# 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 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, T2/T1 = P2/P1

T2 = (6/2) x (273 +28) = 903 K

For a constant volume process, W = 0 and DE = Q

CV = 0.718 kJ/kg.oC (for air; data)

Q = mCV(T2 - T1) = 0.718 x (903 - 301) = 432.24 kJ/kg.

DE = 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 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