minor loss of hydraulics

7/30/2019 minor loss of hydraulics

1/37

Prof. Dr. Atl BULU1

Chapter 3

Local Energy (Head) Losses

3.1. Introduction

Local head losses occur in the pipes when there is a change in the area of the cross-

section of the pipe (enlargement, contraction), a change of the direction of the flow

(bends), and application of some devices on the pipe (vanes). Local head losses are alsonamed as minor losses. Minor losses can be neglected for long pipe systems.

Minor losses happen when the magnitude or the direction of the velocity of the flowchanges. In some cases the magnitude and the direction of the velocity of the flow may

change simultaneously. Minor losses are proportional with the velocity head of the flowand is defined by,

g

VhL

2

2

= (3.1)

Where is the loss coefficient.

3.2. Abrupt Enlargement

There is certain amount of head loss when the fluid moves through a sudden

enlargement in a pipe system such as that shown in Fig. (3.1)

Impuls-momentum equation is applied to the 1221 control volume along the horizontalflow direction. The forces acting on to the control volume are,

a) Pressure force on the 1-1 cross-section, 111 ApF =r b) Pressure force on the 2-2 cross-section, 222 ApF =s c) Pressure force on the 3-3 cross-section, ( )1233 AApF =r Laboratory experiments have verified that 13 pp = , ( )1213 AApF =

r

d) Momentum on the 1-1 cross-section, 11 QVM =r e) Momentum on the 2-2 cross-section, 22 QVM =r


2/37

Prof. Dr. Atl BULU2

Fig. 3.1.

Neglecting the shearing force on the surface of the control volume,

( )( ) ( )

( )122

21

12221

222121111 0

VVgA

Qpp

g

VVQAppQVApAApApQV

=

=

==++

1A1QV12A2 QV2

p1(A2 -A1)

1

3 2

23

Energy Line ( )g

VVhL

2

2

21 =

g

V

2

21

1p

gV2

2

2

2p

Flow

3 2

H.G.L


3/37

Prof. Dr. Atl BULU3

Applying the continuity equation,

2

112

2211

A

AVV

AVAVQ

=

==

Substituting this,

=

111

2

1111

2

21 1 VAVA

AVAV

gA

pp

=

2

1

2

2

1

2

121

A

A

A

A

g

Vpp

(3.2)

Applying Bernoulli equation between cross-sections 1-2 for the horizontal pipe flow,

L

L

hg

V

g

Vpp

hg

Vp

g

Vp

+=

++=+

22

22

2

1

2

221

2

22

2

11

Substituting Equ. (3.2),

g

V

g

VV

g

Vh

gVV

gV

gV

gVh

V

V

V

V

g

V

g

V

g

Vh

V

V

A

AAVAV

hg

V

g

V

A

A

A

A

g

V

L

L

L

L

22

22

22

222

2

221

2

1

21

2

2

2

2

2

1

1

2

2

1

2

2

1

2

2

2

1

1

2

2

12211

2

1

2

2

2

1

2

2

1

2

1

+=

+=

+=

==

+=


4/37

Prof. Dr. Atl BULU4

( )g

VVhL

2

2

21 = (3.3)

This equation is known as Borda-Carnot equation. The loss coefficient can be derived

as,

+=

+=

2

2

2

1

2

1

2

1

2

1

2

2

1

2

2

1

212

212

A

A

A

A

g

Vh

V

V

V

V

g

Vh

L

L

2

2

1

2

1 12

=A

A

g

VhL (3.4)

The loss coefficient of Equ. (3.1) is then,

2

2

11

=A

A (3.5)

The loss coefficient is a function of the geometry of the pipe and is not dependent on tothe flow characteristics.

Special case: Connecting to a reservoir,

Fig. 3.2.

A2

A1

V12/2g

Energy Line

H.G.L.


5/37

Prof. Dr. Atl BULU5

1 a

2

1 a

2

A1A2V1 V2a

Dead Zone

Since the reservoir cross-sectional area is too big comparing to the area pipe cross-

section, 12 AA >> , 02

1 A

A, the loss coefficient will be equal to zero, =0. The head loss is

then,

g

VhL

2

21= (3.6)

3.3. Abrupt Contraction

Flow through an abrupt contraction is shown in Fig. 3.3.

