---
title: "Percentage increase with parallel pipe at midpoint"
subject: "Hydraulics"
chapter: 2
key: "semester_4/hydraulics:parallel-pipe-added-midpoint-increase"
source: https://www.bctengineer.com/q/semester_4/hydraulics/parallel-pipe-added-midpoint-increase
site: BCT Engineering
---

# Percentage increase with parallel pipe at midpoint

Hydraulics · Chapter 2: Simple pipe flow problems and solution

- Asked in: 2073 Bhadra, 8 marks
- Video: [Problem 3 on flow through parallel pipes/ fluid mechanics](https://www.youtube.com/watch?v=FmGhE2_dG_0) — Edutech Guru Engineering Learning-By Priyanka ma'am
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/parallel-pipe-added-midpoint-increase
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-2

## Question

A single uniform pipe joins two reservoirs. Calculate the percentage increase of flow rate obtainable if, from the mid-point of this pipe, another of the same diameter is added in parallel to it. Assume equal friction factor for both pipes and neglect minor losses.

## Short answer

Adding a parallel pipe of same diameter from midpoint increases flow rate by 26.5%. This ratio is independent of pipe diameter, length, friction factor and head difference.

## Answer

**Setting up.** Let the single pipe have length $L$, diameter $D$, friction factor $f$, joining two reservoirs with a level difference $H$. Minor losses are neglected, so the head is lost in friction only. Use $h_f = \dfrac{fLV^2}{2gD}$.

#### Before
$$
H = \frac{fLV_1^2}{2gD} \;\Rightarrow\; V_1 = \sqrt{\frac{2gDH}{fL}}, \qquad Q_1 = AV_1
$$

#### After
From the mid-point, a second pipe of the same diameter and length $L/2$ runs parallel to the second half. The first half carries all the flow ($V$); the second half has two identical pipes, so the flow divides equally and each carries velocity $V/2$.
$$
H = \frac{f(L/2)V^2}{2gD} + \frac{f(L/2)(V/2)^2}{2gD} = \frac{fLV^2}{2gD}\left(\frac12 + \frac18\right) = \frac{fLV^2}{2gD}\left(\frac58\right)
$$
$$
V_2 = \sqrt{\frac{8}{5}}\sqrt{\frac{2gDH}{fL}} = \sqrt{1.6}\,V_1 = 1.2649\,V_1
$$
The discharge in the first half is $Q = AV$, so
$$
\frac{Q_2}{Q_1} = \sqrt{1.6} = 1.2649
$$

$$
\boxed{\text{Increase in flow} = 26.5\%}
$$
(This value does not depend on $D$, $L$, $f$ or $H$.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Beyond the mid-point, how does the flow behave in the two identical pipes?
   A. All flow goes in the original pipe
   B. It divides equally so each carries velocity V/2
   C. Each carries velocity V
   D. It divides in the ratio of lengths
   Answer: B. Two identical parallel pipes share the discharge equally, halving the velocity in each. (Key point: Flow split)

2. What is the total head loss after the pipe is added?
   A. f L V^2/2gD x 1/2
   B. f L V^2/2gD x (1/2 + 1/8) = 5/8 of the original form
   C. f L V^2/2gD x 2
   D. f L V^2/2gD x 1/4
   Answer: B. First half gives 1/2, second half gives (1/2)(1/4) = 1/8. (Key point: Head loss with parallel pipe)

3. What is the percentage increase of flow?
   A. 60%
   B. 25%
   C. 26.5%
   D. 12.5%
   Answer: C. Q2/Q1 = sqrt(8/5) = 1.265, an increase of 26.5%. (Key point: Resulting increase)

4. Does the 26.5% increase depend on D, L, f or H?
   A. Yes, on L only
   B. Yes, on f only
   C. Yes, on H and D
   D. No, the ratio is independent of all of them
   Answer: D. All these quantities cancel in the ratio, leaving only the 5/8 geometry factor. (Key point: Independence of the result)

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