Question

A sealed rigid tank has 0.1 kg saturated vapor R-410a at 0°C that is cooled to 20°C by a 20°C heat sink. Thermodynamic properties of R410a at initial and final states are given as: u1 = 253.02 kJ/kg, 51 -1.0368 kJ/kg.k ; u2 - 136.5 kJ/kg; 52 - 0.5946 kJ/kg.K .. Find the heat transfer during the cooling process. b. Calculate the change in entropy of the R-410a c. Determine the change in entropy of heat sink, a. Estimate the total entropy generation.

Answer 1

solution, Note - In the 1 1 kg of saturated vapor R-410 a cooled from o to -20°c question Temperature is of state 2 is by a - 20°c heat sink > tos tate - 1 4 253.02 KJ/kg 1.0368 kJ/kg k 25 c which wro wrong it should be - 20 (varifieed from properties given ie u the s, for state 2 ų = 136.5 kJ/kg KJ/kg k 82 0.5946 since the tank is rigid de = 0 - SW1-2 from first law of thermodynamics δφ SW + LU 89 0 + m (4-4) 0 + 1 [138.5 - 253:02] - 11. 652 KJ Heat transfer during cooling process - 11.652 KS

(6) Change in Entropy to of the R-40 40 a m (4, - %) 0.1(-5946 - 1.0368) 0.04422 KJ/K (C) Heat Gain by heat sink a heat lost by R-410 a 11.625 KI It is done at - 20°C 273-20 2535 Change in Entropy for heat sink Δς 11 8 T 11.625 253 0.04594 KJ/K (d) we know that Entropy change of the system is given by OS Temperature of heat sink +fo Algen To -> -0.04422 = Asgen + -11.625 253 11.625 -0.044 22+ = 1.7286X10 KS/ 7 Asgen 253 D :11 1.7286 0J/K or 1:7286X1633 KJ/k