Constant Volume Heating & Constant Pressure Cooling

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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

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
[Index]

Last Modified on: 01-May-2024

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