Water Supply Engineering Solved Problems Pdf !!hot!! May 2026

Q_avg = 75,000 × 200 = 15,000,000 L/day = 15,000 m³/day (173.6 L/s)

After one iteration, flows are adjusted typically by 2–5 L/s until loop head loss sum ≈ zero. Problem 6.1 A town has a daily demand pattern with maximum hourly demand of 400 m³/h and average 200 m³/h. Supply from pumping is constant at 220 m³/h for 20 hours. Determine balancing storage required. water supply engineering solved problems pdf

Growth rates: r1 = (55-45)/45 = 0.2222 (22.22%) r2 = (68-55)/55 = 0.2364 r3 = (84-68)/68 = 0.2353 Average r = 0.2313 (23.13%) P2030 = 84,000 × (1+0.2313)² = 84,000 × 1.515 = 127,260 Q_avg = 75,000 × 200 = 15,000,000 L/day

Q_peak hourly = 2.7 × 15,000 = 40,500 m³/day (468.75 L/s) Determine balancing storage required

Increases: 10, 13, 16 Increments of increases: 3, 3 Average increment = 3 Average increase = 13 P2030 = 84,000 + (2×13) + [2×3×(2+1)/2] = 84,000 + 26 + 9 = 110,035 2. Problem Set 2: Water Demand & Fire Flow Problem 2.1 A town of 75,000 people has a per capita water supply of 200 L/day. Calculate: (a) Average daily demand (m³/day) (b) Maximum daily demand (assume factor = 1.8) (c) Peak hourly demand (assume factor = 2.7) (d) Fire demand using Kuichling’s formula

Using h_f = K × Q^1.852 with K = 10.67×L / (C^1.852×D^4.87) D=0.25m → D^4.87 = 0.25^4.87 = 0.25^4 × 0.25^0.87 = 0.003906 × 0.305 = 0.001191 C^1.852 = 100^1.852 = 5120 (approx)

Average increase per decade = [(55-45)+(68-55)+(84-68)] / 3 = (10+13+16)/3 = 13,000 per decade P2030 = P2010 + 2 × 13,000 = 84,000 + 26,000 = 110,000