How is the actual compressor delivery temperature found?

T2 = T1 + (T2s - T1)/eta_c = 528.7 K

What is the turbine expansion ratio?

5.7, since turbine inlet pressure is 0.95 x 600 kPa

Which gas properties are used for the turbine?

Gas values cp = 1.148, gamma = 1.333

Air values cp = 1.005, gamma = 1.4

What is the work ratio and its meaning?

0.345; net work over turbine work, small because compressor takes much

0.655; net work over compressor work

Practice · 8 marks

Actual gas turbine cycle with component efficiencies

Short answer

Considering compressor and turbine efficiencies and 5% pressure loss, T2=528.7 K, T4=751.4 K; net specific work=127.3 kJ/kg; thermal efficiency=20.4%; work ratio=0.345; power=1910 kW.

A gas turbine plant takes in air at 100 kPa and 288 K. The compressor pressure ratio is 6 and its isentropic efficiency is 80%. The pressure loss in the combustion chamber is 5% of the compressor delivery pressure and the turbine inlet temperature is 1073 K. The turbine isentropic efficiency is 85% and it exhausts to 100 kPa. For air $c_p = 1.005$ kJ/kg K, $\gamma = 1.4$ ; for the gases $c_p = 1.148$ kJ/kg K, $\gamma = 1.333$ . Neglect the mass of fuel. Find (a) the compressor delivery temperature, (b) the turbine exit temperature, (c) the net specific work, (d) the thermal efficiency, (e) the work ratio, and (f) the power for a flow of 15 kg/s.

Data: $T_1 = 288$ K, $r_p = 6$ , $\eta_c = 0.80$ , $\eta_t = 0.85$ , $T_3 = 1073$ K, $\dot m = 15$ kg/s.

(a) Compressor

T_{2s} = 288 \times 6^{0.2857} = 480.5\ \text{K}

w_c = \frac{c_p (T_{2s} - T_1)}{\eta_c} = \frac{1.005 \times 192.5}{0.80} = 241.9\ \text{kJ/kg}, \qquad T_2 = 288 + \frac{241.9}{1.005} = 528.7\ \text{K}

(b) Turbine

Pressure at turbine inlet $p_3 = 0.95 \times 600 = 570$ kPa, so the expansion ratio is $570/100 = 5.70$ .

\frac{\gamma - 1}{\gamma} = 0.2498, \quad T_{4s} = \frac{T_3}{r_t^{0.2498}} = \frac{1073}{1.5446} = 694.7\ \text{K}

w_t = \eta_t c_p (T_3 - T_{4s}) = 0.85 \times 1.148 \times 378.3 = 369.2\ \text{kJ/kg}

T_4 = T_3 - \frac{w_t}{c_p} = 1073 - \frac{369.2}{1.148} = 751.4\ \text{K}

(c) Net work

w_{net} = w_t - w_c = 369.2 - 241.9 = 127.3\ \text{kJ/kg}

(d) Thermal efficiency

q_{in} = c_{p,g} (T_3 - T_2) = 1.148 \times (1073 - 528.7) = 624.9\ \text{kJ/kg}

\eta_{th} = \frac{w_{net}}{q_{in}} = \frac{127.3}{624.9} = 0.204

(e) Work ratio and (f) power

\text{Work ratio} = \frac{w_{net}}{w_t} = \frac{127.3}{369.2} = 0.345 \quad (\text{back work ratio} = 0.655)

P = \dot m w_{net} = 15 \times 127.3 = 1910\ \text{kW}

Answer: $T_2 = 528.7$ K; $T_4 = 751.4$ K; $w_{net} = 127.3$ kJ/kg; $\eta_{th} = 20.4\%$ ; work ratio = 0.345; $P = 1910$ kW.

✓ Answer checked by Claude Opus 5.5 (high effort)

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