Air is compressed from 2 atm absolute and 28^{o}C to 6 atm absolute and 28^{o}C by heating at constant volume followed by cooling at constant pressure. Calculate the heat and work requirements and DE and DH of the air.
Calculations:
PV = nRT
We have to rise the pressure to 6 atm by heating at constant volume.
At constant volume, P a T.
Therefore, T_{2}/T_{1} = P_{2}/P_{1}
T_{2} = (6/2) x (273 +28) = 903 K
For a constant volume process, W = 0 and DE = Q
C_{V} = 0.718 kJ/kg.^{o}C (for air; data)
Q = mC_{V}(T_{2}  T_{1}) = 0.718 x (903  301) = 432.24 kJ/kg.
DE = 432.24 kJ/kg
For cooling at constant pressure, heat removed = mC_{P}(T_{2}  T_{1})
= 1.005 x (903  301) = 605.01 kJ/kg (C_{p}  C_{v} = R for ideal gas)
Internal energy change for the total process consisting the above two steps is zero (since internal energy is a function of temperature alone).
Work done on the gas during constant pressure cooling = 605.01  432.24 = 172.77 kJ/kg.
Summary:
Heat required 
Work required 
DE 
DH 

Constant Volume heat addition 
432.24 kJ/kg 
432.24 kJ/kg 
605.01 kJ/kg 

Constant Pressure heat removal 
605.01 kJ/kg 
172.77 kJ/kg 
432.24 kJ/kg 
605.01 kJ/kg 
Over All 
172.77 kJ/kg 
172.77 kJ/kg 
0 
0 
Last Modified on: 11Sep2014
Chemical Engineering Learning Resources  msubbu
email: msubbu.in[AT]gmail.com
Web: http://www.msubbu.in