Question 1 of 8: Steam Injected Gas Turbine

Question 1: Steam Injected Gas 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. An open-cycle gas turbine whose exhaust raises steam that is returned to the combustion chamber, so that gas and steam expand together through one turbine.

Find. The cycle on a T-s diagram, the steam enthalpies needed to close the energy balances, the fuel mass flow rate, and the net power output and thermal efficiency of the steam-injected machine.

Approach. Apply the steady-flow energy equation to the compressor, then to the combustion chamber (which must heat both the gas and the injected steam to 1100 °C), then to the turbine, treating the combustion gas with the given specific heat and the steam with tabulated enthalpies.

1. Part (a) — fix the cycle pressures so the diagram can be drawn. With atmospheric pressure at the compressor inlet and a pressure ratio of 12, $$p_1 = p_4 = 100\ \text{kPa}, \qquad p_2 = p_3 = r\,p_1 = 12 \times 100 = 1200\ \text{kPa}.$$ The gas follows 1–2 (compression), 2–3 (combustion at 1200 kPa), 3–4 (expansion back to 100 kPa); it is an open cycle, so 4–1 is shown dotted as the atmospheric rejection. The steam enters the combustion chamber at state a (540 °C, 1.40 MPa), is heated to state 3 along with the gas, and expands to state b at the exhaust temperature and pressure.
2. Part (b) — read the three steam enthalpies the balances need. The steam is superheated at every state, so the enthalpies come straight from the superheated tables: at injection (1.40 MPa, 540 °C), at the combustor exit (1.20 MPa, 1100 °C, the turbine inlet), and at the turbine exhaust (0.10 MPa, 560 °C). $$h_a = 3562\ \text{kJ/kg},\qquad h_{s3} = 4891\ \text{kJ/kg},\qquad h_{s4} = 3618\ \text{kJ/kg}.$$ A useful check on the extrapolated 1100 °C value: the mean specific heat it implies between 540 and 1100 °C is $(4891-3562)/(1100-540) = 2.37$ kJ/kg·K, which is exactly what is expected of superheated steam at these temperatures.
3. Part (c) — energy balance on the combustion chamber. The fuel must raise the air and the fuel itself from 346 °C to 1100 °C, and separately raise the injected steam from 540 °C to 1100 °C: $$M_f\,CV = (M_{air}+M_f)\,c_{p,g}\,(T_3-T_2) + M_{steam}\,(h_{s3}-h_a).$$ Substituting, with $c_{p,g}(T_3-T_2) = 1.148 \times 754 = 865.6$ kJ/kg, $$M_f (40\,000 - 865.6) = 122 \times 865.6 + 20 \times (4891-3562) = 105\,602 + 26\,580.$$ Solving for the fuel flow, $$\boxed{M_f = \frac{132\,182}{39\,134} = 3.38\ \text{kg/s}}$$ so the combustion-gas flow through the turbine is $M_g = 122 + 3.38 = 125.4$ kg/s.
4. Cross-check the gas flow against the paper itself. Question 2 states the gas mass flow through the heat recovery steam generator as 125 kg/s (air plus fuel). The 125.4 kg/s obtained here reproduces that figure to within a third of a percent, which confirms both the combustion model and the enthalpy of the 1100 °C steam.
5. Part (d) — turbine power. Gas and steam expand together from 1100 °C to 560 °C, the gas handled with its specific heat and the steam with its enthalpies: $$P_T = M_g\,c_{p,g}\,(T_3-T_4) + M_{steam}\,(h_{s3}-h_{s4}).$$ $$P_T = 125.4 \times 1.148 \times 540 + 20 \times (4891-3618) = 77\,738 + 25\,460 = 103\,198\ \text{kW}.$$ The steam therefore contributes about a quarter of the gross output.
6. Compressor work and net output. Only air passes through the compressor, and it is handled with the specific heat of air given on Page 19: $$P_C = M_{air}\,c_{p,a}\,(T_2-T_1) = 122 \times 1.005 \times (346-15) = 40\,584\ \text{kW}.$$ Subtracting gives the net shaft power delivered to the generator, $$\boxed{P_{net} = 103\,198 - 40\,584 = 62\,614\ \text{kW} = 62.6\ \text{MW}}$$
7. Cycle efficiency. All the heat charged to the cycle arrives with the fuel, so $$\eta = \frac{P_{net}}{M_f\,CV} = \frac{62\,614}{3.3777 \times 40\,000} = \frac{62\,614}{135\,108}$$ $$\boxed{\eta = 0.463 = 46.3\ \%}$$ Running the same machine dry — identical temperatures, no steam — would need only 2.70 kg/s of fuel but would deliver 36.7 MW at 34.0  %. Steam injection therefore buys about 26 MW of extra output and roughly twelve percentage points of efficiency, because the steam carries exhaust heat that would otherwise go up the stack back into the expansion.