TITLE.PM5

FIRST LAW OF THERMODYNAMICS 201

dharm

/M-therm/Th4-6.pm5

Q = 0
∴ 0 = ∆u + W
∴ W = – ∆u = – (u 2 – u 1 ) = (u 1 – u 2 )
i.e., W = (u 1 – u 2 ) ...(4.80)
Adiabatic process (not reversible) is also a polytropic process with an index n. The appropri-
ate value of n for adiabatic compression of steam are
n = 1.13 for wet steam
n = 1.3 for superheated steam
When the initial condition and end condition are both in wet region then p 1 v 1 n = p 2 v 2 n
reduces to :
pxv 11 () ( ).g 12 n=p xv2 2g n
As p 1 , x 1 , n and p 2 are specified the value of x 2 can be calculated.
When the end condition is superheated, then
pxv 11 () ().g 12 n=pv 2 sup n
Solving for v 2 , then using
v
v

T

g T

sup

s

2

2

2

2

=

Tsup 2 can be calculated. Knowing Ts 2 and Tsup all properties at the end condition can be

calculated.

+Example 4.58. In a steam engine the steam at the beginning of the expansion process is

at 7 bar, dryness fraction 0.98 and expansion follows the law pv1.1 = constant, down to a pressure

of 0.34 bar. Calculate per kg of steam :

(i)The work done during expansion ;

(ii)The heat flow to or from the cylinder walls during the expansion.

Solution. Initial pressure of steam, p 1 = 7 bar = 7 × 10^5 N/m^2

Dryness fraction, x 1 = 0.98

Law of expansion, pv1.1 = constant

Final pressure of steam, p 2 = 0.34 bar = 0.34 × 10^5 N/m^2.

At 7 bar : vg = 0.273 m^3 /kg

∴ v 1 = x 1 vg = 0.98 × 0.273 = 0.267 m^3 /kg

Also, p 1 v 1 n = p 2 v 2 n

i.e.,

v

v

p

p

n

2

1

1

2

1

=F

HG

I

KJ

/

∴ v^2

1

1

0

7

.267 0 .34

=F 1.

HG

I

KJ or^ v^2 = 0.267

7

0

1

1

.34

F 1.

HG

I

KJ = 4.174 m

(^3) /kg.
(i)Work done by the steam during the process :
W pv pv
n
= −
−
= ×× − ××
−
11 22
55
1
7100 0 104
11
.267 .34 .174
()1.