TITLE.PM5

AVAILABILITY AND IRREVERSIBILITY 335

DHARM
M-therm\Th6-2.PM5

Reversible/theoretical work which must be available,
Wrev = 210.58 – 29.09 = 181.49 kJ. (Ans.)
Actual work developed can be calculated by using the energy equation for adiabatic steady
flow process as follows :

Wactual = (h 1 – h 2 ) +

CC 12 22

2

F −

H

G

I

K

J

= cp (T 1 – T 2 ) +

CC 12 22

2

F −

H

G

I

K

J

= 1.005(453 – 293) +

80 40

210

22

3

−

×

F

HG

I

KJ

= 160.8 + 2.4 = 163.2 kJ/kg. (Ans.)

(ii)Irreversibility and effectiveness :

Irreversibility, I = Wrev – Wactual

= 181.49 – 163.2 = 18.29 kJ/kg. (Ans.)

Effectiveness, ∈ =

W

W

actual

rev

=

163 2

181 49

.

.

= 0.899 or 89.9%. (Ans.)

Example 6.22. Steam expands adiabatically in a turbine from 20 bar, 400°C to 4 bar,

250 °C. Calculate :

(i)The isentropic efficiency of the process ;

(ii)The loss of availability of the system assuming an atmospheric temperature of 20°C ;

(iii)The effectiveness of the process ;

The changes in K.E. and P.E. may be neglected.

Solution. Initial pressure of steam, p 1 = 20 bar

Initial temperature of steam, t 1 = 400°C

Final pressure of steam, p 2 = 4 bar

Final temperature of steam, t 2 = 250°C

Atmospheric temperature, = 20°C (= 293 K).

Initial state 1 : 20 bar, 400°C ; From steam tables,

h 1 = 3247.6 kJ/kg ; s 1 = 7.127 kJ/kg K

Final state 2 : 4 bar 250°C ; From steam tables,

h 2 ′ = 2964.2 kJ/kg, s 2 ′ = 7.379 kJ/kg K

The process is shown as 1 to 2′ in Fig. 6.14

s 1 = s 2 = 7.127 kJ/kg K

Hence, interpolating,

h 2 = 2752.8 +

7127 6 930

7171 6 930

..

..

−

−

F

HG

I

KJ (2860.5 – 2752.8)

= 2752.8 +

0197

0241

.

.

× 107.7 = 2840.8 kJ/kg.