1Chapter 5. Pipe System

Learning Outcomes:1. Understand the concept of hydraulic grade line and

energy grade line,2. Analyse flow in single pipe, pipe is series and pipe in

parallel.3. Analyse loop pipe network using Hardy-Cross method.

25.1 Hydraulic Grade Line and Energy Grade LineHydraulic grade line (HGL) or piezometric line shows the level of piezometric head, i.e.

zp

+

Energy grade line (EGL) shows the total energy head, i.e. gV

zp

2

2

++

Az

Bz

3Entrance loss

Friction loss

Discharge loss

4Entrance lossFriction loss

Discharge loss

Expansion loss

5Entrance loss

Friction loss

Discharge lossContraction loss

6Example 5.1In a fire fighting system, a pipeline with a pump leads to a nozzle as shown in Figure. Find the flow rate when the pump develops a head of 80 ft, given that we may express the friction head loss in the 6-in diameter pipe by hf6 = 5V62/2g, and the friction head loss in the 4-in diameter pipe by hf4 = 12V42/2g. Neglect minor losses.(a) Sketch the energy line and hydraulic grade line,(b) Find the pressure head at the suction side of the pump,(c) Find the power delivered to the water by the pump, and (d) Compute the power of the jet.

7Energy equation between A and C:

C

gV

zphhh

gV

zp C

CC

fpfA

AA

22

2

46

2

++=+++

pA = 0; pC = 0; VA = 0; zA = 70 ft; zC = 80 ft; hp = 80 ft

gVhf 2

5 266

=

gVhf 2

12 244

=

ft 80=ph

gV

gV

gV C

2800

21280

250700

224

26 ++=+++

gV

gV

gV C

2212

2570

224

26

=

Express V4 and V6 in terms of VC using the continuity equation:CCVAVA =44

CCC V.V

DDV 562502

4

2

4 ==

CCVAVA =66

CCC V.V

DDV 2502

6

2

6 ==

8Energy equation between A and C: gV

gV

gV C

2212

2570

224

26

=

CV.V 562504 = CV.V 2506 =Substitute and

( ) ( )g

Vg

V.gV. CCC

225625012

2250570

222

=

702

109452

=

gV

.

C

ft/s 7029.VC =where g = 32.2 ft/s2

/sft 458170294123

3

2

..VAQ CC =

==

pi

(a) Sketch the energy line and hydraulic grade line.

gVhf 2

5 266

= gVhf 2

12 244

=

( )232270292505 2

6.

..hf

=

ft 28146

.hf =

( )2322

70295625012 24

.

..hf

=

ft 02524

.hf =

ft 856302

26

.

gV

=ft 3354

2

24

.

gV

=

ft 70132

2

.

gVC

=

9ft 28146

.hf = ft 02524 .hf =ft 80=ph

ft 856302

26

.

gV

= ft 33542

24

.

gV

= ft 70132

2

.

gVC

=

EGL

70 ft65.72 ft

145.72 ft

93.70 ft

64.86 ft

HGL

141.38 ft

89.36 ft

10

(b) Find the pressure head at the suction side of the pump

Energy equation between A and B:

gV

zph

gV

zp B

BB

fA

AA

22

2

6

2

++=++

ft 856302

ft; 50 ft; 2814 ;0 ft; 70 ;02

6.

gV

z.hVzp BBfAAA =====

856305028140700 .p. B ++=++

856305028140700 .p. B ++=++

ft 8614.pB =

11

(c) Find the power delivered to the water by the pump

ft 80=ph

pQhP =( )( )804581462 ..P =

ft.lb/s 37278.P =

(d) Compute the power of the jetjetQhP =( )( )70134581462 ...P =

ft.lb/s 41246.P =

12

Considering only friction losses, the elevation of P must lie between the surfaces of reservoirs A and C.

If P is level with the surface of reservoir B, then hf2 = h2 = 0 and Q2 = 0.

If P is above the surface of reservoir B, then water must flow in B and

If P is below the surface of reservoir B, then the flow must be out of B and321 QQQ +=

321 QQQ =+

5.2 Branching Pipes

13

5.3 Pipes in Series

According to continuity and the energy equations, the following relations apply to the pipes in series:

L==== 321 QQQQ

L+++= 321 LLLL hhhh

14

Example 5.2Pipes 1, 2 and 3 are 300 m of 300 mm diameter, 150 m of 200 mm diameter, and 250 m of 250 mm diameter, respectively, of new cast iron and are conveying 15C water. If z = 10 m, find the rate of flow from A to B. Assume fully turbulent flow and neglect minor losses.

