Question 1 of 7: Question 1 (Part A, Question A1): Scaling a Fan — Flow, Pressure Rise and Shaft Power at a New Size, Speed and Air Density

Question 1 (Part A, Question A1): Scaling a Fan — Flow, Pressure Rise and Shaft Power at a New Size, Speed and Air Density (16 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The volume flow rate, fan total pressure and shaft power delivered by the 1 m fan at 500 rpm in the second air state.

Approach. Geometric similarity plus unchanged efficiency means the three dimensionless turbomachine coefficients — flow, head (pressure) and power — take the same value in both machines, so each dimensional quantity scales as a fixed power of the speed ratio, the diameter ratio and (for pressure and power) the air-density ratio, which must first be computed from the two barometric states.

1. Fix the air density in each state from the perfect-gas law. The two barometric readings convert to absolute pressures, and $$\rho=\frac{p}{RT},\qquad R_{\text{air}}=287\ \text{J/kg}\cdot\text{K}.$$ For the test state, $p_1=772\times133.322=102\,925\ \text{Pa}$ and $T_1=283.15\ \text{K}$, so $\rho_1=102\,925/(287\times283.15)=1.2665\ \text{kg/m}^3$. For the second state, $p_2=760\times133.322=101\,325\ \text{Pa}$ and $T_2=289.15\ \text{K}$, giving $\rho_2=101\,325/(287\times289.15)=1.2210\ \text{kg/m}^3$. The density ratio that carries through the rest of the question is $$\boxed{\ \frac{\rho_2}{\rho_1}=\frac{1.2210}{1.2665}=0.9640\ }$$ — the second fan handles air that is 3.6 per cent lighter, partly because it is warmer and partly because the barometer is lower.
2. Write the three similarity coefficients. For a family of geometrically similar fans operating at the same efficiency, dimensional analysis (White Ch. 11) gives three groups that must match: $$\begin{aligned} \text{flow coefficient}\quad \phi&=\frac{Q}{ND^{3}}\\ \text{head coefficient}\quad \psi&=\frac{\Delta p}{\rho N^{2}D^{2}}\\ \text{power coefficient}\quad \Pi_P&=\frac{P}{\rho N^{3}D^{5}} \end{aligned}$$ Note that the flow coefficient contains no density: volume flow scales purely kinematically, whereas pressure rise and power both carry one power of density because they are momentum and energy quantities. Because the speed appears only as a ratio, $N$ may be left in rpm throughout provided the same unit is used top and bottom.
3. Scale the volume flow rate. Equating flow coefficients, $$Q_2=Q_1\left(\frac{N_2}{N_1}\right)\left(\frac{D_2}{D_1}\right)^{3} =0.7\times\frac{500}{970}\times\left(\frac{1.0}{0.4}\right)^{3} =0.7\times0.51546\times15.625.$$ The larger fan turns almost half as fast but sweeps a volume nearly sixteen times larger, and the size term wins decisively: $$\boxed{\ Q_2=5.64\ \text{m}^3/\text{s}\ }$$
4. Scale the fan total pressure. Equating head coefficients and carrying the density ratio, $$\Delta p_2=\Delta p_1\left(\frac{\rho_2}{\rho_1}\right) \left(\frac{N_2}{N_1}\right)^{2}\left(\frac{D_2}{D_1}\right)^{2} =245.17\times0.9640\times0.26570\times6.2500.$$ Here the speed and size effects nearly cancel — the tip speed $\pi ND$ rises only from 20.3 m/s to 26.2 m/s — so the pressure rise grows by only about 60 per cent: $$\boxed{\ \Delta p_2=392.5\ \text{Pa}=40.0\ \text{mm H}_2\text{O}\ }$$
5. Scale the shaft power. Equating power coefficients, $$P_2=P_1\left(\frac{\rho_2}{\rho_1}\right)\left(\frac{N_2}{N_1}\right)^{3} \left(\frac{D_2}{D_1}\right)^{5} =250\times0.9640\times0.13696\times97.656.$$ The fifth power of the diameter ratio is what dominates, and the result is $$\boxed{\ P_2=3.22\ \text{kW}\ }$$ about thirteen times the test power for a machine only 2.5 times larger in diameter.
6. Confirm the answer set is self-consistent through the efficiency. The statement "efficiency unchanged" is not an extra assumption we may forget — it is a check. Total-to-total efficiency is $\eta=Q\,\Delta p/P$, so for the tested machine $\eta_1=(0.7\times245.17)/250=0.686$ and for the scaled machine $\eta_2=(5.6379\times392.49)/3223.5=0.686$. The two agree to three figures, which they must, because $\phi\psi/\Pi_P$ is itself a dimensionless group; any slip in one of the three exponents above would have shown up here as a mismatch.