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Design pipes AB, BC; 160 lpcd, peak factor 3

Short answer

Use Hazen-Williams equation with C=100 and per capita demand 160 lpcd, peak factor 3. Pipe AB (44.44 lps, 200 m) adopts 150 mm and pipe BC (24.44 lps, 500 m) adopts 125 mm. Velocities are 2.52 m/s and 1.99 m/s.

Design pipelines AB and BC for the water distribution network shown below. Take per capita demand of water 160 lpcd. Assume peak factor = 3, Hazen William's Constant C = 100. The residual head at any point in the distribution system should not be less than 10 m.

[Figure: Service reservoir at A, RL = 1000 m. A to B: L = 200 m, B has RL = 950 m, population = 3600 supplied along this section. B to C: L = 500 m, population = 4400, C has RL = 900 m.]

Per capita demand 160 lpcd, peak factor 3, C = 100, minimum residual head 10 m. HGL at A = 1000 m. The population along each section is taken as supplied at its end node: AB carries the demand of 3600 + 4400 = 8000 people, and BC carries 4400 people.

  • QAB=8000×160×3/86400=44.44Q_{AB} = 8000 \times 160 \times 3/86400 = 44.44 lps
  • QBC=4400×160×3/86400=24.44Q_{BC} = 4400 \times 160 \times 3/86400 = 24.44 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=44.444Q = 44.444 lps, length L=200L = 200 m
  • HGL at A = 1000.00 m; ground RL at B = 950 m; residual head required = 10 m
  • Available head H=1000.00−950−10=40.00H = 1000.00 - 950 - 10 = 40.00 m
  • Gradient S=H/L=40.00/200=0.20000S = H/L = 40.00/200 = 0.20000
  • From hf=10.67 L Q1.852C1.852D4.87h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}: Dcalc=120.2D_{calc} = 120.2 mm, adopt commercial size 150 mm (raised from the head-based size because velocity would exceed 3 m/s)
  • Actual hf=13.595h_f = 13.595 m, HGL at B = 986.40 m, residual head = 36.40 m (≥ 10 m, OK)
  • Velocity v=Q/A=2.52v = Q/A = 2.52 m/s

Pipe BC

  • Discharge Q=24.444Q = 24.444 lps, length L=500L = 500 m
  • HGL at B = 986.40 m; ground RL at C = 900 m; residual head required = 10 m
  • Available head H=986.40−900−10=76.40H = 986.40 - 900 - 10 = 76.40 m
  • Gradient S=H/L=76.40/500=0.15281S = H/L = 76.40/500 = 0.15281
  • From hf=10.67 L Q1.852C1.852D4.87h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}: Dcalc=101.2D_{calc} = 101.2 mm, adopt commercial size 125 mm
  • Actual hf=27.296h_f = 27.296 m, HGL at C = 959.11 m, residual head = 59.11 m (≥ 10 m, OK)
  • Velocity v=Q/A=1.99v = Q/A = 1.99 m/s

Summary

PipeQ (lps)L (m)Calc. D (mm)Adopted D (mm)h_f (m)Residual (m)v (m/s)
AB44.4444200120.215013.6036.402.52
BC24.4444500101.212527.3059.111.99

Answer: AB = 150 mm and BC = 125 mm. For AB the head-based size is 125 mm, but its velocity would be 3.6 m/s, which is too high, so 150 mm is adopted (v = 2.5 m/s). Residual heads are above 10 m at B and C.

✓ Answer checked by Claude Opus 5.5 (high effort)

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