The R-134a exit temperature and the change in enthalpy are 75.77°F and [tex]10.556\;Btu/Ibm[/tex] respectively.
Given the following data:
How to calculate the R-134a exit temperature.
First of all, we would determine the specific entropy of the saturated R-134a vapor at 6°F from the thermodynamic table (Table A-11E) while interpolating it with the specific entropy of the vapor at 5°F and 10°F as follows:
[tex]\frac{0.22434-s_1}{0.22434-0.22485} =\frac{10-6}{10-5} \\\\\frac{0.22434-s_1}{-0.00051} =\frac{4}{5} \\\\s_1=0.224748 \;Btu/IbmR[/tex]
Note: The specific entropy of the saturated R-134a vapor at the initial and final state would remain the same.
[tex]s_1=s_2=0.224748 \;Btu/IbmR[/tex]
For the specific enthalpy:
Also, we would determine the specific enthalpy of the saturated R-134a vapor at 6°F from the thermodynamic table (Table A-11E) while interpolating it with the specific enthalpy of the vapor at 5°F and 10°F as follows:
[tex]\frac{104.54-h_1}{104.54-103.82} =\frac{10-6}{10-5} \\\\\frac{104.54-h_1}{0.72} =\frac{4}{5} \\\\104.54-h_1=0.72 \times 0.8\\\\h_1=103.964 \;Btu/Ibm[/tex]
For the R-134a exit temperature:
[tex]\frac{80-T_2}{80-65.89} =\frac{0.22661-0.224748}{0.22661-0.22040} \\\\\frac{80-T_2}{14.11} =\frac{0.001862}{0.00621} \\\\\frac{80-T_2}{14.11} =0.2998\\\\80-T_2=4.2302\\\\T_2=80-4.2302\\\\[/tex]
Exit temperature = 75.77°F.
For the final specific enthalpy:
[tex]\frac{115.51-h_2}{115.51-112.20} =\frac{80-75.77}{80-65.89} \\\\\frac{115.51-h_2}{3.31} =\frac{4.23}{14.11} \\\\115.51-h_2=3.31 \times 0.2998\\\\h_2=115.51-0.9923\\\\h_2=114.52 \;Btu/Ibm[/tex]
For the change in the enthalpy of R-134a:
[tex]\Delta h = h_2-h_1\\\\\Delta h =114.520-103.964\\\\\Delta h =10.556\;Btu/Ibm[/tex]
Read more on exit temperature here: https://brainly.com/question/15878720
