Answer Happy

INSTRUCTIONS For your own benefit, write clearly and legibly. To receive full credit, follow instructions, write the answers in the appropriate boxes, and show all steps including equations, calculations, and assumptions. Problem 1 (75%) Consider a steam power plant that operates on a regenerative Rankine cycle with two open feedwater heaters. The mass flow rate of steam through the boiler is 45 kg/s. A schematic of the system and the conditions/properties at each state are provided on the next page. The turbine (all sections) and pumps are insulated. a. Sketch the cycle on a T-s diagram, including the saturation lines (dome). Be sure to place states 7, 8, 9, and 10 in their correct relationship with the saturation dome. b. Write out the conservation of mass and the first law (conservation of energy) for each feed water heater and show how to use these equations to calculate the flow fractions y and z. c. Determine the mass flow rate at each of the turbine exits and through each pump, in kg/s. d. Determine the work input to each pump, in kW. e. Determine the heat addition in the boiler, in kW. f. Determine the work output of each turbine section, in kW. g. Calculate the thermal efficiency. h. Determine the net power output of the power plant, in MW. i. Is the high-pressure section of the turbine operating isentropically? If yes, explain why, if no, calculate the isentropic efficiency. Summarize your answers in the table below. b. flow fractions y = z = Pump II: Pump III: c. flow rates, kg/s Pump I: HP Turbine(8): MP Turbine(9): LP Turbine(10): d. work for pumps, in kW Pump I: Pump II: Pump III: e. heat addition in boiler, in kW f. work for turbines, in kW HP Turbine (7-8): MP Turbine(8-9): LP Turbine(9-10): g. thermal efficiency h.net power, in MW i. isentropic? Yes/No efficiency, if applicable:
7 Turbine Boiler 8 у 10 f1-y-z y 9 Z fwh II fwh I Condenser) PIII PII PI+ P. 0.08 bar °C saturated liquid T. = 41.51 h, 173.88 V, 0.0010084 kJ/kg m/kg 0.2 2.00 bar 174.07 MPa = k/kg MPa = 2.00 bar °C P = 0.2 T:- 120.20 ha 504.70 V- 0.0010605 0.7 saturated liquid 7.00 bar kJ/kg m/K5 MPa = k/kg MPa = °C 505.23 Ps 0.7 7.00 bar 165.00 Ts- hs - Vse saturated liquid 697.22 0.001108 10 kJ/kg m/ MPa 100.00 bar he 707.52 k/kg MPa = = ܕܡ 10 100.00 bar 600.0 °C Ty= h = 3625.30 6.9029 0.7 kJ/kg kJ/kg/K MPa kJ/kg MPa = 7.00 bar hs = 2852.9 0.2 Pa 2.00 bar h = 2618.49 k/kg kPa - P10 0.08 bar 8 2159.8 k/kg