TITLE.PM5

AVAILABILITY AND IRREVERSIBILITY 329

DHARM
M-therm\Th6-2.PM5

= 0.6[50.25 – 42.38] = 4.722 kJ/kg

∴ Effectiveness =

Increase of availability of system

Loss of availability of surroundings

= 1.465

4.722

= 0.31 or 31%. (Ans.)
The low figure for the effectiveness is an indication of the highly irreversible nature of the
mixing process.

+Example 6.17. 2.5 kg of air at 6 bar, 90°C expands adiabatically is a closed system until
its volume is doubled and its temperature becomes equal to that of the surroundings which is at
1 bar, 5°C. For this process determine :
(i)The maximum work ;
(ii)The change in availability ;
(iii)The irreversibility.
For air take : cv = 0.718 kJ/kg K, R = 0.287 kJ/kg K.
Solution. Mass of air, m = 2.5 kg
Initial pressure of air, p 1 = 6 bar = 6 × 10^5 N/m^2

Ratio of final to initial volume,

V

V

2

1

= 2

Initial temperature of air, T 1 = 90 + 273 = 363 K

Final pressure of air, p 2 = 1 bar = 1 × 10^5 N/m^2

Final temperature of air, T 2 = T 0 = 5 + 273 = 278 K

From the property relation

TdS = dU + pdV

or dS =
dU
T

pdV

T

+

dS = mc dTTv +mRdVV QucdT pV mRT

p

T

mR

v V

L ===

NM

O

QP

and or

∴ The entropy change of air between the initial and final states is

dS mc dT

T

mRdV

V

v

1

2

1

2

1

2

zz=+z

or S 2 – S 1 = mcv loge T

T

2

1

+ mR loge V

V

2

1

(i)The maximum work, Wmax :

Also, Wmax = (U 1 – U 2 ) – T 0 (S 1 – S 2 )

= mcT T T c

T

T R

v

vv() log log12 0 eev

2

1

2

1

−+ +

F

HG

I

KJ

L

N

M

M

O

Q

P

P

= 2.5 0 718 363 278 278 0 718^278

363

.(−+). log F 0 287.log 2

HG

I

KJ

F +

HG

I

KJ

L

N

M

M

O

Q

P

P

ee

= 2.5[61.03 + 278(– 0.1915 + 0.1989)] = 157.7 kJ

Hence, maximum work = 157.7 kJ. (Ans.)