Introduction to Compressible
Flow
Dρ
≠0
Dt
The density of a gas changes significantly along a streamline
Compressible Flow
Definition of Compressibility: the fractional change in
volume of the fluid element per unit change in pressure
p + dp
p
p
p
v
p + dp
v − dv
p + dp
p + dp
p
Compressible Flow
1. Mach Number:
M =
V
local velocity
=
c speed of sound
2. Compressibility becomes important for High Speed
Flows where M > 0.3
• M < 0.3 – Subsonic & incompressible
• 0.3 <M < 0.8 – Subsonic & compressible
• 0.8 <M < 1.2 – transonic flow – shock waves appear
mixed subsonic and sonic flow regime
• 1.2 <M < 3.0 - Supersonic – shock waves are present
but NO subsonic flow
• M > 3.0 – Hypersonic Flow, shock waves and other
flow changes are very strong
1
Compressible Flow
3. Significant changes in velocity and pressure result
in density variations throughout a flow field
4. Large Temperature variations result in density
variations.
As a result we now have two new variables we must solve for:
T&ρ
We need 2 new equations.
We will solve: mass, linear momentum, energy and an equation of state.
Important Effects of Compressibility on Flow
1. Choked Flow – a flow rate in a duct is limited by
the sonic condition
2. Sound Wave/Pressure Waves – rise and fall of
pressure during the passage of an acoustic/sound
wave. The magnitude of the pressure change is
very small.
3. Shock Waves – nearly discontinuous property
changes in supersonic flow. (Explosions, high
speed flight, gun firing, nuclear explosion)
4. A pressure ratio of 2:1 will cause sonic flow
Applications
1. Nozzles and Diffusers and converging
diverging nozzles
2. Turbines, fans & pumps
3. Throttles – flow regulators, an obstruction
in a duct that controls pressure drop.
4. One Dimensional Isentropic Flow –
compressible pipe flow.
2
Approach
• Control volume approach
• Steady, One-dimension, Uniform Flow
• Additional Thermodynamics Concepts are
needed
• Restrict our analysis to ideal gases
Thermodynamics
• Equation of State – Ideal Gas Law
p = ρ RT
R=
Universal Gas Constant
8314 J/(kmol ⋅ K)
Ru
=
=
= 287 J/(kg ⋅ K)
Molecular mass of air
28 .97kg/kmol
Mm
Temperature is absolute and the specific volume is
(volume per unit mass):
1
v=
ρ
Thermodynamics – Internal Energy &
Enthalpy
• Internal Energy – individual particle kinetic energy.
Summation of molecular vibrational and rotational energy.
u~ = u~ (v , T )
 ∂ u~ 
 ∂ u~ 
d u~ = 
 dT + 
 dv
 ∂T  v
 ∂v  T
• For an ideal gas u~ = u~ (T )
d u~ = cv dT
• Recall from our integral form of the Energy Equation for
Enthalpy of an ideal gas: h = u~ + pv
h = h (T )
dh = c p dT
3
Thermodynamics – Internal Energy &
Enthalpy
h = u~ + pv
h = u~ + RT
dh = d u~ + RdT
p
ρ
= RT
Substituting:
dh = c p dT
d u~ = c v dT
dh = d u~ + RdT
c p dT = c v dT + RdT
c p = cv + R
c p − c v = R = const
Thermodynamics – Internal Energy &
Enthalpy
Define the ratio of specific heats:
Then,
k≡
cp
cv
= const
kR
k −1
R
cv =
k −1
cp =
For Air:
cp = 1004 J/kg-K
k = 1.4
The 2nd Law of Thermodynamics & Isentropic
Processes
We define entropy by:
 δQ 
ds = 

 T  rev
Combining the 1st and 2nd Laws gives us Gibb’s Equation
Tds = dh −
dp
ρ
Tds = c p dT −
2
2
∫ ds = c ∫
p
1
1
dh = c p dT
dp
ρ
2
dT
dp
− R∫
T
p
1
1
ρT
=
R
p
4
The 2nd Law of Thermodynamics & Isentropic
Processes
s 2 − s1 = c p ln
p
T2
− R ln 2
p1
T1
For an Isentropic process: adiabatic and reversible
We get the following power law relationship
p 2  T2
=
p1  T1
k
 k −1  ρ 2
 = 
 ρ1




