--- title: "Branch diameters equal to a 450 mm direct pipe" subject: "Hydraulics" chapter: 3 key: "semester_4/hydraulics:three-reservoir-abc-equal-discharge-equivalent-pipe" source: https://www.bctengineer.com/q/semester_4/hydraulics/three-reservoir-abc-equal-discharge-equivalent-pipe site: BCT Engineering --- # Branch diameters equal to a 450 mm direct pipe Hydraulics · Chapter 3: Three reservoirs problem and Pipe networks - Asked in: 2069 Bhadra, 10 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/three-reservoir-abc-equal-discharge-equivalent-pipe - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-3 ## Question A reservoir A discharges through a pipe 450 mm in diameter and 900 m long which is connected to two pipes, one 1200 m long leading to reservoir B 36 m below A and the other 1500 m long leading to reservoir C 45 m below A. Calculate the diameters of these two pipes if they have equal discharges which together equal that of a 450 mm diameter pipe of length 2100 m connected directly from reservoir A to reservoir B. Neglect all losses except those due to friction and assume that the friction factor f is the same for all pipes. ## Short answer Direct pipe gives unit head loss 0.017143 m/m. Main AJ loses 15.43 m; available: B 20.57 m, C 29.57 m. Each branch carries Q/2; D⁵ = (kQ²L)/(4h) → D_B ≈ 341 mm, D_C ≈ 332 mm. ## Answer **Data.** Main pipe AJ: 900 m, 450 mm. Branch JB: 1200 m to B, 36 m below A. Branch JC: 1500 m to C, 45 m below A. The branches carry equal discharges, $Q/2$ each; their sum $Q$ equals the discharge of a 450 mm pipe, 2100 m long, running directly from A to B (36 m head). Friction only, same $f$ everywhere. Use $h_f = k\dfrac{LQ^2}{D^5}$ with $k = \dfrac{8f}{\pi^2g}$ (the unknown $f$ and $Q$ cancel). #### Step 1: total discharge from the direct pipe $$ 36 = k\frac{2100\,Q^2}{(0.45)^5} \;\Rightarrow\; \frac{kQ^2}{(0.45)^5} = \frac{36}{2100} = 0.017143 $$ That is, a 450 mm pipe carrying $Q$ loses 0.017143 m of head per metre of length. #### Step 2: head loss in the main pipe AJ It has the same diameter and carries the same $Q$: $$ h_{AJ} = 0.017143(900) = 15.43\ \text{m} $$ So the head at J is 15.43 m below the level of A. #### Step 3: head available for each branch - To B: $36 - 15.43 = 20.57$ m - To C: $45 - 15.43 = 29.57$ m #### Step 4: branch diameters (each carries $Q/2$) $$ h = k\frac{L(Q/2)^2}{D^5} \;\Rightarrow\; D^5 = \frac{kQ^2L}{4h} = \frac{(0.017143)(0.45)^5\,L}{4h}\quad\text{with } (0.45)^5 = 0.018453 $$ $$ D_B^5 = \frac{0.017143(0.018453)(1200)}{4(20.57)} \;\Rightarrow\; D_B = 0.3410\ \text{m} \approx 341\ \text{mm} $$ $$ D_C^5 = \frac{0.017143(0.018453)(1500)}{4(29.57)} \;\Rightarrow\; D_C = 0.3316\ \text{m} \approx 332\ \text{mm} $$ **Answer:** The branch to B should be about 341 mm and the branch to C about 332 mm in diameter. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the 450 mm, 2100 m pipe under 36 m head give? A. Loss per metre of 36/900 B. Loss per metre of 45/2100 C. Loss of 36 m per Q D. Loss per metre of 36/2100 = 0.017143 for discharge Q Answer: D. kQ^2/(0.45)^5 = 36/2100 = 0.017143. (Key point: Total discharge from the direct pipe) 2. Why does main pipe AJ lose 0.017143 x 900 = 15.43 m? A. Its friction factor is zero B. It is 2100 m long C. Same diameter and same total discharge Q D. Half the discharge flows in it Answer: C. It is 450 mm and carries Q, so the loss per metre is the same. (Key point: Head loss in main pipe) 3. What head is available for the branch to C? A. 15.43 m B. 45 - 15.43 = 29.57 m C. 36 - 15.43 = 20.57 m D. 45 m Answer: B. C is 45 m below A, minus the loss in the main pipe. (Key point: Head for branch C) 4. Why does D^5 = kQ^2 L/(4h) for each branch? A. Each branch carries Q/2, so Q^2 becomes Q^2/4 B. Each branch carries 2Q C. The branch has no friction D. h is quartered Answer: A. The two branches share Q equally, so each loses h = kL(Q/2)^2/D^5. 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