What does the Hardy-Cross method do?

Applies a correction repeatedly until the algebraic sum of head losses around a loop is zero

Finds only the pressure at nodes

Assumes equal flow in every pipe

With clockwise flow positive, how is pipe AD (flow A to D) treated?

Negative, as it flows anticlockwise

2069 Asar · 10 marks

Hardy-Cross network discharge AB, BC, AD, CD

Short answer

Use Hardy-Cross method with Hazen-Williams equation. After iterations, final discharges: AB = 20.97 lps, BC = 0.97 lps, AD = 34.03 lps, CD = 9.03 lps from D to C.

Calculate the discharge in pipes AB, BC, AD and CD for the water distribution network given below by using Hardy-cross method. The available data of network are as follows:

Pipe	Length (in meter)	Diameter (in mm)
AB	400	300
BC	300	200
AD	500	400
CD	500	300

Hazen Williams coefficient as 100 for all pipes. Assume other necessary data suitably.

[Figure: loop ABCD; 55 lps enters at A, 20 lps leaves at B, 10 lps leaves at C, 25 lps leaves at D.]

Hardy-Cross method balances the flows in a loop by repeatedly applying a correction until the algebraic sum of head losses around the loop is zero.

Data

Hazen-Williams $h = r\,Q^{1.852}$ with $r = 10.67L/(C^{1.852}D^{4.87})$ , C = 100. Inflow 55 lps at A; outflow 20 lps at B, 10 lps at C and 25 lps at D. Correction:

\Delta Q = -\frac{\sum h}{1.852\sum |h/Q|}

Take clockwise flow (A→B→C→D) as positive.

Trial flows (satisfying continuity at every node)

AB = 35 lps (A→B), BC = 15 lps (B→C), CD = 5 lps (C→D), AD = 20 lps (A→D, anticlockwise, so negative). Check: A: 35 + 20 = 55 ✓; B: 35 − 20 = 15 ✓; C: 15 − 10 = 5 ✓; D: 5 + 20 = 25 ✓.

Iteration 1

Pipe	Q (lps)	h = rQ^1.852 (m)	h/Q
AB	+35.00	+0.597	0.0171
BC	+15.00	+0.672	0.0448
CD	+5.00	+0.020	0.0041
AD	-20.00	-0.065	0.0033

$\sum h = +1.224$ m, $\sum|h/Q| = 0.0692$ , so $\Delta Q = -1.224/(1.852 \times 0.0692) = -9.555$ lps. Corrected flows (the correction is added to clockwise pipes and subtracted from AD): AB = 25.45, BC = 5.45, CD = -4.55, AD = 29.55 lps.

Iteration 2

Pipe	Q (lps)	h = rQ^1.852 (m)	h/Q
AB	+25.45	+0.331	0.0130
BC	+5.45	+0.103	0.0189
CD	-4.55	-0.017	0.0038
AD	-29.55	-0.134	0.0046

$\sum h = +0.282$ m, $\Delta Q = -3.791$ lps.

Further iterations

Corrections: -0.662, -0.026, -0.000, -0.000 lps; the process stops when ΔQ is negligible (< 0.01 lps).

Final discharges

Pipe	Discharge (lps)	Direction
AB	20.97	A → B
BC	0.97	B → C
AD	34.03	A → D
CD	9.03	D → C

Checks: at B, 20.97 = 20 + 0.97 ✓; at C, 0.97 + 9.03 = 10 ✓; at D, 34.03 = 25 + 9.03 ✓.

Answer: AB = 20.97 lps, BC = 0.97 lps, AD = 34.03 lps, CD = 9.03 lps (from D to C).

✓ Answer checked by Claude Opus 5.5 (high effort)

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