Question

Water vapor initially at 240 °C, 1.0 MPa expands in a piston-cylinder assembly isothermally and without internal irreversibilities to a final pressure of 0.1 MPa. Evaluate the (a)work done, in kJ/kg, (b) the change of entropy,, in kJ/kg - K and (c) the change of internal energy, in kJ/kg. Assume the system can be treated as a Van der Waals model.. [Given: the specific volume of the water vapor can be determined from the superheated steam table and they are: Vi = 0.2275 m®/kg and V2 = 2.359 m2/kg. The system can be treated as a single-phase condition.) Integration Equations: dx In x-a x-a and X-a 1

please answer a,b, and c

Answer 1

Calculate the work done. RT dV = RT In w = [pdv = [(*7 =(0.4615) (240 +273)In (2.359) (0.2275) = 553.72 kJ/kg Use superheated steam table to specific entropy and specific internal energy at 1.0 MPa and 240 oC. S7 = 6.8836 kJ/kg-K u = 2693.3 kJ/kg Use superheated steam table to specific entropy and specific internal energy at 0.1 MPa and 240 oC. S2 = 7.9962 kJ/kg-K 1/2 = 2718.7 kJ/kg Calculate the change in entropy. As = 52 - 5 = 7.9962 -6.8836=1.1126 kJ/kg-K Calculate the change in internal energy. Au = 12 – 14 = 2718.7–2693.3 = 25.4 kJ/kg