TITLE.PM5

190 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-5.pm5

6.55 = 2.0457 + x 1 × 4.6139

∴ x 1 6 0457

4

=

.55 2.−

.6139

= 0.976

Fig. 4.61
Now, initial enthalpy (at 8 bar),
hh xh 11 =+f 11 fg
= 720.9 + 0.976 × 2046.5 = 2718.28 kJ/kg
Final enthalpy, h 2 : At state 2 the steam is at 200°C at 8 bar and is therefore, superheated.
From superheated tables, h 2 = 2839.3 kJ/kg
Now, Q = h 2 – h 1 = 2839.3 – 2718.28 = 121.02 kJ/kg
i.e., Heat supplied = 121.02 kJ/kg. (Ans.)
The T-s diagram showing the process is given in Fig. 4.61, the shaded area representing the
heat flow.

3. Constant Temperature or Isothermal Expansion. Fig. 4.62 (a), (b) and (c) shows
   the constant temperature or isothermal expansion on p-v, T-s and h-s diagrams respectively.
   In the wet region, the constant temperature process is also a constant pressure process
   during evaporation and as well as condensation. When the steam becomes saturated it behaves
   like a gas and constant temperature process in superheated region becomes hyperbolic (pv = constant).
   When the wet steam is heated at constant temperature till it becomes dry and saturated,
   then the heat transfer (Q) is given by :
   Q = h 2 – h 1
   and work done, W = p 1 ()vxvgg 21 − 1
   = pvg 1 (1 – x 1 )
   [Q vg 2 = vg 1 as pressure remains constant during this process]