COMPRESSIBLE FLOW 859

dharm

\M-therm\Th16-1.pm5

Dividing both sides by g, we get

p

ρg loge p +

V

g

2

2 + z = constant ...(16.6)

Eqn. (16.6) is the Bernoulli’s equation for compressible flow undergoing isothermal process.

(b) Bernoulli’s equation for adiabatic process :

In case of an adiabatic process,

pvγ = constant or

p

ργ

= constant = c 2 (say)

∴ργ =

p

c 2 or ρ =

p

c 2

F 1/

HG

I

KJ

γ

Hence z

dp

ρ =

dp

pc

c

p

dp c p dp

(/ )

() ()

2

1/^2

1/

1/^2

(^1) 1/ 1/
γ
γ
γ
== = γγ
zzz
−
= ()c 2 1/ pcp() () () ()cp
(^11)
2
1/
1
2 1/
1
1 1 1 1
γ γγ
γ
γ γ γγ
γ
γ
γ
γ
γ
−+ FHG − IKJ F −
HG
I
KJ
F−+
HG
I
KJ
L
N
M
M
M
M
M
O
Q
P
P
P
P
P

F −
HG
I
KJ

−
= γ
γ ργ
γ γ
γ
−
F
HG
I
KJ
F
HG
I
KJ
F −
HG
I
KJ
1
1/^1
p ()p FQ c 2 = p
HG
I
ργKJ

γ
γ
ρ
γ
γ γ
γ
γ
−
F
HG
I
KJ
F
H
G
G
G
I
K
J
J
× J
F −
HG
I
KJ
1
1/
1
1
p ()p

γ
γρ
γ
γρ
γ
γ
γ
−
F
HG
I
KJ

−
F
HG
I
KJ
FHG + − IKJ
11
11
()pp
Substituting the value of z
dp
p in eqn. (16.6), we get
γ
γρ−
F
HG
I
KJ

• 12
  pV^2


• gz = constant
  Dividing both sides by g, we get
  γ
  γρ−
  F
  HG
  I
  KJ


• 
  

  12
  p^2
  g
  V
  g + z = constant ...(16.7)
  Eqn. (16.7) is the Bernoulli’s equation for compressible flow undergoing adiabatic process.
  Example 16.1. A gas with a velocity of 300 m/s is flowing through a horizontal pipe at a
  section where pressure is 78 kN/m^2 absolute and temperature 40°C. The pipe changes in diam-
  eter and at this section, the pressure is 117 kN/m^2 absolute. Find the velocity of the gas at this
  section if the flow of the gas is adiabatic. Take R = 287 J/kg K and γ = 1.4.
  (Punjab University)