UNIT-II FMM-Flow Through Circular Conduits

Unit II
-Flow Through Circular
Conduits

1. Unit II -Flow Through Circular
Conduits

4. Findtheheadlostduetofrictioninapipeofdiameter300mmand
length50m,throughwhichwaterisflowingatavelocityof3m/s
using(i)Darcyformula,(ii)Chezy’sformulaforwhichC=60.
Given:
D = 300 mm = 0.3 m
L = 50 m
V = 3 m/s
C = 60
??????= 0.01 stoke = 0.01 cm
2
/s = 0.01 * 10
-4
m
2
/s
To find:
h
f
(i)Darcy formula
(ii)Chezy’sFormula
??????�??????�2
1
2

5. Darcy Eqn:
h
f =
4�????????????
2
2��
Re =
??????�
??????
=
3∗0.3
0.01∗10
−4= 90 x 10
4
> 2000
Hence it is Turbulent flow
So, f =
0.079
��
0
.
25=
0.079
(90??????104)0.25
= 2.56 x 10
-3
h
f =
4∗2.56x10−3∗50∗3
2
2∗9.81∗0.3
=>h
f =0.7828 m
Chezy’sEqn
V = C ????????????���
�=
�×�×??????
�
??????
�
�
V = C
�
4
ℎ
�
??????
3 = 60
0.3
4
ℎ�
50
h
f = 1.665 m

6. Findthediameterofapipeoflength2000mwhentherateofflowofwater
throughthepipeis200litres/sandtheheadlostduetofrictionis4m.Takethe
valueofC=50inchezy’sformulae.
Given
L = 2000 m
Q = 200 lit/s = 0.2 m
3
/s →Q =A V
h
f= 4 m
C = 50
To find:
d
Solution:
V = C????????????

7. (or)ℎ
�=
4×??????×??????
2
�
2
�
ℎ
�=
4??????2000×??????
2
�
2
�
Discharge, Q = A V
0.2 =
�
4
d
2
* V
V =
0.8
�d
2
ℎ
�=
4??????2000×??????
2
�
2
�
4 =
4??????2000×
0.8
??????d
2
2
50
2
�
d = 0.553 m = 553 mm

8. Loss of Energy in pipes:
When a fluid is flowing through a pipe, the fluid experiences some resistances due to
which some of the energy of fluid is lost. This loss of energy is classified a
Energy Losses
Major Energy Losses
This is due to friction and its is calculated
by the following formulae:
(a) Darcy –WeisbachFormula
(b) Chezy’s Formula
Minor Energy Losses
This is due to
(a) Sudden expansion of pipe,
(b) Sudden contraction of pipe,
(c) Bend in pipe,
(d) Pipe fittings, etc.,
(e) An obstruction in pipe.

9. MAJOR LOSS
❖Themajorlossofenergyisduetofriction
❖Thelossduetofrictionismuchmoreincaseoflongpipe
lines.
❖Itdependsonroughnessofpipe,length,velocityand
diameterofpipe.
�
�=
�×�×�×??????
�
���
FromDARCY’SWEISBACHEQUATION
�
�=
�×�×??????
�
??????
�
�
FromChezy′sEQUATION

10. MINOR LOSS
❖Thelossesduetodisturbanceinflowpatterniscalled
asminorloss
❖Minorlossoccursdueto
❖SuddenExpansion
❖SuddenContraction
❖Valves
❖Fittings
❖Bends

11. LOSS OF ENERGY DUE SUDDEN ENLARGEMENT
❖Considerahorizonalpipeof
areaA
1issuddenlyenlargedto
theareaA
2
�
�??????
�
�
�??????
�
??????
�??????
�
①
②
❖Consideratwosectionof①and
② isbeforeandafter
expansion.
❖Let�
�??????
�??????��??????
�isthePressure
Intensity,VelocityatareaA
1
❖Let�
�??????
�??????��??????
�isthePressureIntensity,VelocityatareaA
2
❖Let�′istheintensityofpressure,oftheliquidEddiesonAreaA
2–A
1

12. LOSS OF ENERGY DUE SUDDEN ENLARGEMENT
❖Theresultantforcebetweensection①&②
??????=�
�??????
�−�
�??????
�+�
′
(??????
�−??????
�)
❖Butithasbeenfoundbyexperimentthat
�
′
=�
�
❖Theresultantforce
??????=�
�??????
�−�
�??????
�+�
�(??????
�−??????
�)
??????=�
�??????
�−�
�??????
�+�
�??????
�−�
�??????
�
??????=�
�−�
�??????
�

