Question 1 of 6: Industrial Gas Turbine — cycle temperatures, efficiency and output

Question 1: Industrial Gas Turbine — cycle temperatures, efficiency and output (15 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. An open-cycle industrial gas turbine driving a generator, with constant specific heats throughout.

Find. The four cycle temperatures (compressor inlet and outlet, turbine inlet and outlet), the thermodynamic cycle efficiency and the electrical power delivered by the generator, and then the direction and size of the change in each when both machine efficiencies fall.

Approach. Work round the cycle in sequence: the isentropic relation gives the ideal compressor delivery temperature, the compressor efficiency converts it to the real one, a steady-flow energy balance on the combustor gives the turbine inlet temperature, and the isentropic relation with the turbine efficiency gives the exhaust; the two work terms then give the efficiency and the output.

1. Part (b), step 1 — fix the gas properties from the given specific heats. The specific heat ratio and the isentropic exponent follow directly from $c_p$ and $c_v$, and it is worth computing the exponent once because every compression and expansion in the question uses it:$$k=\dfrac{c_p}{c_v}=\dfrac{1.005}{0.718}=1.3997 \qquad \dfrac{k-1}{k}=1-\dfrac{c_v}{c_p}=0.28557$$so the temperature ratio across a pressure ratio of 12 is $r_p^{(k-1)/k}=12^{0.28557}=2.0332$.
2. Part (b), step 2 — ideal and actual compressor delivery temperature. For an isentropic compression from state 1 to state 2s, $T_{2s}=T_1\,r_p^{(k-1)/k}$, and the isentropic efficiency is defined as the ideal work divided by the actual work, which for constant $c_p$ is a ratio of temperature rises, $\eta_C=(T_{2s}-T_1)/(T_2-T_1)$. Substituting $T_1=293.15$ K:$$T_{2s}=293.15\times 2.0332=596.0\ \text{K}=322.9^\circ\text{C}$$$$T_2=T_1+\dfrac{T_{2s}-T_1}{\eta_C}=293.15+\dfrac{302.9}{0.85}=649.5\ \text{K}=\boxed{376.3^\circ\text{C}}$$The compressor therefore delivers air 53 K hotter than an ideal machine would, and that extra temperature is paid for in shaft work.
3. Part (b), step 3 — turbine inlet temperature from a combustor energy balance. The paper quotes a gas mass flow of 290 kg/s and a fuel flow of 6 kg/s, so the compressor handles $M_a=290-6=284$ kg/s of air and 290 kg/s of combustion products leave the combustor. Releasing the fuel energy into that stream at constant pressure gives$$Q_{in}=M_f\,CV=6\times 40\,000=240\,000\ \text{kW}$$$$\Delta T_{comb}=\dfrac{Q_{in}}{M_g c_p}=\dfrac{240\,000}{290\times 1.005}=823.5\ \text{K}$$$$T_3=T_2+\Delta T_{comb}=649.5+823.5=1472.9\ \text{K}=\boxed{1199.8^\circ\text{C}}$$A firing temperature of about 1200 °C is exactly what a machine of this class (pressure ratio 12, 17 compressor stages, 3600 rev/min for 60 Hz) is designed for, which is a useful check that the flow interpretation is the right one.
4. Part (b), step 4 — turbine exhaust temperature. The turbine expands back to 100 kPa, so it works across the same pressure ratio. The ideal end state is $T_{4s}=T_3/r_p^{(k-1)/k}$ and the turbine efficiency is the ratio of the actual to the ideal temperature drop, $\eta_T=(T_3-T_4)/(T_3-T_{4s})$:$$T_{4s}=\dfrac{1472.9}{2.0332}=724.5\ \text{K}=451.3^\circ\text{C}$$$$T_4=T_3-\eta_T\,(T_3-T_{4s})=1472.9-0.90\times 748.5=799.3\ \text{K}=\boxed{526.1^\circ\text{C}}$$
5. Part (b), step 5 — the two work terms and the net output. Compressor work is charged on the air flow and turbine work is earned on the gas flow, both through $\dot W=M c_p \Delta T$:$$W_C=M_a c_p (T_2-T_1)=284\times 1.005\times 356.3=101\,706\ \text{kW}$$$$W_T=M_g c_p (T_3-T_4)=290\times 1.005\times 673.6=196\,338\ \text{kW}$$$$W_{net}=W_T-W_C=196\,338-101\,706=94\,632\ \text{kW}$$The compressor absorbs 51.8 % of the turbine output. That very high back-work ratio is the defining feature of a gas turbine and is why machine efficiencies matter so much here, as part (c) shows.
6. Part (b), step 6 — cycle efficiency and electrical output. The thermodynamic cycle efficiency compares the net shaft work with the fuel energy released, and the generator then takes its own 2 %:$$\eta_{th}=\dfrac{W_{net}}{Q_{in}}=\dfrac{94\,632}{240\,000}=\boxed{0.394\ (39.4\ \%)}$$$$P_e=\eta_G\,W_{net}=0.98\times 94\,632=\boxed{92\,739\ \text{kW}\approx 92.7\ \text{MW}}$$

Part (c) is answered by re-running the same arithmetic with both machine efficiencies reduced by five points, to 0.80 and 0.85, with the fuel flow, the air flow and the pressure ratio unchanged. The slopes on the sketch above are drawn to that calculation rather than freehand.

1. Part (c), step 1 — the turbine inlet temperature RISES. A less efficient compressor delivers hotter air for the same pressure ratio, $T_2=293.15+302.9/0.80=671.8$ K, which is 22.3 K above the as-new value. The combustor still adds the same 823.5 K because the fuel and gas flows are unchanged, so the whole rise is carried through to the firing temperature:$$T_3^{\prime}=671.8+823.5=1495.2\ \text{K}=1222.1^\circ\text{C}\qquad (\Delta T_3=+22.3\ \text{K})$$This is counter-intuitive on first reading — the machine gets worse and the metal gets hotter — and it is the reason a deteriorating compressor is a blade-life problem and not only an output problem.
2. Part (c), step 2 — the exhaust temperature rises about twice as much. The expansion starts hotter and recovers a smaller fraction of the available drop:$$T_4^{\prime}=1495.2-0.85\,(1495.2-735.4)=849.4\ \text{K}=576.2^\circ\text{C}\qquad (\Delta T_4=+50.1\ \text{K})$$On the T-s diagram the expansion line 3 to 4 leans further to the right, and its lower end sits visibly higher. A rising exhaust temperature at constant fuel flow is the classic field symptom of fouled compressor blading.
3. Part (c), step 3 — the electrical output falls sharply. Recomputing both work terms with the degraded temperatures,$$W_C^{\prime}=284\times 1.005\times 378.6=108\,063\ \text{kW}\qquad W_T^{\prime}=290\times 1.005\times 645.8=188\,234\ \text{kW}$$$$W_{net}^{\prime}=80\,171\ \text{kW}\qquad \eta_{th}^{\prime}=33.4\ \%\qquad P_e^{\prime}=\boxed{78\,568\ \text{kW}\approx 78.6\ \text{MW}}$$so the output drops by 14.2 MW, or $\boxed{15.3\ \%}$. Ten points of machine efficiency have cost fifteen per cent of the power — the amplification comes straight from the back-work ratio, because the compressor bill goes up at the same time as the turbine income goes down.