---
title: "Fan scaling by similarity laws"
subject: "Turbo Machines"
chapter: 1
key: "semester_7/turbo_machines:fan-laws-similarity-numerical"
source: https://www.bctengineer.com/q/semester_7/turbo_machines/fan-laws-similarity-numerical
site: BCT Engineering
---

# Fan scaling by similarity laws

Turbo Machines · Chapter 1: Introduction

- Asked in: year not given, 6 marks
- Video: [Turbomachinery Similarity Laws](https://www.youtube.com/watch?v=17QKPwBygEA) — Postcard Professor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/turbo_machines/fan-laws-similarity-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/turbo_machines/chapter-1

## Question

A fan running at 1450 rpm delivers 4 m³/s of air at a pressure rise of 300 Pa and absorbs 1.8 kW. Using the similarity laws (air density unchanged), find the flow rate, pressure rise and power (a) when the same fan runs at 1750 rpm, and (b) for a geometrically similar fan of 1.25 times the diameter running at 1450 rpm.

## Short answer

Use similarity: Q∝ND³, Δp∝N²D², P∝N³D⁵. (a) Same D, N ratio 1.207: Q=4×1.207=4.83 m³/s, Δp=300×1.207²=437 Pa, P=1.8×1.207³=3.16 kW. (b) N constant, D ratio 1.25: Q=4×1.25³=7.81 m³/s, Δp=300×1.25²=469 Pa, P=1.8×1.25⁵=5.49 kW.

## Answer

For geometrically similar machines with the same fluid and equal $\phi$, $\psi$, $\lambda$:

$$
\frac{Q_2}{Q_1} = \frac{N_2}{N_1}\left(\frac{D_2}{D_1}\right)^3, \quad
\frac{\Delta p_2}{\Delta p_1} = \left(\frac{N_2}{N_1}\right)^2\left(\frac{D_2}{D_1}\right)^2, \quad
\frac{P_2}{P_1} = \left(\frac{N_2}{N_1}\right)^3\left(\frac{D_2}{D_1}\right)^5
$$

#### (a) Same fan at 1750 rpm
$N_2/N_1 = 1750/1450 = 1.207$, $D_2/D_1 = 1$.

| Quantity | Working | Result |
|---|---|---|
| Flow | $4 \times 1.207$ | 4.83 m³/s |
| Pressure rise | $300 \times 1.207^2$ | 437 Pa |
| Power | $1.8 \times 1.207^3$ | 3.16 kW |

#### (b) Larger fan, same speed
$D_2/D_1 = 1.25$, $N_2/N_1 = 1$.

| Quantity | Working | Result |
|---|---|---|
| Flow | $4 \times 1.25^3$ | 7.81 m³/s |
| Pressure rise | $300 \times 1.25^2$ | 469 Pa |
| Power | $1.8 \times 1.25^5$ | 5.49 kW |

The efficiency is assumed unchanged (equal $\phi$, $\psi$, ignoring Reynolds number effects). Power rises with the cube of speed, so a small speed increase demands a large motor.

**Answer:** (a) 4.83 m³/s, 437 Pa, 3.16 kW; (b) 7.81 m³/s, 469 Pa, 5.49 kW.

## Check yourself

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

1. For the same fan at a higher speed, flow rate scales as:
   A. N^2
   B. N^3
   C. N
   D. 1/N
   Answer: C. With D fixed, Q is proportional to N, giving 4 x 1.207 = 4.83 m3/s. (Key point: Flow scaling)

2. What is the pressure rise at 1750 rpm?
   A. 437 Pa
   B. 362 Pa
   C. 300 Pa
   D. 527 Pa
   Answer: A. Pressure rise scales as N^2: 300 x 1.207^2 = 437 Pa. (Key point: Pressure scaling)

3. For the 1.25 times larger fan at the same speed, power scales as:
   A. D^2
   B. D^3
   C. D^4
   D. D^5
   Answer: D. P is proportional to N^3 D^5, so P = 1.8 x 1.25^5 = 5.49 kW. (Key point: Power scaling with diameter)

4. What does the cube law of power with speed imply?
   A. Power is almost unaffected by speed
   B. A small speed increase needs a much larger motor
   C. Power falls as speed rises
   D. Power rises linearly with speed
   Answer: B. P is proportional to N^3, so 20% more speed gives about 75% more power (1.8 to 3.16 kW). (Key point: Power with speed)

## Related in Turbo Machines

- [Fuel-air ratio in a gas turbine combustor](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Acombustion-chamber-fuel-air-ratio-numerical) — Turbo Machines, chapter 3
- [Convergent nozzle: choking and mass flow](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Aconvergent-nozzle-choking-mass-flow) — Turbo Machines, chapter 4
- [Ideal Brayton cycle work and efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Aideal-brayton-numerical) — Turbo Machines, chapter 3
- [Gas turbine power from isentropic efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Agas-turbine-power-isentropic-efficiency) — Turbo Machines, chapter 1
- [Compressor work and polytropic efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Acompressor-exit-temperature-polytropic-efficiency) — Turbo Machines, chapter 1
- [Parsons stage: work, power and efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Aparsons-stage-numerical) — Turbo Machines, chapter 2

## Related in other subjects

- [Series motor voltage to drive fan at 300 rpm](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aseries-motor-fan-400v-300rpm) — Electrical Machines I, chapter 4
- [Pump specific speed and affinity laws](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Apump-specific-speed-affinity-numerical) — Fluid Machines, chapter 4
- [Solid shaft 300 kW, 120 rpm with twist limit](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Asolid-shaft-300kw-120rpm-twist-limit) — Strength of Materials, chapter 6
- [Torque at 1450 rpm from starting torque](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aim-4pole-150nm-1450rpm-0p5-2ohm) — Electrical Machines I, chapter 5