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Air is compressed in a reversible, isothermal, steady- flow process from 15 psia, 100°F to 100 psia. Calculate the work of compression per pound, the change of entropy, and the heat transfer per pound of air compressed.

Respuesta :

Answer:

|W|=169.28 KJ/kg

ΔS = -0.544 KJ/Kg.K

Explanation:

Given that

T= 100°F

We know that

1 °F = 255.92 K

100°F = 310 .92 K

[tex]P _1= 15 psia[/tex]

[tex]P _1= 100 psia[/tex]

We know that work for isothermal process  

[tex]W=mRT\ln \dfrac{P_1}{P_2}[/tex]

Lets take mass is 1 kg.

So work per unit mass

[tex]W=RT\ln \dfrac{P_1}{P_2}[/tex]

We know that for air R=0.287KJ/kg.K

[tex]W=RT\ln \dfrac{P_1}{P_2}[/tex]

[tex]W=0.287\times 310.92\ln \dfrac{15}{100}[/tex]

W= - 169.28 KJ/kg

Negative sign indicates compression

|W|=169.28 KJ/kg

We know that change in entropy at constant volume

[tex]\Delta S=-R\ln \dfrac{P_2}{P_1}[/tex]

[tex]\Delta S=-0.287\ln \dfrac{100}{15}[/tex]

ΔS = -0.544 KJ/Kg.K

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