Fig. 3.3

Derivation of the loss coefficient may be done by following these steps.

a) Bernoulli equation between 1, C, and 2 cross-sections by neglecting the head lossbetween 1 and C cross-sections is for the horizontal pipe system,

L

CC hg

Vp

g

Vp

g

Vp++=+=+

222

2

22

22

11

b) Continuity equation of the system,

2211 AVAVAVQ CC ===


6/37

Prof. Dr. Atl BULU6

c) Impuls-Momentum equation between C and 2 cross-sections,

( ) ( )CC VVQApp = 222

Solving these 3 equations gives us head loss equation as,

( )

g

V

A

Ah

g

VVh

C

L

C

L

21

2

2

2

2

2

2

2

=

=

(3.7)

Where,

2A

AC CC = = Coefficient of contraction (3.8)

The loss coefficient for abrupt contraction is then,

2

11

=CC

(3.9)

g

VhL

2

2

2=

CC contraction coefficients, and loss coefficients have determined by laboratoryexperiments depending on the A2/A1 ratios and given in Table 3.1.


7/37

Prof. Dr. Atl BULU7

Table 3.1.

A2/A1

CC

0 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00

0.62 0.62 0.63 0.64 0.66 0.68 0.71 0.76 0.81 0.89 1.00

0.50 0.46 0.41 0.36 0.30 0.24 0.18 0.12 0.06 0.02 0

Example: At an abrupt contraction in a pressured pipe flow, Q = 1 m3/sec, D1 = 0.80 m,

D2 = 0.40 m. Calculate the local head when contraction coefficient is CC = 0.60.

Solution: Since discharge has been given,

sec26.130754.0

1

0754.01257.060.0

1257.04

40.0

4

sec96.740.0

144

2

2

222

2

2

22

2

2

mA

QV

mACA

mD

A

mD

QV

C

C

cC

===

===

=

==

=

==

( ) ( ) mgVVh CL 43.1

62.1996.726.13

2

22

2 ===

Or,

mh

V

Ch

L

C

L

44.162.19

96.71

60.0

1

62.191

1

22

2

2

2

=

=

=

Special case: Pipe entrance from a reservoir,

Loss coefficient depends on the geometry of the pipe connection to the reservoir. The

loss coefficient is 0.50 for square-edge entrance. Fig. (3.4)


8/37

Prof. Dr. Atl BULU8

Fig. 3.4

g

VhL

25.0

2

= (3.10)

The head loss at the pipe fittings, bends and vanes is calculated with the general local loss

equation (Equ. 3.1). The loss coefficients can be taken from the tables prepared for thebends and vanes.

Example 3.1. Two reservoirs is connected with a pipe length L. What will the elevationdifference between the water surfaces of the reservoir? Draw the energy and hydraulic

grade line of the system.

Fig. 3.5

H1

H2

H=?

V2/2g

0.5V2/2g

E.L.

H.G.L.

V

L= Pipe Length

V2/2g

0.5V2/2g

R.E.L.

H.G.L.

V


9/37

Prof. Dr. Atl BULU9

Solution: Bernoulli equation between reservoir (1) and (2),

Lhg

Vpz

g

Vpz +++=++

22

2

22

2

2

11

1

Since the velocities in the reservoirs are taken as zero, V1 = V2 =0, and the pressure on

the surface of reservoirs is equal to the atmospheric pressure,

Lhzz = 21

The head losses along the pipe system are,

1) Local loss to the entrance of the pipe,g

VhL

25.0

2

1=

2) Head loss along the pipe, Lg

V

D

fJLhL

2

2

2==

3) Local loss at the entrance to the reservoir,g

VhL

2

2

3=

The total loss is then will be equal to the elevation difference between the reservoir

levels,

Lg

V

D

f

g

V

zz

g

VL

g

V

D

f

g

Vzz

hzz L

225.1

2225.0

22

21

222

21

21

+=

++=

=


10/37

Prof. Dr. Atl BULU10

Numerical Application: If the physical characteristics of the system are given as below,

what will be difference between the reservoir levels? Calculate the ratio of local losses to

the total head loss and give a physical explanation of this ratio.

