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Design pipes AB and BC, 75 lpcd, tank 6 m high

Short answer

Pipe AB: 65 mm, residual head at B 26.9 m, velocity 1.04 m/s. Pipe BC: 40 mm, residual at C 8.4 m, velocity 0.99 m/s. Both meet minimum 5 m head, velocities acceptable.

Design pipes; AB and BC having 400 m and 500 m length respectively. Assume Hazen Williams coefficient C as 100 and maximum demand as 3 times the average demand. The average water consumption is 75 lpcd and population is distributed within the two blocks of 850 in AB section and 480 in BC section. The elevated storage tank is fixed at point A above 6 m height. The R.L. of points A, B and C are 2160, 2125 and 2115 meters respectively. The minimum pressure head of water is to be 5.0 m. Check velocity in the pipes also.

Data: C = 100, peak (maximum) demand = 3 × average, 75 lpcd, minimum residual head 5.0 m. The tank at A stands 6 m high, so the HGL at A = 2160 + 6 = 2166 m. The population of each section is taken as drawn at its end point (B for AB, C for BC), so pipe AB carries the demand of 850 + 480 = 1330 people and BC carries 480 people.

  • QAB=1330×75×3/86400=3.464Q_{AB} = 1330 \times 75 \times 3/86400 = 3.464 lps
  • QBC=480×75×3/86400=1.250Q_{BC} = 480 \times 75 \times 3/86400 = 1.250 lps

Method

Use the Hazen-Williams equation in SI units:

hf=10.67 L Q1.852C1.852D4.87h_f = \frac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}

with QQ in m³/s, LL and DD in m. For each pipe the available head is (HGL at upstream end − ground RL of the tap − required residual head). The gradient S=H/LS = H/L gives the calculated diameter, the next larger commercial size is adopted, and the real head loss and residual head are then checked. Commercial sizes (nominal bore): 15, 20, 25, 32, 40, 50, 65, 80, 100, 125, 150, 200 mm and so on. Constant C = 100.

Pipe AB

  • Discharge Q=3.464Q = 3.464 lps, length L=400L = 400 m
  • HGL at A = 2166.00 m; ground RL at B = 2125 m; residual head required = 5 m
  • Available head H=2166.00−2125−5=36.00H = 2166.00 - 2125 - 5 = 36.00 m
  • Gradient S=H/L=36.00/400=0.09000S = H/L = 36.00/400 = 0.09000
  • From hf=10.67 L Q1.852C1.852D4.87h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}: Dcalc=53.7D_{calc} = 53.7 mm, adopt commercial size 65 mm
  • Actual hf=14.143h_f = 14.143 m, HGL at B = 2151.86 m, residual head = 26.86 m (≥ 5 m, OK)
  • Velocity v=Q/A=1.04v = Q/A = 1.04 m/s

Pipe BC

  • Discharge Q=1.250Q = 1.250 lps, length L=500L = 500 m
  • HGL at B = 2151.86 m; ground RL at C = 2115 m; residual head required = 5 m
  • Available head H=2151.86−2115−5=31.86H = 2151.86 - 2115 - 5 = 31.86 m
  • Gradient S=H/L=31.86/500=0.06371S = H/L = 31.86/500 = 0.06371
  • From hf=10.67 L Q1.852C1.852D4.87h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}: Dcalc=39.1D_{calc} = 39.1 mm, adopt commercial size 40 mm
  • Actual hf=28.483h_f = 28.483 m, HGL at C = 2123.37 m, residual head = 8.37 m (≥ 5 m, OK)
  • Velocity v=Q/A=0.99v = Q/A = 0.99 m/s

Summary

PipeQ (lps)L (m)Calc. D (mm)Adopted D (mm)h_f (m)Residual (m)v (m/s)
AB3.4635440053.76514.1426.861.04
BC1.2550039.14028.488.370.99

Velocities are between 0.6 m/s and 3 m/s, so the sizes are acceptable.

Answer: AB = 65 mm and BC = 40 mm; residual heads 26.9 m at B and 8.4 m at C (both more than 5 m); velocities 1.04 m/s and 0.99 m/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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