Unit6 energy loss in pipelines

LOSSES ENERGY IN PIPELINES J3008/6/1
ENERGY LOSS IN PIPELINES
OBJECTIVES
General Objective : To know, understand and apply Bernoulli’s equation to pipeline
systems
Specific Objectives : At the end of the unit, you should be able to :
 sketch the velocity profile in circular pipe system
 explain and calculate energy loss in pipeline system
 calculate and apply energy loss equation from reservoir
 solve problem related to the pipeline system
UNIT 6

6.0 INTRODUCTION
A pipe is defined as a closed conduit of circular section through which the
fluid flows, filling the complete cross-section. The fluid in the pipe has no free
surface. It will be at a pressure which may vary along the pipe. Losses of energy in a
pipeline cannot be ignored. When the shock losses and friction loss have been
determined, they are inserted in Bernoulli’s equation in the usual way.
hz
g
v
g
p
z
g
v
g
p
++=++ 2
2
22
1
2
11
22 ρρ
Velocity profile in circular pipe system (refer Figure 6.1)
Losses of energy in pipe line are due to :
a) shock loss at sudden enlargement
b) shock loss at sudden contraction
c) frictional resistance to flow
d) loss at entry
e) loss at rounded exit
INPUTINPUT
Figure 6.1
rough pipe wall
smooth pipe wall

Loss of head at enlargement, hL
( )
g
vv
2
2
21 −
=
6.1 SHOCK LOSS AT SUDDEN ENLARGEMENT
Figure 6.2 shows the loss of head when a pipe undergoes a sudden increase in
diameter. To calculate the loss the following equation is given.
When,
v1 = velocity in the smaller pipe upstream of the enlargement
v2 = velocity in the larger pipe
If hL = head lost at the enlargement, then by Bernoulli’s theorem,
Lh
g
vp
g
vp
++=+
22
2
22
2
11
ωω
Special case: When a pipe discharges into a large reservoir through a sharp
exit, conditions are equivalent to a sudden enlargement (refer Figure 6.2).
v1 = pipe velocity = v
v2 = reservoir velocity = 0
Figure 6.2

Loss of head at sharp exit into reservoir, hL
g
v
2
2
=
If the exit is rounded, this loss is greatly reduced and is usually negligible
Example 6.1
A pipe carrying 1800 l/min of water increases suddenly from 10 cm to 15 cm diameter.
Find
a) the head loss due to the sudden enlargement
b) the difference in pressure in kN/m2
in the two pipes
Solution to Example 6.1
a) 1 liter = 0.001 m3
1800 l = 1.8 m3
smmQ /03.0min/8.1 33
==
QA = AA vA = AB vB
So that
vA =Q/AA
( )
4
1.0
03.0
2
π
=
sm /8917.3=
Figure 6.3

vB =Q/AB
( )
4
15.0
03.0
2
π
=
sm /697.1=
Head loss of enlargement, hL =
( )
g
vv BA
2
2
−
( )
( )81.92
697.18197.3
2
−
=
waterofm2294.0=
b) Difference in pressure;
LB
BB
A
AA
hz
g
v
g
p
z
g
v
g
p
+++=++
22
22
ρρ
zA=zB
( )
L
ABBA
h
g
vv
g
pp
+
−
=
−
2
22
( ) ( )9810596.0
62.19
8197.3697.1 22
+
−
=− BA pp
2
/56.3602 mN=
6.2 SHOCK LOSS AT SUDDEN CONTRACTION

