Answer:
0.4058
Explanation:
From the steam table
At 325 K, saturated liquid
Enthalpy, Hf = 217 kJ/kg
Entropy Sf = 0.7274 kJ/kg-K
Specific volume Vf = 1.013 cm3/gm
Saturated pressure Psat = 12.87 kPa
For compressed liquid
P1 = 8000 kPa
Temperature T = 325 K
Thermal expansion coefficient B = 460 x 10^{-6} K^{-1}
Enthalpy at initial conditions
H1 = Hf + Vf (1-BT)(P1 - P_sat)
= 217 + 1.013*10^-3 (1 - 460*10^{-6})(8000 - 12.87)
= 223.881 kJ/kg
Entropy at initial conditions
S1 = Sf - BVf (P1 - Psat)
= 0.7274 - 460×10^{-6}*1.013*10^-3 (8000 - 12.87)
= 0.724 kJ/kg-K
At 8000 kPa, saturated vapor
H2 = 2759.9 kJ/kg
S2 = 5.7471 kJ/kg-K
T = 300 K
Heat added Q = H_2 - H_1
= 2759.9 - 223.881
= 2536 kJ/kg
Maximum work
W = (H1 - H2) - T (S1 - S2)
= 223.881 - 2759.9 - 300(0.724 - 5.7471)
= - 1029 kJ/kg
Fraction of heat added = W/Q
= 1029/2536
= 0.4058
