TITLE.PM5

FIRST LAW OF THERMODYNAMICS 213

dharm

/M-therm/Th4-7.pm5

∴ Mass of air which left the receiver,
m = m 1 – m 2 = 97.34 – 79.3 = 18.04 kg. (Ans.)
Example 4.65. A 1.6 m^3 tank is filled with air at a pressure of 5 bar and a temperature of
100 °C. The air is then let off to the atmosphere through a valve. Assuming no heat transfer,
determine the work obtainable by utilising the kinetic energy of the discharge air to run a frictionless
turbine.
Take : Atmospheric pressure = 1 bar ;
cp for air = 1 kJ/kg K ;
cv for air = 0.711 kJ/kg K.
Solution. Initial volume of air, V 1 = 1.6 m^3
Initial pressure of air, p 1 = 5 bar = 5 × 10^5 N/m^2
Initial temperature of air, T 1 = 100 + 273 = 373 K
Final pressure of air, p 2 = 1 bar = 1 × 10^5 N/m^2
Now, initial quantity of air in the tank before discharge,
m 1 pVRT^11
1

5

3

510 16

0 287 10 373

==×× 747

××

. =
(. )
.kg.
Assuming that system undergoes a reversible adiabatic expansion

T

T

p

1 p

(^2) =F
HG
I
KJ
−
2
1
γ 1
γ
where T 2 is the final temperature of air in the tank.
∴ 373 T^2 =FH IK =
−
1
5 0 631
14 1
14
.

..

T 2 = 373 × 0.631 = 235.4 K (i.e., finally in the line)

The final quantity of air remaining in the tank is

m

pV

(^2) RT
22
2
5
3
110 16
0 287 10 235 4
==×× 2 368
××

. =
(. ).

.kg.

With Q = 0, kinetic energy is found from,

(m 1 – m 2 ) ′+

F ′

HG

I

KJ

h C

2

2 = m 1 u 1 – m 2 u 2

or (m 1 – m 2 ) h′ + (m 1 – m 2 ) C′

2

2 = m^1 u^1 – m^2 u^2

∴ Kinetic energy,

(m 1 – m 2 ) C′

2

2 = (m^1 u^1 – m^2 u^2 ) – (m^1 – m^2 ) h′

= m 1 cvT 1 – m 2 cvT 2 – (m 1 – m 2 ) cpT 2

= 7.47 × 0.771 × 373 – 2.368 × 0.711 × 235.4 – (7.47 – 2.368) × 1 × 235.4

= 2148.24 – 396.33 – 1201 = 550.9 kJ. (Ans.)

+Example 4.66. A frictionless piston is free to move in a closed cylinder. Initially there is

0.035 m^3 of oxygen at 4.5 bar, 60°C on one side of the piston and 0.07 m^3 of methane at 4.5 bar