Chapter 5 : Thermodynamic Properties of Real Fluids
Section 10 : Applications to real fluid processes in process plant equipments
Example 5.7
A certain gas is compressed adiabatically from 293 K and 135 KPa to 550 KPa. What is the work needed? What is the final T2 ? Assume ideal gas behavior. Compressor = 0.8.
For the gas:
For rev. process S = S2-S1 = 0
Flow process thus lead to:
S =
(1.65 + 8.9 x 10-3 T – 2.2 x 10-6 T2) = 8.314 ln
Þ T2 reversible (by iteration) ~ 395 K
By first Law Q (=0) + Ws = H Ws(isentropic, reversible) = H2 – H1 = dT ~ 3960 J/mol. Thus actual work needed =
= 3960/0.8 = 4950 J/mol 4950 =
Þ T2 (actual) ~ 420oK Þ 147oC
= 0.8.
S = S2-S1 = 0
S = 
(1.65 + 8.9 x 10-3 T – 2.2 x 10-6 T2) 
= 8.314 ln 
H
Ws(isentropic, reversible) = H2 – H1 = 
dT
~ 3960 J/mol.
Thus actual work needed = 
4950 = 
