Unit II -Flow Through Circular Conduits

1. Unit II - Flow Through Circular
Conduits

4. Find the head lost due to friction in a pipe of diameter 300 mm and
length 50 m, through which water is flowing at a velocity of 3 m/s
using (i) Darcy formula, (ii) Chezy’s formula for which C = 60.
Given:
D = 300 mm = 0.3 m
L = 50 m
V = 3 m/s
C = 60
𝛾 = 0.01 stoke = 0.01 cm2/s = 0.01 * 10-4 m2/s
To find:
hf
(i) Darcy formula
(ii) Chezy’s Formula
𝛾 𝜋 𝜇 𝜌 2
1
2

5. Darcy Eqn:
hf =
4 𝑓 𝐿 𝑉2
2 𝑔𝑑
Re =
𝑣 𝑑
𝛾
=
3 ∗ 0.3
0.01 ∗ 10−4 = 90 x 104 > 2000
Hence it is Turbulent flow
So, f =
0.079
𝑅𝑒0
.
25 =
0.079
(90 𝑥 104) 0.25
= 2.56 x 10-3
hf =
4 ∗2.56 x 10−3∗50 ∗ 32
2 ∗9.81 ∗0.3
=> hf = 0.7828 m
Chezy’s Eqn
V = C 𝑚𝑖 𝒐𝒓 𝒉𝒇 =
𝟒×𝑳×𝑽𝟐
𝑪𝟐
× 𝒅
V = C
𝑑
4
ℎ𝑓
𝐿
3 = 60
0.3
4
ℎ𝑓
50
hf = 1.665 m

6. Find the diameter of a pipe of length 2000 m when the rate of flow of water
through the pipe is 200 litres/s and the head lost due to friction is 4 m. Take the
value of C = 50 in chezy’s formulae.
Given
L = 2000 m
Q = 200 lit/s = 0.2 m3/s → Q = A V
hf = 4 m
C = 50
To find:
d
Solution:
V = C 𝑚𝑖

7. (or) ℎ𝑓 =
4×𝐿×𝑉2
𝐶2
× 𝑑
ℎ𝑓 =
4 𝑥 2000×𝑉2
𝐶2
× 𝑑
Discharge, Q = A V
0.2 =
𝜋
4
d2 * V
V =
0.8
𝜋 d2
ℎ𝑓 =
4 𝑥 2000×𝑉2
𝐶2
× 𝑑
4 =
4 𝑥 2000×
0.8
𝜋 d2
2
502
× 𝑑
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 – Weisbach Formula
(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
❖ The major loss of energy is due to friction
❖ The loss due to friction is much more in case of long pipe
lines.
❖ It depends on roughness of pipe, length, velocity and
diameter of pipe.
𝒉𝒇 =
𝟒 × 𝒇 × 𝑳 × 𝑽𝟐
𝟐 × 𝒈 × 𝒅
From DARCY’S WEISBACH EQUATION
𝒉𝒇 =
𝟒 × 𝑳 × 𝑽𝟐
𝑪𝟐 × 𝒅
From Chezy′s EQUATION

10. MINOR LOSS
❖ The losses due to disturbance in flow pattern is called
as minor loss
❖ Minor loss occurs due to
❖ Sudden Expansion
❖ Sudden Contraction
❖ Valves
❖ Fittings
❖ Bends

11. LOSS OF ENERGY DUE SUDDEN ENLARGEMENT
❖ Consider a horizonal pipe of
area A1 is suddenly enlarged to
the area A2
𝑷𝟏𝑨𝟏
𝑷𝟐𝑨𝟐
𝑽𝟏 𝑽𝟐
① ②
❖ Consider a two section of ①and
② is before and after
expansion.
❖ Let 𝑷𝟏𝑨𝟏𝒂𝒏𝒅 𝑽𝟏 is the Pressure
Intensity, Velocity at area A1
❖ Let 𝑷𝟐𝑨𝟐𝒂𝒏𝒅 𝑽𝟐is the Pressure Intensity, Velocity at area A2
❖ Let 𝑷′is the intensity of pressure, of the liquid Eddies on Area A2 – A1

