TITLE.PM5

212 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-7.pm5

When the tank is fully insulated and thus no heat transfer takes place,

Q = 0

and (m 2 – m 1 ) ′+

F ′

HG

I

KJ

h C

2

2 = m 2 u 2 – m 1 u 1 ...(4.85)

Also, if the tank is empty initially and fully insulated for heat transfer,

m 1 = 0

Thus h′ + C′

2

2 = u^2 ..(4.86)

Also, if kinetic energy in the pipe line is neglected

h′ = u 2 ...(4.87)

2. Emptying a tank :
   Analogous to the filling of the tank, the equation can be written as

(m 1 – m 2 ) ′+

F ′

HG

I

KJ

h C

2

2 – Q = m 1 u 1 – m 2 u 2 ...(4.88)

where h′ = Specific enthalpy of leaving fluid, and
C′ = Velocity of leaving fluid.
For fully emptying the tank and no heat transfer and negligible exit velocity,
h′ = u 1 ...(4.89)
Example 4.64. An air receiver of volume 5.5 m^3 contains air at 16 bar and 42°C. A valve is
opened and some air is allowed to blow out to atmosphere. The pressure of the air in the receiver
drops rapidly to 12 bar when the valve is then closed.
Calculate the mass of air which has left the receiver.
Solution. Initial volume of air, V 1 = 5.5 m^3
Initial pressure of air, p 1 = 16 bar
Initial temperature of air, T 1 = 42 + 273 = 315 K
Final volume of air, V 2 = V 1 = 5.5 m^3
Final pressure of air, p 2 = 12 bar
Mass of air which left the receiver :
Mass of air in the initial condition,

m

pV

(^1) RT
11
1
5
3
16 10 5 5
0 287 10 315
==×× 97 34
××

. =
(. )

.kg.

Assuming that the mass in the receiver undergoes a reversible adiabatic process, then

T

T

p

p

2

1

2

1

1

=FHG IKJ

−γ

γ

T 2

14 1

14 0 286

315

12

16

12

= 16

F

H

I

K =

F

H

I

K

. −
.. or T
2 = 315 ×

12

16

0 286

F

H

I

K

.

= 290 K

Now mass of air in the receiver in final condition,

m 2 pVRT^22

2

5

3

12 10 55

0287 10 290

==×× 793

××

. =
(. )
.kg.