--- title: "Valve opening to double discharge into C" subject: "Hydraulics" chapter: 3 key: "semester_4/hydraulics:three-reservoir-valve-n-discharge-c-twice-b" source: https://www.bctengineer.com/q/semester_4/hydraulics/three-reservoir-valve-n-discharge-c-twice-b site: BCT Engineering --- # Valve opening to double discharge into C Hydraulics · Chapter 3: Three reservoirs problem and Pipe networks - Asked in: 2070 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-valve-n-discharge-c-twice-b - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-3 ## Question Three reservoirs A, B and C are interconnected by three pipes which all meet at junctions J. The water surface of reservoir B is 20 m above the surface of C whilst the surface of A is 40 m above the surface of B. A flow control valve is fitted just before junction J in pipe AJ. The head loss $h_L$ through pipes and components can be written as $h_L = rQ^2$ where r is the resistance coefficient. The value of r for the valve and the pipes are $r_{AJ}$ = 150, $r_{BJ}$ = 200, $r_{CJ}$ = 300, $r_{valve} = (400/n)^2$. Where n is the percentage valve opening. Find the value of n which will make the discharge into reservoir C twice into reservoir B. ## Short answer From continuity Q_A = 3Q_B, Q_C = 2Q_B. Solve head balance at junction using h = rQ²: Q_B = 0.1414 m³/s, H_J = 24 m. Head loss A–J: 60−24 = (150+r_v)(0.4243)² gives r_v = 50; (400/n)² = 50 → n = 56.6%. ## Answer **Data.** Take the datum at the surface of C: $Z_C = 0$, $Z_B = 20$ m, $Z_A = 20 + 40 = 60$ m. Head losses $h = rQ^2$: $r_{AJ} = 150$, $r_{BJ} = 200$, $r_{CJ} = 300$, and the valve in AJ has $r_v = (400/n)^2$. The discharge into C is twice the discharge into B, so **B receives water**: $Q_C = 2Q_B$, and by continuity at J, $$ Q_A = Q_B + Q_C = 3Q_B $$ Let $H_J$ be the head at the junction. #### Heads at J from the lower reservoirs - J to B: $H_J - 20 = r_{BJ}Q_B^2 = 200\,Q_B^2$ - J to C: $H_J - 0 = r_{CJ}Q_C^2 = 300(2Q_B)^2 = 1200\,Q_B^2$ Subtracting: $1200Q_B^2 - 200Q_B^2 = 20 \Rightarrow Q_B^2 = 0.02$, so $$ Q_B = 0.1414\ \text{m}^3/\text{s},\quad Q_C = 2Q_B = 0.2828\ \text{m}^3/\text{s},\quad Q_A = 3Q_B = 0.4243\ \text{m}^3/\text{s} $$ $$ H_J = 1200(0.02) = 24.0\ \text{m}\quad(\text{check: } 20 + 200(0.02) = 24\ \text{m}) $$ #### Valve opening Head loss from A to J: $60 - 24.0 = 36.0$ m $= (r_{AJ} + r_v)Q_A^2$: $$ 150 + r_v = \frac{36.0}{(0.4243)^2} = \frac{36.0}{0.18} = 200 \;\Rightarrow\; r_v = 50 $$ $$ \left(\frac{400}{n}\right)^2 = 50 \;\Rightarrow\; n = \frac{400}{\sqrt{50}} = 56.57\ \% $$ **Answer:** The valve must be about $56.6\%$ open ($Q_A = 424$ l/s, $Q_B = 141$ l/s, $Q_C = 283$ l/s, $H_J = 24$ m above C). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With Q_C = 2 Q_B, what is Q_A? A. Q_B B. Q_B/2 C. 3 Q_B D. 2 Q_B Answer: C. B receives water, so Q_A = Q_B + Q_C = 3 Q_B. (Key point: Flow relation at J) 2. Which equation gives Q_B? A. 300 Q_B^2 = 200(2Q_B)^2 B. 300(2Q_B)^2 - 200 Q_B^2 = 20 C. 300 Q_B^2 + 200 Q_B^2 = 20 D. 300(2Q_B)^2 + 200 Q_B^2 = 60 Answer: B. H_J = 1200 Q_B^2 from C and H_J = 20 + 200 Q_B^2 from B; subtracting gives Q_B^2 = 0.02. (Key point: Finding Q_B) 3. The head loss from A to J is 36 m with Q_A^2 = 0.18. What is r_v? A. 50 B. 150 C. 200 D. 36 Answer: A. (150 + r_v) = 36/0.18 = 200, so r_v = 50. (Key point: Valve resistance) 4. From r_v = (400/n)^2 = 50, what is n? A. About 8 % B. About 28 % C. About 80 % D. About 56.6 % Answer: D. n = 400/sqrt(50) = 56.57 %. 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