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GASES AND VAPOUR MIXTURES 439

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+Example 9.19. 4 kg of carbon dioxide at 40°C and 1.4 bar are mixed with 8 kg of
nitrogen at 160°C and 1.0 bar to form a mixture at a final pressure of 0.7 bar. The process occurs
adiabatically in a steady flow apparatus. Calculate :
(i)The final temperature of the mixture ; (ii)The change in entropy.
Take value of cp : for CO 2 = 0.85 kJ/kg K and N 2 = 1.04 kJ/kg K.
Solution. Refer Fig. 9.10.

CO 2

N 2

Mixture of

CO + N 22

1

2

1

2

Fig. 9.10

(i)Final temperature, T 2 :

In this process, W = 0, Q = 0

∴ The steady flow equation may be written as

H 1 = H 2

i.e., ()mh 1 CO 2 + ()mh 1 N 2 = ()mh 2 mixture = ()mh 2 CO 2 + ()mh 2 N 2

or [(mh h 12 − )]CO 2 + [(mh h 12 − )]N 2 = 0

or [(mc Tp 12 −T)]CO 2 + [(mc Tp 12 −T)]N 2 = 0

or 4 × 0.85(40 – T 2 ) + 8 × 1.04(160 – T 2 ) = 0
or 136 – 3.4T 2 + 1331.2 – 8.32T 2 = 0
or 1467.2 – 11.72T 2 = 0
∴ T 2 = 125.2°C = 398.2 K. (Ans.)
(ii)Change in entropy :

Now nCO 2 =^4

44

= 0.0909

nN 2 =^8

28

= 0.2857

∴ n = nnCO 22 + N = 0.0909 + 0.2857 = 0.3766

Again,

p

p

(CO ) 22

2

= xCO 2 [p 2 = pressure of the mixture]