Expansion of Refrigerant-134a in a Refrigerator Refrigerant-134a enters the capillary tube of a refrigerator as saturated liquid at 0.8 MPa and is throttled to a pressure of 0.12 MPa. Determine the quality of the refrigerant at the final state and the temperature drop during this process.

Question

Expansion of Refrigerant-134a in a Refrigerator
Refrigerant-134a enters the capillary tube of a refrigerator as saturated liquid at 0.8 MPa and is throttled to a pressure of 0.12 MPa. Determine the quality of the refrigerant at the final state and the temperature drop during this process.

Accepted Answer

Refrigerant-134a that enters a capillary tube as saturated liquid is throttled to a specified pressure. The exit quality of the refrigerant and the temperature drop are to be determined.

Assumptions 1 Heat transfer from the tube is negligible. 2 Kinetic energy change of the refrigerant is negligible.

Analysis A capillary tube is a simple flow-restricting device that is commonly used in refrigeration applications to cause a large pressure drop in the refrigerant. Flow through a capillary tube is a throttling process; thus, the enthalpy of the refrigerant remains constant (Fig. 5-31).

[latex]\begin{array}{ll} ext { At inlet: } \left.\begin{array}{l}P_{1}=0.8 \mathrm{MPa} \ ext { sat. liquid }\end{array} ight\} &\begin{array}{l}T_{1}=T_{\mathrm{sat@ 0.8 \mathrm{MPa}}} =31.31^{\circ} \mathrm{C} \h_{1}=h_{f @ 0.8 \mathrm{MPa}}=95.47 \mathrm{~kJ} / \mathrm{kg}\end{array}\end{array}[/latex] (Table A-12)

At exit: [latex]\begin{array}{c}P_{2}=0.12 \mathrm{MPa} \quad \longrightarrow \quad\qquad h_{f}=22.49 \mathrm{~kJ} / \mathrm{kg} \quad T_{ ext {sat }}=-22.32^{\circ} \mathrm{C}\\quad\\left(h_{2}=h_{1} ight) \quad\qquad\qquad h_{g}=236.97 \mathrm{~kJ} / \mathrm{kg}\end{array}[/latex]

Obviously [latex]h_{f}

[latex]x_{2}=\frac{h_{2}-h_{f}}{h_{f g}}=\frac{95.47-22.49}{236.97-22.49}=0.340[/latex]

Since the exit state is a saturated mixture at 0.12 MPa, the exit temperature must be the saturation temperature at this pressure, which is -22.32°C. Then the temperature change for this process becomes

[latex]\Delta T=T_{2}-T_{1}=(-22.32-31.31)^{\circ} \mathrm{C}=-53.63^{\circ} \mathrm{C}[/latex]

Discussion Note that the temperature of the refrigerant drops by 53.63°C during this throttling process. Also note that 34.0 percent of the refrigerant vaporizes during this throttling process, and the energy needed to vaporize this refrigerant is absorbed from the refrigerant itself.

Question 5.8: Expansion of Refrigerant-134a in a Refrigerator Refrigerant-...

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