Question

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.

Answer 1

|W|=169.28 KJ/kg

ΔS = -0.544 KJ/Kg.K

Step-by-step explanation:

Given that

T= 100°F

We know that

1 °F = 255.92 K

100°F = 310 .92 K

`P _1= 15 psia`

`P _1= 100 psia`

We know that work for isothermal process

`W=mRT\ln (P_1)/(P_2)`

Lets take mass is 1 kg.

So work per unit mass

`W=RT\ln (P_1)/(P_2)`

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

`W=RT\ln (P_1)/(P_2)`

`W=0.287* 310.92\ln (15)/(100)`

W= - 169.28 KJ/kg

Negative sign indicates compression

|W|=169.28 KJ/kg

We know that change in entropy at constant volume

`\Delta S=-R\ln (P_2)/(P_1)`

`\Delta S=-0.287\ln (100)/(15)`

ΔS = -0.544 KJ/Kg.K