TITLE.PM5

196 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-5.pm5

Fig. 4.65

(i)Heat supplied :

Now, heat supplied, Q = shaded area = T(s 2 – s 1 )

= 584(7.163 – 5.619) = 901.7 kJ/kg. (Ans.)

(ii)Work done :

To find work done, applying non-flow energy equation,

Q = (u 2 – u 1 ) + W

or W = Q – (u 2 – u 1 )

From steam tables at 100 bar, dry saturated,

u 1 = ug = 2545 kJ/kg

At 10 bar 311°C, interpolating,

u 2 = 2794 +

311 300

350 300

−

−

F

H

I

K (2875 – 2794)

i.e., u 2 = 2811.8 kJ/kg
Then, W = Q – (u 2 – u 1 )
= 901.7 – (2811.8 – 2545) = 634.9 kJ/kg
Hence, work done by the steam = 634.9 kJ/kg. (Ans.)

4. Reversible Adiabatic or Isentropic Process. Fig. 4.66 (a), (b) and (c) shows the
   isentropic process on p-v, T-s and h-s diagrams respectively.
   Let us consider that the process is non-flow reversible adiabatic. Now applying first law
   energy equation, we have

Q = ∆u + 1 pdv.

2

z = (u^2 – u^1 ) + W

As for adiabatic process, Q = 0

∴ W = (u 1 – u 2 ) ...(4.77)