12. LOSS OF ENERGY DUE SUDDEN ENLARGEMENT
❖ The resultant force between section ① & ②
𝑭 = 𝑷𝟏𝑨𝟏 − 𝑷𝟐𝑨𝟐 + 𝑷′
(𝑨𝟐 − 𝑨𝟏)
❖ But it has been found by experiment that
𝑷′
= 𝑷𝟏
❖ The resultant force
𝑭 = 𝑷𝟏𝑨𝟏 − 𝑷𝟐𝑨𝟐 + 𝑷𝟏(𝑨𝟐 − 𝑨𝟏)
𝑭 = 𝑷𝟏𝑨𝟏 − 𝑷𝟐𝑨𝟐 + 𝑷𝟏𝑨𝟐 − 𝑷𝟏𝑨𝟏
𝑭 = 𝑷𝟏 − 𝑷𝟐 𝑨𝟐

13. LOSS OF ENERGY DUE SUDDEN ENLARGEMENT
❖ Momentum of liquid / sec at section ①
= 𝑴𝒂𝒔𝒔 × 𝑽𝒆𝒍𝒐𝒄𝒊𝒕𝒚
= 𝝆𝑨𝟏𝑽𝟏 × 𝑽𝟏
= 𝝆𝑨𝟏𝑽𝟏
𝟐
❖ Momentum of liquid / sec at section ②
= 𝝆𝑨𝟐𝑽𝟐 × 𝑽𝟐
= 𝝆𝑨𝟐𝑽𝟐
𝟐

14. LOSS OF ENERGY DUE SUDDEN ENLARGEMENT
❖ Change in Momentum of liquid / sec
= 𝝆𝑨𝟐𝑽𝟐
𝟐
− 𝝆𝑨𝟏𝑽𝟏
𝟐
❖ From Continuity Equation
𝑨𝟏𝑽𝟏 = 𝑨𝟐𝑽𝟐 𝑨𝟏 =
𝑨𝟐𝑽𝟐
𝑽𝟏
= 𝝆𝑨𝟐𝑽𝟐
𝟐
− 𝝆
𝑨𝟐𝑽𝟐
𝑽𝟏
𝑽𝟏
𝟐
= 𝝆𝑨𝟐𝑽𝟐
𝟐
− 𝝆𝑨𝟐𝑽𝟏𝑽𝟐

15. LOSS OF ENERGY DUE SUDDEN ENLARGEMENT
❖ From Newton’s Second Law of motion
𝑭𝒐𝒓𝒄𝒆 = 𝑹𝒂𝒕𝒆 𝒐𝒇 𝒄𝒉𝒂𝒏𝒈𝒆 𝒐𝒇 𝒎𝒐𝒎𝒆𝒏𝒕𝒖𝒎 𝒐𝒇 𝒕𝒉𝒆 𝒎𝒂𝒔𝒔
𝑷𝟏 − 𝑷𝟐 𝑨𝟐 = 𝝆𝑨𝟐𝑽𝟐
𝟐
− 𝝆𝑨𝟐𝑽𝟏𝑽𝟐
𝑷𝟏 − 𝑷𝟐 𝑨𝟐 = 𝝆𝑨𝟐(𝑽𝟐
𝟐
− 𝑽𝟏𝑽𝟐)
𝑷𝟏 − 𝑷𝟐
𝝆
= (𝑽𝟐
𝟐
− 𝑽𝟏𝑽𝟐)

16. LOSS OF ENERGY DUE SUDDEN ENLARGEMENT
Applying Bernoulli's equation between section ① & ②
𝑷𝟏
𝝆𝒈
+
𝑽𝟏
𝟐
𝟐𝒈
+ 𝒛𝟏 =
𝑷𝟐
𝝆𝒈
+
𝑽𝟐
𝟐
𝟐𝒈
+ 𝒛𝟐 + 𝒉𝒆
𝑷𝟏
𝝆𝒈
−
𝑷𝟐
𝝆𝒈
= +
𝑽𝟐
𝟐
𝟐𝒈
+ 𝒛𝟐 −
𝑽𝟏
𝟐
𝟐𝒈
− 𝒛𝟏 + 𝒉𝒆
𝑷𝟏−𝑷𝟐
𝝆𝒈
= +
𝑽𝟐
𝟐
𝟐𝒈
+ 𝒛𝟐 −
𝑽𝟏
𝟐
𝟐𝒈
− 𝒛𝟏 + 𝒉𝒆

