Compressor work and polytropic efficiency

Data: $T_{01} = 300$ K, $r = 8$, $\eta_c = 0.82$, $\dot m = 10$ kg/s, $c_p = 1.005$, $\gamma = 1.4$, so $(\gamma-1)/\gamma = 0.2857$.

(a) Exit temperature

Isentropic exit temperature:

$$T_{02s} = T_{01}\, r^{(\gamma-1)/\gamma} = 300 \times 8^{0.2857} = 543.4\ \text{K}$$

Actual temperature rise:

$$T_{02} - T_{01} = \frac{T_{02s} - T_{01}}{\eta_c} = \frac{243.4}{0.82} = 296.9\ \text{K}$$ $$T_{02} = 300 + 296.9 = 596.9\ \text{K}$$

(b) Work and power

$$w = c_p (T_{02} - T_{01}) = 1.005 \times 296.9 = 298.4\ \text{kJ/kg}$$ $$P = \dot m w = 10 \times 298.4 = 2984\ \text{kW}$$

(c) Polytropic efficiency

$$\eta_p = \frac{\frac{\gamma-1}{\gamma}\ln r}{\ln (T_{02}/T_{01})} = \frac{0.2857 \times \ln 8}{\ln (596.9/300)} = \frac{0.5941}{0.6879} = 0.864$$

The polytropic efficiency (86.4%) is higher than the isentropic efficiency (82%) because of the preheat effect: the losses in each small stage heat the air entering the next stage, which then needs more work to compress.

Answer: $T_{02} \approx 596.9$ K; $w \approx 298.4$ kJ/kg; $P \approx 2984$ kW; $\eta_p \approx 86.4\%$.