COMPRESSIBLE FLOW 879

dharm

\M-therm\Th16-1.pm5

or V 2 =^2
1

1

1

2

2

γ

()γρρ−

−

F

HG

I

KJ

pp

or V 2 =^2
1

(^11)
1
2
2
1
1
γ
γρ ρ
ρ
()−
−−
F
HG
I
KJ
pp
p
...(1)
For 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
...(i)
Substituting the value of
ρ
ρ
1
2
in eqn. (1), we get
V 2 =
2
1
(^11)
1
2
1
1
2
1
γ
γρ
γ
()−
−×
F
HG
I
KJ
L
N
M
M
M
O
Q
P
P
P
pp
p
p
p =
2
1
(^11)
1
2
1
1 1
γ
γρ
γ
()−
−
F
HG
I
KJ
L
N
M
M
M
O
Q
P
P
P
−
pp
p
or V 2 =
2
1
(^11)
1
2
1
1
γ
γρ
γ
γ
()−
−F
HG
I
KJ
L
N
M
M
M
O
Q
P
P
P
−
pp
p ...(16.27)
The mass rate of flow of the compressible fluid,
m = ρ 2 A 2 V 2 , A 2 being the area of the nozzle at the exit
= ρ 2 A 2
2
1
(^11)
1
2
1
1
γ
γρ
γ
γ
()−
−
F
HG
I
KJ
L
N
M
M
M
O
Q
P
P
P
−
pp
p , [substituting V^2 from eqn. (16.27)]
or m = A 2
2
1
(^11)
1
2
(^22)
1
1
γ
γρ
ρ
ρ
γ
γ
()−
×−F
HG
I
KJ
L
N
M
M
M
O
Q
P
P
P
−
pp
From eqn. (i), we have ρ 2 =
ρ ρ
γ
1 γ
12 1/
1 2
1
1
(/)pp
p
p

F
HG
I
KJ
∴ρ 22 = ρ 12 p
p
2
1
2
F
HG
I
KJ
γ
Substituting this value in the above equation, we get
m = A 2
2
1
(^11)
1
12 2
1
2
2
1
1
γ
γρ
ρ
γ
γ
γ
−
×
F
HG
I
KJ
−
F
HG
I
KJ
L
N
M
M
M
O
Q
P
P
P
−
pp
p
p
p
= A 2 2
1 11
2
1
2
2
1
12
γ
γ
ρ
γ γγγ
−
F
HG
I
KJ
−
F
HG
I
KJ
L
N
M
M
M
O
Q
P
P
P
− +
p p
p
p
p
/