In a steam power plant, 1MW heat is added in the boiler at 700 o C, 0.58 MW heat is taken out in the condenser at 40 o C, and the pump work is 0.02 MW. a) What is the work (in MW) of the turbine in

a) The work of the turbine can be found by using the First Law of Thermodynamics, which states that the energy input equals the energy output plus any work done. Therefore, the work done by the turbine is: Work = Input Heat - Output Heat - Pump Work Work = 1 MW - 0.58 MW - 0.02 MW Work = 0.4 MW So, the work of the turbine in this power plant is 0.4 MW. b) The thermal efficiency of the power plant can be found using the formula: Thermal Efficiency = Output Work / Input Heat The output work has already been calculated in part (a) to be 0.4 MW. The input heat is 1 MW. Therefore, Thermal Efficiency = 0.4 MW / 1 MW Thermal Efficiency = 0.4 or 40% So, the thermal efficiency of the power plant is 40%. c) The coefficient of performance (COP) of the system as a refrigerator is given by: COP = Desired Output / Required Input The desired output of a refrigerator is the amount of heat removed from the inside of the refrigerator. In this case, that is the amount of heat removed in the condenser, which is 0.58 MW. The required input is the amount of work required to remove that heat, which is the amount of work done by the pump (0.02 MW). Therefore, COP = 0.58 MW / 0.02 MW COP = 29 So, the coefficient of performance of the system as a refrigerator is 29.