
1=17.658N/cm
2
=17.658x10
4
N/m
2
=18mofwater(ρforwater=1000kg/m
3
)
=-30cmofmercury
=-0.3mofmercury=-0.3x13.6=-4.08m
Required:dischargeofwaterthroughventurimeter
Solution:
10
2
.0
4

100081.9
10
4
658.171



g
p

g
p

2

36. Example 2
Differential head : = 18-(-4.08) = 22.08 m of water
Discharge :
gg
h
pp

21

gh
aa
aa
Cd
Q 2
2
2
2
1
21



08.2281.92
007854.0031416.0
007854.0031416.0
98.0
22





=0.165 m
3
/s = 165.55lit/s
Answer : Discharge of water through venturimeter = 165.55lit/s

37. Problem -3
In a vertical pipe conveying oil of specific gravity of 0.8, two pressure gauges have been
installed at A and B where the diameters are 16 cm and 8 cm respectively. A is 2 meters
above B. The pressure gauges readings have shown that the pressure at B is greater
than A by 0.981N/cm2. Neglecting all losses, calculate the flow rate. If the gauges are
replaced by tubes containing same liquid and connected to a U-tube containing
mercury. Calculate the difference of level of mercury in the two limbsof U-tube.
Given Data:
Specific gravity of oil = 0.8Specific gravity of oil = 0.8
Area at section A (a
1)= Πd
1
2
/4= 3.14 x 0.16
2
/4=0.02 m
2
Area at section B (a
2)=Πd
2
2
/4= 3.14 x 0.08
2
/4=0.005 m
2
Datum head at A Z
1= 2m
Datum head at B Z
2= 0m
Difference in pressure p
2–p
1= 0.981 N/cm
2
= 0.981 x 10
4
= N/m
2
Difference in pressure head (p
2–p
1)/ρg = (0.981 x 10
4
) / (800 x 9.81)
= 1.25 m
But we need, (p
2–p
1)/ρg = -1.25m

38. Step 1:Find h:
So h = -1.25 + 2 = 0.75 m
Step 2: Find
Q
th= 0.0198m
3
/s
Step 3 : Second Part : TO FIND DIFFERENCE OF MERCURY LEVELS
IN U TUBE MANOMETER ie x

















 z
p
z
p
gg
h
2
2
1
1

gh
aa
aa
Q
th
2
2
2
2
1
21


IN U TUBE MANOMETER ie x
0.75 = x ((13.6/0.8)-1)
So x = 0.04687 m = 4.687 cm
1
So
Sh
xh

39. Syllabus
•Second Half of Unit -2
»Bernoulli Equation(Energy equation)
»Applications:
Venturimeter( C
d= 0.96 to 0.98)
Orificemeter (C
d= 0.62 to 0.68)
Pitot tube (C= 0.98)Pitot tube (C
v = 0.98)
»Momentum Equation
Forces on pipe bends (Momentum eqn)
F= ρQ (V
2-V
1)
Forces on sprinkler system
(Moment of Momentum)
F= ρQ (V
2r
2–V
1r
1)

40. Momentum equation
Itisbasedonthelawofconservationofmomentumor
onthemomentumprinciple,whichstatesthatthenet
forceactingonafluidmassisequaltothechangein
momentumofflowperunittimeinthatdirection.
Theforceactingonafluidmass‘m’isgivenbytheTheforceactingonafluidmass‘m’isgivenbythe
Newton’ssecondlawofmotion.
F=mxa
where‘a’istheaccelerationactinginthesame
directionasforceF.

41. Momentum equation
dt
dv
mF
dt
dv
a


F = m x a
Net Force F =massx acceleration
F = Density x volume x acceleration
F = Density x volume x velocity
time
F = Density x Dischargex change

dt
mvd
F
dt

Thisisthemomentumprinciple
F.dt=d(mv)whichisknownastheimpulse-momentumequationand
statesthattheimpulseofaforce‘F’actingonafluidmass‘m’ina
shortintervaloftime‘dt’isequaltothechangeofmomentumd(mv)
inthedirectionofforce.
in velocity
F = ρQv

42. Momentum equation
Force exerted by a flowing fluid on a pipe-bend
By using Impulse-momentum equation:
LetV
1=velocity of flow at section (1)
P
1=pressure intensity at section (1)
A
1=area of cross-section of pipe at section(1)
V
2,P
2,A
2= are for section (2)

43. v
2
pA
p
2A
2F
Y
p
2A
2cosθ
v
2Sinθ
v
2Cosθ
p
2A
2sinθ
Net Force F = ρQ V
Along x direction,
p
1A
1–p
2A
2cosθ–F
x=ρQ (V
2cosθ-V
1)
Along y direction
-p
2A
2sinθ–F
y= ρ Q V
2sinθ
Resultant Force
Direction of Force tan θ
1= F
y/F
x
+ve Direction
+ve Direction
v
1
θ
p
1A
1
F
x
 FFF yxR
22


44. Problems
•A45
o
reducingbendisconnectedtoapipeline,the
diametersattheinletandoutletofbendbeing600mm
and300mmrespectively.Findforceexertedbywater
onbendifthepressureintensityatinletis8.829N/cm
2
andpressureintensityatoutletis5.45N/cm
2.
TheRate
offlowis600litres/secofflowis600litres/sec
Given Data : Angle of bend θ= 45
Diameter at sec.1 d
1= 0.6 mFind A
1
Diameter at sec. 2 d
2 = 0.3 mFind A
2
Pressure intensity at sec 1 p
1 = 8.829 x 10
4
N/m
2
Pressure intensity at sec 2 p
2 = 5.45 x 10
4
N/m
2
Discharge Q = 0.6 m
3
/s

45. •Use continuity Equation:
Q = A
1V
1= A
2V
2
Find V
1and V
2
•Force along x direction
p
1A
1–p
2A
2cosθ–F
x=ρQ (V
2cosθ-V
1)
–Find F
x
Answers: V1 = 2.122m/s , V2 = 8.488 m/s
Answers:–Find F
x
•Force along x direction
-p
2A
2sinθ–F
y= ρ Q V
2sinθ
–Find F
y
•Resultant Force
 FFF yxR
22

Answers:
Fx= 19.911kN
Fy = -6.322 kN
Fr = 20.89 kN

46. Example
250litresofwaterisflowinginapipehavinga
diameterof300mm.Ifthepipeisbentby135,find
themagnitudeanddirectionoftheresultantforce
onthebend.Thepressureofwaterflowingis39.24
N/cm
2
.N/cm.
figure

48. Example
Given : Since pipe diameter is same, V
1= V
2,
Substituting, V
1= V
2and Z
1= Z
2we get,
Pressure P
1=P
2= 39.24 N/cm
2
= 39.24 x 10
4
N/m
2
Discharge : Q = 250 litres/s = 0.25 m3/s
D
1= D
2=300 mm = 0.3m
A
1= A
2=
V
1= V
2= 3.54 m/s
Required: magnitude and direction of the resultant force on
07068.0
25.0

A
Q

49. Example
Solution :
Force along x-axis:
P
1A
1 + P
2A
2cos 45-F
x = Q (-V
1+V
2cos 45)
F= PA+ PAcos 45+Q (-V+Vcos 45)F
x = P
1A
1 + P
2A
2cos 45+Q (-V
1+V
2cos 45)
= 48855.4 N
  45cos537.3537.325.01000)45cos1(07068.024.3910
4

50. Example
Force along y-axis :
-F
y -P
2A
2sin 45= Q (V
2 sin 45)
F
y = -Q(V
2sin 45) -P
2A
2cos 45
= -20236.3 N
   45sin07068.024.3945sin537.325.01000 10
4
= -20236.3 N
= 52880.6 N
  3.202364.48855
2222
 FFF yxR
'30224142.0
4.48855
3.20236
tan 
F
F
x
y


