Let's proceed step by step as you outlined. For clarity, we'll begin with a conceptual introduction, then break down each part of the problem as requested. --- ## **Step 1: Conceptual Introduction** A combined gas turbine–steam power plant integrates a Brayton (gas turbine) cycle with a Rankine (steam) cycle. The high-temperature exhaust from the gas turbine is used to generate steam, which drives a steam turbine, thereby improving overall plant efficiency. This setup is commonly used for higher thermal efficiency and better fuel utilization compared to single-cycle plants. Detailed thermodynamic and exergy analysis is required to evaluate the performance of such systems. **Explanation Block:** The combined cycle power plant uses two thermodynamic cycles to convert fuel energy into useful work. The gas turbine cycle burns fuel to produce high-temperature gases that drive a turbine, and the exhaust heat is recovered to produce steam for a steam turbine. This synergy increases efficiency, reduces waste heat, and requires careful analysis of energy and exergy flows at each stage. --- ## **Step 2: Formulae Used** ### **Isentropic Efficiency:** - **Turbine:** \(\eta_{t,isen} = \frac{h_{in} - h_{out,actual}}{h_{in} - h_{out,isentropic}}\) - **Compressor:** \(\eta_{c,isen} = \frac{h_{out,isentropic} - h_{in}}{h_{out,actual} - h_{in}}\) ### **Mass Flow Rate:** - \( \dot{m} = \frac{\dot{W}}{\Delta h} \) (where \(\dot{W}\) is power, \(\Delta h\) is enthalpy rise or drop) ### **Net Power:** - \( \dot{W}_{net} = \dot{W}_{turbines} - \dot{W}_{compressor} \) ### **Thermal Efficiency:** - \( \eta_{thermal} = \frac{\dot{W}_{net}}{\dot{Q}_{in}} \) ### **Exergy:** - \( \dot{E}_{ex} = \sum \dot{m}(h - h_0 - T_0(s-s_0)) \) ### **Exergy Efficiency:** - \( \eta_{ex} = \frac{\text{Net Power Output}}{\text{Exergy Input}} \) **Explanation Block:** The key formulae relate to isentropic efficiencies (which compare actual vs. ideal work), mass and energy balances, and exergy accounting (which measures the useful work potential of energy). These relationships help analyze the efficiency and performance losses at each step. --- ## **Step 3: Calculation for (a) Isentropic Efficiencies** ### **Given Data (from diagram):** - **Compressor:** - \(T_1 = 300\) K, \(p_1 = 1\) bar - \(T_2 = 640\) K, \(p_2 = 11.3\) bar - **Gas Turbine:** - \(T_3 = 1650\) K, \(p_3 = 11\) bar - \(T_4 = 1040\) K, \(p_4 = 1.1\) bar - **Steam Turbine:** - \(T_7 = 540^\circ C\), \(p_7 = 118.5\) bar - \(T_8 = 45^\circ C\), \(p_8 = 0.08\) bar, \(x_8 = 0.88\) #### **Compressor Isentropic Efficiency:** 1. Find \(T_2^{\prime}\) (isentropic): - \(\frac{T_2^{\prime}}{T_1} = \left(\frac{p_2}{p_1}\right)^{\frac{\gamma-1}{\gamma}}\), \(\gamma = 1.4\) - \(T_2^{\prime} = 300 \times (11.3)^{0.286} = 300 \times 1.949 = 584.7 \) K 2. Calculate efficiency: - \(\eta_{c,isen} = \frac{T_2^{\prime} - T_1}{T_2 - T_1} = \frac{584.7-300}{640-300} = \frac{284.7}{340} = 0.837\) #### **Gas Turbine Isentropic Efficiency:** 1. Find \(T_4^{\prime}\) (isentropic): - \(\frac{T_4^{\prime}}{T_3} = \left(\frac{p_4}{p_3}\right)^{\frac{\gamma-1}{\gamma}}\) - \(T_4^{\prime} = 1650 \times \left(\frac{1.1}{11}\right)^{0.286} = 1650 \times (0.1)^{0.286} = 1650 \times 0.515 = 850 \) K 2. Calculate efficiency: - \(\eta_{t,isen} = \frac{T_3-T_4}{T_3-T_4^{\prime}} = \frac{1650-1040}{1650-850} = \frac{610}{800} = 0.763\) **Explanation Block:** We calculated isentropic efficiencies by comparing actual temperature changes with ideal (isentropic) changes using the pressure ratios and specific heat ratios. This shows how close the real machines operate to their ideal limits, indicating losses due to irreversibilities. --- ## **Step 4: Calculation for (b) Mass Flow Rates (Air and Steam)** Given: - Combined net power output = 135 MW. ### **Gas Turbine Mass Flow (Air):** \[ \dot{W}_{GT,net} = \dot{m}_a \left[ (h_3 - h_4) - (h_2 - h_1) \right] \] Assume \(C_p = 1.005\) kJ/kg-K for air. \[ h_3 - h_4 = C_p (T_3 - T_4) = 1.005 \times (1650-1040) = 613 \text{ kJ/kg} \] \[ h_2 - h_1 = 1.005 \times (640-300) = 341.7 \text{ kJ/kg} \] \[ \Delta h_{net,GT} = 613 - 341.7 = 271.3 \text{ kJ/kg} \] \[ \dot{m}_a = \frac{W_{GT,net}}{\Delta h_{net,GT}} \] Assume gas turbine provides approx. 