Question

A centrifugal air compressor has inlet root diameter $D_1=0.25\,\text{m}$ and outlet impeller diameter $D_2=0.6\,\text{m}$. Pressure ratio $\pi_c=p_{02}/p_{01}=5.0$. Air at rotor inlet: $p_{01}=1\,\text{atm}$, $T_{01}=25^\circ\text{C}=298\,\text{K}$. Polytropic efficiency $\eta_p=0.8$, slip factor $\sigma=0.92$. Take $C_p=1.004\,\text{kJ/kg-K}$ and $\gamma=1.4$. Find the impeller speed (RPM). (round off to the nearest integer)

Hint:

For centrifugal compressors with negligible inlet whirl, \( \Delta h_0 \approx \sigma U_2^2 \). Combine this with the polytropic \(T_0\)-rise \(T_{02}/T_{01}=\pi^{(\gamma-1)/(\gamma\eta_p)}\) to get the tip speed, then RPM from \(U_2=\pi D_2 N/60\).

Answer 1

Step 1: Stagnation temperature rise from the given pressure ratio and polytropic efficiency.
For a compressor with polytropic efficiency $\eta_p$, \[ \frac{T_{02}}{T_{01}}=\pi_c^{\frac{\gamma-1}{\gamma\,\eta_p}} =5^{\frac{0.4}{1.4\times 0.8}} =5^{0.35714}\approx 1.776. \] Hence \[ \Delta T_0=T_{02}-T_{01}=298(1.776-1)=231.5\ \text{K}. \]

Step 2: Actual specific work.
\[ w = C_p\,\Delta T_0 = 1004\times 231.5 \approx 2.324\times 10^{5}\ \text{J/kg}. \]

Step 3: Euler head and slip.
With zero pre-whirl at the eye and slip factor $\sigma$, the ideal work from the impeller is \[ \Delta h_0 = U_2 V_{\theta2}= \sigma U_2^2 = w. \] So \[ U_2 = \sqrt{\frac{w}{\sigma}} = \sqrt{\frac{2.324\times10^{5}}{0.92}} \approx 5.03\times10^{2}\ \text{m/s}. \]

Step 4: Convert to RPM.
\[ U_2=\frac{\pi D_2 N}{60} \Rightarrow N= \frac{60U_2}{\pi D_2} =\frac{60(502.6)}{\pi(0.6)}\approx 1.60\times 10^{4}\ \text{RPM}. \] \[ \boxed{N \approx 16000\ \text{RPM}} \]