TITLE.PM5

FIRST LAW OF THERMODYNAMICS 199

dharm

/M-therm/Th4-6.pm5

Final pressure of steam, p 2 = 38 bar

From superheat tables, at 120 bar and 400°C

h 1 = 3051.3 kJ/kg and v 1 = 0.02108 m^3 /kg

Now, using the equation :

u = h – pv

∴ u 1 = 3051.3 –

120 10 0 02108

10

5

3

××.
= 2798.34 kJ/kg
Also, u 1 = ug at 38 bar = 2602 kJ/kg.
Since the cylinder is perfectly thermally insulated then no heat flows to or from the steam
during the expansion, the process therefore is adiabatic.
∴ Work done by the steam, W = u 1 – u 2
= 2798.34 – 2602 = 196.34 kJ/kg. (Ans.)
The process is shown on p-v diagram in Fig. 4.67, the shaded area representing the work
done.

5. Polytropic process. In this process, the steam follows the law pvn = constant. This
   process on p-v, T-s and h-s diagrams is shown in Fig. 4.68 (a), (b) and (c).
   The work done during this process is given by

W =

pv pv

n

11 2 2

1

−

−

F

HG

I

KJ N-m/kg

Applying the first law energy equation to non-flow process, we have

Q = ∆u + W

= (u 2 – u 1 ) +

pv pv

n

11 2 2

1

−

−

F

HG

I

KJ

= (h 2 – p 2 v 2 ) – (h 1 – p 1 v 1 ) +

pv p v

n

11 2 2

1

−

−

F

HG

I

KJ

(a)