unit 6-fluid mechanics

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.

hzg

vg

pzg

vg

p++=++ 2

222

1

211

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

INPUT

Figure 6.1

rough pipe wall

smooth pipe wall


Loss of head at enlargement, hL ( )

gvv

2

221 −=

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,

Lhg

vpg

vp++=+

22

222

211

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 gv2

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

( )

41.0

03.02

π= sm /8917.3=

Figure 6.3


vB =Q/AB

( )

415.003.0

2

π=

sm /697.1=

Head loss of enlargement, hL = ( )

gvv BA

2

2−

( )( )81.92

697.18197.3 2−=

waterofm2294.0=

b) Difference in pressure;

LBBB

AAA hz

gv

gp

zg

vg

p+++=++

22

22

ρρ

zA=zB

( )L

ABBA hgvv

gpp

+−

=−

2

22

( ) ( )9810596.0

62.198197.3697.1 22

+−

=− BA pp

2/56.3602 mN=


Loss of head at sudden contraction, hC g

vCc 2

11 22

2

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

6.2 SHOCK LOSS AT SUDDEN CONTRACTION

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 = ( )

gvvC

2

22−

For continuity of flow CC vava =22

If the coefficient of contraction = C

CC v

aC =

C

C ava

v 22=

21 v

CC

=


Loss of head at sharp entrance, hC ⎟⎟⎠

⎞⎜⎜⎝

⎛=

gv22

1 2

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. Solution to Example 6.2 Inserting this expression for the loss of head in Bernoulli’s equation,

222

222

211 11

222 ⎟⎟⎠

⎞⎜⎜⎝

⎛−++=+

CCgv

gv

gp

gv

gp

ρρ

g

vCg

vgpp

C 2111

2

21

22221 −

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

−ρ

Given, 655.021 =−gpp

ρ

Using the continuity of flow Q = Av for velocity v1 and v2.

1

1 AQv =

( )22.0

406.0π

×=

sm /91.1=

2

2 AQv =

( )215.0

406.0π

×=

sm /4.3=


Thus,

( ) ( )81.92

91.111181.92

4.3655.0222

×−

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+

×=

CC

65.31116.1186.122

−⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

CC

39.16.1151.16111

2

==⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+

CC

39.0112

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

CC

625.011=−⎟⎟

⎠

⎞⎜⎜⎝

⎛

CC

625.11=

CC

coefficient of contraction, 615.0=CC


Loss of head due to friction, hf gv

dfL

24 2

=

6.3 FRICTION LOSS

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.

Solution to Example 6.3

g

vdfLh f 2

4 2

=

( )( )( )81.92

5.12

1405.04 2

×=

oilofm16.0=

Figure 6.5

A


ACTIVITY 6A

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

( )

gvv

hL 2

221 −=

using continuity of flow, Q1 = Q2 A1v1 = A2v2

2

112 A

vAv =

22

12

1

dvd

=

( ) ( )( )2

2

3.04.215.0

=

sm /6.0=

( )

gvv

hL 2

221 −=

( )( )81.92

6.04.2 2−= m165.0=


loss of head at sudden contraction, hC

gv

Ch

CC 2

11 22

2

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

( )( )81.92

4.2162.01 22

⎟⎠⎞

⎜⎝⎛ −=

m110.0=


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

lossfrictionallossshockzg

vg

pzg

vg

p+++=++ 2

222

1

211

22 ρρ

6.4.1 Discharge to atmosphere

g

vdfLpipeinFriction

24 2

=

gventryatLoss22

1 2

=

To understand the discharge to atmosphere, let’s look at Example 6.4.

INPUT

Figure 6.6


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)


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

gv

dfL

gv

gvH

24

221

2

222

++=

Putting H = 12 m, f = 0.01, L = 450 m, d = 100 mm = 0.1 m

⎟⎠⎞

⎜⎝⎛ ++=

dfL

gvH 4

211

22

⎟⎠⎞

⎜⎝⎛ ××

+=1.0

45001.045.12

122

gv

g

v2

5.1812

=

smv /14.1=

Discharge vd4

2π=

( ) ( )14.141.0 2π

=

sm /1096.8 33−×=


6.4.2 Pipe in series

( )

gvv

ementenlatLoss2

arg2

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



22

212

1 41

41 vdvdQ ππ ==

21

14dQv

π=

( )2

3

05.0108.24

π

−××=

sm /426.1=

22

24dQv

π=

( )2

3

075.0108.24

π

−××=

sm /634.0=

Applying Bernoulli’s equation to A and B at which pA = pB = atmospheric pressure and vA = 0, for unit weight

CDinlossfrictionalCatlossshockBCinlossfrictionalBatlossshockDatenergytotalAatenergyTotal

++++=

There is no shock loss at D as discharge is to atmosphere.

gv

DatenergyTotal2

22=

( )( )81.92634.0 2

=

m020.0=


gv

BentryatLoss22

1,2

1=

( )( )81.92426.1

21 2

=

m052.0=

g

vdfLBCinlossFrictional

24 2

1

1

1=

( )( ) ( )( )81.92426.1

050.0150048.04 2

=

m597.0=

( )

gvv

CatlossShock2

221 −=

( )( )81.92

634.0426.1 2−=

m032.0=

gv

dfLCDinlossFrictional

24 2

2

2

2=

( )( ) ( )( )81.92634.0

075.0240058.04 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. Solution to Example 6.6 Since d1 = 150 mm and d2 = 225 mm

22

2

1 49

150225 vvv ⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=


The losses are

g

ventryatLoss

221 2

1=

g

v24

921 2

22

⎟⎠⎞

⎜⎝⎛=

g

v2

53.22

2=

gv

dfLpipeminlossFrictional

246

21

1

1=

( )( )g

v215.0

601.04 21=

g

v2

6.12

1=

g

v24

96.12

22

⎟⎠⎞

⎜⎝⎛=

g

v2

1.82

2=

( )

gvv

ementenlatlossShock2

arg2

21 −=

22

2 149

2⎟⎠⎞

⎜⎝⎛ −=

gv

g

v2

56.12

2=


gv

dfLpipeminlossFrictional

2415

22

2

2=

( )( )g

v2225.0

1501.04 22=

g

v2

67.22

2=

g

vexitatlossShock

2

22=

g

v2

00.12

2=

Total loss of head = g

v2

53.22

2

gv2

1.82

2+g

v2

56.12

2+g

v2

67.22

2+g

v2

00.12

2+

g

v2

86.152

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

v2

86.1562

2=

so

86.1526

2gv ×

=

sm /72.2=


Discharge vd4

2π=

( ) ( )72.24225.0 2π

=

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

v22

1 21=

2. Friction loss g

vdfL

24 2

1

1

1=

3. Loss at sudden enlargement ( )

gvv

2

221 −=

4. Friction loss g

vdfL

24 2

2

2

2=

5. Loss at exit g

v2

22=


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

unit 6-fluid mechanics

Documents