Saturated water vapor enters a heat exchanger at a pressure

Saturated water vapor enters a heat exchanger at a pressure of 125 kPa and rate of 5 kg/s. The water exits at the same pressure with a quality of x = 0.65. Heat is transferred between the water and air (you determine the direction), which remains at a constant temperature of 25°C. Determine the rate of heat transfer, the rate of change of the entropy of the water, and the rate of change of the entropy of the air. What is the net rate of entropy generation?

Solution

At 125 kPa for sat. vapor, we get T1 = 106 deg C, v1 = 1.37 m^3/kg, u1 = 2510 kJ/kg, h1 = 2680 kJ/kg, s1 = 7.28 kJ/kg-K

At P2 = P1 = 125 kPa for x = 0.65, we get T2 = 106 deg C, v2 = 0.894 m^3/kg, u2 = 1790 kJ/kg, h2 = 1900 kJ/kg, s2 = 5.22 kJ/kg-K

Q - W = (u2 - u1)

Q - P(v2 - v1) = (u2 - u1)

Q - 125*(0.894 - 1.37) = (1790 - 2510)

q = -779.5 kJ/kg

Q = 5*(-779.5) = -3897.5 kW

Change of entropy of water = m*(s2 - s1)

= 5*(5.22 - 7.28)

= -10.3 kW/K

Change of entropy of air = -Q/T

= 3897.5 / (25+273)

= 13.079 kW/K

Rate of entropy generation = 13.079 - 10.3 = 2.779 kW/K