17. LOSS OF ENERGY DUE SUDDEN ENLARGEMENT
𝑽𝟐
𝟐
−𝑽𝟏𝑽𝟐
𝒈
= +
𝑽𝟐
𝟐
𝟐𝒈
+ 𝒛𝟐 −
𝑽𝟏
𝟐
𝟐𝒈
− 𝒛𝟏 + 𝒉𝒆
❖ Datum head of section ① & ② are equal
∴ 𝒛𝟏= 𝒛𝟐
𝑽𝟐
𝟐
−𝑽𝟏𝑽𝟐
𝒈
=
𝑽𝟐
𝟐
𝟐𝒈
−
𝑽𝟏
𝟐
𝟐𝒈
+ 𝒉𝒆 𝒉𝒆 =
𝟐𝑽𝟐
𝟐
−𝟐𝑽𝟏𝑽𝟐−𝑽𝟐
𝟐
−𝑽𝟏
𝟐
𝟐𝒈
𝒉𝒆 =
𝑽𝟐
𝟐
−𝟐𝑽𝟏𝑽𝟐−𝑽𝟏
𝟐
𝟐𝒈
𝒉𝒆 =
𝑽𝟏−𝑽𝟐
𝟐
𝟐𝒈

18. The rate of flow of water through a horizontal pipe is 0.3m3/sec. the diameter of the pipe,
which is 25cm, is suddenly enlarged to 50 cm. the pressure intensity in the smaller pipe is 14
N/cm2. determine the loss of head due to sudden enlargement, pressure intensity in the larger
pipe power lost due to enlargement.
Given
Q = 0.3 m3/sec
D1 = 25 cm = 0.25 m
D2 = 50 cm = 0.5 m
p1 = 14 N/cm2 = 14 * 104 N/m2
To find:
he
p2
P
Solution:
Q = A1 V1 => V1 = Q/A1 = Q/(
𝜋
4
d1
2 ) = 6.11 m/s
Q = A2 V2 => V2 = Q/A2 = Q/(
𝜋
4
d2
2 ) = 1.52 m/s

19. he =
𝑽𝟏−𝑽𝟐
𝟐
𝟐𝒈
=
𝟔.𝟏𝟏 −𝟏.𝟓𝟐 𝟐
𝟐∗𝟗.𝟖𝟏
he = 1.07 m
Applying Bernoulli's Equation,
𝑃1
𝜌𝑔
+
𝑉1
2
2𝑔
=
𝑃2
𝜌𝑔
+
𝑉2
2
2𝑔
+ ℎ𝑒
14 ∗ 104
1000 ∗9.81
+
6.112
2 ∗9.81
=
𝑃2
9.81 ∗1000
+
1.522
2 ∗9.81
+ 1.07
𝑷𝟐 = 𝟏𝟒. 𝟔𝟗 × 𝟏𝟎𝟒 N/m2
Power:
𝑷 = 𝝆 × 𝒈 × 𝑸 × 𝒉𝒆 = 1000 × 9.81 × 0.3 × 1.073
𝑷 = 𝟑. 𝟏𝟓𝟕𝟖 × 𝟏𝟎𝟑
W

20. LOSS OF ENERGY DUE SUDDEN CONTRACTION
𝑷𝟏𝑨𝟏 𝑷𝟐𝑨𝟐
𝑽𝟏
𝑽𝟐
① ②
©
❖ Consider a horizonal pipe of
area A1 is suddenly contracted
to the area A2
❖ As the liquid flows from
larger pipe to smaller pipe
the area of flow decreases
and becomes minimum of ©
❖ Let 𝑷𝟏𝑨𝟏𝒂𝒏𝒅 𝑽𝟏 is the Pressure
Intensity, Velocity at area A1
❖ Let 𝑷𝟐𝑨𝟐𝒂𝒏𝒅 𝑽𝟐is the Pressure Intensity, Velocity at area A2

21. LOSS OF ENERGY DUE SUDDEN CONTRACTION
𝑷𝟏𝑨𝟏 𝑷𝟐𝑨𝟐
𝑽𝟏
𝑽𝟐
① ②
©
❖ Actually the loss of head is due
to enlargement between section
© & ②
𝐡𝐜 =
𝑽𝒄−𝑽𝟐
𝟐
𝟐𝒈
𝐡𝐜 =
𝑽𝟐
𝟐
𝟐𝒈
𝑽𝒄
𝑽𝟐
− 𝟏
𝟐

22. LOSS OF ENERGY DUE SUDDEN CONTRACTION
❖ From Continuity Equation
𝑨𝒄𝑽𝒄 = 𝑨𝟐𝑽𝟐
𝑽𝒄
𝑽𝟐
=
𝑨𝟐
𝑨𝒄
𝑽𝒄
𝑽𝟐
=
𝟏
𝑨𝒄/𝑨𝟐
𝑽𝒄
𝑽𝟐
=
𝟏
𝑪𝒄

