Page 34 - DJJ20063- Thermodynamics 1

DJJ20063- Thermodynamics 1

At point A, x = 0

Sat. steam At point B, x = 1
Sat. steam
Sat. liquid Between point A and B, 0  x  1.0
Sat. liquid
P x = 0.2 x = 0.8

Note that for a saturated liquid, x = 0;
and that for dry saturated steam, x = 1.

A B

ts

vf vg v

Figure 2.2.4-2 P-v diagram showing the location point of the dryness
fraction

2.3 Analyze the state of steam using the properties of pure substances

2.3.1 Calculate saturation temperature and pressure, specific enthalpy, specific

volume, dryness fraction.

Specific volume
For a wet steam, the total volume of the mixture is given by the volume of liquid present

plus the volume of dry steam present.

Therefore, the specific volume is given by,

volume of a liquid + volume of dry steam
v =
total mass of wet steam

Now for 1 kg of wet steam, there are (1 – x) kg of liquid and x kg of dry steam, where x is

the dryness fraction as defined earlier. Hence,

v = vf(1 – x) + vgx

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