144 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-3.pm5

In case of a polytropic process,

T

T

p

p

n

2 n

1

2

1

(^131)
66 3
1

F
HG
I
KJ
=F
HG
I
KJ
− −
.

1. 
   


2. 
   


3. 
   

   = 1.5119
   ∴ T 2 = 300 × 1.5119 = 453.6 K
   Now, work done, W
   RT T
   n
   = −
   −
   = −
   −
   () 12 .)
   1
   0 300 453 6
   31
   .277(

   
   


4. 
   

   = – 141.8 kJ/kg
   To find heat flow, using the relation,
   Q= −n W
   −
   F
   HG
   I
   KJ
   = −
   −
   F
   HG
   I
   KJ
   γ
   γ 1
   188 1 3
   188 1

   
   


5. 
   


6. 
   

   .
   × – 141.8 = + 84.5 kJ/kg
   i.e., Heat supplied = 84.5 kJ/kg. (Ans.)
   Example 4.31. 0.1 m^3 of an ideal gas at 300 K and 1 bar is compressed adiabatically to 8
   bar. It is then cooled at constant volume and further expanded isothermally so as to reach the
   condition from where it started. Calculate :
   (i)Pressure at the end of constant volume cooling.
   (ii)Change in internal energy during constant volume process.
   (iii)Net work done and heat transferred during the cycle. Assume
   cp = 14.3 kJ/kg K and cv = 10.2 kJ/kg K.
   Solution. Given : V 1 = 0.1 m^3 ; T 1 = 300 K ; p 1 = 1 bar ; cp = 14.3 kJ/kg K ;
   cv = 10.2 kJ/kg K.
   Refer to Fig. 4.24.
   (i)Pressure at the end of constant volume cooling, p 3 :
   γ =
   c
   c
   p
   v

   
   

   14 3
   10 2
   .
   .
   = 1.402
   Characteristic gas constant,
   R = cp – cv = 14.3 – 10.2 = 4.1 kJ/kg K
   Considering process 1-2, we have :
   p 1 V 1
   γ
   = p 2 V 2
   γ
   V 2 = V 1 × p
   p
   1
   2
   1
   F
   HG
   I
   KJ
   γ
   = 0.1 ×
   1
   8
   1
   F
   HG
   I
   KJ
   1.402
   = 0.0227 m^3
   Also, T
   T
   p
   p
   2
   1
   2
   1
   (^111)
   8 1
   1
   =F
   HG
   I
   KJ
   =F
   HG
   I
   KJ
   γ− −
   γ
   .402
   .402 = 1.815
   or T 2 = T 1 × 1.815 = 300 × 1.815 = 544.5 K
   Considering process 3–1, we have
   p 3 V 3 = p 1 V 1
   ∴ p 3 =
   pV
   V
   11
   3
   101
   0 0227
   = ×.
   .
   = 4.4 bar. (Ans.) (Q V 3 = V 2 )
   (ii)Change in internal energy during constant volume process, (U 3 – U 2 ) :
   Mass of gas, m =
   pV
   RT
   11
   1
   110 01^5
   4 1 1000 300
   = ××
   ××
   ().
   (. )
   = 0.00813 kg
   8 2
   p 3
   V=C
   1
   p(bar)
   V=V 23 0.1 V(m )^3
   pV = Cg
   pV=C
   3
   1
   Fig. 4.24