dinThe steam power plant diagram below is assumed to follow a Carnotcycle principle. The inlet and outlet of the boiler (2 and 3) aresaturated liquid and saturated vapor, respectively. The temperaturesin the boiler and in the condenser are 200°C and 20°C, respectively.Boilerpump,inWturb,outAssume the mass flow rate is 2 kg/s and use the table below to listthe needed properties.PumpTurbineComputea. power output by the turbineb. heat supplied to the boilerheat wasted by the condenserd. required pump powerthermal efficiency of the power plantCondenserс.Toute.Use the table to state the properties – you DO NOT NEED to fill up all of the cells – state only what youneedNodeQuality(X)PressureTemperatureEnthalpy (h)Entropy (s)(kPa)C)124

Given: Inlet temperature of boiler is 20°C Boiler outlet temperature is 200 °C Mass flow rate is 2 kg/s Steam power plant diagram is shown below:Saturated steam properties: At 200°C (473 K) Temperature, K Pressure, bar Enthalpy, kJ/kg Entropy, kJ/kg K Condensed Vapor Condensed Vapor 470 14.55 838.2 2789 2.301 6.451 473 15.55 851.76 2790.8 2.329 6.429 480 17.90 883.4 2795 2.395 6.377 At 20°C (293 K) Temperature, K Pressure, bar Enthalpy, kJ/kg Entropy, kJ/kg K Condensed Vapor Condensed Vapor 290 0.01917 70.7 2532 0.251 8.740 293 0.02127 76.97 2534.7 0.2726 8.706 295 0.02617 91.6 2541 0.323 8.627a) Power output by the turbine: Q=m×∆H=m×H3-H4=2kgs×2790.8-851.76kJkg=3878.08 kW b) Heat supplied to boiler: Q=m×H3-H2=2kgs×2790.8-76.97kJkg=5427.66 kW c) Heat wasted by the condenser: Q=m×H4-H1=2kgs×851.76-76.97kJkg=1549.58 kW d) Pump power required: Power=m×∆P3.6×106=7.2m3h×1552873 Pa3.6×106=3.1057 kW e) Thermal efficiency of the power plant: η=QBoiler-QCondenserQBoiler×100=5427.66-1549.585427.66×100=0.7145×100=71.45 %