Figure 3.6

Q = 0.2 m3/sec, f = 0.02, L = 2000 m, D = 0.30 m

The velocity in the pipe,

sec83.2

3.0

2.044

22

mV

D

QV

=

==

The velocity head,

mg

V41.0

62.19

83.2

2

22

==

The elevation difference is then,

mzz

zz

zz

Lg

V

D

f

g

Vzz

62.82

0.8262.0

20002.0

02.041.041.05.1

225.1

21

21

21

22

21

=

+=

+=

+=

z1

z2

H=82.62

V2/2g=0.41m

0.5V2/2g=0.20m

E.L.

H.G.L.

V

L= 2000m


11/37


The ratio of local losses to the total head loss is,

0075.062.82

62.0=

As can be seen from the result, the ratio is less than 1%. Therefore, local losses aregenerally neglected for long pipes.

3.4 Conduit SystemsThe other applications which are commonly seen in practical applications for conduitsystems are,

a) The Three-Reservoir Problems,b) Parallel Pipes,c) Branching Pipes,d) Pump Systems.

3.4.1 The Three-Reservoir ProblemsConsider the case where three reservoirs are connected by a branched-pipe system. Theproblem is to determine the discharge in each pipe and the head at the junction point D.

There are four unknowns (VAD, VBC, VDC and pD/ ), and the solution is obtained by

solving the energy equations for the pipes (neglecting velocity heads and including onlypipe losses and not the minor losses) and the continuity equation. The physical

characteristics of the system such as lengths, diameters and f friction factors of the pipes,

the geometric elevations of the water surfaces at the reservoirs and the piezometric head

at the junction D are given. The reservoirs are located as, zA> zB> zC. There are threepossible solutions of this problem,

a) AD

zp

z =

+

Figure 3.7

A

zA

zB

zC

B

C

Dp

D

zD

E.L


12/37


Since AD

zp

z =

+

, there will be no flow from or to reservoir A, and therefore,

0,0,0 === ADADAD JVQ

DCBD QQ =

The relations between the reservoir levels B, and D and the piezometric head at the

junction D are,

CDBD LC

D

DLB hzp

zhz +=+=

Energy line slope (JBD) of the BD pipe can be found as,

BD

L

BD

AB

BD

D

DB

BDL

h

L

zz

L

pzz

J BD=

=

=

(3.11)

Using the Darcy-Weisbach equation,

gD

fVJ

2

2

=

The flow velocity in the BD pipe can be calculated by,

21

2

=

BD

BDBD

BDf

JgDV (3.12)

And the discharge is,

BDBDBD VAQ =


13/37


The energy line slope, the velocity and the discharge of the CD pipe are then,

BDCdCDCD

CD

CDCD

CD

CD

L

CD

CA

CD

QVAQ

f

JgDV

L

h

L

zzJ CD

==

=

=

=

21

2

b) BD

AC zp

zzz


14/37


The energy line slopes of the pipes are,

AD

L

AD

A

D

AD

BD

L

BD

D

B

BD

L

h

L

zp

z

J

L

h

L

pzz

J

AD

BD

=

+

=

=

+

=

CD

L

CD

C

D

CDL

h

L

zp

z

JCD=

+

=

c) BAD

C zzpzz


15/37


The energy line slopes of the pipes are,

CD

L

CD

C

D

CD

AD

L

AD

D

A

AD

BD

L

BD

D

B

BD

L

h

L

zp

z

J

L

h

L

pzz

J

L

h

L

pzz

J

CD

AD

BD

=

+

=

=

+

=

=

+

=

Example: The water surface levels at the A and C reservoirs are respectively zA=100 mand zC= 70 for the given three-reservoir system. Reservoirs A and B are feeding reservoir

C and QA = QB . Calculate the water surface level of the reservoir B. Draw the energyline of the system. The physical characteristics of the pipes are,

Figure 3.9

Pipe Length (m) Diameter (mm) f

AD 500 150 0.03

BD 1000 200 0.02

DC 1500 250 0.03

A

zA=100m

zB=89.47m

zC=70m

B

C

Dp

D

zD

E.L.


