Why does velocity, not head, govern the choice of diameters here?

The available head (135 m) is far more than needed, so the pipes are sized to keep velocity about 1 to 2.5 m/s

The Hardy-Cross correction for the loop is:

dQ = -sum h / (1.852 sum |h/Q|)

The final residual pressure at B and C is about 140 m. What does the answer recommend?

Use a pressure-reducing valve or high-class pipe

Do nothing since more pressure is better

2066 Jestha (old course)

Design pipes AB, BC and AC, C = 110

Short answer

Use Hardy-Cross method with Hazen-Williams equation. Adopt AB = 125 mm, AC = 80 mm, BC = 80 mm. Final flows: AB = 26.06 lps, BC = 1.06 lps, AC = 6.94 lps. Velocities below 3 m/s.

Design the pipes AB, BC and AC for the water distribution network given below. Take Hazen William's co-efficiency as 110 for all the pipes and assume suitable data as required.

[Figure: triangle network; 33 lps enters at A; AB length = 200 m, BC length = 200 m, AC length = 270 m; 25 lps leaves at B and 8 lps leaves at C.] Consider RL of points A, B and C are same. The pressure available at A is 15 kg/cm $^2$ and the minimum pressure required at B and C is 1.5 kg/cm $^2$ .

Given data

Triangle A–B–C at the same RL. Inflow 33 lps at A; outflow 25 lps at B and 8 lps at C. $L_{AB} = 200$ m, $L_{BC} = 200$ m, $L_{AC} = 270$ m, $C = 110$
Pressure at A = 15 kg/cm² = 150 m of water; minimum pressure required at B and C = 15 m
Head that can be lost between A and B (or C) $= 150 - 15 = 135$ m

Step 1: Choose diameters

The head available (135 m) is much more than what is needed, so the velocity (about 1–2.5 m/s) governs the choice. The first estimate of flows: AB = 25 lps (to B), AC = 8 lps (to C), and BC just a connection. Adopt (commercial sizes): AB = 125 mm, AC = 80 mm, BC = 80 mm.

Step 2: Flow distribution by Hardy-Cross

Loop A→B→C→A. Resistance $r = 10.67L/(C^{1.852}D^{4.87})$ and $h = rQ^{1.852}$ , $\Delta Q = -\sum h/(1.852\sum |h/Q|)$ . Trial flows: AB = 25, BC = 0, AC = 8 lps (A to B, B to C and A to C).

Iteration	ΔQ (lps)	AB (lps)	BC, B→C (lps)	AC (lps)
1	+1.076	26.08	1.08	6.92
2	-0.016	26.06	1.06	6.94
3	-0.000	26.06	1.06	6.94

Final flows: AB = 26.06 lps, BC = 1.06 lps (B to C), AC = 6.94 lps. Continuity check: at B, 26.06 − 25 = 1.06 ✓; at C, 1.06 + 6.94 = 8 ✓.

Step 3: Check heads and velocities

Head loss in AB $= 10.30$ m, in BC $= 0.24$ m and in AC $= 10.54$ m; the loop closes because AB + BC $= 10.54$ m $=$ AC
Pressure head at B $= 150 - 10.30 = 139.7$ m; at C $= 150 - 10.54 = 139.5$ m; both are far more than 15 m
Velocities: AB $= 2.12$ m/s, AC $= 1.38$ m/s, BC $= 0.21$ m/s (all below 3 m/s)

Answer: AB = 125 mm, AC = 80 mm, BC = 80 mm. The pressure at B and C (about 140 m) is high, so a pressure-reducing valve or high-class pipe must be used.

✓ Answer checked by Claude Opus 5.5 (high effort)

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All Water Supply Engineering chapter 7 questions →

Design pipes AB, BC and AC, C = 110

Given data

Step 1: Choose diameters

Step 2: Flow distribution by Hardy-Cross

Step 3: Check heads and velocities

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