Steam is the working fluid in a heat engine. During steady-state operation, the
ID: 1850448 • Letter: S
Question
Steam is the working fluid in a heat engine. During steady-state operation, the steam enters the condenser as a saturated vapor at a pressure of 200 kPa, transfers heat to a coolant liquid, and exits with a quality of 0.10 at the same pressure. The entropy of the cooling liquid increases during the process by 5.9 kJ/K per kg of steam. a. Is the process reversible of irreversible? b. What is the temperature of the cooling liquid?Explanation / Answer
At 200 kPa, s1 = s_g = 7.13 kJ/kg-K At 200 kPa and x = 0.1, s2 = 2.09 kJ/kg-K a) Entropy change of steam = s2 - s1 = 2.09 - 7.13 = -5.04 kJ/kg-K Entropy gain of coolant = 5.9 kJ/kg-K Total entropy change of universe = 5.9 + (-5.04) = 0.86 kJ/kg-K Since total entropy change > 0, process is irreversible. b) At 200 kPa, saturation T = 120 deg C = 120+273 K = 393 K Heat transfer from steam = T ds = 393*(-5.04) = -1980.7 kJ/kg Thus, temperature of coolant = 1980.7/5.9 = 335.7 K = 62.7 deg C