COMPRESSIBLE FLOW 883

dharm

\M-therm\Th16-2.pm5

∴ρ 1 = ρ 2 p

p

2

1

1

F

HG

I

KJ

−γ

or ρ 2 2

1

1

1

γ

γ

γγ

+

F

HG

I

KJ

− ×

or ρ 2 2

1

1

1

γ

γ

+

F

HG

I

KJ

−

Substituting the values of p 1 and ρ 1 in the above eqn. (16.31), we get

V 2 =

2

1

2

1

12

(^21)
1
2
1
γ^1
γγ ργ
γ
γγ

• F
  HG
  I
  KJ
  ×


• 
  

  F
  HG
  I
  KJ
  ××

  
  


• 
  

  F
  HG
  I
  KJ
  R
  S
  |
  T
  |
  U
  V
  |
  W
  |
  −−
  p
  =^2
  1
  2
  1
  2
  2
  1
  1
  γ^1
  γργ
  γ
  γγ

  
  


• 
  

  F
  HG
  I
  KJ
  ××

  
  


• 
  

  F
  HG
  I
  KJ
  p − + −

  
  

  2
  1
  2
  1
  2
  2
  1
  γ
  γργ+
  F
  HG
  I
  KJ
  ×

  
  


• 
  

  F
  HG
  I
  KJ
  −
  p
  or V 2 =
  2
  1
  1
  2
  2
  2
  γ
  γρ
  γ

  
  


• 
  

  F
  HG
  I
  KJ
  × F +
  HG
  I
  KJ
  p

  
  

  γ
  ρ
  p 2
  2
  = C 2
  i.e., V 2 = C 2 ...(16.33)
  Hence the velocity at the outlet of nozzle for maximum flow rate equals sonic velocity.

  
  

16.9. Variables of Flow in Terms of Mach Number

In order to obtain relationship involving change in velocity, pressure, temperature and
density in terms of the Mach number use is made of the continuity, perfect gas, isentropic flow and
energy equations.
For continuity equation, we have
ρAV = constant
Differentiating the above equation, we get
ρ[AdV + VdA] + AVdρ = 0
Dividing throughout by ρAV, we have
dV
V

dA

A

++dρ

ρ

= 0

From isentropic flow, we have p

ργ

= constant or dp

p

= γ dp

p

For perfect gas, we have p = ρRT or dp

p

=

dρ

ρ

+

dT

T

From energy equation, we have cpT +

V^2

2 + constant

Differentiating throughout, we get

cpdT + VdV = 0 or

γ

γ

R

−

F

HG

I

(^1) KJ
dT + VdV = 0 Q c
R
p= −
F
HG
I
KJ
γ
γ 1
or,
γ
γ
R
− 1
dT
V^2

• dV
  V = 0 ...(i)
  Also, sonic velocity, C = γRT ∴ γR =
  C
  T
  2