TITLE.PM5

288 ENGINEERING THERMODYNAMICS

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/M-therm/th5-4.pm5

Temperature fall when expanded reversibly and adiabatically = 165 K

Temperature fall when expanded into an evacuated space = 25 K

Fig. 5.44
Change in entropy :
Path 1-2 : Reversible adiabatic process.
Change in entropy, (s 2 – s 1 ) = 0.
Path 1-3 : Adiabatic process such that v 3 = 0.186 m^3 /kg (= v 2 )
(States 2 and 3 lie on the same constant volume line on T-s diagram)
and T 1 – T 3 = 25 K.
Change in entropy during this adiabatic process = s 3 – s 1.
To calculate (s 3 – s 1 ) a reversible path has to be selected joining 3 and 1.
This is achieved by selecting the reversible adiabatic path 1-2 and the reversible constant
volume process 2-3.
s 3 – s 1 = (s 3 – s 2 ) + (s 2 – s 1 )
= (s 3 – s 2 ) + 0 = (s 3 – s 2 )

= cv loge TT^3

2

= 1.25 loge

530 25

530 165

−

−

F

H

I

K

= 1.25 loge (^) HF 365505 IK = 0.4058 kJ/kg K. (Ans.)
Example 5.43. A heat pump operates between two identical bodies which are at tempera-
ture T 1 and cools one of the bodies to a temperature T 2 (T 2 < T 1 ). Prove that for this operation the
minimum work required by the heat pump is given by
WcT
T
p^1 T2T
2
22
=+− 21
F
HG
I
KJ
where cp is the specific heat which is same for both the bodies.