TITLE.PM5

274 ENGINEERING THERMODYNAMICS

dharm

/M-therm/th5-4.pm5

Note that if in this problem sA – s 2 happened to be greater than sA – s 1 , this would mean that

s 1 was greater than s 2 , and the process should appear as in Fig. 5.36.

Note. The change of entropy can also be found by using the following relation :

s 2 – s 1 = cv

n

n

−

−

F

HG

I

KJ

γ

1 loge^

T

T

2

1

= 0.718

25 399

25 1

−

−

F

HG

I

KJ loge^

597.3

873

F

HG

I

KJ Qγ= = =

L

N

M

O

Q

P

c

c

p

v

1. (^005) 1.
   0718
   399
   .
   = 0.718 × (– 0.596) × (– 0.3795) = 0.1626 kJ/kg K (increase).
   Example 5.28. In an air turbine the air expands from 7 bar and 460°C to 1.012 bar and
   160 °C. The heat loss from the turbine can be assumed to be negligible.
   (i)Show that the process is irreversible ;
   (ii)Calculate the change of entropy per kg of air.
   Solution. Refer Fig. 5.37.
   Initial pressure, p 1 = 7 bar = 7 × 10^5 N/m^2
   Initial temperature, T 1 = 460 + 273 = 733 K
   Final pressure, p 2 = 1.012 bar = 1.012 × 10^5 N/m^2
   Final temperature, T 2 = 160 + 273 = 433 K
   Fig. 5.37
   (i)To prove that the process is irreversible :
   Since the heat loss is negligible, the process is adiabatic.
   For a reversible adiabatic process for a perfect gas, using the following equation, we have :
   T
   T
   2
   1 =
   p
   p
   2
   1
   1
   F
   HG
   I
   KJ
   −γ
   γ
   T 2
   733 =


2. 
   

   1.4
   012 1.4
   7
   1
   F
   HG
   I
   KJ
   F −
   HG
   I
   KJ