16/37


Solution: The head loss along the pipes 1 and 3 is,

mzzhh CALL 307010031 ===+ (1)

Since,

QQ

QQQ

23

21

=

==

Using the Darcy-Weisbach equation for the head loss,

5

2

2

2

2

2

8

4

22

D

LQ

g

fh

D

Q

g

fL

gD

LfVh

L

L

=

==

Using Equ. (1),

( )

sec8.30sec0308.0

3156630

25.0

21500

15.0

500

81.9

03.08

30

8

3

2

5

2

5

2

2

5

2

2

222

5

1

2

111

2

ltmQ

Q

QQ

D

QLf

D

QLf

ghL

==

=

+

=

+=

sec74.115.0

0308.044

sec0308.0

22

1

1

1

3

1

mD

QV

mQ

=

==

=

sec26.125.0

0616.044

sec0616.00308.022

22

3

3

3

3

13

mD

QV

mQQ

=

==

===


17/37


Head losses along the pipes are,

mhh

mgD

LVfh

mgD

LVfh

Ll

L

L

00.3057.1443.15

57.1425.062.19

150026.103.0

2

43.1515.062.19

50074.103.0

2

31

3

1

2

3

3

2

33

2

1

1

2

11

=+=+

=

==

=

==

For the pipe 2,

sec0308.03

21 mQQ ==

mgD

LVfh

m

D

QV

L 90.420.062.19

100098.002.0

2

sec98.0

20.0

0308.044

2

2

2

2

22

222

2

2

2=

==

=

==

Water surface level of the reservoir B is,

mhp

zz

mhzp

z

mhzp

z

L

D

B

LC

D

LA

D

47.8990.457.84

57.8457.1400.70

57.8443.1500.100

2

3

1

=+=+

+=

=+=+=

+

===

+

3.4.2 Parallel Pipes

Consider a pipe that branches into two parallel and then rejoins as in Fig.. . In pipelinepractice, looping or laying a pipeline parallel to an existing pipeline is a standard method

of increasing the capacity of the line. A problem involving this configuration might be to

determine the division of flow in each pipe given the total flow rate. In such problems,

velocity heads, and minor losses are usually neglected, and calculations are made on thebasis of coincident energy and hydraulic grade lines.


18/37


Figure 3.10

Evidently, the distribution of flow in the branches must be such that the same head loss

occurs in each branch; if this were not so there would be more than one energy line for

the pipe upstream and downstream from the junctions.

Application of the continuity principle shows that the discharge in the mainline is equal

to the sum of the discharges in the branches. Thus the following simultaneous equations

may be written.

21

21

QQQ

hh LL

+=

=(3.13)

Using the equality of head losses,

gD

VLf

gD

VLf

22 2

2

222

1

2

111 =

21

211

122

2

1

=

DLf

DLf

V

V (3.14)

If f1 and f2 are known, the division of flow can be easily determined.

Q1, L1, D1, f1

Q2, L2, D2, f2

Q Q

A B

hL1 = hL2

PB/

PA/

E.L


19/37


Expressing the head losses in terms of discharge through the Darcy-Weisbach equation,

42

22 16

22 D

Q

gD

fL

gD

LfVhL

=

2

2

522

16

KQh

QgD

fLh

L

L

=

=

(3.14)

Substituting this to the simultaneous Equs (3.12),

21

2

22

2

11

QQQ

QKQK

+=

=(3.15)

Solution of these simultaneous equations allows prediction of the division of a discharge

Q into discharges Q1 and Q2 when the pipe characteristics are known. Application ofthese principles allows prediction of the increased discharge obtainable by looping an

existing pipeline.

