TITLE.PM5

COMPRESSIBLE FLOW 893

dharm

\M-therm\Th16-2.pm5

This equation is ‘known as Rankine Line Equation.

Now combining continuity and energy equations [refer eqns. (i) and (iii)], we get

γ

γ− 1

p 1

ρ 1

F

HG

I

KJ

+

()ρ

ρ

112

212

V

=

γ

γ− 1

p 2

ρ 2

F

HG

I

KJ

+

()ρ

ρ

22

2

222

V

...(16.43)

This equation is called Fanno Line Equation.

Further combining eqns. (i), (ii) and (iii) and solving for

p

p

2

1

, we get

p

p

2

1

=

γ

γ

ρ

ρ

γ

γ

ρ

ρ

+

−

F

HG

I

KJ

−

+

−

F

HG

I

KJ

−

1

1

1

1

1

2

1

2

1

...(16.44)

Solving for density ratio

ρ

ρ

2

1

, the same equations yield

ρ

ρ

2

1

= V

V

1

2

=

1 1

1

1

1

2

1

2

1

+ +

−

F

HG

I

KJ

+

−

F

HG

I

KJ

+

γ

γ

γ

γ

p

p

p

p

...(16.45)

The eqns. (16.44) and (16.45) are called Ranking-Hugoniot equations.

One can also express p

p

2

1

, V

V

2

1

, ρ

ρ

2

1

and

T

T

2

1

in terms of Mach number as follows :

p

p

2

1

=

21

1

1

γγ^2

γ

M −−

+

()

() ...(16.46)

V

V

1

2

= ρ

ρ

2

1

=

()

()

γ

γ

+

−+

1

12

1

2

12

M

M ...(16.47)

T

T

2

1

=

[( ) ] [ ( )]

()

γγγ

γ

−+ −−

+

122 1

1

1

2

1

2

2 12

MM

M ...(16.48)

By algebraic manipulation the following equation between M 1 and M 2 can be obtained.

M 22 =

()

()

γ

γγ

−+

−−

12

21

12

12

M

M

...(16.49)

Example 16.18. For a normal shock wave in air Mach number is 2. If the atmospheric
pressure and air density are 26.5 kN/m^2 and 0.413 kg/m^3 respectively, determine the flow condi-
tions before and after the shock wave. Take γ = 1.4.
Sol. Let subscripts 1 and 2 represent the flow conditions before and after the shock wave.
Mach number, M 1 = 2
Atmospheric pressure, p 1 = 26.5 kN/m^2
Air density, ρ 1 = 0.413 kg/m^3