---
title: "Second parallel pipe to increase flow"
subject: "Hydraulics"
chapter: 2
key: "semester_4/hydraulics:two-reservoirs-second-pipe-increase-flow"
source: https://www.bctengineer.com/q/semester_4/hydraulics/two-reservoirs-second-pipe-increase-flow
site: BCT Engineering
---

# Second parallel pipe to increase flow

Hydraulics · Chapter 2: Simple pipe flow problems and solution

- Asked in: 2074 Bhadra, 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 Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/two-reservoirs-second-pipe-increase-flow
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-2

## Question

Difference in level between two reservoir is 100 m and distance between them is 10 km. The reservoir is connected by a single pipe to carry 200 lps. Calculate the diameter of the pipe and length of second pipe, which is connected to increase the rate of flow by $5\times10^6$ lit/day with same diameter pipe. Take friction factor for all pipes 0.03.

## Short answer

Single pipe diameter from Darcy–Weisbach: D=0.397 m. With additional 0.0579 m³/s, total 0.2579 m³/s. A parallel identical pipe must be laid for 5313 m (5.31 km) to maintain 100 m head.

## Answer

**Data.** $H = 100$ m, $L = 10$ km, $f = 0.03$, $Q = 200$ l/s $= 0.2$ m³/s. Only friction loss is considered.

#### Diameter of the single pipe
$$
H = \frac{8fLQ^2}{\pi^2gD^5} \;\Rightarrow\; D^5 = \frac{8(0.03)(10\,000)(0.2)^2}{\pi^2(9.81)(100)} = 0.00992 \;\Rightarrow\; D = 0.397\ \text{m}
$$
($D \approx 397$ mm; check $V = 1.61$ m/s.) For this pipe the head loss over a length $L$ at discharge $Q$ is $h_f = kLQ^2$, where
$$
k = \frac{8f}{\pi^2gD^5} = 0.2500\ \text{s}^2/\text{m}^6 \text{ per metre of pipe}
$$
(so that $100 = k(10\,000)(0.2)^2$).

#### Second pipe
The extra flow is $5\times10^6$ l/day $= 5000$ m³/day $= 0.05787$ m³/s, so the new total discharge is
$$
Q' = 0.2 + 0.05787 = 0.25787\ \text{m}^3/\text{s}
$$
A second pipe of the same diameter ($D$, same $f$) is laid in parallel along a length $x$, measured from one end. Over this length the flow divides equally ($Q'/2$ in each pipe); over the remaining length $(L - x)$ the single pipe carries $Q'$.
$$
H = k(L - x)Q'^2 + k\,x\left(\frac{Q'}{2}\right)^2 = kQ'^2\left(L - 0.75x\right)
$$
$$
100 = 0.2500(0.25787)^2(10\,000 - 0.75x) \;\Rightarrow\; 10\,000 - 0.75x = 6015.3
$$
$$
x = \frac{10\,000 - 6015.3}{0.75} = 5313\ \text{m}
$$

**Answer:** Diameter of the pipe $D = 0.397$ m (about 397 mm). The second pipe, of the same diameter, must be laid in parallel for a length of about $5313$ m ($5.31$ km, about 53% of the line).

## Check yourself

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

1. How is the diameter of the single pipe found?
   A. D^5 = 8 f L Q^2/(pi^2 g H) from Darcy-Weisbach
   B. D = Q/V with V = 1 m/s
   C. D^2 = 8 f L Q/(pi g H)
   D. D^5 = 8 f Q^2/(pi^2 g H L)
   Answer: A. H = 8fLQ^2/(pi^2 g D^5) rearranged gives D = 0.397 m. (Key point: Pipe diameter)

2. What is the new total discharge after adding 5x10^6 litres/day?
   A. 0.2 + 5.787 m^3/s
   B. 0.2 + 0.05787 = 0.2579 m^3/s
   C. 0.05787 m^3/s
   D. 0.2 - 0.05787 m^3/s
   Answer: B. 5x10^6 l/day = 0.05787 m^3/s is added to the 0.2 m^3/s. (Key point: Extra discharge)

3. Over length x with a second equal pipe in parallel, what is the head loss?
   A. k x Q'^2
   B. k x (Q'/2)^2, a quarter of the single-pipe value
   C. k x Q'^2/2
   D. 2 k x Q'^2
   Answer: B. Each pipe carries Q'/2, and loss varies with the square of Q. (Key point: Parallel section head loss)

4. Which equation gives the length x of the second pipe?
   A. H = k Q'^2 (L - 0.75 x)
   B. H = k Q'^2 (L - x)
   C. H = k Q'^2 (L - 0.5 x)
   D. H = k Q'^2 (L + 0.75 x)
   Answer: A. Single part (L - x) at full loss plus x at one quarter gives L - 0.75x. (Key point: Equation for x)

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