---
title: "Pipe diameter to halve pumping cost"
subject: "Hydraulics"
chapter: 1
key: "semester_4/hydraulics:petrol-pipe-diameter-pumping-cost"
source: https://www.bctengineer.com/q/semester_4/hydraulics/petrol-pipe-diameter-pumping-cost
site: BCT Engineering
---

# Pipe diameter to halve pumping cost

Hydraulics · Chapter 1: Pipe flow

- Asked in: 2074 Bhadra, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/petrol-pipe-diameter-pumping-cost
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-1

## Question

Petrol of kinematic viscosity 0.6 mm²/s is to be pumped at the rate of 0.8 m³/s through a horizontal pipe 500 mm diameter. However, to reduce pumping costs a pipe of different diameter is suggested. Assuming that the absolute roughness of the walls would be the same for a pipe of slightly different diameter, and that, for Re > $10^6$, f is approximately proportional to the cube root of the roughness, determine the diameter of pipe for which the pumping costs would be halved. Neglect all head losses other than pipe friction. How are the running costs altered if n pipes of equal diameter are used in parallel to give the same total flow rate at the same Reynolds number as for a single pipe?

## Short answer

Pumping power is proportional to head loss, which varies as D^(−16/3) for rough turbulent flow with f ∝ D^(−1/3). Halving the cost requires D' ≈ 569 mm. Using n parallel pipes at the same Re increases running cost by a factor of n^(10/3).

## Answer

**Data.** $\nu = 0.6\times10^{-6}$ m²/s, $Q = 0.8$ m³/s, $D = 0.5$ m, same absolute roughness $k$ for the new pipe.
$$
V = \frac{4Q}{\pi D^2} = 4.074\ \text{m/s},\qquad Re = \frac{VD}{\nu} = 3395305 > 10^6
$$
So $f \propto (k/D)^{1/3}$ may be used.

#### Part 1: Diameter for half the pumping cost
Pumping cost is proportional to the power $P = \rho g Q h_f$. With $Q$ fixed, cost $\propto h_f$ (neglecting all losses except friction).

$$
h_f = \frac{fLV^2}{2gD}, \qquad V = \frac{4Q}{\pi D^2} \;\Rightarrow\; h_f \propto \frac{f}{D^5}
$$
For constant $k$: $f \propto \left(\dfrac{k}{D}\right)^{1/3} \propto D^{-1/3}$, so
$$
h_f \propto D^{-1/3}\,D^{-5} = D^{-16/3}
$$
Cost halved: $\left(\dfrac{D'}{D}\right)^{-16/3} = \dfrac{1}{2}$ gives
$$
\frac{D'}{D} = 2^{3/16} = 1.1388 \;\Rightarrow\; D' = 0.5(1.1388) = 0.569\ \text{m} \approx 569\ \text{mm}
$$
Check: new velocity $V' = 3.142$ m/s and $Re' = 2981506 > 10^6$, so the assumption holds.

#### Part 2: n equal pipes in parallel at the same Reynolds number
The same $Re = VD/\nu$ for the same fluid means $V'D' = VD$. Each of the $n$ pipes carries $Q/n$:
$$
\frac{Q}{n} = \frac{\pi}{4}D'^2V' \;\Rightarrow\; \frac{\pi}{4}D^2V = n\frac{\pi}{4}D'^2V' = n\frac{\pi}{4}D'\,(VD)\;\Rightarrow\; D' = \frac{D}{n},\quad V' = nV
$$
Relative roughness rises to $k/D' = n(k/D)$, so $f' = n^{1/3}f$. Head loss of each pipe (all pipes have the same $h_f$ in parallel):
$$
h_f' = \frac{f'LV'^2}{2gD'} = \frac{n^{1/3}f\,L\,n^2V^2}{2g(D/n)} = n^{3+1/3}h_f = n^{10/3}h_f
$$
The discharge is the same, so the power and the running cost increase by the factor
$$
\boxed{\frac{\text{cost}'}{\text{cost}} = n^{10/3}}
$$
For example, two pipes ($n = 2$) cost $2^{10/3} = 10.1$ times as much. Using one large pipe is much cheaper than several small ones.

**Answer:** $D' \approx 569$ mm (about 570 mm); running cost rises by $n^{10/3}$ for $n$ parallel pipes.

## Check yourself

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

1. With Q fixed, what is the pumping cost proportional to?
   A. Head loss h_f
   B. Diameter
   C. Velocity squared only
   D. Reynolds number
   Answer: A. Power = rho g Q h_f, so for fixed Q cost is proportional to h_f. (Key point: Cost proportional)

2. For constant k, how does h_f scale with D?
   A. D^(-5)
   B. D^(-16/3)
   C. D^(-4)
   D. D^(-7/3)
   Answer: B. h_f is proportional to f/D^5 and f to D^(-1/3), giving D^(-16/3). (Key point: Scaling of h_f)

3. What is the diameter for half the cost?
   A. About 0.53 m
   B. About 0.569 m
   C. About 0.71 m
   D. About 0.45 m
   Answer: B. D'/D = 2^(3/16) = 1.1388, so D' = 569 mm. (Key point: New diameter)

4. By what factor does cost change with n equal pipes at the same Re?
   A. n^(10/3)
   B. n
   C. 1/n
   D. n^2
   Answer: A. D' = D/n, V' = nV, f' = n^(1/3) f, giving h_f' = n^(10/3) h_f. (Key point: Parallel pipes)

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