TITLE.PM5

108 ENGINEERING THERMODYNAMICS

dharm

M-therm/th4-1.pm5

In a constant pressure process, the work done by the fluid,

W = p(V 2 – V 1 )

= mR(T 2 – T 1 )

Q pV mRT

pV mRT

pp p

11 1

22 2

12

=

=

==

L

N

M

M

M

O

Q

P

P

in this caseP

On substituting

Q = mcv (T 2 – T 1 ) + mR (T 2 – T 1 ) = m(cv + R) (T 2 – T 1 )

But for a constant pressure process,

Q = mcp (T 2 – T 1 )

By equating the two expressions, we have

m(cv + R)(T 2 – T 1 ) = mcp(T 2 – T 1 )

∴ cv + R = cp

or cp – cv = R ...(4.23)
Dividing both sides by cv, we get
c
c

p

v

• 1 =

R

cv

∴ cv =
R
γ− 1 ...[4.23 (a)]
(where γ = cp/cv)
Similarly, dividing both sides by cp, we get

cp =

γ

γ

R

− 1 ...[4.23 (b)]

In M. K.S. units

In SI units the value of is unity.

:;

()

,

()

ccR

J

c R

J

c R

J

J

p−=vv= − p= −

L

N

M

M

M

O

Q

P

P

P

γ

γ

11 γ

4.8.5. Enthalpy

— One of the fundamental quantities which occur invariably in thermodynamics is the

sum of internal energy (u) and pressure volume product (pv). This sum is called

Enthalpy (h).

i.e., h = u + pv ...(4.24)

— The enthalpy of a fluid is the property of the fluid, since it consists of the sum of a

property and the product of the two properties. Since enthalpy is a property like inter-

nal energy, pressure, specific volume and temperature, it can be introduced into any

problem whether the process is a flow or a non-flow process.

The total enthalpy of mass, m, of a fluid can be

H = U + pV, where H = mh.

For a perfect gas,

Referring equation (4.24),

h = u + pv