23. LOSS OF ENERGY DUE SUDDEN CONTRACTION
𝐡𝐜 =
𝑽𝟐
𝟐
𝟐𝒈
𝟏
𝑪𝒄
− 𝟏
𝟐
𝐡𝐜 =
𝒌𝑽𝟐
𝟐
𝟐𝒈 𝒔𝒊𝒏𝒄𝒆 𝒌 =
𝟏
𝑪𝒄
− 𝟏
𝟐
❖ If the valve of CC is not given, then
𝐡𝐜 =
𝟎.𝟓𝑽𝟐
𝟐
𝟐𝒈

24. In Fig shown below, when a sudden contraction is introduced in a horizontal pipe line from 50 cm to 25
cm, the pressure changes from 10,500 kg/m2 (103005 N/m2) to 6900 kg/m2 (67689 N/m2). Calculate the rate
of flow. Assume co-efficient of contraction of jet to be 0.65. Following this if there is a sudden enlargement
from 25 cm to 50cm and if the pressure at the 25 cm section is 6900 kg/m2 (67689 N/m2) what is the
pressure at the 50 cm enlarged section?
Given:
D1 = 50 cm = 0.5 m
D2 = 25 cm = 0.25 m
p1 = 103005 N/ m2
p2 = 67689 N/ m2
p3 = 67689 N/ m2
CC = 0.65
D4 = 50 cm = 0.5 m
D3= 25 cm = 0.25 m
To find:
Q
p4

25. Solution:
From formula, hc =
𝑘𝑉2
2
2𝑔
,
where, 𝑘 =
1
𝐶𝑐
− 1
2
=
1
0.65
− 1
2
= 0.289
Applying Bernoulli’s eqn
𝑃1
𝜌𝑔
+
𝑉1
2
2𝑔
=
𝑃2
𝜌𝑔
+
𝑉2
2
2𝑔
+ ℎ𝑐
103005
9.81 ∗1000
+
𝑉1
2
2 ∗9.81
=
67689
9.81 ∗1000
+
𝑉2
2
2 ∗9.81
+
0.289𝑉2
2
2𝑔
From Continuity Equation, A1 V1 = A2 V2
V1 = (A2 V2) / A1
V1 = (d2
2 V2) / d1
2
V1 = 0.25 V2

26. 103005
9.81 ∗1000
+
0.25 𝑉2 2
2 ∗9.81
=
67689
9.81 ∗1000
+
𝑉2
2
2 ∗9.81
+
𝟎.𝟐𝟖𝟗𝑽𝟐
𝟐
𝟐𝒈
V2 = 8.3 m/s
V1 = 0.25 V2 = 2.07 m/s
Discharge, Q = A2 V2 =
𝜋
4
d2
2 * V2
Q = 0.40 m3/s
Applying Bernoulli’s Equation for section 3 - 4
𝑃3
𝜌𝑔
+
𝑉3
2
2𝑔
=
𝑃4
𝜌𝑔
+
𝑉4
2
2𝑔
+ ℎ𝑒
Here, V2 = V3 = 8.3 m/s
V1 = V4 = 2.07 m/s
he = (V3 – V4)2/2g = (8.3 – 2.07)2 / 2 * 9.81 = 1.98 m
67689
9.81 ∗1000
+
8.32
2 ∗9.81
=
𝑃4
9.81 ∗1000
+
2.072
2 ∗9.81
+ 1.98
P4 = 80587.37 N/m2

27. LOSS OF ENERGY AT ENTRANCE AND EXIT
𝐡𝐢 =
𝟎.𝟓𝑽𝟏
𝟐
𝟐𝒈
❖ Loss of energy at Entrance
❖ Loss of energy at Exit
𝐡𝐨 =
𝑽𝟐
𝟐
𝟐𝒈

28. LOSS OF ENERGY DUE TO GRADUAL CONTRACTION AND EXPANSION
𝐡𝐠 =
𝒌 𝑽𝟏−𝑽𝟐
𝟐
𝟐𝒈
𝐡𝐛 =
𝒌𝑽𝟐
𝟐𝒈
LOSS OF ENERGY DUE TO BEND IN A PIPE

29. LOSS OF ENERGY DUE TO VARIOUS PIPE FITTINGS
𝐡𝐯 =
𝒌𝑽𝟐
𝟐𝒈
𝐡𝐨𝐛 =
𝑽𝟐
𝟐𝒈
𝑨
𝑪𝑪(𝑨−𝒂)
− 𝟏
LOSS OF ENERGY DUE TO SUDDEN OBSTRUCTION
𝑪𝒄 =
𝑨𝑪
(𝑨−𝒂)