Example: A 300 mm pipeline 1500 m long is laid between two reservoirs having adifference of surface elevation of 24. The maximum discharge obtainable through this

line is 0.15m3/sec. When this pipe is looped with a 600 m pipe of the same size and

material laid parallel and connected to it, what increase of discharge may be expected?

Solution: For the single pipe, using Equ. (3.14),

1067

15.0242

2

=

=

=

K

K

KQhL


20/37


Figure 3.11

K for the looped section is,

427

10671500

6002

=

==

A

A

K

KL

LK

For the unlooped section,

640

10671500

6001500

2

=

=

=

K

K

K

L

LLK

For the looped pipeline, the headloss is computed first in the common section plus branch

A as,

22

22

42764024 A

AAL

QQ

QKQKh

+=

+=(1)

And then in the common section plus branch B as,

22 42764024 BQQ += (2)

24m1

1A

B, LB=600m


21/37


In which QA and QB are the discharges in the parallel branches. Solving these by

eliminating Q shows that QA=QB (which is expected to be from symmetry in thisproblem). Since, from continuity,

2QQ

QQQ

A

BA

=

+=

Substituting this in the Equ (1) yields,

( )

sec18.0

sec09.0

298724

427264024

3

3

2

22

mQ

mQ

Q

QQ

A

A

AA

=

=

=

+=

Thus the gain of discharge capacity by looping the pipe is 0.03 m3/sec or 20%.

Example: Three pipes have been connected between the points A and B. Total dischargebetween A and B is 200 lt/sec. Physical characteristics of the pipes are,

mDmDmD

mLLL

fff

30.0,20.0,10.0

1000

02.0

321

321

321

===

===

===

Calculate the discharges of each pipe and calculate the head loss between the points A

and B. Draw the energy line of the system.

Figure

A B

1

2

3

hL= 13.33m


22/37


Solution: Head losses through the three pipes should be the same.

321 LLLhhh ==

332211 LJLJLJ ==

Since the pipe lengths are the same for the three pipes,

321 JJJ ==

5

2

242

22 816

2

1

2 D

Q

g

f

D

Q

gD

f

g

V

D

fJ ===

And also roughness coefficients f are the same for three pipe, the above equation takes

the form of,

5

3

2

3

5

2

2

2

5

1

2

1

D

Q

D

Q

D

Q==

Using this equation,

21

5.225

2

1

2

1

177.0

177.020.0

10.0

QQ

D

D

Q

Q

=

=

=

=

23

5.225

2

3

2

3

76.2

76.220.0

30.0

QQ

D

D

Q

Q

=

=

=

=

Substituting these values to the continuity equation,

sec051.0200.0937.3

200.076.2177.0

200.0

3

2

2

222

321

mQQ

QQQ

QQQ

==

=++

=++

sec140.0051.076.2

sec009.0051.0177.0

3

3

3

1

mQ

mQ

==

==

sec200.0140.0051.0009.0 3321 mQQQ =++=++


23/37


Calculated discharges for the parallel pipes satisfy the continuity equation.

The head loss for the parallel pipes is,

mh

LD

Q

g

fh

AB

AB

L

L

33.13100030.0

14.0

81.9

02.08

8

5

2

2

353

2

3

2

=

=

=

3.4.2 Branching Pipes

Figure 3.12

Consider the reservoir A supplying water with branching pipes BC and BD to a city.

Pressure heads (p/) are required to be over 25 m to supply water to the top floors in

multistory buildings.

mp

mp DC 25,25

Using the continuity equation,

BDBCAB QQQ +=

A

zA

B

C

D

minppc

minppD

zC

zD

Bp


24/37


Two possible problems may arise for these kinds of pipeline systems;

a) The physical characteristics of the system such as the lengths, diameters andfriction factors, and also the discharges of each pipe are given. Water surface

level in the reservoir is searched to supply the required (p/)min pressure head at

points C and D. The problem can be solved by following these steps,

1) Velocities in the pipes are calculated using the given discharges and diameters.

2

4

D

Q

A

QV

==

2) Head losses for the BC and BD pipes are calculated.

