Question 1 of 8: Compressor and Turbine

Question 1: Compressor and Turbine (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A simple open-cycle, single-shaft gas turbine:

Find. The T-s diagram with every state numbered, the temperature at each numbered state, the net specific work in kW·s per kilogram of gas leaving the turbine, and the specific fuel consumption in kilograms of fuel per kilowatt-hour.

Approach. Fix the two isentropic end states from the pressure ratio, open them out to the real end states with the two component efficiencies, then take a work balance per kilogram of gas passing through the turbine — the compressor handles only the air part of that gas, which is where the air/fuel ratio enters.

1. Part (a) — number the cycle and identify each process. State 1 is ambient air at the compressor intake; 1→2 is the real (irreversible) compression against the delivery pressure $p_2 = r_p p_1 = 1200\ \text{kPa}$, with 2s the ideal end point on the same pressure line; 2→3 is constant-pressure combustion to the turbine inlet temperature; 3→4 is the real expansion back to 100 kPa, with 4s its ideal counterpart; and 4→1 closes the diagram as the atmospheric rejection of the exhaust. Figure 1.1 is the required sketch. Both irreversible processes lean to the right, because entropy is generated in each.
2. Part (b) — ideal compressor delivery temperature. Along an isentrope between the two pressure levels, $$\frac{T_{2s}}{T_1}=\left(\frac{p_2}{p_1}\right)^{(k-1)/k}=12^{0.2857}=2.034$$ so $T_{2s}=288.15\times 2.034$, giving $$\boxed{T_{2s}=586.1\ \text{K}\ (313.0\ ^\circ\text{C})}$$
3. Real compressor delivery. The isentropic efficiency is the ratio of ideal to actual temperature rise, so the real rise is the larger of the two: $$T_2=T_1+\frac{T_{2s}-T_1}{\eta_c}=288.15+\frac{297.9}{0.86}=634.6\ \text{K}$$ $$\boxed{T_2=634.6\ \text{K}\ (361.4\ ^\circ\text{C})}$$
4. Ideal turbine exhaust temperature. The expansion returns to ambient pressure through the same ratio, so $$T_{4s}=\frac{T_3}{12^{0.2857}}=\frac{1350.15}{2.034}=663.8\ \text{K}$$
5. Real turbine exhaust. Here the efficiency multiplies the ideal drop, because the real machine delivers less than the isentropic work: $$T_4=T_3-\eta_t\,(T_3-T_{4s})=1350.15-0.89\times 686.4=739.3\ \text{K}$$ $$\boxed{T_4=739.3\ \text{K}\ (466.2\ ^\circ\text{C})}$$ State 3 is given directly as $T_3=1350.15$ K and state 1 as $T_1=288.15$ K, which completes part (b).
6. Part (c) — split the mass flow between air and gas. The question asks for work per kilogram of gas, and the gas leaving the combustor is air plus fuel. With $A/F=50$, one kilogram of gas contains $$m_{air}=\frac{50}{51}=0.9804\ \text{kg air},\qquad m_{fuel}=\frac{1}{51}=0.01961\ \text{kg fuel}$$ The turbine passes the whole kilogram; the compressor only ever handled the air fraction.
7. Turbine and compressor specific works. Treating the products as air with the page-13 specific heat, $$w_t=c_p\,(T_3-T_4)=1.005\times 610.9=613.9\ \text{kJ/kg of gas}$$ $$w_c=\frac{A/F}{A/F+1}\,c_p\,(T_2-T_1)=0.9804\times 1.005\times 346.4=341.3\ \text{kJ/kg of gas}$$
8. Net specific work output. The shaft efficiency debits the mechanical losses between turbine and compressor: $$w_{net}=\eta_m w_t-w_c=0.98\times 613.9-341.3=260.3\ \text{kJ/kg}$$ Because one kilojoule per kilogram is one kilowatt-second per kilogram, $$\boxed{w_{net}=260.3\ \text{kW}\cdot\text{s/kg of gas}}$$
9. Part (d) — specific fuel consumption. Each kilogram of gas carries 0.01961 kg of fuel and yields 260.3 kJ of shaft work; one kilowatt-hour is 3600 kJ, so $$\text{SFC}=\frac{m_{fuel}}{w_{net}}\times 3600=\frac{0.01961}{260.3}\times 3600$$ $$\boxed{\text{SFC}=0.271\ \text{kg/kWh}}$$
10. Check the answer against the cycle efficiency. The heat added is $q_{in}=c_p(T_3-T_2)=1.005\times 715.6=719.2$ kJ/kg, so the thermal efficiency is $260.3/719.2=36.2\ \%$. A fuel of calorific value about 45 MJ/kg burnt at an air/fuel ratio of 50 releases roughly $45000/51=882$ kJ per kilogram of gas, and 36 % of that is close to the computed net work — the two routes agree, which confirms the mass bookkeeping.