30. S.No. Loss Of Energy Formulae
1 Loss Of Energy Due Sudden
Enlargement 𝒉𝒆 =
𝑽𝟏 − 𝑽𝟐
𝟐
𝟐𝒈
2 Loss Of Energy Due Sudden Contraction
𝐡𝐜 =
𝒌𝑽𝟐
𝟐
𝟐𝒈
Where, 𝒌 =
𝟏
𝑪𝒄
− 𝟏
𝟐
Cc ==
𝑨𝒄
𝑨𝟐
3 Loss Of Energy At Entrance
𝐡𝐢 =
𝟎. 𝟓𝑽𝟏
𝟐
𝟐𝒈
4 Loss Of Energy At Exit
𝐡𝐨 =
𝑽𝟐
𝟐
𝟐𝒈
5 Loss Of Energy Due To Gradual
Contraction And Expansion 𝐡𝐠 =
𝒌 𝑽𝟏 − 𝑽𝟐
𝟐
𝟐𝒈
6 Loss Of Energy Due To Bend In A Pipe
𝐡𝐛 =
𝒌𝑽𝟐
𝟐𝒈
7 Loss Of Energy Due To Various Pipe
Fittings 𝐡𝐯 =
𝒌𝑽𝟐
𝟐𝒈
8 Loss Of Energy Due To Sudden
Obstruction
𝐡𝐨𝐛 =
𝑽𝟐
𝟐𝒈
𝑨
𝑪𝑪(𝑨−𝒂)
− 𝟏 where, 𝑪𝒄 =
𝑨𝑪
(𝑨−𝒂)

31. Water is flowing through a horizontal pipe of diameter 200 mm at a velocity of 3 m/s.
A circular solid plate of diameter 150 mm is placed in the pipe to obstruct the flow.
Find the loss of head due to obstruction in the pipe if Cc = 0.62.
Given:
D = 200 mm = 0.2 m
A =
𝜋
4
D2 =
𝜋
4
0.22 =0.031 m2
V = 3 m/s
d = 150 mm = 0.15 m
a =
𝜋
4
d2 =
𝜋
4
0.152 =0.017 m2
Cc = 0.62
To find:
hob
Solution:
hob =
𝑉2
2𝑔
𝐴
𝐶𝐶(𝐴−𝑎)
− 1
hob =
32
2 ∗9.81
0.031
0.62(0.037−0.017)
− 1
hob = 3.315 m

32. A horizontal pipe line 40 m long is connected to a water tank at one end and discharges freely
into the atmosphere at the other end. For the first 25 m of its length from the tank, the pipe is
150 mm diameter and its diameter is suddenly enlarged to 300 mm. The height of water level in
the tank is 8 m above the center of the pipe. Considering all losses if head which occur,
determine the rate of flow. Take f = 0.01 for both sections of the pipe
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

33. Solution:
𝑃1
𝜌𝑔
+
𝑉1
2
2𝑔
+ 𝑍1 =
𝑃2
𝜌𝑔
+
𝑉2
2
2𝑔
+ 𝑍2 + ℎ𝑖 + ℎ𝑓1 +ℎ𝑒 + ℎ𝑓2
0 +0 + 8 = 0 +
𝑉2
2
2𝑔
+ 0 + ℎ𝑖 + ℎ𝑓1 +ℎ𝑒 + ℎ𝑓2
8 =
𝑉2
2
2𝑔
+ ℎ𝑖 + ℎ𝑓1 +ℎ𝑒 + ℎ𝑓2 --------1
Using continuity eqn, A1 V1 = A2 V2
V1 = (A2 V2) / A1
V1 = (d2
2 V2) / d1
2
V1 = 4 V2

34. 𝐡𝐢 =
𝟎.𝟓𝑽𝟏
𝟐
𝟐𝒈
. =
8 V2
2
𝟐𝒈
𝒉𝒆 =
𝑽𝟏−𝑽𝟐
𝟐
𝟐𝒈
=
𝟒 𝑽𝟐−𝑽𝟐
𝟐
𝟐𝒈
=
9 V2
2
𝟐𝒈
𝒉𝒇𝟏 =
𝟒×𝒇×𝑳𝟏×𝑽𝟏
𝟐
𝟐×𝒈×𝒅
= 106.66 V2
2
𝒉𝒇𝟐 =
𝟒×𝒇×𝑳𝟐×𝑽𝟐
𝟐
𝟐×𝒈×𝒅
= 2 V2
2
Multiply 2g
16𝑔 =
𝑉2
2
2𝑔
+ 8 V2
2 + 106.66V2
2+ 9 V2
2 + 2V2
2
V2 = 1.1132 m/s
Q = A2 V2 =
𝜋
4
d2
2 * V2
Q = 0.0786 m3/s.
𝛾 𝜋 𝜇 𝜌
1
2