Lg

V

D

fhL =2

2

Piezometric head at junction B is calculated by using the given geometric elevations of

the points C and D.

BCLC

B

hzpp

z +

=

+

min

(1)

BDLD

B

hzpp

z +

=

+

min

(2)

Since there should be only one piezometric head (z+p/) at any junction, the largest value

obtained from the equations (1) and (2) is taken as the piezometric head for the junctionB. Therefore the minimum pressure head requirement for the points C and D has been

achieved.

3) The minimum water surface level at the reservoir to supply the required pressurehead at points C and D is calculated by taking the chosen piezometric head for thejunction,

ABL

B

A hp

zz +

+=


25/37


b) The physical characteristics and the discharges of the pipes are given. Thegeometric elevations of the water surface of the reservoir and at points C and D

are also known. The minimum pressure head requirement will be checked,

1)

VAB , VBC and VBD velocities in the pipes are calculated.

2) Head losses along the pipes AB, BC and BD are calculated.3) Piezometric heads at points B, C and D are calculated.

min

min

+=

+

+=

+

=

+

ph

pz

pz

ph

pz

pz

hzp

z

BD

BC

AB

L

BD

L

BC

LA

B

4) If the minimum pressure head requirement is supplied at points C and D, thepipeline system has been designed according to the project requirements. If theminimum pressure head is not supplied either at one of the points or at both

points,

a) The reservoir water level is increased up to supply the minimum pressure head atpoints C and D by following the steps given above.

b) Head losses are reduced by increasing the pipe diameters.

Example: Geometric elevation of the point E is 50 m for the reservoir system given. Therequired minimum pressure at point E is 300 kPa. The discharge in the pipe DE is 200

lt/sec. The physical characteristics of the reservoir-pipe system are given as,

Pipe Length(m) Diameter(mm) fABD 1000 200 0.03

ACD 1000 300 0.03

DE 1500 400 0.03


26/37


Figure 3.13

Calculate the minimum water surface level to supply the required pressure at the outlet E.Draw the energy line of the system. water= 10 kN/m

3.

Solution: The velocity and the head loss along the pipe DE are,

mgD

LfV

h

mD

QV

DEDE

L

De

DE 50.144.062.19

150059.103.0

2

sec59.14.0

2.044

22

22

=

==

=

==

Since,

ACDABD

ACDABD

ACDABD

ACD

ACDACD

ABD

ABDABD

LL

VV

VV

g

V

g

V

gD

LfV

gD

LfV

hhACDABD

816.0

2.03.0

3.02

100003.0

2.02

100003.0

22

22

22

22

=

=

=

=

=

A

B

CD E

50m

80m

94.50m

116.59m


27/37


Applying the continuity equation,

( )

sec70.108.2816.0

sec08.2

255.009.0816.004.0

255.009.004.0

200.03.02.04

222

mV

mV

VV

VV

VV

QQQ

ABD

ACD

ACDACD

ACDABD

ACDABD

ACDABD

==

=

=+

=+=+

+=

The discharges of the looped pipes are,

sec053.0

70.14

2.0

4

3

2

2

mQ

Q

VD

Q

ABD

ABD

ABD

ABD

ABD

=

=

=

sec147.0

08.24

3.0

4

3

2

2

mQ

Q

VD

Q

ACD

ACD

ACD

ACD

ACD

=

=

=

The head loss along the looped pipes is,

mh

h

gD

LfVhh

L

L

ABD

ABDABD

LL ABDACD

09.22

2.062.19

100070.103.0

2

2

2

=

=

==


28/37


Piezometric head at the outlet E is,

mp

z

Ew

8010

30050 =+=

+

The required minimum water surface level at the reservoir is,

ACDDe LL

E

A hhp

zz ++

+=

mz

z

A

A

59.116

09.2250.1480

=

++=

Example: The discharges in the AB and AC pipes are respectively Q1=50 lt/sec andQ2=80 lt/sec for the pipe system given. The required pressure at the B and C outlets is200 kPa and the geometric elevations for these points are zB= 50 m and zc= 45 m. The

physical characteristics of the pipe system are,


RA 2000 300 0.02

AB 1000 350 0.02

AC 1500 400 0.02

Calculate the minimum water surface level of the reservoir R to supply the requiredpressure at the outlets. Draw the energy line of the system. water= 10 kN/m

3.

