1. **Problem statement:** A 12,000 kW gas turbine operates with exhaust gas flow of 80 kg/s at 440°C entering a waste heat recovery boiler and leaving at 120°C. The gas specific heat is 0.895 kJ/kg-°C. The boiler produces steam at 4.0 MPa and 300°C from feed water at 90°C. The system's overall thermal efficiency is 30%. The steam turbine-generator set has a combined steam rate of 6 kg/kW-hr. Calculate: a) Mass flow rate of steam (kg/hr) b) Additional kW capacity obtained c) Total heat generated in the combustor (kJ/sec) 2. **Formulas and important rules:** - Heat lost by exhaust gas to boiler: $$Q = \dot{m}_{gas} \times c_p \times (T_{in} - T_{out})$$ - Mass flow rate of steam: $$\dot{m}_{steam} = \text{steam rate} \times \text{power output}$$ - Thermal efficiency: $$\eta = \frac{\text{power output}}{\text{heat input}}$$ - Power output of gas turbine: 12,000 kW - Steam rate is in kg/kW-hr, so convert power to kW-hr for steam mass flow 3. **Calculate heat recovered by boiler:** $$Q = 80 \times 0.895 \times (440 - 120) = 80 \times 0.895 \times 320 = 22,912 \text{ kJ/s}$$ 4. **Calculate total heat generated in combustor:** Using thermal efficiency $$\eta = 0.30$$ and power output $$P = 12,000 \text{ kW}$$ $$Q_{in} = \frac{P}{\eta} = \frac{12,000}{0.30} = 40,000 \text{ kW} = 40,000 \text{ kJ/s}$$ 5. **Calculate mass flow rate of steam:** Steam rate = 6 kg/kW-hr Power output from steam turbine is from heat recovered by boiler, so first find additional power from steam turbine: Additional power $$P_{steam} = \frac{Q}{\text{heat rate}}$$ but heat rate is not given, so use steam rate: Convert power to kW-hr for 1 hour: $$P_{steam} = \frac{Q \times 3600}{\text{heat rate}}$$ Alternatively, mass flow rate of steam: $$\dot{m}_{steam} = \text{steam rate} \times P_{steam} \text{ (kW-hr)}$$ Since steam rate is 6 kg/kW-hr, and power from steam turbine is: Power from steam turbine $$P_{steam} = \frac{Q}{\text{heat input per kW}}$$ but heat input per kW is not given, so use steam rate directly: Mass flow rate of steam: $$\dot{m}_{steam} = 6 \times P_{steam}$$ But we need $P_{steam}$, which is the additional power from steam turbine. 6. **Calculate additional kW capacity from steam turbine:** Heat recovered by boiler is $$Q = 22,912 \text{ kJ/s} = 22,912 \text{ kW}$$ Assuming all heat recovered is converted to power with same efficiency 30%: $$P_{steam} = 0.30 \times 22,912 = 6,873.6 \text{ kW}$$ 7. **Calculate mass flow rate of steam:** Steam rate = 6 kg/kW-hr Convert $P_{steam}$ to kW-hr for 1 hour: $$P_{steam} = 6,873.6 \text{ kW} \times 1 \text{ hr} = 6,873.6 \text{ kW-hr}$$ Mass flow rate of steam: $$\dot{m}_{steam} = 6 \times 6,873.6 = 41,241.6 \text{ kg/hr}$$ **Final answers:** a) Mass flow rate of steam = 41,241.6 kg/hr b) Additional kW capacity = 6,873.6 kW c) Total heat generated in combustor = 40,000 kJ/s