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) andc)  compute sum of total head loss

in pipes having clockwise & anticlockwise direction of flow,call it hf.

f) Calculate hf / Qa for each pipe and sum for loop

hf/ Qa.  g)  Calculate correction q= hf /(1.85hf/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

200mm,500m

350mm,330m

200mm,330m

200mm,500m

12 m3/min0.5 m3/min

1.5 m3/min

D

A B

C

10m3/min

Problem 1

200mm,500m

350mm,330m

200mm,330m

200mm,500m

1m3/min

12 m3/min0.5 m3/min

0.5m3/min

1m3/min

11m3/min

10m3/min

1.5 m3/min

D

A B

C

Solution 1(trial 1)LineDia D Length Assumed flow Q1 C H H/Q

(mm) (m) m3/min) (m3/sec) (m of water)

AB 200 500 1 0.0167 100

BD 200 330 0.5 0.00833 100

         

AC 350 330 11 0.18333 100

CD 200 500 1 0.01667 100

q1= (m3/s) - hf /(1.85hf/Qa).  (m3/

min)

Solution 1(trial 1)Line Dia D Length Assumed flow Q1 C H H/Q

(mm) (m) m3/min) (m3/sec) (m of water)

AB 200 500 1 0.01666667 100 1.382686 82.9612

BD 200 330 0.5 0.00833333 100 0.253141 30.3769

         

AC 350 330 11 -0.18333333 100 5.049505 27.5428

CD 200 500 1 -0.01666667 100 1.382686 82.9612

223.842

q1= (m3/s) 0.011582

  (m3/min)

0.694944

q= hf /(1.85hf/Qa).

Solution 1(trial2)

Line Dia D Length Assumed flow Q2 C H H/Q (mm) (m) (m3/sec) (m of water)

AB 200 500   0.01667+0.011582=0.02824867

100 3.669862 129.913

BD 200 330   0.01991533 100 1.268658 63.7026          H1= 4.93852 AC 350 330   -0.1833+

0.011582= 0.1717513

100 4.475251 26.0566

CD 200 500   0.00508467 100 0.153776 30.2431H2= 4.629027 249.915

q2= (m3/s) -0.00067

  (m3/min)

-0.04016

Problem 2

pipe L D

1 305m 150mm

2 305m 150mm

3 610m 200mm

4 457m 150mm

5 153m 200mm

1

234

5

37.8 L/s

25.2 L/s

63 L/s

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

Problem 2

pipe L D

1 305m 150mm

2 305m 150mm

3 610m 200mm

4 457m 150mm

5 153m 200mm

1

234

5

37.8 L/s

25.2 L/s

63 L/s

24

39 11.4

12.6

25.2

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

Solution 2(loop1,trial1)

Line Dia D

Length Assumed flow Q1

C H H/Q

(mm) (m) L/sec) (m3/sec) (m of water)

1 150 305 24 0.024 100 6.721649 280.0692 150 305 11.4 0.0114 100 1.695739 148.749               3 200 610 39 0.039 100 -8.13079 208.482               

Hf= 0.286595 637.3q1= (m3/s) -0.00024  (m3/min) -0.01458

Solution 2(loop2,trial 1)

Line Dia D Length Assumed flow Q1

C H H/Q

(mm) (m) m3/min) (m3/sec) (m of water)

4 150 457 12.6 0.0126 100 3.057644 242.67   

               2 150 305 11.4 0.0114 100 -1.69574 148.7495 200 153 25.2 0.0252 100 -0.90911 36.0758

H2= 0.452796 427.495q1= (m3/s) 0.000573  (m3/min) 0.034352

Pipe

Dia Length

1st adjst

2nd

adjst

3rd

adjst

head

(mm)

(m)

Q hf hf/Q

Q hf hf/Q

Q hf hf/Q

(m)

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

300mm,660m

200mm,330m 200mm,330m

200mm,660m

Q=0.085m3/s0.025m3/s

0.025m3/s

150mm,330m

150mm,330m

150mm,660m

0.004m3/s

0.023m3/s0.008m3/s

A B

DC

FE

300mm,660m

200mm,330m 200mm,330m

200mm,660m

0.057m3/s

Q=0.085m3/s0.025m3/s

0.032m3/s

0.014m3/s

0.028m3/s

0.025m3/s

150mm,330m

150mm,330m

150mm,660m

0.004m3/s

0.021m3/s

0.002m3/s

0.01m3/s

0.023m3/s0.008m3/s

A B

DC

FE

Solution 3(loop 1 trial 1)

Line Dia D Length Assumed flow Q1

C H H/Q

(mm) (m) (m3/sec) (m of water)

AB 300 660   0.057 100 2.464232 43.2321BD 200 330   0.032 100 3.050527 95.329           AC 200 330   -0.028 100 -2.382812 85.1004CD 200 660   -0.014 100 -1.321948 94.4248

318.086q1= (m3/s) -0.00308

Solution 3(loop 2, trial 1)LineDia D Length Assumed

flow Q1C H H/Q

(mm) (m) (m3/sec) (m of water)

CD 200 660   0.014 100 1.321948 94.4

DF 150 330   0.021 100 5.680739 271

           

CE 150 330   -0.01 100 -1.43979 144

EF 150 660   -0.002 100 -0.146634 73.3

582

q1= (m3/s) -0.00503

Solution 3(loop 1 trial 2)

Line Dia D Length Assumed flow Q2C H H/Q

(mm) (m) (m3/sec) (m of water)

AB 300 660   0.05392 100 2.223567 41.2383BD 200 330   0.02892 100 2.529673 87.4714           AC 200 330   -0.03108 100 -2.890263 92.9943CD 200 660   -0.014-

0.00308+0.005=-0.01205

100 -1.001621 83.1221

304.826q2= (m3/s) -0.00153

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

h Assumed flow Q2

C H H/Q

(mm) (m) (m3/sec) (m of water)

CD 200 660   0.014-0.00503+0.00308=0.01205

100 1.001621 83.1

DF 150 330   0.01597 100 3.42305 214           CE 150 330   -0.01503 100 -3.05966 204EF 150 660   -0.00703 100 -1.500363 213

714q2= (m3/s) 0.000102

Class problem 1

Example: Obtain the flow rates in the network shown below. 90 l/s A 55 600 m B 45 35 600 m 254 mm 600 m C C 152 mm 15 15 60l/s 66600 600 m E 600 m 5 D 152 mm 152 mm

254 mm 10 +ve 600 152 mm

Correct the flows as shown below: 90 l/s A 63 B 49

27 C 60 //s 11 E 3 D 30 l/s

14

Correct flows again for the third trial 90 l/s 65 A B 52 25 C 60 l/s 8 E 5 D 30 l/s

13

Final Water Flows

Final Water Flows 90 l/s 66 l/s 53 l/s 24 l/s 60 l/s 7 30 l/s 6 l/s

Note: A computer programme exists for analysis using the Hardy Cross Method

13 l/s

Problem 1