## P-h Diagram

Refrigerant HFC-134a is used for a vapor-compression refrigeration cycle. If the evaporator temperature is 40°C, and the condenser temperature is 60°C, what is the evaporator cooling in kJ/kg?

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__**Hint**

**Hint**

For this specific problem, the pressure vs enthalpy diagram is:

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__**Hint 2**

**Hint 2**

For Boilers, Condensers, and Evaporators:

$$$h_{in}+q=h_{exit}$$$

where
$$h$$
is specific enthalpy, and
$$q$$
is the heat transfer per unit mass.

For this specific problem, the pressure vs enthalpy diagram is:

For Boilers, Condensers, and Evaporators:

$$$h_{in}+q=h_{exit}$$$

where
$$h$$
is specific enthalpy, and
$$q$$
is the heat transfer per unit mass. Per the problem statement’s P-h diagram:

$$$q_{evaporator}=h_{1}-h_{4}$$$

In the P-h diagram for Refrigerant HFC-134a, the enthalpy associated with the evaporator’s 40°C temperature intersecting with the saturated vapor curve (1) is
$$h_1$$
(2). The enthalpy associated with condenser’s 60°C intersecting with the saturated liquid curve (3) is
$$h_3$$
(4).

Because
$$h_3=h_4$$
based on the vapor-compression refrigeration cycle:

$$$q_{evaporator}=430\frac{kJ}{kg}-290\frac{kJ}{kg}=140\:\frac{kJ}{kg}$$$

140 kJ/kg