A centrifugal compressor impeller rotates at $N = 18000\text{ RPM}$ with tip radius $r_{tip} = 0.25\text{ m}$. With radial blades (slip factor $\sigma = 0.90$) and zero inlet pre-whirl, what is the specific work input $w$ in kJ/kg? (Take $\pi \approx 3.1416$).

Answer Key:

199.9

### Solution: Tip speed $U_2 = \frac{2\pi N r_{tip}}{60} = \frac{2 \times 3.1416 \times 18000 \times 0.25}{60} = 471.24\text{ m/s}$. Specific work $w = \sigma U_2^2 = 0.90 \times (471.24)^2 = 0.90 \times 222066 = 199859\text{ J/kg} \approx 199.9\text{ kJ/kg}$. Correct: 199.9 (range 198 to 202).

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