13. LOSS OF ENERGY DUE SUDDEN ENLARGEMENT
❖Momentumofliquid/secatsection①
=�??????��×??????�������
=????????????
�??????
�×??????
�
=????????????
�??????
�
�
❖Momentumofliquid/secatsection②
=????????????
�??????
�×??????
�
=????????????
�??????
�
�

14. LOSS OF ENERGY DUE SUDDEN ENLARGEMENT
❖ChangeinMomentumofliquid/sec
=????????????
�??????
�
�
−????????????
�??????
�
�
❖FromContinuityEquation
??????
�??????
�=??????
�??????
�??????
�=
??????
�??????
�
??????
�
=????????????
�??????
�
�
−??????
??????
�??????
�
??????
�
??????
�
�
=????????????
�??????
�
�
−????????????
�??????
�??????
�

15. LOSS OF ENERGY DUE SUDDEN ENLARGEMENT
❖FromNewton’sSecondLawofmotion
??????����=�??????������??????�������������������??????��
�
�−�
�??????
�=????????????
�??????
�
�
−????????????
�??????
�??????
�
�
�−�
�??????
�=????????????
�(??????
�
�
−??????
�??????
�)
�
�−�
�
??????
=(??????
�
�
−??????
�??????
�)

16. LOSS OF ENERGY DUE SUDDEN ENLARGEMENT
ApplyingBernoulli'sequationbetweensection①&②
�
�
??????�
+
??????
�
�
��
+�
�=
�
�
??????�
+
??????
�
�
��
+�
�+�
�
�
�
??????�
−
�
�
??????�
=+
??????
�
�
��
+�
�−
??????
�
�
��
−�
�+�
�
��−��
??????�
=+
??????
�
�
��
+�
�−
??????
�
�
��
−�
�+�
�

17. LOSS OF ENERGY DUE SUDDEN ENLARGEMENT
??????
�
�
−??????
�??????
�
�
=+
??????
�
�
��
+�
�−
??????
�
�
��
−�
�+�
�
❖Datumheadofsection①&②areequal
∴�
�=�
�
??????
�
�
−??????
�??????
�
�
=
??????
�
�
��
−
??????
�
�
��
+�
��
�=
�??????
�
�
−�??????
�??????
�−??????
�
�
−??????
�
�
��
�
�=
??????
�
�
−�??????
�??????
�−??????
�
�
��
�
�=
??????
�−??????
�
�
��

18. Therateofflowofwaterthroughahorizontalpipeis0.3m
3
/sec.thediameterofthepipe,
whichis25cm,issuddenlyenlargedto50cm.thepressureintensityinthesmallerpipeis14
N/cm
2
.determinethelossofheadduetosuddenenlargement,pressureintensityinthelarger
pipepowerlostduetoenlargement.
Given
Q = 0.3 m
3
/sec
D
1= 25 cm = 0.25 m
D
2= 50 cm = 0.5 m
p
1= 14 N/cm
2
= 14 * 10
4
N/m
2
To find:
he
p
2
P
Solution:
Q = A
1V
1=> V
1= Q/A
1= Q/(
�
4
d
1
2
) = 6.11 m/s
Q = A
2 V
2=> V
2= Q/A
2= Q/(
�
4
d
2
2
) = 1.52 m/s

19. he =
??????�−??????�
�
��
=
�.��−�.��
�
�∗�.��
he =1.07 m
Applying Bernoulli's Equation,
�1
��
+
??????
1
2
2�
=
�2
��
+
??????
2
2
2�
+ℎ
�
14∗10
4
1000∗9.81
+
6.11
2
2∗9.81
=
�
2
9.81∗1000
+
1.52
2
2∗9.81
+1.07
�
�= ��.����
�
N/m
2
Power:
�=??????���
�=1000×9.81×0.3×1.073
�= �.������
�
W

20. LOSS OF ENERGY DUE SUDDEN CONTRACTION
�
�??????
� �
�??????
�
??????
�
??????
�
① ②©
❖Considerahorizonalpipeof
areaA
1issuddenlycontracted
totheareaA
2
❖Astheliquidflowsfrom
largerpipetosmallerpipe
theareaofflowdecreases
andbecomesminimumof©
❖Let�
�??????
�??????��??????
�isthePressure
Intensity,VelocityatareaA
1
❖Let�
�??????
�??????��??????
�isthePressureIntensity,VelocityatareaA
2

