R-134a vapor enters into a turbine at 250 psia and 175°F. The temperature of R-134a is reduced to 20°F in this turbine while its specific entropy remains constant. Determine the change in the enthalpy of R-134a as it passes through the turbine. Use the tables for R-134a.

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MrRoyal MrRoyal · Answer 1 · 2020-03-16T02:47:19+01:00

Answer:

∆h = −22.79Btu/Ibm

Explanation:

The entropy and the initial enthalpy can be obtained from table A-13E with the given temperature and pressure using interpolation.

a = 0.2328Btu/IbmR

h1 = 129.7 Btu/Ibm

The final enthalpy can be obtained from final temperature and entropy from table A-13E

h2 = 106.91 Btu/Ibm

The enthalpy change is then calculated as follows

∆h = h2 - h1

∆h = 106.91 - 129.7

∆h = −22.79Btu/Ibm

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abiolataiwo2015 abiolataiwo2015 · Answer 2 · 2022-03-29T17:12:30+02:00

The change in the enthalpy of R-134a as it passes through the turbine is: −22.79Btu/Ibm.

Temperature (t1) of R-134a vapor at state 1= 175°F

Pressure (p1) of R-134a vapor at state 1 =250 psia

Specific Enthalpy (h1) of R-134a vapor at state 1= 129.7 Btu/Ibm

Specific Entropy (s1) of R-134a vapor at state 1 =0.2328Btu/IbmR

Final enthalpy (h2) = 106.91 Btu/Ibm

Hence:

Enthalpy change (∆h) = h2 - h1

Enthalpy change (∆h) = 106.91 - 129.7

Enthalpy change (∆h) = −22.79Btu/Ibm

Inconclusion the change in the enthalpy of R-134a as it passes through the turbine is: −22.79Btu/Ibm.

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