FIRST LAW OF THERMODYNAMICS 209

dharm

/M-therm/Th4-6.pm5

(ii) Change in enthalpy = h 2 – h 1

= 2559.29 – 2692.64 = – 133.35 kJ/kg. (Ans.)

(–ve sign indicates decrease).

(iii) Change in internal energy :

Internal energy at point 1,

u 1 = h 1 – pxv 11 g 1

= 2692.64 – 15 × 10^5 × 0.95 × 0.132 × 10–3 = 2504.54 kJ/kg

Internal energy at point 2,

u 2 = h 2 – pxv 22 g 2

= 2559.29 – 7.5 × 10^5 × 0.9 × 0.255 × 10–3 = 2387.16 kJ/kg

∴ Change in internal energy

= u 2 – u 1 = 2387.16 – 2504.54 = – 117.38 kJ/kg

(–ve sign indicates decrease)

Consider the throttling expansion 2-3 :

Entropy at point 2,

s 2 = (s 1 ) = 6.2343 kJ/kg K

Entropy at point 3,

ss xs s 33 =+f 333 ()g −f

The pressure at point 3 can be read from

chart ( 0.06 bar) and the corresponding

values of and from steam tables.

hs

p

sf hfg

• 3

(^33)

Condition 3. At 0.06 bar, x 3 = 1. From steam tables,
sf 3 = 0.521 kJ/kg K, sg 3 = 8.330 kJ/kg K
∴ s 3 = 0.521 + 1 × (8.330 – 0.521) = 8.330 kJ/kg K
Change in entropy = s 3 – s 2
= 8.330 – 6.2343 = 2.0957 kJ/kg K
Change in enthalpy = 0
i.e., h 2 = h 3
Change in internal energy = 0
i.e., u 3 = u 2
Combining the results obtained from isentropic and throttling expansion, we get during the
entire process :
(i)Change in entropy = 2.0957 kJ/kg K (increase). (Ans.)
(ii)Change in enthalpy = 133.35 kJ/kg K (decrease). (Ans.)
(iii)Change in internal energy = 117.38 kJ/kg (decrease). (Ans.)
Only the expansion of steam from point 1 to 2 (i.e., isentropic expansion) is reversible
because of unresisted flow whereas the expansion from point 2 to point 3 (i.e., throttling expansion)
is irreversible because of frictional resistance to flow. Increase of entropy also shows that expan-
sion from point 2 to point 3 is irreversible.