Figure 3.14

A

B

C

R

70.79m

mpc 22.24=

50m

mpB 20=

45m

93.79m


29/37


Solution: Since the discharges in the pipes are given,

mgD

LfVh

m

D

QV

AB

ABAB

L

AB

AB

AB

AB79.0

35.062.19

100052.002.0

2

sec52.0

35.0

050.044

22

22

=

==

=

==

ABL

BA

hp

zp

z +

+=

+

mpzA

79.7079.01020050 =+ +=

+

mgD

LfVh

mD

QV

AC

ACAC

L

AC

AC

AC

AC57.1

40.062.19

150064.002.0

2

sec64.040.0

080.044

22

22

=

==

=

==

mp

z

hp

zp

z

A

L

CA

AC

57.6657.110

20045 =++=

+

+

+=

+

If we take the largest piezometric head of the outlets 70.79 m, the pressure will be

200kPa at the outlet B, and the pressure at outlet C is,

kPakPap

mp

zhp

zp

w

C

CL

AC

AC

2002.24222.241022.24

22.2400.4557.179.70

>===

==

+=


30/37


Water surface level at the reservoir R is,

mhp

zz

mgD

LfVh

mD

QV

mQQQ

RA

RA

L

A

R

RA

RARA

L

RA

RA

RA

ACABRA

79.9300.2379.70

00.2330.062.19

200084.102.0

2

sec84.130.0

13.044

sec13.008.005.0

22

22

3

=+=+

+=

=

==

===

=+=+=

3.4.4. Pump Systems

When the energy (head) in a pipe system is not sufficient enough to overcome the headlosses to convey the liquid to the desired location, energy has to be added to the system.

This is accomplished by a pump. The power of the pump is calculated by,

QHN = (3.16)

Where = Specific weight of the liquid (N/m3), Q = Discharge (m

3/sec), H = Head to

be supplied by pump (m), N = Power of the pump (Watt). Calculated pump power should

be divided by = Efficiency factor of the pump.

Example: Reservoir D is fed by a pump and pipe system. Pressure heads in the pipe flow

at the entrance and at the outlet from the pump are respectively 5 and 105 m. The

geometric of the pipe before the pump is 0.50 m. The efficient power of the pump is 75kW. The friction factor at all the pipe system is f = 0.03. (Minor losses will be neglected).

a) Calculate the discharge in pipe BC,b) What will be the discharges in the parallel C1D and C2D pipes?c) Calculate the water surface level of the reservoir D.d) Draw the energy line of the system.


BC 1000 300 0.03

C1D 3000 400 0.03

C2D 1500 200 0.03


31/37


Figure 3.

Solution:

