TITLE.PM5

200 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-6.pm5

(b)

(c)

Fig. 4.68. Polytropic process.

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

1

+ − 1

F

H

I

n K

= (h 2 – h 1 ) +
n
n− 1 (p^1 v^1 – p^2 v^2 ) ...(4.79)
In adiabatic process Q = 0 and if ∆s ≠ 0 then the process behaves like adiabatic process and
not isentropic. Such a process with steam will be a particular case of the law pvn = constant. The
index n in this case will be that particular index which will satisfy the condition :