860 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th16-1.pm5

Sol. Section 1 : Velocity of the gas, V = 300 m/s

Pressure, p 1 = 78 kN/m^2

Temperature, T 1 = 40 + 273 = 313 K

Section 2 : Pressure, p 2 = 117 kN/m^2

R = 287 J/kg K, γ = 1.4

Velocity of gas at section 2, V 2 :

Applying Bernoulli’s equations at sections 1 and 2 for 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 [Eqn. (16.7)]

But z 1 = z 2 , since the pipe is horizontal.

∴

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 sides, we get

γ

γρρ−

F

HG

I

KJ

−

F

HG

I

KJ

=−

122

1

1

2

2

2

2

1

pp V V^2

or,

γ

γρ ρ

ρ

−

F

HG

I

KJ

−×

F

HG

I

KJ

=−

1

1

22

1

1

2

2

1

1

2

2

1

pp^2

p

VV

∴

γ

γρ

ρ

− ρ

F

HG

I

KJ

−×

F

HG

I

KJ

=−

1

1

22

1

1

2

1

1

2

pp 22 12

p

VV

...(i)

For an adiabatic flow :

p 1

ρ 1

γ =

p 2

ρ 2 γ

or

p

p

1

2

1

2

=

F

HG

I

KJ

ρ

ρ

γ

or

ρ

ρ

1 γ

2

1

2

1

=

F

HG

I

KJ

p

p

Substituting the value of

ρ

ρ

1

2

in eqn (i), we get

γ

γρ

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
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
−
or,
γ
γρ
γ
γ
−
F
HG
I
KJ
−
F
HG
I
KJ
R
S
|
T
|
U
V
|
W
|
= −
−
1
1
2
1
1
2
1
1
pp 22 12
p
VV
...(ii)
At section 1 :
p 1
ρ 1
= RT 1 = 287 × 313 = 89831
p
p
2
1

117
78 = 1.5, and V^1 = 300 m/s