35. WHEN PIPE ARE CONNECTED SERIES
𝑸 = 𝑸𝟏 = 𝑸𝟐 = 𝑸𝟑
𝑯 = 𝒉𝒇𝟏 + 𝒉𝒇𝟐 + 𝒉𝒇𝟑
Neglecting Minor loss
𝑯 = 𝒉𝒊 + 𝒉𝒇𝟏 + 𝒉𝒄 + 𝒉𝒇𝟐 + 𝒉𝒆 + 𝒉𝒇𝟑 + 𝒉𝒐
Considering Minor loss

36. The difference in water surface levels in two tanks, which are connected by three pipes in series
of lengths 300 m, 170 m and 210 m and of diameters 300 mm, 200 mm and 400 mm respectively
is 12 m. Determine the rate of flow of water if co-efficient of friction are 0.005, 0.0052 and
0.0048 respectively, considering: (i) minor losses also (ii) neglecting minor losses.
Given:
L1 = 300 m; L2 = 170 m; L3 = 210 m
D1 = 300 mm = 0.3m
D2 = 200 mm = 0.2 m
D3 = 400 mm = 0.4 m
H = 12 m
f1 = 0.005
f2 = 0.0052
f3 = 0.0048
To find:
Q
(i) All losses
(ii) Neglect Minor losses

37. From continuity eqn, A1 V1 = A2 V2 = A3 V3
A1 V1 = A2 V2
V1 = (A2 V2) / A1
V1 = (d2
2 V2) / d1
2 = (0.22 V2) / 0.3
V1 = 0.4442 V2
A2 V2 = A3 V3
V3 = (A2 V2) / A3
V3 = (d2
2 V2) / d3
2 = (0.22 V2) / 0.4
V3 = 0.25 V2

38. Solution:
(i) Considering All Losses
H = ℎ𝑖 + ℎ𝑓1 +ℎ𝑐 + ℎ𝑓2 + ℎ𝑒 + ℎ𝑓3 +ℎ𝑜
12 = ℎ𝑖 + ℎ𝑓1 +ℎ𝑐 + ℎ𝑓2 + ℎ𝑒 + ℎ𝑓3 +ℎ𝑜
Losses:
𝐡𝐢 =
𝟎.𝟓𝑽𝟏
𝟐
𝟐𝒈
. =
𝟎.𝟓 ∗ (𝟎.𝟒𝟒𝟒 𝑽𝟐
)𝟐
𝟐𝒈
= 0.098 V2
2 / 2g
𝒉𝒆 =
𝑽𝟐−𝑽𝟑
𝟐
𝟐𝒈
=
𝑽𝟐
−𝟎.𝟐𝟓𝑽𝟐
𝟐
𝟐𝒈
= 0.5625 V2
2 / 2g
𝐡𝐨 =
𝑽𝟑
𝟐
𝟐𝒈
= 0.0625 V2
2 / 2g
𝐡𝐜 =
𝒌𝑽𝟐
𝟐
𝟐𝒈
=
𝟎.𝟓 𝑽𝟐
𝟐
𝟐𝒈
𝒉𝒇𝟏 =
𝟒×𝒇𝟏
×𝑳𝟏
×𝑽𝟏
𝟐
𝟐×𝒈×𝒅𝟏
=
𝟒×𝟎.𝟎𝟎𝟓 ×𝟑𝟎𝟎×(𝟎.𝟒𝟒𝟒 𝑽𝟐
)𝟐
𝟐×𝒈×𝟎.𝟑
= 3.942 V2
2 / 2g
𝒉𝒇𝟐 =
𝟒×𝒇𝟐
×𝑳𝟐
×𝑽𝟐
𝟐
𝟐×𝒈×𝒅𝟐
=
𝟒×𝟎.𝟎𝟎𝟓𝟐 ×𝟏𝟕𝟎×𝑽𝟐
𝟐
𝟐×𝒈×𝟎.𝟐
= 17.68 V2
2 / 2g
𝒉𝒇𝟑 =
𝟒×𝒇𝟑
×𝑳𝟑
×𝑽𝟑
𝟐
𝟐×𝒈×𝒅𝟑
=
𝟒×𝟎.𝟎𝟎𝟒𝟖 ×𝟐𝟏𝟎×(𝟎.𝟐𝟓𝑽𝟐
)𝟐
𝟐×𝒈×𝟎.𝟒
= 0.63 V2
2 / 2g
𝛾 𝜋 𝜇 𝜌
1
2

