Refrigerant 134a (R134a) enters the coils of the evaporator of a refrigeration system as a saturated liquid-vapour mixture at a pressure of 160 kPa. The refrigerant absorbs 180 kJ of heat from the cooled space, which is maintained at -5oC, and leaves as saturated vapour at the same pressure. (a) What is the temperature of the R134a liquid-vapour mixture as it enters the system? (b) What is the specific enthalpy of the R134a liquid-vapour mixture as it exits the system? (c) What is the entropy change of the R134a? (d) What is the total entropy change for this process?

Part a: The temperature of the mixture at 160kPA is 257.68K.

Part b:The specific enthalpy of the mixture as it exits the system is 237.97 kJ/kg.

Part c:The entropy change for the mixture  is 0.6999kJ/K

Part d: The total entropy change is 0.027 kJ/K

Explanation:

Part a

The temperature is as viewed from the the table of Saturated properties for the Refrigerant 134a for Pressure of 160kPa.

The temperature of the mixture at 160kPA is 257.68K

Part b

For specific enthalpy from the same table of Saturated properties for the Refrigerant 134a for Pressure of 160kPa for the vapour phase which is given as

The specific enthalpy of the mixture as it exits the system is 237.97 kJ/kg.

Part c

The entropy change for the mixture is given as

Here

ΔQ is the heat in which is given as 180kJ

T is the temperature of the mixture which is calculated in part a as 257.28K

so the equation becomes.

The entropy change for the mixture  is 0.6999kJ/K

Part d

For overall entropy change, the entropy change in the cooled space is also to be calculated such that

Q is the heat in which is given as 180kJ

T is the temperature of the cooled space which is given as -5°C or 268K

so the equation becomes.

So the total entropy change is

The total entropy change is 0.027 kJ/K

answer:

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Explanation: