IDEAL AND REAL GASES 399

dharm

\M-therm\Th8-2.pm5

∴

453

T 2

=^10

105

1

.

F ()/

HG

I

KJ

γγ−

cv = 0.837 kJ/kg K (given)

R = R

M

(^0) = 8 314
44

. (Molecular weight of CO
2 = 44)
= 0.1889 kJ/kg K
Also cp – cv = R
∴ cp – 0.837 = 0.1889
cp = 1.0259 kJ/kg K

∴γ =

c

c

p

v

=

1 0259

0 837

.

. = 1.23
Substituting for γ in equation (ii)
453
T 2
=^10
105

123 1 123

.

F (. )/.

HG

I

KJ

−

∴ T 2 = 297 K

Final temperature = 297 – 273 = 24 °C. (Ans.)

p 2 v 2 = RT 2

∴ 1.05 × 10^5 × v 2 = (0.1889 × 1000) × 297

∴ v 2 =

(. )

.

01889 1000 297

105 10^5

××

× = 0.5343 m

(^3) /kg
i.e., Final specific volume = 0.5343 m^3 /kg. (Ans.)
As the process is reversible and adiabatic
∆s = 0
i.e., Increase in entropy = 0. (Ans.)
Since the process is adiabatic, therefore, heat transfer rate from turbine = 0. (Ans.)
Applying steady flow energy equation (S.F.E.E.) on unit time basis,
mh& 1 C^1 ZQ&
2
++ 2 1
L
N
M
M
O
Q
P
P

• = mh& 2 C^2 ZW
  2
  ++ 2 2
  L
  N
  M
  M
  O
  Q
  P
  P


• 
  

  By data changes in velocity and elevation are negligible, and Q = 0.
  ∴ S.F.E.E. reduces to
  i.e., W = m&(h 1 – h 2 )
  = m&cp(T 1 – T 2 ) as
  dh
  dT
  L =−= −ch h cT Tp p
  NM
  O
  QP
  ,() 12 12

  
  

  65
  60
  .
  × 1.0259 (453 – 297) = 17.34 kW
  Hence power delivered by the turbine = 17.34 kW. (Ans.)
  Example 8.8. A certain quantity of air initially at a pressure of 8 bar and 280°C has a
  volume of 0.035 m^3. It undergoes the following processes in the following sequence in a cycle :
  (a) Expands at constant pressure to 0.1 m^3 ,