Your company is installing a battery pack into an existing heat pump system so the compressor can be powered using renewable energy from a nearby wind turbine. Your team needs to know the power input required by the compressor for sizing of the battery pack. An on-site engineer has gathered the necessary data and compiled it into the following report, with a schematic and a description of the heat pump system: On-Site Report Condenser Expansion valve Compressor Air source heat pump. Evaporator Working fluid: Refrigerant R-134a, mass flow rate of 0.05 kgs. Fluid enters evaporator at 22C with quality of 32 %. Fluid enters compressor at 1.40 bar and 60C. Isentropic compressor efficiency of 75 %. Fluid leaves compressor at 1.20 MPa and enters condenser. Fluid experiences a pressure drop across the condenser and exits in a compressed liquid state. A colleague performed the initial calculations for this task, presented on the next page. However, concerns have been raised by your supervisor, who suspects there are errors in these calculations. You have been tasked with reviewing and correcting the calculations, which involves the following steps: (a) Redraw a correct T-s diagram, clearly indicating the areas where errors occurred in the previous version. [4 marks] (b) Identify the errors in the initial calculations and provide an explanation for why they are incorrect. [8 marks] (c) Present the correct solutions for determining specific enthalpy at Point 1, Point 2, and Point 3 and ultimately calculate the power input requirement. [8 marks] Parts (b) and (c) can be presented together or separately. Marks will be awarded equally based on both your reasoning and calculations. It is essential to be explicit in explain- ing scientifically using your knowledge of thermodynamic systems why the previous calculations were flawed, what the correct methodology should entail, and ultimately, provide the correct solutions. POINT 1 R-134a before the evaporator: xsaturated liquid T = 22C Data from Table A-11: POINT 1 h =hye22c -82.14 kJ kg- P = 12 bar POINT 2 R-134a before the compressor: X2-saturated vapour POINT 2 P = 140 kPa R-134a is a saturated vapour at given pressure, read the properties from Table A-12: POINT 3 h hg@140 kPa 239.16 kJ kg" S2 Sp@140 kPa 0.94456 kJ kg- K-1 R-134a after the compressor: xy-superheated S3 82 0.94456 kJ kg K-POINT 3 P = 0.12 MPa Isentropic compression of fluid in compressor (S-52) Interpolate to find h, data from A-13: T-s diagram for the heat pump system. s (kJ/kg K) h(kl/kg) 0.9267 0.9614 278.27 289.64 Following interpolation: POWER REQUIRED h = 284.12 kJ kg W-ma(h-hs) W = 0.05 kg s (239.16-284.12)k) kg = -2.25 kW