Question

The gas-turbine cycle of a combined gas-steam power plant has a pressure ratio of \(12 .\) Air enters the compressor at \(310 \mathrm{K}\) and the turbine at \(1400 \mathrm{K}\). The combustion gases leaving the gas turbine are used to heat the steam at \(12.5 \mathrm{MPa}\) to \(500^{\circ} \mathrm{C}\) in a heat exchanger. The combustion gases leave the heat exchanger at \(247^{\circ} \mathrm{C}\). Steam expands in a high-pressure turbine to a pressure of \(2.5 \mathrm{MPa}\) and is reheated in the combustion chamber to \(550^{\circ} \mathrm{C}\) before it expands in a low-pressure turbine to \(10 \mathrm{kPa} .\) The mass flow rate of steam is \(12 \mathrm{kg} / \mathrm{s}\). Assuming all the compression and expansion processes to be isentropic, determine (a) the mass flow rate of air in the gas-turbine cycle, ( \(b\) ) the rate of total heat input, and ( \(c\) ) the thermal efficiency of the combined cycle.

Short answer

Answer: The thermal efficiency of the combined cycle in the given gas-steam power plant is approximately 86.79%.

Step-by-step solution

Determine mass flow rate of air in the gas-turbine cycle.

We'll begin by finding the temperatures at the compressor and the turbine inlets and outlets using the isentropic relations. For the compressor, we have: \(T_2 = T_1 \left(\frac{P_2}{P_1}\right)^{(\gamma - 1)/\gamma} = 310 \times \left(\frac{12}{1}\right)^{(\frac{1.4 - 1}{1.4})} = 665.8 \mathrm{K}\) For the turbine, we have: \(T_4 = T_3 \left(\frac{P_4}{P_3}\right)^{(\gamma - 1)/\gamma} = 1400 \times \left(\frac{1}{12}\right)^{(\frac{1.4 - 1}{1.4})} = 734.8 \mathrm{K}\) Now we can use the heat exchanger's outlet temperatures to find the mass flow rate of air in the gas-turbine cycle. \(\dot{m}_{air} = \dot{m}_{steam} \frac{(h_5-h_4)-(h_7-h_6)}{(h_2-h_1)-(h_4-h_3)} = 12 \mathrm{kg/s} \frac{(3204.3-3009.5) - (3425.9-3123.4)}{(663.3-206.79) - (1005.55-939.83)} = 1.006 \mathrm{kg/s}\) The mass flow rate of air in the gas-turbine cycle is 1.006 kg/s.

Calculate the rate of total heat input.

The rate of total heat input is the sum of the heat input in the gas-turbine cycle and the steam cycle. For the gas-turbine cycle, we have: \(\dot{Q}_{in_{1}}= \dot{m}_{air}(h_3-h_2) = 1.006 \times (1400-665.8) = 737.7 \mathrm{kW}\) For the steam cycle, we have: \(\dot{Q}_{in_{2}}= \dot{m}_{steam}[(h_1'-h_2')+(h_3'-h_4')] = 12[(3477.8 - 3293.5) + (3501 - 3204.3 + 3009.5 - 3425.9)] = 6619.7 \mathrm {kW}\) Therefore, the rate of total heat input in the combined cycle is: \(\dot{Q}_{in_{total}} = \dot{Q}_{in_{1}} + \dot{Q}_{in_{2}} = 737.7 + 6619.7 = 7357.4 \mathrm{kW}\)

Calculate the thermal efficiency of the combined cycle.

To determine the thermal efficiency of the combined cycle, we'll find the work done in each cycle and the total work output for both cycles. In the gas-turbine cycle: \(\dot{W}_{gt} = \dot{m}_{air}((h_3-h_4)-(h_2-h_1)) = 1.006[(939.83-665.8)-(1005.55-663.3)] = 201.37 \mathrm {kW}\) In the steam cycle: \(\dot{W}_{steam} = \dot{m}_{steam}((h_1'-h_2')+(h_3'-h_4')-(h_6-h_7)) = 12[(-13)+(3123.4 - 3425.9 +1926)] = 6187.2 \mathrm {kW}\) The total work output is: \(\dot{W}_{total} = \dot{W}_{gt} + \dot{W}_{steam} = 201.37 + 6187.2 = 6388.5 \mathrm{kW}\) Finally, we can find the combined cycle's thermal efficiency: \(\eta_{combined} = \frac{\dot{W}_{total}}{\dot{Q}_{in_{total}}} = \frac{6388.5}{7357.4} = 0.8679 = 86.79 \%\) The thermal efficiency of the combined cycle is approximately 86.79%