k
Control Volume Analysis of a Finite Strength
Pressure Wave
Moving Wave of Frontal Area A
Stationary Wave
Reference frame moving with wave
c
p
p + ∆p
ρ
ρ + ∆ρ
T
V =0
T + ∆T
1
p
2
p + ∆p
ρ
ρ + ∆ρ
T
T + ∆T
V = c − ∆V
V =c
∆V
Steady State Continuity Equation (Solve for the induced velocity ∆V):
0=
∫ ρ (V • nˆ )dA = − ∫ ρcdA + ∫ (ρ + ∆ρ )(c − ∆V )dA
r
CS
1
2
ρ cA = ( ρ + ∆ ρ )(c − ∆ V )A
ρ c = c ( ρ + ∆ ρ ) − ∆ V (ρ + ∆ ρ )
∆ρ
∆V = c
(A)
ρ + ∆ρ
The Speed of sound (c) is the rate of propagation of a pressure wave of infinitesimal
strength through a still fluid.
Control Volume Analysis of a Finite Strength
Pressure Wave
p
Steady State Momentum Equation:
(Find ∆p and c)
∑F
x
=
∫ ρV (V • nˆ )dA = m& (V
ρ
r
x
2
− V1 )
T
V =c
1
2
p + ∆p
ρ + ∆ρ
T + ∆T
V = c − ∆V
CS
pA − ( p + ∆ p ) A = ρ cA (c − ∆ V − c )
∆ p = ρ c∆ V
(B)
Now combine A & B and solve for the speed of sound:
c2 =
∆ p ρ + ∆ ρ ∆p
=
ρ
∆ρ
∆ρ
c2 =
∂p
∂ρ

∆ρ 
 1 +

ρ 

in the limit of ∆ ρ → 0
Small Amplitude moderate frequency waves are
isentropic and
p
ρk
= const
5
Control Volume Analysis of a Finite Strength
Pressure Wave
Calculating the Speed of Sound for an ideal gas:
p
ρk
= const
∂p
p
=k
∂ρ
ρ
c=
c=
k
p
ρ
=
kRT
Typical Speeds of Sound
Fluid
c (m/s)
kRT
Gases:
For Air:
k=
H2
Air
cp
≈ 1 .4
cv
R = 287 J/(kg ⋅ K)
1,294
340
Liquids:
Water
Ethyl Alcohol
1,490
1,200
Data From White 2003
Example 1: Speed of sound calculation
Determine the speed of sound in Argon (Ar) at 120 oC. MW = 40
kg/kmol:
c=
R=
k=
kRT
Ru
8314 J/(kmol ⋅ K)
=
= 207 .9 J/(kg ⋅ K)
Mm
4 0kg/kmol
cp
cv
≈ 1 .668
c = 1 .668 (207 .9 J/kgK )(393 K ) = 318 .8 ms -1
Movement of a sound source
and wave propagation
Source moves to the right at a speed V
Zone of silence
Mach cone
α
3 c ∆t
V>c
V<c
V=0
V ∆t
V ∆t
sin α =
V ∆t
V
1
=
c M
6
Example 2: a needle nose projectile traveling at a
speed of M=3 passes 200m above an observer. Find
the projectiles velocity and determine how far
beyond the observer the projectile will first be heard
M =3
200 m
α
x
Example 2: a needle nose projectile traveling at a
speed of M=3 passes 200m above an observer. Find
the projectiles velocity and determine how far
beyond the observer the projectile will first be heard
c = kRT = 1.4(287 )(300 ) = 347.2m/s
V = Mc = 3(347.2 ) = 1041.6m/s
 1 
−1  1 
o
 = sin   = 19.5
M 
 3
200m
tan α =
x
200m
x=
= 565m
tan 19.5
α = sin −1 
Steady Isentropic Flow – Control Volume
Analysis
Applications where the assumptions of steady,
uniform, isentropic flow are reasonable:
1. Exhaust gasses passing through the blades
of a turbine.
2. Diffuser near the front of a jet engine
3. Nozzles on a rocket engine
4. A broken natural gas line
7
Steady Isentropic Flow
1
2
h + dh
ρ + dρ
T + dT
p + dp
h
V
ρ
T
p
dx
Steady State Continuity Equation:
(
V + dV
)
r
0 = ∫ ρ V • nˆ dA = − ρ1V1 A1 + ρ 2V 2 A2
CS
ρ VA = ( ρ + d ρ )(V + dV )( A + dA )
ρ VA = ρ AV + ρ VdA + VAd ρ + Vd ρ dA + ρ AdV + ρ dAdV + Ad ρ dV + d ρ dAdV
Steady Isentropic Flow
1
2
h + dh
ρ + dρ
T + dT
p + dp
h
V
ρ
T
p
dx
Steady State Continuity Equation:
0=
∫ ρ (V • nˆ )dA = − ρ V A
r
1 1
1
V + dV
+ ρ 2V 2 A2
CS
ρ VA = ( ρ + d ρ )(V + dV )( A + dA ) ~ 0
~0
~0
~0
ρ VA = ρ AV + ρ VdA + VAd ρ + Vd ρ dA + ρ AdV + ρ dAdV + Ad ρ dV + d ρ dAdV
0=
dA d ρ dV
+
+
A
ρ
V
Only retain 1st order differential terms & divide
By ρVA
Steady Isentropic Flow
1
2
h + dh
ρ + dρ
T + dT
p + dp
h
V
ρ
T
p
V + dV
dx
Steady State Energy Equation with
1 inlet & 1 exit:
Q& − W& s V 22 − V12
=
+ g ( z 2 − z1 ) + (u~ + pv )2 − (u~ + pv )1
m&
2
Neglecting potential energy and recalling: h = u~ + pv
Q& − W& s V 22 − V12
=
+ h2 − h1
m&
2
Assuming and ideal gas:
Q& − W& s V 22 − V12
=
+ c p (T2 − T1 )
m&
2
8
Steady Isentropic Flow
1
2
h + dh
ρ + dρ
T + dT
p + dp
h
V
ρ
T
p
Steady State Energy Equation with 1 inlet
& 1 exit, neglecting potential energy &
assuming Isentropic duct flow:
V + dV
dx
V22
V2
+ h2 = 1 + h1
2
2
Assuming and ideal gas:
V22
V2
+ c p T2 = 1 + c p T1
2
2
V22
k
V2
k
RT1
+
RT 2 = 1 +
2
k −1
2
k −1
Stagnation Conditions
Insolated
walls
2
1
Assume the area A2is so big V2 ~ 0, then
h2 =
V12
+ h1 = ho
2
Stagnation enthalpy
Similarly, as we adiabatically bring a fluid parcel to zero velocity
there is a corresponding increase in temperature
V22
V2
+ c p T2 = 1 + c p T1
2
2
To =
V2
+T
2c p
Stagnation Temperature
Stagnation Conditions – maximum velocity
To =
V2
+T
2c p
(+)
If the temperature, T is taken taken down to absolute zero,
then (+) can be solved for the maximum velocity:
V max =
2 c p To
No higher velocity is possible unless energy is added to the
flow through heat transfer or shaft work.
9
Stagnation Conditions – Mach number relations
M =
Recall, that the Mach number is defined as:
To =
V2
+T
2c p
V
c
For Ideal gases:
1
 kR 
c pT = 
T = kRT
k −1
 k −1
c2
cp
k −1 2
To k − 1 V 2
M +1
=
+1 =
2
T
2 c
2
2
To
V
=
+1
T
T 2c p
To k − 1 2
=
M +1
T
2
Stagnation Conditions – Isentropic pressure &
density relationships
To k − 1 2
=
M +1
T
2
k
k
1
1
p o  To  k −1  k − 1 2
 k −1
=  =
M + 1
p T 
 2

