858 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th16-1.pm5

Differentiating the above equation, we get

d(ρAV) = 0 or ρd(AV) + AVdρ = 0

or ρ(AdV + VdA) + AVdρ = 0 or ρAdV + ρVdA + AVdρ = 0

Dividing both sides by ρAV and rearranging, we get

ddA

A

dV

V

ρ

ρ

++ = 0 ...(16.2)

Eqn. (16.2) is also known as equation of continuity in differential form.

16.2.2. Momentum equation

The momentum equation for compressible fluids is similar to the one for incompressible
fluids. This is because in momentum equation the change in momentum flux is equated to force
required to cause this change.
Momentum flux = mass flux × velocity = ρAV × V
But the mass flux i.e., ρAV = constant ...By continuity equation
Thus the momentum equation is completely independent of the compressibility effects and
for compressible fluids the momentum equation, say in X-direction, may be expressed as :
ΣFx = (ρAVVx) 2 – (ρAVVx) 1 ...(16.3)

16.2.3. Bernoulli’s or energy equation

As the flow of compressible fluid is steady, the Euler equation is given as :

dp

ρ + VdV + gdz = 0 ...(16.4)

Integrating both sides, we get

z^

dp

ρ + z^ VdV + z^ gdz = constant

or z dp

ρ

+ V

2

2

+ gz = constant ...(16.5)

In compressible flow since ρ is not constant it cannot be taken outside the integration sign.

In compressible fluids the pressure (p) changes with change of density (ρ), depending on the type of

process. Let us find out the Bernoulli’s equation for isothermal and adiabatic processes.

(a) Bernoulli’s or energy equation for isothermal process :

In case of an isothermal process,

pv = constant or

p

ρ = constant = c^1 (say)

(where v = specific volume = 1/ρ)

∴ρ =

p

c

Hence (^) z
dp
ρ = zzz

dp
pc
cdp
p
c
dp
/ 1 p
(^11) = c
1 loge p =
p
ρ loge p^
Q c
p
1 =
F
HG
I
ρKJ
Substituting the value of z
dp
p in eqn. (16.5), we get
p
ρ loge p +
V^2
2

• gz = constant