In: Other
Overall process can be divided into 2 process:
one is isothermal process to reach the pressure at vapor pressure of 175 DegC
2nd process is condensation of vapor to liquid stream
Now for 1st process, it is isothermal, final temperature will be same as 175 DegC
Steam is fully condensed at this temperature, so pressure in system = vapor pressure of water at 175 DegC = 892.5 kPa
Now Work done for isothermal process = W1 = nRTln(P1/P2) =
(4000/18)*8.314*(273+175)*ln(400/892.5) = -664.3J
Now for 2nd process W2 = P*total mass * change in specific volume
change in specific volume = specific value of saturated liquid-specific vol of saturated vapor
= 0.0011-0.2166 = -0.2155 m3/Kg ( reffered steam table)
W2 = 892.5*1000*4*(-0.2155) = 769335 J = -769.335 kJ
Now W = W1 + W2 = -664.3 + -769.335 = -1433.6 kJ
work done on the system = -W = 1433.6 kJ (please note answer slightly differs than given in question which might be bacause of difference in steam table data of mine and the one used in question)
Now Q1 = Delta U1 + W1 = W1 = -664.3 (because for isothermal process Delta U1 = 0)
Now Q2 = Mass *Delta Hcondensation + W2 at 892.5 Pa = -2031.55*4 = -8126-769.335= -8895.5
Q= Q1+ Q2 = --664.3-8895.5 = -9559.8 (please note answer slightly differs than given in question which might be bacause of difference in steam table data of mine and the one used in question)