21. LOSS OF ENERGY DUE SUDDEN CONTRACTION
�
�??????
� �
�??????
�
??????
�
??????
�
① ②©
❖Actuallythelossofheadisdue
toenlargementbetweensection
©&②
�
�=
??????
�−??????
�
�
��
�
�=
??????
�
�
��
??????
�
??????�
−�
�

22. LOSS OF ENERGY DUE SUDDEN CONTRACTION
❖FromContinuityEquation
??????
�??????
�=??????
�??????
�
??????
�
??????
�
=
??????
�
??????
�
??????
�
??????
�
=
�
??????
�/??????
�
??????
�
??????
�
=
�
??????
�

23. LOSS OF ENERGY DUE SUDDEN CONTRACTION
�
�=
??????
�
�
��
�
??????
�
−�
�
�
�=
�??????
�
�
��������=
�
??????
�
−�
�
❖IfthevalveofC
Cisnotgiven,then
�
�=
�.�??????
�
�
��

24. InFigshownbelow,whenasuddencontractionisintroducedinahorizontalpipelinefrom50cmto25
cm,thepressurechangesfrom10,500kg/m
2
(103005N/m
2
)to6900kg/m
2
(67689N/m
2
).Calculatetherate
offlow.Assumeco-efficientofcontractionofjettobe0.65.Followingthisifthereisasuddenenlargement
from25cmto50cmandifthepressureatthe25cmsectionis6900kg/m
2
(67689N/m
2
)whatisthe
pressureatthe50cmenlargedsection?
Given:
D
1= 50 cm = 0.5 m
D
2= 25 cm = 0.25 m
p
1= 103005 N/m
2
p
2= 67689 N/m
2
p
3 = 67689 N/m
2
C
C= 0.65
D
4= 50 cm = 0.5 m
D
3= 25 cm = 0.25 m
To find:
Q
p
4

25. Solution:
From formula, h
c=
????????????
2
2
2�
,
where, ??????=
1
�??????
−1
2
=
1
0.65
−1
2
= 0.289
Applying Bernoulli’s eqn
�1
��
+
??????
1
2
2�
=
�2
��
+
??????
2
2
2�
+ℎ
�
103005
9.81∗1000
+
??????
1
2
2∗9.81
=
67689
9.81∗1000
+
??????
2
2
2∗9.81
+
0.289??????
2
2
2�
From Continuity Equation, A
1V
1= A
2V
2
V
1= (A
2V
2) / A
1
V
1= (d
2
2
V
2) / d
1
2
V
1= 0.25 V
2

26. 103005
9.81∗1000
+
0.25??????2
2
2∗9.81
=
67689
9.81∗1000
+
??????
2
2
2∗9.81
+
�.���??????
�
�
��
V
2= 8.3 m/s
V
1= 0.25 V
2= 2.07 m/s
Discharge, Q = A
2V
2=
�
4
d
2
2
* V
2
Q = 0.40 m
3
/s
Applying Bernoulli’s Equation for section 3 -4
�
3
��
+
??????
3
2
2�
=
�
4
��
+
??????
4
2
2�
+ℎ
�
Here, V
2= V
3= 8.3 m/s
V
1= V
4= 2.07 m/s
h
e= (V
3–V
4)
2
/2g = (8.3 –2.07)
2
/ 2 * 9.81 = 1.98 m
67689
9.81∗1000
+
8.3
2
2∗9.81
=
�
4
9.81∗1000
+
2.07
2
2∗9.81
+1.98
P
4= 80587.37 N/m
2

27. LOSS OF ENERGY AT ENTRANCE AND EXIT
�
�=
�.�??????
�
�
��
❖LossofenergyatEntrance
❖LossofenergyatExit
�
??????=
??????
�
�
��

28. LOSS OF ENERGY DUE TO GRADUAL CONTRACTION AND EXPANSION
�
�=
�??????�−??????�
�
��
�
�=
�??????
�
��
LOSS OF ENERGY DUE TO BEND IN A PIPE

29. LOSS OF ENERGY DUE TO VARIOUS PIPE FITTINGS
�
??????=
�??????
�
��
�
??????�=
??????
�
��
??????
??????
??????(??????−??????)
−�
LOSS OF ENERGY DUE TO SUDDEN OBSTRUCTION
??????
�=
??????
??????
(??????−??????)

