A onestage adiabatic compressor is taking gas k 125 at stan

A one-stage adiabatic compressor is taking gas (k = 1.25) at standard conditions and discharging it at 279.8 psia. When it is replaced by a three-stage compressor with cooling back to standard temperature between stages, the rate, total power-consumption and individual efficiencies remain the same. What will be the new outlet pressure in psia?

Solution

Inlet pressure P1 = 14.7 psia

Inlet temperature T1 = 68 F = 528 R

T2 / T1 = (P2 / P1)^(k-1)/k

T2 / 528 = (279.8 / 14.7)^{(1.25-1)/1.25}

T2 = 951.8 R

Power consumption P = Cp (T2 - T1) = Cp*(951.8 - 528) = 423.8*Cp

Now for 3-stage compression for states 1-2, 2\'-3, 3\'-4,

T2 / T1 = (P2 / P1)^(k-1)/k

T2 / 528 = r^{(1.25-1)/1.25}..........Here is the pressure ratio of each stage

T3 / 528 = r^{(1.25-1)/1.25}.

T4 / 528 = r^{(1.25-1)/1.25}.

This gives T2 = T3 = T4

Also, T2\' = T3\' = T1

Power input = [Cp (T2 - T1) - R (T2\' - T2)] + [Cp (T3 - T2\') - R (T3\' - T3)] + Cp (T4 - T3\')

= [Cp (T2 - T1) - R (T1 - T2)] + [Cp (T2 - T1) - R (T1 - T2)] + Cp (T2 - T1)

= (3Cp + 2R)(T2 - T1)

Also, Cp - Cp / k = R

Power input = (3Cp + 2Cp (1 - 1/k))(T2 - T1)

= (5 - 2/k) Cp (T2 - T1)

= (5 - 2/k) Cp [528*r^{(1.25-1)/1.25} - 528]

= 3.4 Cp [528*r^{(1.25-1)/1.25} - 528]

We get, 423.8*Cp = 3.4 Cp [528*r^{(1.25-1)/1.25} - 528]

Solving this, r = 2.8855

Hence, Outlet presure = 2.8855*2.8855*(2.8855*14.7)

= 353.17 psia