L1 = 300 m

D1 = 0.3 m L2 = 150 m

D2 = 0.2 m L3 = 250 m

D3 = 0.25 m

= 10 m

15

L1 = 300 m

D1 = 0.3 m L2 = 150 m

D2 = 0.2 m L3 = 250 m

D3 = 0.25 m

= 10 m

Energy equation between A and B:g

Vz

phhhg

Vz

p BB

Bfff

AA

A

22

22

321++=++

0.0200.0210.019f0.001000.001250.00083e/D

0.250.20.3D (m)250150300L (m)321Pipe

m 000250 iron Cast .e =

00222

01002

3

3

332

2

2

222

1

1

11 +=++g

VDLf

gV

DLf

gV

DLf

From continuity:112

2

21

2 252 V.VDDV == 112

3

21

3 441 V.VDDV ==

( )( ) ( )( ) ( ) ( )( ) ( ) 02441

250250020

2252

201500210

230300019010

21

21

21

=

gV.

.

.

gV.

.

.

gV

.

.

m/s 18311 .V =/sm 08360 3.Q =

102

211402

1=

gV

.

16

5.4 Pipes in Parallel

For the parallel or looping pipes of the following figure, the continuity and energy equations provide the following relations:

L+++= 321 QQQQ

L==== 321 LLLL hhhh

This is because pressures at A and B are common to all pipes.

17

Example 5.3

Three pipes A, B and C are interconnected as in figure below. The pipe characteristicsare as follows:

0.02440008C0.03216004B0.02020006A

fL (ft)D (in)Pipe

Find the rate at which water will flow in each pipe. Find also the pressure at point P. All pipe lengths are much greater than 1000 diameters, so neglect minor losses.

18

1

2

Energy equation between 1 and 2:g

Vz

phhg

Vz

pCA ff 22

22

22

21

11 ++=++

gV

gV

DLf

gV

DLf CC

C

CCA

A

AA

2500

2202000

222

++=++

From continuity equation: CBA QQQ =+CCBBAA VAVAVA =+

CCBBAA VDVDVD 222 =+

CBA V.V.V. 4444011110250 =+

For pipes in parallel:BA ff hh =

gV

DLf

gV

DLf B

B

BBA

A

AA

22

22

=

22 615380 BA V.V =

AB V.V 72170=

( ) CAA V.V..V. 444407217011110250 =+CA V.V 3461=

19

1

2

Energy equation between 1 and 2:g

Vg

VDLf

gV

DLf CC

C

CCA

A

AA

222150

222

=

( ) ( )( )( ) g

Vg

V.gV. CCC

2212840000240

2346180150

222

=

CA V.V 3461=

gV

gV

gV

.

CCC

22144

295144150

222

=

( ) 1502

1449514412

=++g

V.

C

ft/s 7725.VC =

( ) ft/s7697 77253461 ...VA ==( ) ft/s6075 .769772170 ..VB ==

/sft 0152 3.QC =/sft 5261 3.QA =/sft 4890 3.QB =

Check the continuity: CBA QQQ =+48905261 ..QQ BA +=+CBA Q.QQ ==+ 0152

20

Pressure at point P

Energy equation between 1 and P:g

Vz

phg

Vz

p PP

PfA 22

221

11 ++=++

1

2

gVp

gV

.

CPC

2120

29514402000

22

++=++

( )( )

Pp.

.

. =

232277259514580

2

ft/s 7725.VC =

ft 4964.pP =

21

5.5 Pipe Networks

Three simple methods to solve for pipe networks in loop configuration are:a. Hardy Cross (using either Darcy-Weisbach or Hazen-Williams equation),b. Linear theory, andc. Newton-Raphson.

The most popular is the Hardy Cross method which involves a series of successive approximations and corrections to flows in individual pipes.

22

According to the Darcy-Weisbach equation,g

VdfLhf 2

2=

gQ

dfL

216 2

52pi=

nf KQh =

51012 d.fLK =where,

The sum of head losses around any closed loop is zero, i.e.

0= fh

According to the Hazen-Williams equation, 8518748517274

.

..f QdC

L.h =

8748517010

.. dCL.K =

and n = 2 (Darcy-Weisbach in S.I. unit)

and n = 1.85 (Hazen-Williams in S.I. unit)

23

12020.0 104Wood stave13040.0 104Concrete11060.0 104Riveted steel1201.5 104Commercial and welded steel1.5 104Wrought iron4.0 104Asphalted iron

1205.0 104Galvanized iron100Cast iron (old)1308.0 104Cast iron (new)150SmoothPVC, plastic140SmoothBrass, copper, aluminium

Hazen-Williams coefficient C

Equivalent roughness e (ft)Pipe material

Roughness values for pipes

24

Q (m3/s), L (m), d (m), hf (m)

Q (gpm), L (ft), d (in), hf (ft)

Q (cfs), L (ft), d (ft), hf (ft)Darcy-Weisbach

Q (m3/s), L (m), d (m), hf (m)

Q (gpm), L (ft), d (in), hf (ft)

Q (cfs), L (ft), d (ft), hf (ft)Hazen-Williams

KUnits of MeasurementFormula

874851734

.. dCL.

