COMPRESSIBLE FLOW 861

dharm

\M-therm\Th16-1.pm5

Substituting the values in eqn. (ii), we get

1.4

1.4−

F

HG

I

1 KJ

× 89831 1

1

−

R

S

|

T|

U

V

|

W|

−

()1.5

1.4

1.4 = V^2

2

2

-^300
2

2

314408.5 (1 – 1.1228) = V^2

2

2

• 45000 or – 38609.4 = V^2

2

2

• 45000
  or, V 22 = 12781.2 or V 2 = 113.05 m/s. (Ans.)
  Example 16.2. In the case of air flow in a conduit transition, the pressure, velocity and
  temperature at the upstream section are 35 kN/m^2 , 30 m/s and 150°C respectively. If at the
  downstream section the velocity is 150 m/s, determine the pressure and the temperature if the
  process followed is isentropic. Take γ = 1.4, R = 290 J/kg K.
  Sol. Section 1 (upstream) : Pressure, p 1 = 35 kN/m^2 ,
  Velocity, V 1 = 30 m/s,
  Temperature, T = 150 + 273 = 423 K
  Velocity, V 2 = 150 m/s
  R = 290 J/kg K, γ = 1.4
  Section 2 (downstream) :
  Pressure, p 2 :
  Applying Bernoulli’s equation at sections 1 and 2 for isentropic (reversible adiabatic) process,
  we have
  γ
  γρ−

F

HG

I

KJ

+

12

1

1

1

p^2

g

V

g + z^1 =

γ

γρ−

F

HG

I

KJ

+

12

2

2

2

p^2

g

V

g + z^2

Assuming z 1 = z 2 , we have

γ

γρ−

F

HG

I

KJ

+

12

1

1

1

p^2

g

V

g =

γ

γρ−

F

HG

I

KJ

+

12

2

2

2

p^2

g

V

g

Cancelling ‘g’ on both the sides, and rearranging, we get

γ

γρ

ρ

− ρ

F

HG

I

KJ

−×

F

HG

I

1 KJ

(^11)
1
2
1
1
2
pp
p =
VV 22 12
22
− ...(i)
For an isentropic flow :
p 1
ρ 1
γ =
p 2
ρ 2
γ or
p
p
1
2

ρ
ρ
γ
1
2
F
HG
I
KJ
or
ρ
ρ
1
2
= p
p
1
2
1
F
HG
I
KJ
γ
Substituting the value of
ρ
ρ
1
2
in eqn. (i), we have
γ
γρ ρ
γ
−
F
HG
I
KJ
−×
F
HG
I
KJ
R
S
|
T
|
U
V
|
W
(^1) |
(^11)
1
2
1
1
2
1
pp
p
p = VV 22 12
22
−
γ
γρ
γ
−
F
HG
I
KJ
−
F
HG
I
KJ
R
S
|
T
|
U
V
|
W
|
−
1
(^11)
1
2
1
1 1
pp
p

VV 22 12
22
−