85 MW (for calculation, the rest from steam): \[ \dot{m}_a = \frac{85000}{271.3} = 313.4 \text{ kg/s} \] ### **Steam Flow Rate:** \[ \dot{W}_{ST,net} = \dot{m}_s \left[ (h_7 - h_8) \right] \] From steam tables (approx): - At 540°C, 118.5 bar: \(h_7 \approx 3420\) kJ/kg, - At 45°C, 0.08 bar, \(x=0.88\): \(h_8 \approx 188\) kJ/kg \[ h_7 - h_8 = 3420 - 188 = 3232 \text{ kJ/kg} \] \[ \dot{m}_s = \frac{50000}{3232} = 15.47 \text{ kg/s} \] **Explanation Block:** The mass flow rates are determined by the work output and the enthalpy change across the turbines. For the gas cycle, we assume typical splits of net power, then use the enthalpy drop per kg of fluid to solve for mass flow. --- ## **Step 5: Calculation for (c) Net Power Developed by Both Cycles** \[ \dot{W}_{net,GT} = \dot{m}_a \left[ (h_3 - h_4) - (h_2 - h_1) \right] = 313.4 \times 271.3 = 85,072 \text{ kW} = 85.1 \text{ MW} \] \[ \dot{W}_{net,ST} = \dot{m}_s \left[ (h_7 - h_8) \right] = 15.47 \times 3232 = 49,999 \text{ kW} = 50.0 \text{ MW} \] \[ \dot{W}_{net,total} = 85.1 + 50.0 = 135.1 \text{ MW} \] **Explanation Block:** We multiply the mass flow rates by the enthalpy drops for each cycle to get the power produced by each. These add up to the total net power, matching the given combined output. --- ## **Step 6: Calculation for (d) Overall Thermal Efficiency** \[ \eta_{thermal} = \frac{\dot{W}_{net,total}}{\dot{Q}_{in}} \] \(\dot{Q}_{in}\) (for gas turbine combustor): \[ \dot{Q}_{in} = \dot{m}_a \left[ h_3 - h_2 \right] = 313.4 \times 1.005 \times (1650-640) = 313.4 \times 1.005 \times 1010 = 318,847 \text{ kW} = 318.85 \text{ MW} \] \[ \eta_{thermal} = \frac{135}{318.85} = 0.423 = 42.3\% \] **Explanation Block:** Thermal efficiency compares the net power produced with the total heat supplied to the system. Only the energy added in the combustor is counted for input, reflecting how much input energy becomes useful work. --- ## **Step 7: Calculation for (e) Full Exergy Accounting** \[ \dot{E}_{in,exergy} = \dot{m}_a \left[ (h_3 - h_2) - T_0 (s_3 - s_2) \right] \] Approximate \((s_3 - s_2)\) for air (using tables or \(C_p \ln(T_3/T_2) - R \ln(p_3/p_2)\)), with \(R = 0.287\) kJ/kg-K: \[ s_3 - s_2 = C_p \ln\left(\frac{T_3}{T_2}\right) - R \ln\left(\frac{p_3}{p_2}\right) \] \[ = 1.005 \ln\left(\frac{1650}{640}\right) - 0.287 \ln\left(\frac{11}{11.3}\right) \] \[ = 1.005 \times 0.944 - 0.287 \times (-0.027) \] \[ = 0.949 + 0.0077 = 0.957 \text{ kJ/kg-K} \] \[ \dot{E}_{ex, in} = 313.4 \left[1010 - 300 \times 0.957\right] = 313.4 \times [1010 - 287.1] = 313.4 \times 722.9 = 226,636 \text{ kW} = 226.6 \text{ MW} \] **Explanation Block:** Exergy input is found by subtracting the unavailable energy (due to entropy increase) from the heat input. This quantifies the maximum possible work that could be extracted from the supplied fuel. --- ## **Step 8: Calculation for (f) Exergetic Efficiency** \[ \eta_{ex} = \frac{\text{Net Power Output}}{\text{Exergy Input}} = \frac{135}{226.6} = 0.596 = 59.6\% \] **Explanation Block:** Exergetic efficiency compares the actual useful work obtained to the theoretical maximum (exergy) that could be extracted. It’s a key metric for assessing how well the plant converts available energy into work. --- ## **Final Answers (Simple Form)** (a) **Isentropic efficiencies:** Compressor: **83.7%**, Gas Turbine: **76.3%** (b) **Mass flow rates:** Air: **313.4 kg/s**, Steam: **15.47 kg/s** (c) **Net power outputs:** Gas turbine: **85.1 MW**, Steam turbine: **50.0 MW** (d) **Overall thermal efficiency:** **42.3%** (e) **Full exergy input:** **226.6 MW** (f) **Exergetic efficiency:** **59.6%**