--- title: "Design pipes AB, BC and AC, C = 110" subject: "Water Supply Engineering" chapter: 7 key: "semester_5/water_supply_engineering:pd-j66" source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pd-j66 site: BCT Engineering --- # Design pipes AB, BC and AC, C = 110 Water Supply Engineering · Chapter 7: Reservoirs and Distribution System - Asked in: 2066 Jestha (old course) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pd-j66 - Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-7 ## Question 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$^2$. ## 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. ## Answer #### 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Pressure at A of 15 kg/cm^2 corresponds to what head, and how much head may be lost to reach B or C? A. 15 m; 1.5 m B. 150 m; 135 m C. 150 m; 15 m D. 135 m; 150 m Answer: B. 15 kg/cm^2 = 150 m; with 15 m required at B and C, 150 - 15 = 135 m can be lost. (Key point: Pressure conversion) 2. Why does velocity, not head, govern the choice of diameters here? A. The pipes are very short B. C is high C. 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 D. Water hammer is absent Answer: C. Plenty of head means head loss will not control; practical velocity limits decide sizes. (Key point: What governs the diameter) 3. The Hardy-Cross correction for the loop is: A. dQ = -sum h / (1.852 sum |h/Q|) B. dQ = sum h / sum Q C. dQ = -sum Q / sum h D. dQ = -sum h / (2 sum Q) Answer: A. For h = rQ^1.852, dQ = -sum h / (1.852 sum |h/Q|). (Key point: Hardy-Cross correction) 4. The final residual pressure at B and C is about 140 m. What does the answer recommend? A. Reduce the diameter of AB B. Use a pressure-reducing valve or high-class pipe C. Do nothing since more pressure is better D. Remove pipe BC Answer: B. Such high pressure is excessive, so a pressure-reducing valve or high-class pipe is used. 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