Loss of head at sudden contraction, hC
g
v
Cc 2
1
1
2
2
2






−=
Loss of head at sharp entrance, hC 





=
g
v
22
1 2
Figure 6.4
In a sudden contraction, the flow converges to form a vena contracta at section
(3) in the smaller pipe. The loss of energy in the convergence from sections (1)
to (3) is small and the main loss occurs in the enlargement from sections (3) to
(2). It is usual to ignore the loss from sections (1) to (3) and treat the loss from
(3) to (2) as if it was due to a sudden enlargement from the area of the vena
contracta ac to the area a2 of the smaller pipe.( Figure 6.4 )
Loss of head =
( )
g
vvC
2
2
2−
For continuity of flow
CC vava =22
If the coefficient of contraction =
C
C
C
v
a
C =
C
C
a
va
v 22
=
2
1
v
CC
=
Special case: if the entrance of a pipe line from a reservoir is sharp (no
rounded or bell-mouthed) it is equivalent to a sudden contraction from a pipe
of infinite size to that of the pipe line. The loss of head at sharp entrance is

v = velocity in the pipe
Example 6.2
A pipe carrying 0.06 m3
/s suddenly contracts from 200 mm to 150 mm diameter.
Assuming that the vena contracta is formed in the smaller pipe, calculate the
coefficient of contraction if the pressure head at a point upstream of the contraction is
0.655 m greater than at a point just downstream of the vena contracta.
Inserting this expression for the loss of head in Bernoulli’s equation,
22
2
2
22
2
11
1
1
222 





−++=+
CCg
v
g
v
g
p
g
v
g
p
ρρ
g
v
Cg
v
g
pp
C 2
1
1
1
2
2
1
22
221
−














−+=
−
ρ
Given, 655.021
=
−
g
pp
ρ
Using the continuity of flow Q = Av for velocity v1 and v2.
1
1
A
Q
v =
( )2
2.0
406.0
π
×
=
sm /91.1=
2
2
A
Q
v =
( )2
15.0
406.0
π
×
=
sm /4.3=
Thus,
( ) ( )
81.92
91.1
1
1
1
81.92
4.3
655.0
222
×
−














−+
×
=
CC

65.31
1
16.1186.12
2
−














−+=
CC
39.1
6.11
51.16
1
1
1
2
==














−+
CC
39.01
1
2
=





−
CC
625.01
1
=−





CC
625.1
1
=
CC
coefficient of contraction, 615.0=CC
6.3 FRICTION LOSS

Loss of head due to friction, hf
g
v
d
fL
2
4 2
=
When there is a loss of head due to friction in a pipeline in terms of the
velocity head; we assume that the frictional resistance per unit area of the pipe
wall is proportional to the square of the mean velocity of flow.
Consider a cylinder of fluid of length L completely filling the pipe of cross
sectional area A and moving with a mean velocity v ( Figure 6.5).
The force acting on the cylinder is the force due to pressure difference and the
force due to frictional resistance. Since the velocity is constant and there is no
acceleration, the resultant of these two forces in the direction of motion must
be zero.
Example 6.3
Determine the loss of head due to friction in a pipe 14 m long and 2 m diameter which
carries 1.5 m/s oil. Take into consideration f = 0.05.
g
v
d
fL
hf
2
4 2
=
( )( )
( )81.92
5.1
2
1405.04 2
×=
oilofm16.0=
ACTIVITY 6A
Figure 6.5
A

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT
INPUT…!
6.1 Water flows vertically downwards through a 150 mm diameter pipe with a
velocity of 2.4 m/s. The pipe suddenly enlarges to 300 mm in diameter. Find
the loss of head. If the flow is reversed, find the loss of head, assuming the
coefficient of contraction now being 0.62.
FEEDBACK ON ACTIVITY 6A

6.1
loss of head at sudden enlargement, hL
( )
g
vv
hL
2
2
21 −
=
using continuity of flow,
Q1 = Q2
A1v1 = A2v2
2
11
2
A
vA
v =
2
2
1
2
1
d
vd
=
( ) ( )
( )2
2
3.0
4.215.0
=
sm /6.0=
( )
g
vv
hL
2
2
21 −
=
( )
( )81.92
6.04.2
2
−
= m165.0=
loss of head at sudden contraction, hC

