76 ENGINEERING THERMODYNAMICSDharm
\M-therm/th3-1.p65— A line of constant entropy between two state points 2 and 3 defines the properties at all
points during an isentropic process between the two states.
Example 3.1. Calculate the dryness fraction (quality) of steam which has 1.5 kg of water
in suspension with 50 kg of steam.
Solution. Mass of dry steam, ms = 50 kg
Mass of water in suspension, mw = 1.5 kg
∴ Dryness fraction, x =
Mass of dry steam
Mass of dry steam mass of water in suspension+
= m
mms
sw+=
+50
50 1.5
= 0.971. (Ans.)☞ Example 3.2. A vessel having a volume of 0.6 m^3 contains 3.0 kg of liquid water and water
vapour mixture in equilibrium at a pressure of 0.5 MPa. Calculate :
(i)Mass and volume of liquid ;
(ii)Mass and volume of vapour.
Solution. Volume of the vessel, V = 0.6 m^3
Mass of liquid water and water vapour, m = 3.0 kg
Pressure, p = 0.5 MPa = 5 bar
Thus, specific volume, v = Vm =^0630 .. = 0.2 m^3 /kg
At 5 bar : From steam tables,
vfg = vg – vf = 0.375 – 0.00109 = 0.3739 m^3 /kg
We know that, v = vg – (1 – x) vfg, where x = quality of the vapour.
0.2 = 0.375 – (1 – x) × 0.3739
∴ (1 – x) = (0.375 0.2)
0.3739− = 0.468or x = 0.532
(i)Mass and volume of liquid, mliq. =? Vliq. =?
mliq. = m(1 – x) = 3.0 × 0.468 = 1.404 kg. (Ans.)
Vliq. = mliq. vf = 1.404 × 0.00109 = 0.0015 m^3. (Ans.)
(ii)Mass and volume of vapour, mvap. =? Vvap. =?
mvap. = m.x = 3.0 × 0.532 = 1.596 kg. (Ans.)
Vvap. = mvap. vg = 1.596 × 0.375 = 0.5985 m^3. (Ans.)
☞ Example 3.3. A vessel having a capacity of 0.05 m^3 contains a mixture of saturated water
and saturated steam at a temperature of 245°C. The mass of the liquid present is 10 kg. Find the
following :
(i)The pressure, (ii) The mass,
(iii)The specific volume, (iv)The specific enthalpy,
(v)The specific entropy, and (vi)The specific internal energy.
Solution. From steam tables, corresponding to 245°C :
psat = 36.5 bar, vf = 0.001239 m^3 /kg, vg = 0.0546 m^3 /kg
hf = 1061.4 kJ/kg, hfg = 1740.2 kJ/kg, sf = 2.7474 kJ/kg K
sfg = 3.3585 kJ/kg K.