A steam compressor operates with P1 = 100 kPa and x1 = 1 at the inlet, a pressure ratio of r,Problem= 16, a mass flow rate of m = 5 kg/s, and an isentropic efficiency of 1 = 75%. Complete a thermodynamic analysisof the compressor, using the appropriate data for water, by finding: (A) The temperature at the exit, T2, (B) thework required to operate the compressor, Wc, and (C) the rate of entropy production of the compressor, Šgen: Plotthe process on a T-s diagram and clearly label states 1, 2s, and 2.1

Given, Initial pressure, P1 = 100 kPa mole fraction, x1 = 1 From the steam tables, the thermodynamic properties of steam at 100 kPa are as follows: Enthalpy, H1 = 2675.0 kJ/kg Entropy, S1 = 7.3589 kJ/kg·KGiven, pressure ratio, rp = 16P2P1 = 16 Final pressure, P2 = 16P1 = 1600 kPa Since the process is an isentropic process. The entropy of the state-1 and state-2s are equal. S1 = S2S = 7.3589 kJ/kg·K Enthalpy, H2S = 3349.1 kJ/kg Given the efficiency of the isentropic process, η = 75% Calculate the enthalpy of state-2 as follows: η = H2S-H1H2-H10.75 = 3349.1-2675H2-2675H2-2675 = 674.10.75H2 = 898.8 + 2675H2 = 3573.8 kJ/kg (a) At H2 = 3573.8 kJ/kg and P2 = 1600 kPa Using steam tables, The final temperature, T2 = 546°C Therefore, the required solution is 546 °C.(b) Calculate the work done on the system as follows: W˙C = m˙H2-H1= 5 kg/s×3573.8 -2675 kJ/kg= 4494 kW Therefore, the required solution is 4494 kW.(c) At state-2, H2 = 3573.8 kJ/kg and P2 = 1600 kPa From steam tables, The entropy, S2 = 7.6641 kJ/kg·K Write the expression for the entropy balance as follows: S˙in-S˙out+ S˙gen = 0S˙gen = S˙out-S˙inS˙gen = m˙S2-S1 Substitute the corresponding values in the above equation and calculate S˙gen. S˙gen = m˙S2-S1S˙gen = 5 kg/s7.6641 kJ/kg·K-7.3589 kJ/kg·KS˙gen = 1.526 kW/K Therefore, the required solution is 1.526 kW/K.Draw the T-S graph for the process as follows: