TITLE.PM5

202 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-6.pm5

Fig. 4.69

=^10

0

5

.1

(1.869 – 1.419) = 10^5 × 4.5 N-m/kg

i.e., Work done =^10 4 5

10

5

3

×. = 450 kJ/kg. (Ans.)

At 0.34 bar : vg 2 = 4.65 m^3 /kg, therefore, steam is wet at state 2 (since v 2 < vg 2 ).

Now, v 2 = x 2 vg 2 , where x 2 = dryness fraction at pressure p 2 (0.34 bar)

4.174 = x 2 × 4.65 or x 2 =^4

4

.174
.65
= 0.897
The expansion is shown on a p-v diagram in Fig. 4.69, the area under 1-2 represents the
work done per kg of steam.

(ii)Heat transferred :

Internal energy of steam at initial state 1 per kg,

u 1 = (1 – x 1 )uf 1 + x 1 ug 1 = (1 – 0.98) 696 + 0.98 × 2573 = 2535.46 kJ/kg

Internal energy of steam at final state 2 per kg,

u 2 = (1 – x 2 ) uf 2 + x 2 ug 2

= (1 – 0.897) 302 + 0.897 × 2472 = 2248.49 kJ/kg

Using the non-flow energy equation,

Q = (u 2 – u 1 ) + W

= (2248.49 – 2535.46) + 450 = 163.03 kJ/kg

i.e., Heat supplied = 163.03 kJ/kg. (Ans.)