TITLE.PM5

FIRST LAW OF THERMODYNAMICS 193

dharm

/M-therm/Th4-5.pm5

1

2

165° C

1.5 × 10^5

7 × 10^5

p (N/m )^2

v 1 v 2 v (m /kg)

3

Fig. 4.63

(i)Change of internal energy :

The internal energy at state 1 is found by using the relation :

u 1 = (1 – x) uf + xug

= (1 – 0.95) 696 + (0.95 × 2573)

∴ u 1 = 2479.15 kJ/kg

Interpolating from superheat tables at 1.5 bar and 165°C, we have

u 2 = 2580 +^1550 (2856 – 2580)

= 2602.8 kJ/kg

∴ Gain in internal energy,

u 2 – u 1 = 2602.8 – 2479.15 = 123.65 kJ/kg. (Ans.)

(ii)Change of enthalpy :

Enthalpy at state 1 (7 bar),

h 1 = hxhf 11 + 1 fg

At 7 bar. hf = 697.1 kJ/kg and hfg = 2064.9 kJ/kg

∴ h 1 = 697.1 + 0.95 × 2064.9 = 2658.75 kJ/kg

Interpolating from superheat tables at 1.5 bar and 165°C, we have

h 2 = 2772.6 +^1550 (2872.9 – 2772.6) = 2802.69 kJ/kg

∴ Change of enthalpy

= h 2 – h 1 = 2802.69 – 2658.75 = 143.94 kJ/kg. (Ans.)