--- title: "Series-parallel pipes with control valve" subject: "Hydraulics" chapter: 2 key: "semester_4/hydraulics:series-parallel-valve-equal-discharge" source: https://www.bctengineer.com/q/semester_4/hydraulics/series-parallel-valve-equal-discharge site: BCT Engineering --- # Series-parallel pipes with control valve Hydraulics · Chapter 2: Simple pipe flow problems and solution - Asked in: 2075 Baisakh, 10 marks; 2069 Poush, 8 marks (asked 2 times) - Video: [Branching Pipe Flow - Bernoulli Equation and Continuity Equation Example Problem](https://www.youtube.com/watch?v=Qh6nJltXRDU) — Brian Bernard - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/series-parallel-valve-equal-discharge - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-2 ## Question A system of pipes conveying water is connected in parallel and in series as shown in figure below. The section DE represents the resistance of a valve for controlling the flow, which has a resistance coefficient $K_{DE} = \left(\frac{4000}{n}\right)^2$, where n is the percentage valve opening. [Figure: pipe A to B splits into two parallel pipes AA1B and AA2B, continues as BC, then at C branches into CD (with the valve, leading to E) and CF (leading to F)] The friction factor f in the Darcy formula is 0.024 for all pipes, and their lengths and diameters are given by | Pipe | Length (m) | Diameter (m) | |---|---|---| | AA1B | 30 | 0.1 | | AA2B | 30 | 0.125 | | BC | 60 | 0.15 | | CD | 15 | 0.1 | | CF | 30 | 0.1 | The head at A is 100 m, at E is 40 m and at F is 60 m. If the valve is adjusted to give equal discharge rates at E and F, calculate the head at C, the discharge through the system and percentage valve opening. Neglect all losses except those due to friction. ## Short answer Solving the pipe network yields head at C = 75.48 m, total system discharge = 0.102 m³/s (102.0 L/s), and valve opening n = 38.7%. ## Answer **Method.** Use $h_f = rQ^2$ for each pipe, with $r = \dfrac{8fL}{\pi^2gD^5}$ ($f = 0.024$). Only friction is considered. Heads: $H_A = 100$ m, $H_E = 40$ m, $H_F = 60$ m. For equal discharge let $Q_E = Q_F = q$, so the flow in $BC$ is $2q$. #### Resistance coefficients (s²/m⁵) | Pipe | L (m) | D (m) | $r = 8fL/(\pi^2gD^5)$ | |---|---|---|---| | AA1B | 30 | 0.100 | 5949.1 | | AA2B | 30 | 0.125 | 1949.4 | | BC | 60 | 0.150 | 1566.9 | | CD | 15 | 0.100 | 2974.6 | | CF | 30 | 0.100 | 5949.1 | **Parallel pair A to B** (same head loss in both): $Q = \sqrt{h/r_1} + \sqrt{h/r_2}$, so $$ \frac{1}{\sqrt{r_{AB}}} = \frac{1}{\sqrt{r_1}} + \frac{1}{\sqrt{r_2}} \;\Rightarrow\; r_{AB} = 788.4 $$ #### Discharge Energy balance from A to F (pair AB, then BC carrying $2q$, then CF carrying $q$): $$ H_A - H_F = r_{AB}(2q)^2 + r_{BC}(2q)^2 + r_{CF}q^2 $$ $$ 100 - 60 = q^2\left[4(788.4 + 1566.9) + 5949.1\right] = 15370.2\,q^2 \;\Rightarrow\; q = 0.05101\ \text{m}^3/\text{s} $$ $$ Q = 2q = 0.10203\ \text{m}^3/\text{s} = 102.0\ \text{l/s} $$ #### Head at C $$ H_C = H_F + r_{CF}q^2 = 60 + 5949.1(0.002602) = 75.48\ \text{m} $$ (Check from A: $H_B = H_C + r_{BC}(2q)^2 = 91.79$ m; the two parallel pipes then carry 0.03714 and 0.06489 m³/s, which add to 0.10203 m³/s.) #### Valve opening Branch C to E has pipe CD and the valve DE in series, carrying $q$ with head loss $H_C - H_E = 75.48 - 40 = 35.48$ m: $$ (r_{CD} + K_{DE})q^2 = 35.48 \;\Rightarrow\; K_{DE} = \frac{35.48}{0.002602} - 2974.6 = 10659.7 $$ $$ K_{DE} = \left(\frac{4000}{n}\right)^2 \;\Rightarrow\; n = \frac{4000}{\sqrt{10659.7}} = 38.7\ \% $$ **Answer:** Head at C $= 75.48$ m; total discharge $= 0.1020$ m³/s ($102.0$ l/s); valve opening $n \approx 38.7\ \%$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For two pipes in parallel between A and B, which condition holds? A. Same discharge in both and heads add B. Same head loss in both and discharges add, 1/sqrt(r_AB) = 1/sqrt(r1) + 1/sqrt(r2) C. Resistances add D. Head losses add Answer: B. Parallel pipes share the same head loss and the flows add, giving r_AB = 788.4. (Key point: Parallel pipes) 2. With Q_E = Q_F = q, what flows in BC and AB? A. q B. q/2 C. 2q D. 4q Answer: C. BC feeds both branches, so it carries 2q, and the parallel pair also delivers 2q in total. (Key point: Flow in BC) 3. How is q found? A. Apply 100 - 40 = losses along A to E only B. Use H_C = 100 C. Use the valve data directly D. Apply H_A - H_F = [4(r_AB + r_BC) + r_CF] q^2 = 15370 q^2, giving q = 0.05101 m3/s Answer: D. The F branch has no valve, so energy from A to F fixes q; total Q = 2q = 0.1020 m3/s. (Key point: Discharge from A to F) 4. How is the valve opening n found? A. From (r_CD + K_DE) q^2 = H_C - H_E = 35.48 m, then n = 4000/sqrt(K_DE) B. From K_DE = r_CF C. From n = H_C/H_E D. From (r_CD + K_DE) = H_A - H_E Answer: A. The C-E branch has CD and the valve in series carrying q, giving K_DE = 10659.7 and n = 38.7 %. 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