39. Multiplying with 2g,
24g = 0.098 V2
2 + 3.942 V2
2 + 𝟎. 𝟓 𝑽𝟐
𝟐
+ 17.68 V2
2 + 0.5625 V2
2 + 0.63 V2
2 + 𝟎. 𝟎𝟔𝟐𝟓 𝑽𝟐
𝟐
V2 = 3.167 m/s
Q = A2 V2 =
𝜋
4
d2
2 * V2 =
𝜋
4
0.2 2 * 3.167
Q = 0.09944 m3/s = 99.44 lit/s
(ii) Considering Major Losses
H = ℎ𝑓1 + ℎ𝑓2 + ℎ𝑓3
24g = 3.942 V2
2 + 17.68 V2
2 + 0.63 V2
2
V2 = 3.353 m/s
Q = A2 V2 =
𝜋
4
d2
2 * V2 =
𝜋
4
0.2 2 * 3.353
Q = 0.102 m3/s = 102.18 lit/s

40. EQUIVALENT PIPE
𝐿
𝑑5
=
𝐿1
𝑑1
5 +
𝐿2
𝑑2
5 +
𝐿3
𝑑3
5
where, L = L1 + L2 +L3
The above equation is known as Dupuit’s equation.

41. Three pipes of lengths 800 m, 500 m and 400 m and of diameters 500 mm, 400 mm and
300 mm respectively are connected in series. These pipes are to be replaced by a single
pipe of length 1700 m. find the diameter of the single pipe.
Given:
L1 = 800 m; L2 = 500 m; L3 = 400 m
L = 1700 m
D1 = 500 mm = 0.5 m; D2 = 400 mm = 0.4 m; D3 = 300 mm = 0.3 m
To find:
D
Solution:
𝐿
𝐷5 =
𝐿1
𝐷1
5 +
𝐿2
𝐷2
5 +
𝐿3
𝐷3
5
1700
𝐷5 =
800
0.52 +
500
0.42 +
400
0.32
D = 0.37 mm

42. WHEN PIPE ARE CONNECTED PARALLEL
𝑸 = 𝑸𝟏 + 𝑸𝟐
𝒉𝒇 = 𝒉𝒇𝟏 = 𝒉𝒇𝟐

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
𝑯 =
𝟒×𝒇×𝑳×𝑽𝟐
𝟐×𝒈×𝒅
=> 20 =
𝟒×𝟎.𝟎𝟏𝟓×𝟐𝟎𝟎𝟎×𝑽𝟐
𝟐×𝟗.𝟖𝟏×𝟎.𝟐
=> v = 0.808 m/s
Qold =
𝜋
4
d2 * V => Q = 0.025 m3/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 =
𝜋
4
d2 =
𝜋
4
0.22 = 0.0314 m2
To find:
Qnew – Qold
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)
Q1 = 2Q2 or Q1 = 2Q3
Loss of head:
H=
𝟒×𝒇𝟏
×𝑳𝟏
×𝑽𝟏
𝟐
𝟐×𝒈×𝒅𝟏
+
𝟒×𝒇𝟐
×𝑳𝟐
×𝑽𝟐
𝟐
𝟐×𝒈×𝒅𝟐
(We know, hf2 = hf3)
20 =
𝟒×𝟎.𝟎𝟏𝟓×𝟖𝟎𝟎×𝑽𝟏
𝟐
𝟐×𝟗.𝟖𝟏×𝟎.𝟐
+
𝟒×𝟎.𝟎𝟏𝟓×𝟏𝟐𝟎𝟎×𝑽𝟐
𝟐
𝟐×𝟗.𝟖𝟏×𝟎.𝟐
From Continuity equation,
Q1 = A1 V1 => V1 = Q1/A1
Q2 = A2 V2 => V2 = Q2/A2 = Q1/2A2

