Flow of Compressible Fluid

8.1 Introduction
All fluids are to some degree compressible, compressibility is sufficiently great to
affect flow under normal conditions only for a gas. If the pressure of the gas does not
change by more than about 20%, [or when the change in density more than 5-10 %] it is
usually satisfactory to treat the gas as incompressible fluid with a density equal to that at
the mean pressure.
When compressibility is taken into account, the equations of flow become more
complex than they are for an incompressible fluid.
The flow of gases through orifices, nozzles, and to flow in pipelines presents in all
these cases, the flow may reach a limiting maximum value which independent of the
downstream pressure (P2); this is a phenomenon which does not arise with
incompressible fluids.
8.2 Velocity of Propagation of a Pressure Wave
The velocity of propagation is a function of the bulk modulus of elasticity (ε),
where;
increase of stress within the fluid dP
ε= =
resulting volumetric strain − dυ / υ
dP
⇒ ε = −υ
dυ
where, υ: specific volume (υ = 1/ρ).
Suppose a pressure wave to be transmitted at a velocity uw over a distance dx in a
fluid of cross-sectional area A, from section  to section  as shown in Figure;
Now imagine the pressure wave to be brought to rest by causing the fluid to flow
at a velocity uw in the opposite direction.
From conservation of mass law; dx
m& 1 = m& 2
ρuwA = (ρ+dρ)(uw +duw)A uw (uw + duw)
P (P + dP)
uw ( u w + duw )
⇒ A= A
υ (υ + dυ) uw
u m& m&  
and m& = w A ⇒ uw = υ ⇒ duw = dυ
υ A A
Newton’s 2nd law of motion stated that “The rate of change in momentum of fluid
is equal to the net force acting on the fluid between sections  and .
Thus;
m&
m& [(uw − du w ) − u w ] = A[ P − ( P + dP ) ] ⇒ du = −dP
A w
dP ⎛ m& ⎞
2
m& m& m&
but duw = dυ ⇒ ( dυ) = − dP ⇒− =⎜ ⎟
A A A dυ ⎝ A ⎠
ε ⎛ m& ⎞
2
dP ε
we have − = ⇒ = ⎜ ⎟ = G2
dυ υ υ ⎝ A⎠
 ε ⎛ uw A / υ⎞ ⎛ uw ⎞
2 2

⇒ =⎜ ⎟ =⎜ ⎟ ⇒ uw2 = υε
υ ⎝ A ⎠ ⎝ υ⎠
∴ uw = υε
For ideal gases
Pυκ = const . where, κ = 1.0 for isothermal conditions
cp
κ=γ for isentropic conditions, γ=
cv
υk
dP P
d (Pυk ) = 0 ⇒ υk dP + Pkυk −1 dυ = 0 ⇒ υk dP = − kP dυ ⇒ = −k
υ dυ υ
⎛ dP ⎞
⇒ − υ⎜ ⎟ = kP = ε ∴ u w = kPυ
⎝ dυ ⎠
- For isothermal conditions k = 1 ⇒ uw = Pυ
- For isentropic (adiabatic) conditions k = γ ⇒ u w = γ Pυ
The value of uw is found to correspond closely to the velocity of sound in the
fluid and its correspond to the velocity of the fluid at the end of a pipe uder conditions
of maximum flow.
Mach Number
Is the ratio between gas velocity to sonic velocity,
u
Ma =
uw
where, Ma > 1 supersonic velocity
Ma = 1 sonic velocity
Ma < 1 subsonic velocity
8.3 General Energy Equation for Compressible Fluids
Let E the total energy per unit mass of the fluid where,
E=Internal energy (U)+Pressure energy (Pυ)+Potential energy(zg)+Kinetic energy (u2/2)
Assume the system in the Figure; q
Energy balance
E1 + q = E2 + Ws E1 E2
⇒E –E =q–W
2 1 s
System
⇒ ΔU + Δ(Pυ) + gΔ(z) + Δ(u2/2) = q – Ws
[α = 1 for compressible fluid since it almost in turbulent flow] Ws
but ΔH = ΔU + Δ(Pυ)
⇒ ΔH + gΔ(z) + Δ(u2/2) = q – Ws
For irreversible process
dW = Pdυ – dF -------useful work
dH + gd(z) + ud(u) = dq – dWs dU = dq – dW -------closed system
but, dH = dU + d(Pυ)
dH = dq + dF + υdP = dq – dW + d(Pυ)
where, = dq – (Pdυ – dF) + d(Pυ)
dF: amount of mechanical energy converted = dq – Pdυ + dF + Pdυ + υdP
⇒ dH = dq + dF + υdP
into heat
 ⇒ u du + g dz + υ dP +dWs + dF = 0
P
Δu2 2

+ g Δz + ∫ υdP + Ws + F = 0
∴
General equation of energy apply to any
2 P 1
type of fluid

 For compressible fluid flowing through (dl) of pipe of constant area

L
Substitute these equations into  
equation (*), to give
G υ (G dυ) + g dz + υ dP +dWs + 4Φ (dl/d) (G υ)2 = 0
á For horizontal pipe (dz = 0), and no shaft work (Ws =0)
⇒ G2 υ (dυ) + υ dP + 4Φ (dl/d) (G υ)2 = 0 ----------------------------(**)
Dividing by (υ2) and integrating over a length L of pipe to give;
⎛ υ2 ⎞ P dP2
L General equation of energy apply to
G ln⎜ ⎟ + ∫
2
+ 4φ G 2 = 0
⎝ υ1 ⎠ P υ 1
d compressible fluid in horizontal pipe with
Kinetic Pressure Frictional no shaft work
energy energy energy
8.3.1 Isothermal Flow of an Ideal Gas in a Horizontal Pipe
For isothermal conditions of an ideal gas
P υ = constant ⇒ P υ = P1 υ1 ⇒ 1/υ = P / (P1 υ1)
P2 P2

∫ P dP = 2 P υ (P − P12 ) -----------------------(1)
dP 1 1
⇒ ∫ υ = Pυ 2
2

P1 1 1 P1 1 1

P1 υ1 = P2 υ2 ⇒ υ2 / υ1 = P1/ P2 -----------------------(2)
Substitute equations (1) and (2) into the genral equation of compressibl fluid to
give;
⎛ P ⎞ ( P 2 − P12 ) L
G 2 ln⎜ 1 ⎟ + 2 + 4φ G 2 = 0
⎝ P2 ⎠ 2 P1 υ1 d
Let υm the mean specific volume at mean pressure Pm, where,
Pm =(P1 + P2)/2
Pm υm = P1 υ1 ⇒ Pm =(P1 + P2)/2 = P1 υ1/ υm
( P22 − P12 ) ⎛ P2 + P1 ⎞⎛ P2 − P1 ⎞ ⎛ P1 υ1 ⎞⎛ P2 − P1 ⎞
=⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟
2 P1 υ1 ⎝ 2 ⎠⎝ P1 υ1 ⎠ ⎝ υm ⎠⎝ P1 υ1 ⎠
P22 − P12 P2 − P1
∴ =
2 P1 υ1 υm