Constant Volume Heating & Constant Pressure Cooling
Air is compressed from 2 atm absolute and 28oC to 6 atm absolute and 28oC by heating at constant volume followed by cooling at constant pressure. Calculate the heat and work requirements and ΔU and ΔH of the air.
Calculations:
PV = nRT
We have to rise the pressure to 6 atm by heating at constant volume.
At constant volume, P ∝ T.
Therefore, T2/T1 = P2/P1
T2 = (6/2) x (273 +28) = 903 K
For a constant volume process, W = 0 and ΔU = Q
CV = 0.718 kJ/kg.oC (for air; data)
Q = mCV(T2 - T1) = 0.718 x (903 - 301) = 432.24 kJ/kg.
ΔU = 432.24 kJ/kg
For cooling at constant pressure, heat removed = mCP(T2 - T1)
= 1.005 x (903 - 301) = 605.01 kJ/kg (Cp - Cv = 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 | ΔU | ΔH | |
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 |