TITLE.PM5

188 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-5.pm5

x 1

1

a

T 1

T 2

p= p 12

2

h

s
(c)
Fig. 4.59. Constant pressure process.
If the initial condition of steam is wet and final condition is superheated, then
Q = (u 2 + pvsup 2 ) – (u 1 + p. x 1 vg 1 )
= (h 2 – h 1 ) ...(4.73)
here h 1 and h 2 are the actual enthalpies of steam per kg before and after heating.
The heat added during the constant pressure process is equal to the change in enthalpy of
steam during the process. When the steam is wet before heating and becomes superheated after
heating the work done,

W = p (vxvsup 21 − 1 g) ...(4.74)
Example 4.52. 0.08 kg of dry steam is heated at a constant pressure of 2 bar until the
volume occupied is 0.10528 m^3. Calculate :
(i)Heat supplied ;
(ii)Work done.
Solution. Mass of steam, m = 0.08 kg
Pressure of steam, p = 2 bar
Volume occupied after heating = 0.10528 m^3
Initially the steam is dry saturated at 2 bar, hence
h 1 = hg (at 2 bar) = 2706.3 kJ/kg
Finally the steam is at 2 bar and the specific volume is given by

v 2 0

08

= .10528
0.
= 1.316 m^3 /kg
Hence the steam is superheated finally (since the value of vg at 2 bar = 0.885 m^3 /kg). From
superheat tables at 2 bar and 1.316 m^3 /kg the temperature of steam is 300°C, and the enthalpy,
h 2 = 3071.8 kJ/kg.