82 ENGINEERING THERMODYNAMICS

Dharm

\M-therm/th3-1.p65

0.918 × 0.462 = x 3 × 0.606

∴ x 3 =0918 0462..0 606.× = 0.699. (Ans.)
(iii)Heat lost during cooling :
Heat lost during cooling = m(u 3 – u 2 ), where u 2 and u 3 are the internal energies of steam
before starting cooling or after blowing and at the end of the cooling.
∴ u h22 22=−pxvg 22 = +()hf xh22 222fg −pxvg
= (604.7 + 0.918 × 2133) – 4 × 10^5 × 0.918 × 0.462 × 10–3
= 2562.79 – 169.65 = 2393.14 kJ/kg
u 3 =−h 3 pxv 33 g 33 = +()hf xh3 3fg −pxv33 3g
= (561.4 + 0.669 × 2163.2) – 3 × 10^5 × 0.699 × 0.606 × 10–3
= 2073.47 – 127.07 = 1946.4 kJ/kg
∴ Heat transferred during cooling
= 2.045 (1946.4 – 2393.14) = – 913.6 kJ.
i.e., Heat lost during cooling = 913.6 kJ. (Ans.)
Example 3.11. If a certain amount of steam is produced at a pressure of 8 bar and dryness
fraction 0.8. Calculate :
(i)External work done during evaporation.
(ii)Internal latent heat of steam.
Solution. Pressure of steam, p = 8 bar
Dryness fraction, x = 0.8
At 8 bar. From steam tables,
vg = 0.240 m^3 /kg, hfg = 2046.5 kJ/kg
(i)External work done during evaporation
= pxvg = 8 × 10^5 × 0.8 × 0.24 N-m

= 810 08024 10

5

3

×××.. = 153.6 kJ. (Ans.)

(ii)Internal latent heat = xhfg – external work done

= 0.8 × 2046.5 – 153.6

= 1483.6 kJ. (Ans.)

Example 3.12. A quantity of steam at 10 bar and 0.85 dryness occupies 0.15 m^3. Deter-

mine the heat supplied to raise the temperature of the steam to 300°C at constant pressure and

percentage of this heat which appears as external work.

Take specific heat of superheated steam as 2.2 kJ/kg K.

Solution. Pressure of steam, p 1 = p 2 = 10 bar

Dryness fraction, x 1 = 0.85

Volume of steam, V 1 = 0.15 m^3

Final temperature of steam, tsup 2 = 300°C

Specific heat of superheated steam, cps = 2.2 kJ/kg K

Mass of steam =

V

xvg

1

(^11)

0.15
0.85 0.194×
= 0.909 kg (Q At 10 bar :vg=0.194 m / k^3 g)