51. Moment Of Momentum Equation
MomentofMomentumequationisderivedfrommomentof
momentumprinciplewhichstatesthattheresultingtorqueacting
onarotatingfluidisequaltotherateofchangeofmomentof
momentum.
letV
1=velocityoffluidatsection1
r
1=radiusofcurvatureatsection1
1
Q=rateofflowoffluid
ρ=densityoffluid
V
2&r
2=velocityandradiusofcurvatureatsection2
Momentumoffluidatsection1=massxvelocity
=ρQxV
1/s
momentofmomentumpersecatsec-1=ρQxV
1xr
1

52. Moment Of Momentum Equation
Similarly moment of momentum per sec at sec -2
= ρQ x V
2x r
2
Rate of change of moment of momentum
= ρQV
2r
2-ρQV
1r
1
= ρQ(V
2r
2-V
1r
1)
According to moment of momentum principle :According to moment of momentum principle :
Resultant torque = rate of change of moment of momentum
T = ρQ(V
2r
2-V
1r
1)
This eqis known as moment of momentum equation. It is
applied :
Analysis of flow problems in turbines and centrifugal pumps
For finding torque exerted by water on sprinkler.

53. Example
Alawnsprinklerasshowninfig.has0.8cmdiameternozzleattheend
ofarotatingarmanddischargewaterattherateof10m/svelocity.
Determinethetorquerequiredtoholdtherotatingarmstationary.Also
determinetheconstantspeedofrotationofthearm,iffreetorotate.
Given:
Dia.Ofeachnozzle:0.8cm=0.008m
Areaofeachnozzle= =5.026x10
5
m
2 008.0
2
4

Velocityofflowateachnozzle=10m/s
Required:torque&speedofrotationofthearm
Solution:
Discharge:Q=AreaXvelocity
=5.026x10
5
x10=5.026x10
4
m
3
/s
 
4

54. Example
Torque exerted by water moment of momentum
coming through nozzle = of water through A
= r
Ax ρx Q x V
A
= 0.25 x 1000 x 5.026 x 10
4
x 10
= 1.26 (clockwise)
Torque exerted by water moment of momentum Torque exerted by water moment of momentum
coming through nozzle = of water through B
= r
Bx ρx Q x V
B
= 0.20 x 1000 x 5.026 x 10
4
x 10
= 1.0 (clockwise)
Torque exerted by water on sprinkler :
= 1.26+ 1 = 2.26 Nm

55. Example
Speed of rotation of arm , if free to rotate:
Let ω= speed of rotation of the sprinkler
The absolute velocity of flow of water at the nozzles A & B:
V
1= 10 -0.25 x ω
V
2= 10 -0.2 x ωV
2= 10 -0.2 x ω
Torque exerted by water coming out at A on sprinkler :
= r
Ax ρx Q x V
1
= 0.25 x 1000 x 5.026 x 10
4
x (10 -0.25 x ω)
= 0.125 (10 -0.25 x ω)
Torque exerted by water coming out at B on sprinkler :
= r
Bx ρx Q x V
2
= 0.2 x 1000 x 5.026 x 10
4
x (10 -0.2 x ω)
= 0.1 (10 -0.2 x ω)

56. Example
Torque exerted by water = 0.125 (10 -0.25 x ω) + 0.1 (10 -0.2 x ω)
Since moment of momentum of the flow entering is zero and no external torque
is applied on sprinkler, so the resultant torque on the sprinkler must be zero.
0.125 (10 -0.25 x ω) + 0.1 (10 -0.2 x ω) = 0
ω = 43.9 rad/s
 9.436060 
N
= 419.2 r.p.m
Answer : Torque = 2.26 Nm & speed of rotation of the arm = 43.9
rad/s


2
9.4360
2
60 
N

57. References
1.Bansal,R.K.,“FluidMechanicsandHydraulicsMachines”,
5thedition,LaxmiPublicationsPvt.Ltd,NewDelhi,2008
2.ModiP.NandSeth"HydraulicsandFluidMechanics
includingHydraulicMachines",StandardBookHouseNewincludingHydraulicMachines",StandardBookHouseNew
Delhi.2015.