Hardy cross method • Assuming flow distribution in network and balancing resulting headloss • hf=K Qn • hf= headloss;K=constant (size of pipe,internal conditions,units); Q=discharge;n=1.85(H-W eq. used)

Hardy-Cross Method (Procedure) 1. Divide network into number of closed loops. 2. For each loop: a) Assume discharge Qa and direction for each pipe. Apply Continuity at each node, Total inflow = Total Outflow. Clockwise positive. b) Calculate hf=K Qan for each pipe. Retain sign from step (a) and c) compute sum of total head loss in pipes having clockwise & anticlockwise direction of flow,call it S hf.

f) Calculate hf / Qa for each pipe and sum for loop Shf/ Qa. g) Calculate correction q =-S hf /(1.85Shf/Qa). NOTE: For common members between 2 loops both corrections have to be made. As loop 1 member, q= q1 - q2. As loop 2 member, q= q2 – q1 h) Apply correction to Qa, Qnew=Qa +q. i) Repeat the procedure till q<0.2m3/min or q<10% of flow in that pipe

Problem 1 12

m3/min

A

350mmf, 330m

C

200mmf,500m

B

0.5 m3/min

200mmf,330m 10m3/min

200mmf,500m

D 1.5 m3/min

Problem 1 12

m3/min

A

200mmf,500m

B

0.5 m3/min

1m3/min 350mmf, 330m

C

200mmf,330m

11m3/min 0.5m3/min 10m3/min 1m3/min 200mmf,500m

D 1.5 m3/min

Solution 1(trial 1) Line Dia D Length Assumed flow Q1 C (mm)

(m)

m3/min)

(m3/sec)

H/Q (m of water)

AB 200 BD 200

500 330

1 0.5

0.0167 0.00833

100 100

AC 350 CD 200

330 500

11 1

0.18333 0.01667

100 100

q1=

H

(m3/s) (m3/m in)

-S hf /(1.85Shf/Qa).

Solution 1(trial 1) Line

Dia D Length Assumed flow Q1 C (mm)

(m)

m3/min)

(m3/sec)

H

H/Q (m of water)

AB BD

200 200

500 330

1 0.5

0.01666667 0.00833333

100 100

1.382686 82.9612 0.253141 30.3769

AC CD

350 200

330 500

11 1

-0.18333333 100 -0.01666667 100

5.049505 27.5428 1.382686 82.9612 223.842

q1=

q =-S hf /(1.85Shf/Qa).

(m3/s)

0.011582

(m3/min)

0.694944

Solution 1(trial2) Line Dia D (mm)

Length Assumed flow Q2 C (m)

AB 200

500

BD 200

330

AC 350

330

CD 200

500

(m3/sec)

H

H/Q (m of water)

0.01667+0.01158 100 2=0.02824867 0.01991533 100 H1= -0.1833+ 100 0.011582= 0.1717513 0.00508467 100 H2= (m3/s) q2= (m3/mi n)

3.669862

129.913

1.268658 4.93852 4.475251

63.7026

0.153776 4.629027

30.2431 249.915

-0.00067 -0.04016

26.0566

Problem 2 Solve the following pipe network using Hazen William Method CHW =100

63 L/s

1

3 pipe

L

4 2 37.8 L/s

D

1 305m

150mm

2 305m

150mm

3 610m

200mm

4 457m

150mm

5 153m

200mm

5 25.2 L/s

Problem 2 Solve the following pipe network using Hazen William Method CHW =100

63 L/s

24

3 pipe

L

11.4

1

4 2 37.8 L/s

D

1 305m

150mm

2 305m

150mm

3 610m

200mm

4 457m

150mm

5 153m

200mm

5 25.2 L/s

Solution 2(loop1,trial1) Line Dia Length Assumed flow C D Q1 (mm)

(m)

L/sec)

H

(m3/sec)

H/Q

(m of water)

1 2

150 305 150 305

24 11.4

0.024 0.0114

100 6.721649 280.069 100 1.695739 148.749

3

200 610

39

0.039

100 -8.13079 208.482 Hf=

q1=

(m3/s) (m3/min)

0.286595 -0.00024 -0.01458

637.3

Solution 2(loop2,trial 1) Line Dia D (mm)

Length Assumed C flow Q1 (m)

m3/min)

(m3/sec)

4

150

457

12.6 0.0126

2 5

150 200

305 153

11.4 0.0114 25.2 0.0252 q1=

H

H/Q

(m of water)

100 3.057644

242.67

100 -1.69574 148.749 100 -0.90911 36.0758 H2= 0.452796 427.495 (m3/s) 0.000573 (m3/min) 0.034352

Pip Dia Len 1st adj e gth st Q hf (m m) (m)

adj adj he 3rd st ad 2n st d hf/ Q hf hf/ Q hf hf/ (m) Q Q Q

Problem 3 Calculate the flows in various pipes of the circuit and the residual pressures at all points of the network • Input pressure A=23m,C=100

A

Q=0.085m3/s

300mmf,660m

B

200mmf,330m

0.004m3/s

0.025m3/s

200mmf,330m

C

D

0.025m3/s

200mmf,660m 150mmf, 330m

150mmf,330m

E 0.008m3/s

150mmf,660m

F 0.023m3/s

A

Q=0.085m3/s

300mmf,660m 0.057m3/s

200mmf,330m

0.004m3/s

0.028 m3/s

C

0.025m3/s

B 0.032 m3/s

200mmf,330m

D

0.014m3/s

0.025m3/s

200mmf,660m 150mmf, 330m

0.01 m3/s

150mmf,330m

0.021 m3/s 0.002m3/s

E 0.008m3/s

150mmf,660m

F 0.023m3/s

Solution 3(loop 1 trial 1) Line Dia D

Length Assumed flow C Q1

(mm)

(m)

H

(m3/sec)

H/Q (m of water)

AB 300 BD 200

660 330

0.057 0.032

100 100

2.464232 3.050527

43.2321 95.329

AC 200 CD 200

330 660

-0.028 -0.014

100 100

-2.382812 -1.321948

85.1004 94.4248 318.086

(m3/s)

-0.00308

q1=

Solution 3(loop 2, trial 1) Line Dia D Length Assumed C flow Q1 (mm)

(m)

H

(m3/sec)

H/Q

(m of water)

CD 200 DF 150

660 330

0.014 0.021

100 1.321948 100 5.680739

94.4 271

CE 150 EF 150

330 660

-0.01 -0.002

100 -1.43979 100 -0.146634

144 73.3 582

q1=

(m3/s)

-0.00503

Solution 3(loop 1 trial 2) Line Dia D Length Assumed flow Q2 C (mm)

(m)

H

(m3/sec)

H/Q (m of water)

AB BD

300 200

660 330

0.05392 0.02892

100 100

2.223567 2.529673

41.2383 87.4714

AC CD

200 200

330 660

-0.03108 -0.0140.00308+0.005= -0.01205

100 100

-2.890263 -1.001621

92.9943 83.1221

304.826 q2=

(m3/s)

-0.00153

Solution 3(loop 2 trial 2) Line Dia D (mm)

Leng th (m)

CD

200

660

DF

150

CE EF

150 150

Assumed C flow Q2

H

(m3/sec)

H/Q (m of water)

100

330

0.0140.00503 +0.00308= 0.01205 0.01597

100

3.42305

214

330 660

-0.01503 -0.00703

100 100

-3.05966 -1.500363

204 213 714

q2=

(m3/s)

1.001621 83.1

0.000102