TITLE.PM5

FIRST LAW OF THERMODYNAMICS 185

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/M-therm/Th4-5.pm5

If after cooling the condition of steam remains wet, then the mass fraction be obtained as
follows :

V

xv

V

12 g 12 =xvg

x

xv

v

g

(^2) g
(^11)
2
= ...(4.70)
(c)
Fig. 4.57. Constant volume process.
where vg 2 can be found from the steam tables corresponding to pressure p 2.
Applying the first law of thermodynamics, we have
Q = ∆u + pdv.
1
2
z = ∆u as 1 pdv
2
z = 0
= u 2 – u 1
i.e., Q 1 = [h 2 – pv 2 sup 2 ] – [h 1 – p 1 ()xv 1 g 1 ] ...(4.71)
In case the condition of steam remains wet after heating, then
Q = (u 2 – u 1 ) = [h 2 – p 2 ()xv 2 g 2 ] – [h 1 – p 1 ()xv 1 g 1 ] ...(4.72)
In the cooling process, the same equations are used except that the suffixes 1, 2 are
interchanged.
+Example 4.51. A rigid cylinder of volume 0.028 m^3 contains steam at 80 bar and 350°C.
The cylinder is cooled until the pressure is 50 bar. Calculate :
(i)The state of steam after cooling ;
(ii)The amount of heat rejected by the steam.
Solution. Volume of rigid cylinder = 0.028 m^3
Pressure of steam before cooling, p 1 = 80 bar
Temperature of steam before cooling = 350°C
Pressure of steam after cooling, p 2 = 50 bar