g
v
C
h
C
C
2
1
1
2
2
2






−=
( )
( )81.92
4.2
1
62.0
1
22






−=
m110.0=
INPUTINPUT

6.4 PIPELINE PROBLEMS
All pipeline problems should be solved by applying Bernoulli’s theorem
between points for which the total energy is known and including expressions
for any loss of energy due to shock or to friction, thus
lossfrictionallossshockz
g
v
g
p
z
g
v
g
p
+++=++ 2
2
22
1
2
11
22 ρρ
6.4.1 Discharge to atmosphere
g
v
d
fL
pipeinFriction
2
4 2
=
g
v
entryatLoss
22
1 2
=
To understand the discharge to atmosphere, let’s look at Example 6.4.
Example 6.4
Water from a large reservoir is discharged to atmosphere through a 100 mm
diameter pipe 450 m long. The entry from the reservoir to the pipe is sharp and
the outlet is 12 m below the surface level in the reservoir. Taking f = 0.01 in
the Darcy formula, calculate the discharge.(refer to Figure 6.6)
Figure 6.6

Apply Bernoulli’s theorem to A and B, assuming velocity at A is zero and that
pA = pB = atmospheric pressure
Total energy at A = Total energy at B + loss at entry + frictional loss
g
v
d
fL
g
v
g
v
H
2
4
22
1
2
222
++=
Putting H = 12 m, f = 0.01, L = 450 m, d = 100 mm = 0.1 m






++=
d
fL
g
v
H
4
2
1
1
2
2





 ××
+=
1.0
45001.04
5.1
2
12
2
g
v
g
v
2
5.181
2
=
smv /14.1=
Discharge v
d
4
2
π
=
( ) ( )14.1
4
1.0
2
π
=
sm /1096.8 33−
×=
6.4.2 Pipe in series

( )
g
vv
ementenlatLoss
2
arg
2
21 −
=
To understand the pipe in series, let’s look at Example 6.5.
Example 6.5
Water is discharged from a reservoir into the atmosphere through a pipe 39 m
long. There is a sharp entrance to the pipe and the diameter is 50 mm for the
first 15 m from the entrance. The pipe then enlarges suddenly to 75 mm in
diameter for the remainder of its length. Taking into account the loss of head at
entry and at the enlargement, calculate the difference of level between the
surface of the reservoir and the pipe exit which will maintain a flow of 2.8
dm3
/s. Take f as 0.0048 for the 50 mm pipe and 0.0058 for the 75 mm pipe.
Figure 6.7

2
2
21
2
1
4
1
4
1
vdvdQ ππ ==
2
1
1
4
d
Q
v
π
=
( )2
3
05.0
108.24
π
−
××
=
sm /426.1=
2
2
2
4
d
Q
v
π
=
( )2
3
075.0
108.24
π
−
××
=
sm /634.0=
Applying Bernoulli’s equation to A and B at which pA = pB = atmospheric
pressure and vA = 0, for unit weight
CDinlossfrictionalCatlossshock
BCinlossfrictionalBatlossshockDatenergytotalAatenergyTotal
++
++=
There is no shock loss at D as discharge is to atmosphere.
g
v
DatenergyTotal
2
2
2
=
( )
( )81.92
634.0
2
=
m020.0=
g
v
BentryatLoss
22
1
,
2
1
=
( )
( )81.92
426.1
2
1
2
=

m052.0=
g
v
d
fL
BCinlossFrictional
2
4
2
1
1
1
=
( )( ) ( )
( )81.92
426.1
050.0
150048.04
2
=
m597.0=
( )
g
vv
CatlossShock
2
2
21 −
=
( )
( )81.92
634.0426.1
2
−
=
m032.0=
g
v
d
fL
CDinlossFrictional
2
4
2
2
2
2
=
( )( ) ( )
( )81.92
634.0
075.0
240058.04
2
=
m152.0=
Difference of level = H = 0.02 + 0.052 + 0.597 + 0.032 + 0.152
= 0.853 m of water
6.4.3 Hydraulic Gradient

