---
title: "Pump specific speed and affinity laws"
subject: "Fluid Machines"
chapter: 4
key: "semester_5/fluid_machines:pump-specific-speed-affinity-numerical"
source: https://www.bctengineer.com/q/semester_5/fluid_machines/pump-specific-speed-affinity-numerical
site: BCT Engineering
---

# Pump specific speed and affinity laws

Fluid Machines · Chapter 4: Pump

- Asked in: year not given, 6 marks
- Video: [Pump Affinity Laws + Exercises Applied Fluid Dynamics - Class 055](https://www.youtube.com/watch?v=kATC6hsIi_0) — Chemical Engineering Guy
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/fluid_machines/pump-specific-speed-affinity-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/fluid_machines/chapter-4

## Question

Define the specific speed of a centrifugal pump and state its significance. A centrifugal pump delivers 0.05 m³/s at a head of 25 m when running at 1450 rpm, with an overall efficiency of 72 %. (a) Calculate its specific speed and the shaft power. (b) Using the affinity laws, find the discharge, head and power when the speed is increased to 1750 rpm (assume efficiency unchanged).

## Short answer

Specific speed Ns = N√Q / H^(3/4) classifies pump type: radial (10-70), mixed (70-150), axial (>150). (a) Ns=29.0, shaft power=17.03 kW. (b) At 1750 rpm: Q=0.0603 m³/s, H=36.4 m, P=29.94 kW.

## Answer

#### Specific speed

The **specific speed** of a centrifugal pump is the speed of a geometrically similar pump that delivers **unit discharge (1 m³/s) against unit head (1 m)**.

$$
N_s=\frac{N\sqrt{Q}}{H_m^{3/4}}
$$

($N$ in rpm, $Q$ in m³/s, $H_m$ in m.) It is the type number of the pump:

| Ns (rpm, m³/s, m) | Impeller type |
|---|---|
| About 10 to 70 | Radial flow, high head, low discharge (below 30: narrow, very high head) |
| 70 to 150 | Mixed flow |
| Above 150 | Axial flow, low head, large discharge |

It helps in selecting the pump type for a duty and in predicting performance of similar pumps. (The head $H_m$ is for a single stage in a multistage pump, and $Q$ is for one side for a double suction impeller.)

#### (a) Specific speed and shaft power

$$
N_s=\frac{1450\times\sqrt{0.05}}{25^{0.75}}=\frac{1450\times0.2236}{11.180}=29.0
$$

This lies in the radial-flow (centrifugal) range.

$$
P_1=\frac{\rho gQH}{\eta_o}=\frac{1000\times9.81\times0.05\times25}{0.72}=17.03\ \text{kW}
$$

#### (b) Speed changed to 1750 rpm

Speed ratio $r=\dfrac{1750}{1450}=1.2069$. For the same pump:

$$
Q\propto N,\qquad H\propto N^2,\qquad P\propto N^3
$$

$$
\begin{aligned}
Q_2&=0.05\times1.2069=0.0603\ \text{m}^3/\text{s}\ (60.3\ \text{L/s})\\
H_2&=25\times1.2069^2=36.4\ \text{m}\\
P_2&=17.03\times1.2069^3=29.94\ \text{kW}
\end{aligned}
$$

**Answer:** (a) $N_s=29.0$, $P_1=17.03$ kW; (b) $Q_2=0.0603$ m³/s, $H_2=36.4$ m, $P_2=29.94$ kW.

## Check yourself

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

1. The specific speed of a centrifugal pump is the speed of a geometrically similar pump that:
   A. delivers 1 m^3/s against 1 m head
   B. has unit power and unit head
   C. runs at unit efficiency
   D. has unit diameter and unit speed
   Answer: A. Ns = N sqrt(Q) / Hm^(3/4) corresponds to unit discharge against unit head. (Key point: Definition of specific speed)

2. A pump has Ns about 29. What type of impeller is it?
   A. Axial flow
   B. Mixed flow
   C. Radial flow, high head and low discharge
   D. Cannot be determined
   Answer: C. Ns of about 10 to 70 indicates a radial-flow centrifugal pump. (Key point: Significance of Ns)

3. When the speed of the same pump is raised, how do Q, H and P vary with N?
   A. Q ~ N, H ~ N^2, P ~ N^3
   B. Q ~ N^2, H ~ N, P ~ N^3
   C. Q ~ N, H ~ N^3, P ~ N^2
   D. Q ~ N^3, H ~ N^2, P ~ N
   Answer: A. These affinity laws give 60.3 L/s, 36.4 m and 29.94 kW at 1750 rpm. (Key point: Affinity laws)

## Related in Fluid Machines

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