884 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th16-2.pm5

Substituting the value of γR = C

T

2

in eqn. (i), we get

C

T

dT

V

dV

V

2

()γ− 1 ×+^2 = 0

or,
1
()γ− 1 2

×+

M

dT

T

dV

V = 0

Q M V

C

FHG = IKJ ...(16.34)

From the Mach number relationship

M =

V

γRT

(where γRT = C)

dM

M =

dV

V –

1

2

dT

T ...(16.35)

Substituting the value of

dT

T from eqns. (16.34) in eqn. (16.35), we get

dM

M =

dV

V –

1

2

L−×−

NM

O

QP

dV

V

()γ 1 M^2

=

dV

V +

1

2

dV

V × (γ – 1) M

2

or,

dM

M =

dV

V^

1

1

2

L + − 2

NM

O

QP

γ

M or dV

V =

1

1 1

2

+F −^2

HG

I

KJ

L

NM

O

QP

γ M^

dM

M ...(16.36)

Since the quantity within the bracket is always positive, the trend of variation of velocity
and Mach number is similar. For temperature variation, one can write

dT

T =

−−

+FHG − IKJ

L

N

M

M

M

M

O

Q

P

P

P

P

()γ

γ

1

1 1

2

2

2

M

M

dM

M ...(16.37)

Since the right hand side is negative the temperature changes follow an opposite trend to
that of Mach number. Similarly for pressure and density, we have

dp

p =

−

+ −

L

N

M

M

M

O

Q

P

P

P

γ

γ

M

M

2

1 1 2

2

dM

M ...(16.38)

and, dρ
ρ

−

+F −

HG

I

KJ

L

N

M

M

M

M

O

Q

P

P

P

P

M

M

2

1 1 2

2

γ^

dM

M ...(16.39)

For changes in area, we have

dA

A =

−−

+ −

L

N

M

M

M

O

Q

P

P

P

()1

1 1

2

2

2

M

γ M

dM

M ...(16.40)