Figure 6.8
To understand the hydraulic gradient, let’s look at Example 6.6.
Example 6.6
Two reservoirs are connected by a pipeline which is 150 mm in diameter for
the first 6 m and 225 mm in diameter for the remaining 15 m. The entrance and
exit are sharp and the change of section is sudden. The water surface in the
upper reservoir is 6 m above that in the lower. Tabulate the losses of head
which occur and calculate the rate of flow in m3
/s. Friction coefficient f is 0.01
for both pipes.
Since d1 = 150 mm and d2 = 225 mm
22
2
1
4
9
150
225
vvv 





=





=
The losses are

g
v
entryatLoss
22
1
2
1
=
g
v
24
9
2
1
2
2
2






=
g
v
2
53.2
2
2
=
g
v
d
fL
pipeminlossFrictional
2
4
6
2
1
1
1
=
( )( )
g
v
215.0
601.04
2
1
=
g
v
2
6.1
2
1
=
g
v
24
9
6.1
2
2
2






=
g
v
2
1.8
2
2
=
( )
g
vv
ementenlatlossShock
2
arg
2
21 −
=
22
2
1
4
9
2






−=
g
v
g
v
2
56.1
2
2
=
g
v
d
fL
pipeminlossFrictional
2
4
15
2
2
2
2
=

( )( )
g
v
2225.0
1501.04
2
2
=
g
v
2
67.2
2
2
=
g
v
exitatlossShock
2
2
2
=
g
v
2
00.1
2
2
=
Total loss of head =
g
v
2
53.2
2
2
g
v
2
1.8
2
2
+
g
v
2
56.1
2
2
+
g
v
2
67.2
2
2
+
g
v
2
00.1
2
2
+
g
v
2
86.15
2
2
=
Applying Bernoulli’s Equation to A and B for unit weight
lossesBatenergytotalAatenergyTotal +=
Pressures at A and B are equal and if the reservoirs are large the velocities will be
zero. Taking datum level at B,
H = 0 + losses
g
v
2
86.156
2
2
=
so
86.15
26
2
g
v
×
=
sm /72.2=
Discharge v
d
4
2
π
=
( ) ( )72.2
4
225.0
2
π
=

sm /185.0 3
=
ACTIVITY 6B

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT
INPUT…!
6.2 According to the figure below, list out the losses of head which occur, giving
an expression for each.
FEEDBACK ON ACTIVITY 6B

Answers:
6.1 The losses of head which will occur are as follows:
1. Loss at entry
g
v
22
1
2
1
=
2. Friction loss
g
v
d
fL
2
4
2
1
1
1
=
3. Loss at sudden enlargement
( )
g
vv
2
2
21 −
=
4. Friction loss
g
v
d
fL
2
4
2
2
2
2
=
5. Loss at exit
g
v
2
2
2
=
SELF-ASSESSMENT

You are approaching success. Try all the questions in this self-assessment section
and check your answers with those given in the Feedback on Self-Assessment If you
face any problems, discuss it with your lecturer. Good luck.
6.1 Water is discharged from a reservoir into the atmosphere through a pipe 80 m long.
There is a sharp entrance to the pipe and the diameter is 250 mm for the first 50
m. The outlet is 35 m below the surface level in the reservoir. The pipe then
enlarges suddenly to 450 mm in diameter for the reminder of its length. Take
f = 0.004 for both pipes. Calculate the discharge.
6.2 Two reservoirs have a difference in level of 9 m and are connected by a pipe line,
which is 38 mm in diameter for the first 13 m and 23 mm for the remaining 6
m. Take f = 0.01 for both pipes and CC = 0.66. Calculate the discharge.
6.3 A pipe carrying 0.056 m3
/s suddenly changes diameter from;
a) 200 mm to 150 mm
b) 300 mm to 150 mm
c) 450 mm to 150 mm
Find the loss of head and the pressure difference across the contraction in each
case, given CC = 0.62.
FEEDBACK ON SELF-ASSESSMENT
Answers:

6.1 Q = 0.623 m3
/s
6.2 Q = 0.00345 m3
/s
6.3 (a) 0.19 m, 0.54 m
(b) 0.19 m, 0.673 m
(c) 0.19 m, 0.699 m

Unit6 energy loss in pipelines

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