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Fan scaling by similarity laws

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.

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.

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

Q2Q1=N2N1(D2D1)3,Δp2Δp1=(N2N1)2(D2D1)2,P2P1=(N2N1)3(D2D1)5\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

N2/N1=1750/1450=1.207N_2/N_1 = 1750/1450 = 1.207, D2/D1=1D_2/D_1 = 1.

QuantityWorkingResult
Flow4×1.2074 \times 1.2074.83 m³/s
Pressure rise300×1.2072300 \times 1.207^2437 Pa
Power1.8×1.20731.8 \times 1.207^33.16 kW

(b) Larger fan, same speed

D2/D1=1.25D_2/D_1 = 1.25, N2/N1=1N_2/N_1 = 1.

QuantityWorkingResult
Flow4×1.2534 \times 1.25^37.81 m³/s
Pressure rise300×1.252300 \times 1.25^2469 Pa
Power1.8×1.2551.8 \times 1.25^55.49 kW

The efficiency is assumed unchanged (equal ϕ\phi, ψ\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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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