a) The head to be supplied by the pump is,

mHpump 1005105 ==

The efficient power of the pump is,

QHN

=

sec76sec076.010081.91000

75000 3 ltmH

NQ ==

==

b) The velocity in the pipe BC,

sec08.1

30.0

076.04422

m

D

QV

BC

BC =

==

The head loss is,

mh

Lg

V

D

fh

BC

BC

L

BC

BC

BC

L

94.5100062.19

08.1

30.0

03.0

2

2

2

==

=

0.50mPum

A B

C

1

2

D


32/37


Since,

VVV

VV

g

V

g

V

L

g

V

D

fL

g

V

D

f

hh LL

==

=

=

=

=

21

2

2

2

1

2

2

2

1

2

2

2

2

1

2

1

1

1500220.0

03.03000

240.0

03.022

21

By equation of continuity,

2

2

2

2

1

2

11

21

4

4

VD

Q

VDQ

QQQ

=

=

+=

21

2

2

2

1

2

1

4

4

40.0

4

2.0

4

16.0

4

4.0

QQ

VVQ

VVQ

=

=

=

=

=

sec765

sec76

2

3

21

ltQ

mQQ

=

=+

sec2.151 ltQ = sec8.602 ltQ =

c) The velocity and the head loss in the parallel pipes are,

mLg

V

D

fhh

mVVV

LL 64.2300062.19

48.0

40.0

03.0

2

sec48.020.0

0152.04

2

1

2

1

221

21====

=

===


33/37


Total head loss is,

mh

hhh

L

LLL BC

58.864.294.5

1

=+=

+=

The water surface level of the reservoir D is,

mz

hp

zz

D

L

B

D

92.9658.810550.0 =+=

+=

Example: Reservoir A is feeding reservoirs B and C by a pump and pipe system. Thedischarge to reservoir C is Q2 = 0.10 m

3/sec. If the efficient coefficient of the pump is =

0.70 n, what will be required power of pump? Draw the energy line of the system . Thephysical characteristics of the pipe system are,

Pipe Diameter (mm) Length (m) f

1 300 400 0.020

2 150 300 0.015

3 200 1000 0.025

5.50m

105.50m

99.56m96.92m

p/ =100m

A ump B

C

D

0.50m


34/37


Figure..

Solution: Beginning with pipe 2,

sec66.515.0

10.04

4

22

2

2

2

2

2

2

mV

D

Q

A

QV

=

=

==

mh

Lg

VDfh

L

L

4930062.19

66.5

15.0

015.0

2

2

2

2

2

2

2

2

2

=

=

Energy level at the outlet of the pump is,

mhzp

zH LCP

Poutlet1394990

2=+=+=

+=

Pipe 3:

Head loss along the pipe 3 is,

mzHh BPL outlet 291101393 ===

65m

A

B

C

110m

90m

1

2

3


35/37


sec067.013.24

20.0

4

sec13.2

100062.19200.0

025.029

2

32

3

2

3

3

3

2

3

3

2

3

3

3

3

mVD

Q

mV

V

Lg

V

D

fhL

=

==

=

=

=

Pipe 1:

mLg

V

D

fh

mD

QV

mQ

QQQ

L 57.740062.19

36.2

30.0

020.0

2

sec36.230.0

167.044

sec167.0067.0100.0

2

1

2

1

1

1

22

1

1

1

3

1

321

1===

=

==

=+=

+=

Energy level at the entrance to the pump is,

mhzH LAPent 43.5757.700.651 ===

The energy (head) to be supplied by pump is,

mHHHentoutlet PPp

57.8143.5700.139 ===

The required power of the pump is,

kWWN

HQN

P

P

P

19119095270.0

59.81167.0100081.9=

=

=


36/37


Example: Reservoir B is fed by reservoir A with 600 lt/sec discharge. The water surface

levels in the reservoirs are zA = 100 m and zB = 90 m. If the pipe diameter is D = 0.50m,

the friction factor is f = 0.02, and the efficiency factor of the pump is = 0.70, calculatethe required pump power. The lengths of the pipes are respectively L1 = 500 m and L2 =

1000 m. Draw the energy line of the system.

Figure

Solution: Since the discharge of the system is known, the head loss from reservoir A to

the pump can be calculated as,

sec06.350.0

60.04422

mD

QV =

==

100m

90m

A

BC

pump

1

2

65m

A

B

C

110m

90m

1

2

3

139m

57.43m

Pump


37/37

mh

Lg

V

D

fh

AC

AC

L

ACL

54.950062.19

06.3

50.0

02.0

2

2

2

==

=

The energy level at the entrance to the pump is,

mHentC

46.9054.9100 ==

Since the pipe diameter is same along the pipe, and L2 = 2L1, the head loss along the

pipe 2 is,

mhh LL 08.1954.922 12 ===

The energy level at the outlet of the pump should be,

mHoutC

80.10908.1900.90 =+=

The pumping is then,

mHpump 62.1846.9080.109 ==

The required power of the pump is,

kWW

QH

N 15715656870.0

62.18600.09810

=

==

A

100m

90m

CB

90.46m

109.08m

minor loss of hydraulics

Documents