Question

Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compressor. Determine the compressor power per unit mass flow rate if the device is (a) isentropic, (b) polytropic with n =1.3, (c) isothermal

Answer 1

(a)
`W_(isoentropic)=8.125(kJ)/(mol)`

(b)
`W_(polytropic)=7.579(kJ)/(mol)`

(c)
`W_(isothermal)=5.743(kJ)/(mol)`

Step-by-step explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant
`k` should be computed for air as an ideal gas by:

`(R)/(Cp_(air))=1-(1)/(k) \\\\(8.314)/(29.11) =1-(1)/(k)\\`

`0.2856=1-(1)/(k)\\\\k=1.4`

Next, we compute the final temperature:

`T_2=T_1((p_2)/(p_1) )^(1-1/k)=300K((1000kPa)/(100kPa) )^(1-1/1.4)=579.21K`

Thus, the work is computed by:

`W_(isoentropic)=(kR(T_2-T_1))/(k-1) =(1.4*8.314(J)/(mol*K)(579.21K-300K))/(1.4-1)\\\\W_(isoentropic)=8.125(kJ)/(mol)`

(b) In this case, since
`n` is given, we compute the final temperature as well:

`T_2=T_1((p_2)/(p_1) )^(1-1/n)=300K((1000kPa)/(100kPa) )^(1-1/1.3)=510.38K`

And the isentropic work:

`W_(polytropic)=(nR(T_2-T_1))/(n-1) =(1.3*8.314(J)/(mol*K)(510.38-300K))/(1.3-1)\\\\W_(polytropic)=7.579(kJ)/(mol)`

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

`W_(isothermal)=RTln((p_2)/(p_1) )=8.314(J)/(mol*K)*300K*ln((1000kPa)/(100kPa) ) \\\\W_(isothermal)=5.743(kJ)/(mol)`

Regards.