TITLE.PM5

FIRST LAW OF THERMODYNAMICS 207

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/M-therm/Th4-6.pm5

At 18 bar : hf = 884.6 kJ/kg, hfg = 1910.3 kJ/kg

∴ 2776.4 = 884.6 + x 1 × 1910.3

or x 1 =
2776.4 884
1910

− .6

.3 = 0.99

i.e., Initial, dryness fraction = 0.99. (Ans.)

Fig. 4.75
The process is shown on a p-v diagram in Fig. 4.75. States 1 and 2 are fixed, but the
intermediate states are indeterminate ; the process must be drawn dotted, as shown. No work is
done during the process, and the area under the line 1-2 is not equal to work done.
Example 4.62. Steam at 10 bar and 0.9 dryness fraction is throttled to a pressure of 2 bar.
Determine the exit condition of steam using Mollier chart.
Solution. Refer to Fig. 4.76.

Saturationline

x=0.9 1

1 2

10 bar

2 bar

x = 0.94 2

2573

h (kJ/kg)

s (kJ/kg K)

Fig. 4.76