Page 41 - DJJ20063- Thermodynamics 1

DJJ20063- Thermodynamics 1

Example 2.1

For a steam at 20 bar with a dryness fraction of 0.9, calculate the
a) specific volume
b) specific enthalpy

c) specific internal energy

An extract from the steam tables

p ts vg uf ug hf hfg hg sf sfg sg

20 212.4 0.09957 907 2600 909 1890 2799 2.447 3.893 6.340

a) Specific volume (v),

v = xvg

= 0.9(0.09957)
3
= 0.0896 m /kg

b) Specific enthalpy (h),

h = hf + xhfg
= 909 + 0.9(1890)

= 2610 kJ/kg

c) Specific internal energy (u),
u = uf + x( ug -uf )

= 907 + 0.9(2600 - 907)

= 2430.7 kJ/kg

P

bar
x = 0.9

20

o
ts = 212.4 C

3
v m /kg
uf v vg
hf u ug
sf h hg
s sg
32 | P a g e