TITLE.PM5

138 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-3.pm5

(ii)Index of expansion, n :

If the work done by the polytropic process is the same,

W1–2–3 = W1–3 =

pV pV

n

11 33

1

−

−

85454 =

4 10 0 02 10 0

1

336

1

××− ××^55

−

=^5

−

.2 1. .732

()nn

∴ n= 854545336 + 1

i.e., n = 1.062

Hence, value of index = 1.062. (Ans.)

Example 4.24. The following is the equation which connects u, p and v for several gases

u = a + bpv

where a and b are constants. Prove that for a reversible adiabatic process,

pvγ = constant, where γ =

b1

b

+

.

Solution. Consider a unit mass.

For a reversible adiabatic process, first law gives

0 = du + pdv

∴ dudv = – p ...(i)

Also, u = a + bpv

∴ dudv = da bpv()dv+ = bv dpdv + bp

= b pv

dp

+ dv

F

H

I

. K ...(ii)
Equating (i) and (ii), we get

b pv

dp

+ dv

F

H

I

. K = – p

bp + b. v. dpdv = – p

bp + p + bv. dp

dv

= 0

p(b + 1) + bv. dpdv = 0

Multiplying both sides by dv

bpv

, we get

b

b

dv

v

dp

p

F +

HG

I

KJ +

1

= 0

or
dp
p

b

b

dv

+ v

F +

H

I

K

1

= 0

d(loge p) +

b

b

F +

H

I

K

1

d(loge v) = 0