30. S.No. Loss Of Energy Formulae
1Loss Of Energy Due Sudden
Enlargement
�
�=
??????
�−??????
�
�
��
2Loss Of Energy Due Sudden Contraction
�
�=
�??????
�
�
��
Where, �=
�
??????�
−�
�
C
c==
??????�
??????�
3Loss Of Energy At Entrance
�
�=
�.�??????
�
�
��
4Loss Of Energy At Exit
�
??????=
??????
�
�
��
5Loss Of Energy Due To Gradual
Contraction And Expansion
�
�=
�??????
�−??????
�
�
��
6Loss Of Energy Due To Bend In A Pipe
�
�=
�??????
�
��
7Loss Of Energy Due To Various Pipe
Fittings
�
??????=
�??????
�
��
8Loss Of Energy Due To Sudden
Obstruction
�
??????�=
??????
�
��
??????
????????????(??????−??????)
−�where,??????
�=
??????
??????
(??????−??????)

31. Waterisflowingthroughahorizontalpipeofdiameter200mmatavelocityof3m/s.
Acircularsolidplateofdiameter150mmisplacedinthepipetoobstructtheflow.
FindthelossofheadduetoobstructioninthepipeifCc=0.62.
Given:
D = 200 mm = 0.2 m
A =
�
4
D
2
=
�
4
0.2
2
=0.031 m
2
V = 3 m/s
d = 150 mm = 0.15 m
a =
�
4
d
2
=
�
4
0.15
2
=0.017 m
2
Cc = 0.62
To find:
h
ob
Solution:
h
ob=
??????
2
2�
??????
�
??????(??????−??????)
−1
h
ob=
3
2
2∗9.81
0.031
0.62(0.037−0.017)
−1
h
ob=3.315 m

32. Ahorizontalpipeline40mlongisconnectedtoawatertankatoneendanddischargesfreely
intotheatmosphereattheotherend.Forthefirst25mofitslengthfromthetank,thepipeis
150mmdiameteranditsdiameterissuddenlyenlargedto300mm.Theheightofwaterlevelin
thetankis8mabovethecenterofthepipe.Consideringalllossesifheadwhichoccur,
determinetherateofflow.Takef=0.01forbothsectionsofthepipe
Given:
l = 40 m
l
1= 25 m
d
1= 150 mm =0.15 m
l
2= 15 m
d
2= 300 mm = 0.3 m
z = 8 m
f = 0.01
To find:
Q = AV

33. Solution:
�
1
��
+
??????
1
2
2�
+??????
1=
�
2
��
+
??????
2
2
2�
+??????
2+ℎ
??????+ℎ
�1+ℎ
�+ℎ
�2
0 +0+8=0+
??????
2
2
2�
+0+ℎ
??????+ℎ
�1+ℎ
�+ℎ
�2
8=
??????
2
2
2�
+ℎ
??????+ℎ
�1+ℎ
�+ℎ
�2--------1
Using continuity eqn, A
1V
1= A
2V
2
V
1= (A
2V
2) / A
1
V
1= (d
2
2
V
2) / d
1
2
V
1= 4 V
2

34. �
�=
�.�??????
�
�
��
. =
8V
2
2
��
�
�=
??????�−??????�
�
��
=
�??????�−??????�
�
��
=
9V
2
2
��
�
��=
���
�
×??????
�
�
���
= 106.66 V
2
2
�
��=
���
�
×??????
�
�
���
= 2 V
2
2
Multiply 2g
16??????=
??????
2
2
2??????
+8V
2
2
+106.66V
2
2
+9V
2
2
+2V
2
2
V
2=1.1132m/s
Q = A2 V2 =
�
4
d
2
2
* V
2
Q = 0.0786 m
3
/s.
??????�??????�
1
2

35. WHEN PIPE ARE CONNECTED SERIES
�=�
�=�
�=�
�
??????=�
��+�
��+�
��
NeglectingMinorloss
??????=�
�+�
��+�
�+�
��+�
�+�
��+�
�
ConsideringMinorloss

36. Thedifferenceinwatersurfacelevelsintwotanks,whichareconnectedbythreepipesinseries
oflengths300m,170mand210mandofdiameters300mm,200mmand400mmrespectively
is12m.Determinetherateofflowofwaterifco-efficientoffrictionare0.005,0.0052and
0.0048respectively,considering:(i)minorlossesalso(ii)neglectingminorlosses.
Given:
L
1= 300 m; L
2= 170 m; L
3= 210 m
D
1= 300 mm = 0.3m
D
2= 200 mm = 0.2 m
D
3= 400 mm = 0.4 m
H = 12 m
f
1= 0.005
f
2= 0.0052
f
3= 0.0048
To find:
Q
(i) All losses
(ii) Neglect Minor losses