8748514410

.. dCL.

8748517010

.. dCL.

57039 d.fL

51532 d.fL

51012 d.fL

Equivalent resistance K for pipe

25

Consider that Qa is an assumed pipe discharge that varies from pipe to pipe of a loop to satisfy the continuity of flow. If is the correction made in the assumed flow of all pipes of a loop to satisfy then,0= fh

( ) 0=+ naQK Expansion by binomial theorem and retaining only the first two terms yield,

= 1na

na

KQnKQ

=

a

f

f

Qh

n

h

26

Procedure of Hardy-Cross method:

1. Divide the network into a number of closed loops. Computations are made for one loop at a time.

2. Compute K for each pipe.3. Assume a discharge Qa and its direction in each pipe of the loop. At each joint, the

total flow in should equal to the flow out. Consider the clock-wise flow to be positive and the counterclockwise flow to be negative.

4. Compute hf for each pipe and retain the sign of the flow direction.5. Compute hf/Q for each pipe without regard to the sign.6. Determine the correction . Apply the correction algebraically to the discharge of

each member of the loop.7. For common members among two loops, both corrections should be made, one

for each loop.8. For the adjusted Q, repeat steps 4 to 7 until becomes very small for all loops.

27

Example 5.4

Find the discharge in each pipe of the welded steel pipe network shown in Figure below. All pipes are 4 in. in diameter. The pressure head at A is 50 ft. Determine the pressure at the different nodes.

0.6 cfs

0.8 cfs

50 ft

100 ft

100 ft

50 ft

100 ft

100 ft

I II1.8 cfs

0.4 cfs

A

B

C

D

E

28

0.6 cfs

0.8 cfs

I II1.8 cfs

0.4 cfs

A

B

C

D

E

+ +

1.0 cfs

0.8 cfs

0.2 cfs 0.2 cfs

0.8 cfs

0.6 cfs

Assumed discharges

29

24.62+3.890.1851.800.360.27.12CB0.6859.235.540.614.25EC+0.1151.80+0.36+0.27.12DE

+0.71511.79+9.43+0.814.25BD2

27.84+5.18

0.911.799.430.814.25CA+0.1851.80+0.36+0.27.12BC+0.914.25+14.25+1.014.25AB1

hf = KQa1.85Qa (cfs)KPipelineLoop

(7)(6)(5)(4)(3)(2)(1)1st iteration

Qhf

+= aQQcorrected

For loop 1, ( ) 1008427851185

.

..

.

=

=

For loop 2, ( ) 08506224851893

.

..

.

=

=

874851734

.. dCL.K =

=

a

f

f

Qh

n

h

30

23.86+0.400.1851.680.310.1857.12CB0.68510.347.080.68514.25EC+0.1151.13+0.13+0.1157.12DE

+0.71510.71+7.66+0.71514.25BD2

27.74+0.310.913.0311.730.914.25CA

+0.1851.68+0.31+0.1857.12BC+0.913.03+11.73+0.914.25AB1

hf = KQa1.85Qa (cfs)KPipelineLoop

(7)(6)(5)(4)(3)(2)(1)2nd iteration

Qhf

+= aQQcorrected

For loop 1, ( ) ( )negligible 00607427851310

.

..

.

=

=

For loop 2, ( ) ( )negligible 00908623851400

.

..

.

=

=

31

0.6 cfs

0.8 cfs

0.115 cfs

0.715 cfs

0.685 cfs

0.185 cfs

0.9 cfs

0.9 cfs

I II1.8 cfs

0.4 cfs

A

B

C

D

E

+ +

Final discharges

32

Project Question

A water distribution network is shown as in figure. Elevation of A is 61 m, with pressure head = 45.72 m. Elevation of D is 30.5 m. Using f = 0.015, find:(i) Flow rate through each pipe, and(ii)Pressure head at node D.

0.3 m3/s

0.6 m3/s1000 m

-

500

mm

0.5 m3/s

0.2 m3/s

A

B

C

D

1000 m

- 250 mm

1200 m

- 750 mm

1500

m

-

500

mm

700 m -

500 mm