136 ENGINEERING THERMODYNAMICS

dharm

M-therm/th4-2.pm5

QW==+=+=+=paddle ∆∆∆∆ ∆U pV U () (pV U pV)∆H
Hence paddle work is equal to change in enthalpy. (Ans.)
+Example 4.23. 0.2 m^3 of air at 4 bar and 130°C is contained in a system. A reversible
adiabatic expansion takes place till the pressure falls to 1.02 bar. The gas is then heated at
constant pressure till enthalpy increases by 72.5 kJ. Calculate :
(i)The work done ;
(ii)The index of expansion, if the above processes are replaced by a single reversible polytropic
process giving the same work between the same initial and final states.
Take cp = 1 kJ/kg K, cv = 0.714 kJ/kg K.
Solution. Refer Fig. 4.22.

p (Pressure)

Constant

pressure

heating

(^23)
1
Adiabatic
expansion
V (Volume)
Fig. 4.22
Initial volume, V 1 = 0.2 m^3
Initial pressure, p 1 = 4 bar = 4 × 10^5 N/m^2
Initial temperature, T 1 = 130 + 273 = 403 K
Final pressure after adiabatic expansion,
p 2 = 1.02 bar = 1.02 × 10^5 N/m^2
Increase in enthalpy during constant pressure process
= 72.5 kJ.
(i)Work done :
Process 1-2 : Reversibe adiabatic process :
p 1 V 1 γ = p 2 V 2 γ
VV
p
(^21) p
1
2
1

F
H
G
I
K
J
γ