TITLE.PM5

84 ENGINEERING THERMODYNAMICS

Dharm

\M-therm/th3-2.p65

(i)Dryness fraction, x :
At 7 bar, hg = 2762 kJ/kg, hence since the actual enthalpy is given as 2550 kJ/kg, the steam
must be in the wet vapour state.
Now, using the equation,
h = hf + xhfg
∴ 2550 = 697.1 + x × 2064.9

i.e., x=2550 697.1−
2064.9
= 0.897
Hence, dryness fraction = 0.897. (Ans.)
(ii)Specific volume of wet steam,
v = xvg = 0.897 × 0.273 = 0.2449 m^3 /kg. (Ans.)
(iii)Specific internal energy of wet steam,
u = (1 – x)uf + xug
= (1 – 0.897) × 696 + 0.897 × 2573
= 2379.67 kJ/kg. (Ans.)
Example 3.15. Steam at 120 bar has a specific volume of 0.01721 m^3 /kg, find the tempera-
ture, enthalpy and the internal energy.

Solution. Pressure of steam, p = 120 bar
Specific volume, v = 0.01721 m^3 /kg
(i)Temperature :
First it is necessary to decide whether the steam is wet, dry saturated or superheated.
At 120 bar, vg = 0.0143 m^3 /kg, which is less than the actual specific volume of
0.01721 m^3 /kg, and hence the steam is superheated.
From the superheat tables at 120 bar, the specific volume is 0.01721 m^3 /kg at a tempera-
ture of 350 °C. (Ans.)
(ii)Enthalpy :
From the steam tables the specific enthalpy at 120 bar, 350°C,
h = 2847.7 kJ/kg. (Ans.)
(iii)Internal energy :
To find internal energy, using the equation,
u = h – pv

= 2847.7 –

120 10 001721

10

5

3

××.

= 2641.18 kJ/kg. (Ans.)
Example 3.16. Steam at 140 bar has an enthalpy of 3001.9 kJ/kg, find the temperature,
the specific volume and the internal energy.
Solution. Pressure of steam, p = 140 bar
Enthalpy of steam, h = 3001.9 kJ/kg
(i)Temperature :
At 140 bar, hg = 2642.4 kJ, which is less than the actual enthalpy of 3001.9 kJ/kg, and
hence the steam is superheated.
From superheat tables at 140 bar, h = 3001.9 kJ/kg at a temperature of 400 °C. (Ans.)