Question

Refrigerant-134a enters the coils of the evaporator of a refrigeration system as a saturated liquid-vapor mixture at a pressure of 160 kPa. The refrigerant absorbs 180 kJ/kg of heat from the cooled space, which is maintained at -5°C, and leaves as a saturated vapor at the same pressure. Determine
(a) the entropy change of the refrigerant,

(b) the entropy change of the cooled space, and

(c) the total entropy change for this process.

Answer 1

Answer:

(a) 0.699 kJ/K

(b) -0.671 kJ/K

(c) 0.028 kJ/K

Explanation:

The Refrigerant-134a flows into the evaporator as a saturated liquid-vapor mixture and flows out as a saturated vapor at a saturation pressure of 160 kPa and temperature of -15.64°C (estimated from the Saturated Refrigerant-134a Temperature Table).

(a) The entropy change of the refrigerant (ΔS$_{R-134a}$) = Q/T$_{1}$

Q = 180 kJ

T$_{1}$ = -15.64 + 273.15 = 257.51 K

ΔS$_{R-134a}$ = Q/T$_{1}$ = 180/257.51 = 0.699 kJ/K

(b) The entropy change (ΔS$_{c}$) of the cooled space (ΔS$_{c}$) = -Q/T$_{2}$

Q = -180 kJ

T$_{2}$ = -5 + 273.15 = 268.15 K

ΔS$_{c}$ = Q/T$_{2}$ = -180/268.15 = -0.671 kJ/K

(c) The total entropy change for this process (ΔS$_{t}$) = ΔS$_{R-134a}$ + ΔS$_{c}$ = 0.699 - 0.671 = 0.028 kJ/K