Page 57 - DJJ20063- Thermodynamics 1

DJJ20063- Thermodynamics 1

Example 2.10

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0.9 m of dry saturated steam at 225 kN/m is contained in a rigid cylinder. If it is

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cooled at constant volume process until the pressure drops to180 kN/m ,

determine the following:
a) mass of steam in the cylinder
b) dryness fraction at the final state
Sketch the process in the form of a P-v diagram.

Solution to Example 2.10
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Data: V1 = 0.9 m , P1 = 225 kN/m = 2.25 bar, P2 = 180 kN/m = 1.80 bar

a) Firstly, find the specific volume of dry saturated steam at 2.25 bar. Note
that the pressure 2.25 bar is not tabulated in the steam tables and it is

necessary to use the interpolation method.

From the Steam Tables,
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vg at 2.2 bar = 0.8100 m /kg
3
vg at 2.3 bar = 0.7770 m /kg

vg1 at 2.25 bar,
v − . 0 8100 . 0 7770 − . 0 8100
1 g =
. 2 25 − . 2 20 . 2 30 − . 2 20

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vg1 = 0.7935 m /kg

V
 Mass of steam in cylinder, m = 1 (m x kg/m )
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3
v g 1
0 9
.
=
0 7935
.

= 1.134 kg

48 | P a g e