37. From continuity eqn, A
1V
1= A
2V
2= A
3V
3
A
1V
1= A
2V
2
V
1= (A
2V
2) / A
1
V
1= (d
2
2
V
2) / d
1
2
= (0.2
2
V
2) / 0.3
V
1= 0.4442 V
2
A
2V
2= A
3V
3
V
3= (A
2V
2) / A
3
V
3= (d
2
2
V
2) / d
3
2
= (0.2
2
V
2) / 0.4
V
3= 0.25 V
2

38. Solution:
(i)Considering All Losses
H = ℎ
??????+ℎ
�1+ℎ
�+ℎ
�2+ℎ
�+ℎ
�3+ℎ
??????
12 = ℎ
??????+ℎ
�1+ℎ
�+ℎ
�2+ℎ
�+ℎ
�3+ℎ
??????
Losses:
�
�=
�.�??????
�
�
��
. =
�.�∗(�.���??????
�
)
�
��
= 0.098 V
2
2
/ 2g
�
�=
??????�−??????�
�
��
=
??????
�
−�.��??????�
�
��
= 0.5625 V
2
2
/ 2g
�
??????=
??????
�
�
��
= 0.0625 V
2
2
/ 2g
�
�=
�??????
�
�
��
=
�.�??????
�
�
��
�
��=
��
�
�
�
×??????
�
�
���
�
=
�×�.���×���×(�.���??????
�
)
�
���.�
= 3.942 V
2
2
/ 2g
�
��=
��
�
�
�
×??????
�
�
���
�
=
�×�.����×���×??????
�
�
���.�
= 17.68V
2
2
/ 2g
�
��=
��
�
�
�
×??????
�
�
���
�
=
�×�.����×���×(�.��??????
�
)
�
���.�
= 0.63 V
2
2
/ 2g
??????�??????�
1
2

39. Multiplying with 2g,
24g = 0.098 V
2
2
+3.942V
2
2
+�.�??????
�
�
+17.68V
2
2
+0.5625V
2
2
+0.63V
2
2
+�.����??????
�
�
V
2= 3.167 m/s
Q = A
2V
2=
�
4
d
2
2
* V
2=
�
4
0.2
2
* 3.167
Q = 0.09944 m
3
/s = 99.44 lit/s
(ii) Considering Major Losses
H = ℎ
�1+ℎ
�2+ℎ
�3
24g = 3.942 V
2
2
+ 17.68 V
2
2
+ 0.63 V
2
2
V
2= 3.353 m/s
Q = A
2V
2=
�
4
d
2
2
* V
2=
�
4
0.2
2
* 3.353
Q = 0.102 m
3
/s = 102.18 lit/s

40. EQUIVALENT PIPE
??????
??????
5
=
??????
1
??????
1
5
+
??????
2
??????
2
5
+
??????
3
??????
3
5
where, L = L
1+ L
2+L
3
The above equation is known as Dupuit’sequation.

41. Threepipesoflengths800m,500mand400mandofdiameters500mm,400mmand
300mmrespectivelyareconnectedinseries.Thesepipesaretobereplacedbyasingle
pipeoflength1700m.findthediameterofthesinglepipe.
Given:
L
1= 800 m; L
2= 500 m; L
3= 400 m
L = 1700 m
D
1= 500 mm = 0.5 m; D
2= 400 mm = 0.4 m; D
3= 300 mm = 0.3 m
To find:
D
Solution:
??????
�
5
=
??????1
�
1
5+
??????2
�
2
5+
??????3
�
3
5
1700
�
5
=
800
0.5
2+
500
0.4
2+
400
0.3
2
D = 0.37 mm

42. WHEN PIPE ARE CONNECTED PARALLEL
�=�
�+�
�
�
�=�
��=�
��

43. A pipe of diameter 20 cm and length 2000 m connects two reservoirs, having difference of
water levels as 20 m, Determine the discharge through the pipe. Take f = 0.015 and neglect
minor losses.
Given:
D = 20 cm = 0.2 m
L = 2000 m
H = 20 m
To find:
Q
Solution:
Q = A V
??????=
�×�×�×??????
�
���
=> 20 =
�×�.���×����×??????
�
��.���.�
=> v = 0.808 m/s
Q
old=
�
4
d
2
* V => Q = 0.025 m
3
/s

44. Given:
D1 = 20 cm = 0.2 m
L1 = 800 m
D2 = 20 cm = 0.2 m
L2 = 1200 m
D3 = 20 cm = 0.2 m
L3 = 1200 m
H = 20 m
A1= A2 = A3 =
�
4
d
2
=
�
4
0.2
2
= 0.0314 m
2
To find:
Q
new–Q
old
If an additional pipe of diameter 20 cm and length 1200 m is attached to the last 1200 m
length of the existing pipe, find the increase in the discharge. Take f = 0.015 and neglect
minor losses.