46. 20 =
𝟒×𝟎.𝟎𝟏𝟓×𝟖𝟎𝟎
𝟐×𝟗.𝟖𝟏×𝟎.𝟐
x(
𝑄1
0.0314
)2 +
𝟒×𝟎.𝟎𝟏𝟓×𝟏𝟐𝟎𝟎
𝟐×𝟗.𝟖𝟏×𝟎.𝟐
x(
𝑄1
0.0628
)2
Q1 = 0.076 m3/s
Q2 = Q1 / 2
Qold = 0.025 m3/s
Qnew = Q1 = 0.076 m3/s
Qnew – Qold = 0.076 – 0.025 = 0.051 m3/s

47. HYDRAULIC ENERGY LINE (OR) HYDRAULIC GRADIENT LINE
❖ It is defined as a line which gives the sum of
pressure head and datum head of flowing fluid in
pipe with respect to some reference line. (H.G.L.)
TOTAL ENERGY LINE (OR) TOTAL GRADIENT LINE
❖ It is defined as a line which gives the sum of
pressure head, datum head and kinetic head of
flowing fluid in pipe with respect to some
reference line. (T.E.L)

48. A horizontal pipe line 40 m long is connected to a water tank at one end and discharges freely
into the atmosphere at the other end. For the first 25 m of its length from the tank, the pipe is
150 mm diameter and its diameter is suddenly enlarged to 300 mm. The height of water level in
the tank is 8 m above the center of the pipe. Considering all losses if head which occur,
determine the rate of flow. Take f = 0.01 for both sections of the pipe. Draw the hydraulic
gradient and total energy line.
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
𝛾 𝜋 𝜇 𝜌
1
2

49. Solution:
𝑃1
𝜌𝑔
+
𝑉1
2
2𝑔
+ 𝑍1 =
𝑃2
𝜌𝑔
+
𝑉2
2
2𝑔
+ 𝑍2 + ℎ𝑖 + ℎ𝑓1 +ℎ𝑒 + ℎ𝑓2
0 +0 + 8 = 0 +
𝑉2
2
2𝑔
+ 0 + ℎ𝑖 + ℎ𝑓1 +ℎ𝑒 + ℎ𝑓2
8 =
𝑉2
2
2𝑔
+ ℎ𝑖 + ℎ𝑓1 +ℎ𝑒 + ℎ𝑓2
𝐡𝐢 =
𝟎.𝟓𝑽𝟏
𝟐
𝟐𝒈
. =
8 V2
2
𝟐𝒈
=
8 x1.1132 2
𝟐𝒈
= 0.5 m
𝒉𝒆 =
𝑽𝟏−𝑽𝟐
𝟐
𝟐𝒈
=
𝟒 𝑽𝟐−𝑽𝟐
𝟐
𝟐𝒈
=
9 V2
2
𝟐𝒈
= 0.568 m
𝒉𝒇𝟏 =
𝟒×𝒇×𝑳𝟏×𝑽𝟏
𝟐
𝟐×𝒈×𝒅
= 106.66 V2
2 / 2g = 6.73 m
𝒉𝒇𝟐 =
𝟒×𝒇×𝑳𝟐×𝑽𝟐
𝟐
𝟐×𝒈×𝒅
= 2 V2
2/ 2g = 0.126 m
𝛾 𝜋 𝜇 𝜌
1
2

50. A1 V1 = A2 V2
V1 = (A2 V2) / A1
V1 = (d2
2 V2) / d1
2
V1 = 4 V2
Multiply with 2g
16𝑔 =
𝑉2
2
2𝑔
+ 8 V2
2 + 106.66V2
2+ 9 V2
2 + 2V2
2
V2 = 1.1132 m/s
Q = A2 V2 =
𝜋
4
d2
2 * V2 = 0.0786 m3/s.
𝛾 𝜋 𝜇 𝜌
1
2

51. TEL: 𝒉𝒊 = 𝟎. 𝟓 𝒎 𝒉𝒆 = 𝟎. 𝟓𝟔𝟖 𝒎
𝒉𝒇𝟏 = 𝟔. 𝟕𝟑 𝒎 𝒉𝒇𝟐 = 𝟎. 𝟏𝟐𝟔 𝒎
𝟎. 𝟓 𝒎
𝟔. 𝟕𝟑 𝒎
𝟎. 𝟓𝟔𝟖 𝒎
𝟎. 𝟏𝟐𝟔 𝒎

52. HEL:
𝑽𝟏
𝟐
𝟐𝒈
= 𝟏. 𝟎 𝒎
𝟎. 𝟓 𝒎
𝟔. 𝟕𝟑 𝒎
𝟎. 𝟓𝟔𝟖 𝒎
𝟎. 𝟏𝟐𝟔 𝒎
𝟏. 𝟎 𝒎

53. THANK YOU