Gas turbine power from isentropic efficiency

Data: $T_{01} = 1100$ K, pressure ratio $r = 600/100 = 6$, $\eta_t = 0.88$, $\dot m = 20$ kg/s, $c_p = 1.148$ kJ/kg K, $\gamma = 1.333$. Inlet and exit velocities are taken as negligible (static = stagnation values).

(a) Isentropic exit temperature

$$\frac{\gamma - 1}{\gamma} = \frac{0.333}{1.333} = 0.250$$ $$T_{02s} = T_{01}\left(\frac{1}{r}\right)^{(\gamma-1)/\gamma} = 1100 \times (1/6)^{0.250} = 703.1\ \text{K}$$

Isentropic temperature drop: $T_{01} - T_{02s} = 396.9$ K.

(b) Actual exit temperature

$$T_{01} - T_{02} = \eta_t (T_{01} - T_{02s}) = 0.88 \times 396.9 = 349.3\ \text{K}$$ $$T_{02} = 1100 - 349.3 = 750.7\ \text{K}$$

(c) Work and power

$$w = c_p (T_{01} - T_{02}) = 1.148 \times 349.3 = 401.0\ \text{kJ/kg}$$ $$P = \dot m w = 20 \times 401.0 = 8.02\ \text{MW}$$

Answer: $T_{02s} \approx 703.1$ K, $T_{02} \approx 750.7$ K, $w \approx 401.0$ kJ/kg, power $\approx 8.02$ MW.