45. Solution:
Q1 = Q2 + Q3 (We know, Q2 = Q3)
Q
1= 2Q
2or Q
1= 2Q
3
Loss of head:
H=
��
�
�
�
×??????
�
�
���
�
+
��
�
�
�
×??????
�
�
���
�
(We know, hf
2= hf
3)
20 =
�×�.���×���×??????
�
�
��.���.�
+
�×�.���×����×??????
�
�
��.���.�
From Continuity equation,
Q
1= A
1V
1=> V
1= Q
1/A
1
Q
2 = A
2V
2=> V
2= Q
2/A
2= Q
1/2A
2

46. 20 =
��.������
��.���.�
x(
�
1
0.0314
)
2
+
��.�������
��.���.�
x(
�
1
0.0628
)
2
Q
1= 0.076 m
3
/s
Q
2= Q
1/ 2
Q
old= 0.025 m
3
/s
Q
new= Q
1= 0.076 m
3
/s
Q
new–Q
old= 0.076 –0.025 = 0.051 m
3
/s

47. HYDRAULIC ENERGY LINE (OR) HYDRAULIC GRADIENT LINE
❖Itisdefinedasalinewhichgivesthesumof
pressureheadanddatumheadofflowingfluidin
pipewithrespecttosomereferenceline.(H.G.L.)
TOTAL ENERGY LINE (OR) TOTAL GRADIENT LINE
❖Itisdefinedasalinewhichgivesthesumof
pressurehead,datumheadandkineticheadof
flowingfluidinpipewithrespecttosome
referenceline.(T.E.L)

48. Ahorizontalpipeline40mlongisconnectedtoawatertankatoneendanddischargesfreely
intotheatmosphereattheotherend.Forthefirst25mofitslengthfromthetank,thepipeis
150mmdiameteranditsdiameterissuddenlyenlargedto300mm.Theheightofwaterlevelin
thetankis8mabovethecenterofthepipe.Consideringalllossesifheadwhichoccur,
determinetherateofflow.Takef=0.01forbothsectionsofthepipe.Drawthehydraulic
gradientandtotalenergyline.
Given:
L = 40 m
L1 = 25 m
D1 = 150 mm =0.15 m
L2 = 15 m
D2 = 300 mm = 0.3 m
Z = 8 m
f = 0.01
To find:
Q = AV
??????�??????�
1
2

49. Solution:
�
1
��
+
??????
1
2
2�
+??????
1=
�
2
��
+
??????
2
2
2�
+??????
2+ℎ
??????+ℎ
�1+ℎ
�+ℎ
�2
0 +0+8=0+
??????
2
2
2�
+0+ℎ
??????+ℎ
�1+ℎ
�+ℎ
�2
8=
??????
2
2
2�
+ℎ
??????+ℎ
�1+ℎ
�+ℎ
�2
�
�=
�.�??????
�
�
��
. =
8V
2
2
��
=
8x1.1132
2
��
= 0.5 m
�
�=
??????�−??????�
�
��
=
�??????�−??????�
�
��
=
9V
2
2
��
= 0.568 m
�
��=
���
�
×??????
�
�
���
= 106.66 V
2
2
/ 2g = 6.73 m
�
��=
���
�
×??????
�
�
���
= 2 V
2
2
/ 2g = 0.126 m
??????�??????�
1
2

50. A
1V
1= A
2V
2
V
1= (A
2V
2) / A
1
V
1= (d
2
2
V
2) / d
1
2
V
1= 4 V
2
Multiply with 2g
16??????=
??????
2
2
2??????
+8V
2
2
+106.66V
2
2
+9V
2
2
+2V
2
2
V
2=1.1132m/s
Q = A2 V2 =
�
4
d
2
2
* V
2= 0.0786 m
3
/s.
??????�??????�
1
2

51. TEL:�
�=�.���
�=�.����
�
��=�.����
��=�.����
�.��
�.���
�.����
�.����

52. HEL:
??????
�
�
��
=�.��
�.��
�.���
�.����
�.����
�.��

53. THANK YOU