--- title: "Parallel pipe added in second half" subject: "Hydraulics" chapter: 2 key: "semester_4/hydraulics:parallel-pipe-added-second-half" source: https://www.bctengineer.com/q/semester_4/hydraulics/parallel-pipe-added-second-half site: BCT Engineering --- # Parallel pipe added in second half Hydraulics · Chapter 2: Simple pipe flow problems and solution - Asked in: 2081 Chaitra, 8 marks - Video: [Mastering Parallel Pipe Flow Systems | Fluid Mechanics Explained](https://www.youtube.com/watch?v=B4pysiBSBWM) — Engineering Key Tutorials - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/parallel-pipe-added-second-half - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-2 ## Question A pipeline of 600 mm diameter is 1.5 km long. To increase the discharge, another line of same diameter is introduced parallel to the first in the second half of the length. If f = 0.01 and head at inlet is 300 mm, calculate the increase in discharge. [Figure: reservoir with 0.3 m head at the inlet A; pipe of total length L = 1500 m made of two 750 m halves; in the second half the flow splits at B into two parallel pipes (discharges $Q_1$ and $Q_2$), each $L_1 = L_2$ = 750 m and $D_1 = D_2$ = 0.6 m] ## Short answer Adding a parallel pipe of same diameter in the second half increases discharge by 0.0364 m³/s, a 26.5% rise, from 0.137 to 0.174 m³/s. ## Answer **Data.** $D = 0.6$ m, $L = 1500$ m, $f = 0.01$, head $H = 0.3$ m (as given). Minor losses are neglected, so the head is used up in friction only. Here $Q_{before}$ and $Q_{after}$ are the total discharges in the single pipe and in the modified line (they are not the discharges $Q_1$, $Q_2$ in the two parallel pipes of the figure). #### Before: single pipe of length 1500 m $$ H = \frac{fLV_b^2}{2gD} \;\Rightarrow\; \frac{V_b^2}{2g} = \frac{HD}{fL} = \frac{0.3(0.6)}{0.01(1500)} = 0.01200\ \text{m} $$ $$ V_b = 0.4852\ \text{m/s},\qquad Q_{before} = \frac{\pi}{4}(0.6)^2V_b = 0.13719\ \text{m}^3/\text{s} $$ #### After: first 750 m single, second 750 m has two identical pipes in parallel Let $V_a$ be the velocity in the first half. In the second half the flow divides equally between two pipes of the same size, so the velocity in each is $V_a/2$. $$ H = \frac{f(750)V_a^2}{2gD} + \frac{f(750)(V_a/2)^2}{2gD} = \frac{f(750)}{D}\,\frac{V_a^2}{2g}\,(1 + 0.25) $$ $$ \frac{V_a^2}{2g} = \frac{HD}{f(750)(1.25)} = \frac{0.3(0.6)}{0.01(750)(1.25)} = 0.01920\ \text{m} \;\Rightarrow\; V_a = 0.6138\ \text{m/s} $$ $$ Q_{after} = \frac{\pi}{4}(0.6)^2V_a = 0.17354\ \text{m}^3/\text{s} $$ (each parallel pipe carries 0.08677 m³/s). #### Increase in discharge $$ \Delta Q = Q_{after} - Q_{before} = 0.17354 - 0.13719 = 0.03635\ \text{m}^3/\text{s}, \text{ about } 26.5\%\ \text{increase} $$ (In general $\dfrac{Q_{after}}{Q_{before}} = \sqrt{\dfrac{1500}{937.5}} = \sqrt{1.6} = 1.265$.) **Answer:** The discharge increases by $0.0364$ m³/s, i.e. by about 26.5%. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the original single pipe, which relation gives V1? A. H = f L V^2/(2 g D), so V^2/2g = H D/(f L) B. H = V^2/2g C. H = f D V^2/(2 g L) D. Q = H/(f L) Answer: A. The whole head is friction, so Darcy-Weisbach is solved for the velocity head. (Key point: Single pipe velocity) 2. In the second half with two identical parallel pipes, what is the velocity in each? A. V B. 2V C. V/2 D. V/4 Answer: C. The flow splits equally between two pipes of equal size, so each carries half the discharge and half the velocity. (Key point: Flow division in parallel) 3. What is the friction head with the added parallel pipe? A. f(750)V^2/2gD x (1 + 1) because both halves lose the same B. f(750)V^2/2gD x (1 + 0.25) C. f(1500)V^2/2gD x 0.25 D. f(750)V^2/2gD x (1 + 0.5) Answer: B. The first half has velocity V and the second half V/2, whose loss is a quarter of the first. (Key point: Head equation after addition) 4. What is the ratio Q2/Q1 and the resulting percentage increase? A. sqrt(1.6) = 1.265, about 26.5% B. 1.6, a 60% increase C. 2, a 100% increase D. 1.25, a 25% increase Answer: A. V^2 is proportional to 1/(effective length factor): 1500 against 937.5 gives sqrt(1.6). 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