ρ o  To  k −1  k − 1 2  k −1
=  =
M + 1
ρ T 

 2
Critical Values: conditions when M = 1
T*  2 
=

To  k + 1 
k
p *  2  k −1
=

po  k + 1 
1
ρ *  2  k −1
=

ρo  k +1 
1
c*  2  2
=

co  k + 1 
10
Critical Values: conditions when M = 1
For Air k = 1.4
T*  2 
=
 = 0 .8333
To  k + 1 
k
p *  2  k −1
=
 = 0 .5283
po  k + 1 
1
ρ *  2  k −1
=
 = 0 .9129
ρo  k + 1
1
c*  2  2
=
 = 0 .9129
co  k + 1 
In all isentropic flow, all critical values are constant.
Critical Values: conditions when M = 1
Critical Velocity: is the speed of sound c*
1
c*  2  2
=

co  k + 1 
1
V * = c* =
1
 2  2  2 kRT o  2
kRT * = c o 
 =

 k +1
 k +1 
Example 3: Stagnation Conditions
Air flows adiabatically through a duct. At point 1 the velocity
is 240 m/s, with T1 = 320K and p1 = 170kPa. Compute
(a) To
(b) Po
(c) ro
(d) M
(e) Vmax
(f) V*
11
Steady Isentropic Duct Flow
1
V
2
h
h + dh
ρ
T
ρ + dρ
T + dT
p
p + dp
V + dV
dx
Recall, for Steady isentropic flow Continuity:
(†) 0 =
dA d ρ dV
+
+
A
ρ
V
dA
d ρ dV
=−
−
A
ρ
V
For compressible, isentropic flow the momentum equation is:
2
(*) 0 = dp + dV = dp + VdV
ρ
2
ρ
Bernoulli’s Equation!
neglecting gravity
Substitute (†) into (*)
dA
dρ
dp
dp  1
dρ 


=−
+
=
−
ρ
ρV 2
ρ  V 2 dp 
A
Steady Isentropic Duct Flow
1
V
2
h
h + dh
ρ
T
ρ + dρ
T + dT
p
p + dp
V + dV
dA dp  1
dρ 


=
−
ρ  V 2 dp 
A
2
Recall that the speed of sound is: c =
∂p
∂ρ
dA dp  1
dp  V 2 
1 
 1 − 2 
=
− =

A
ρ  V 2 c 2  ρ V 2 
c 
Substituting the Mach number:
dA
dp
=
(1 − M 2 )
ρV 2
A
M =
V
c
Describes how the pressure
behaves in nozzles and diffusers
under various flow conditions
Nozzle Flow Characteristics
dA
dp
=
(1 − M 2 )
ρV 2
A
1.
2.
3.
4.
Subsonic Flow: M < 1 and dA < 0, then dP < 0:
indicating a decrease in pressure in a converging
channel.
Supersonic Flow: M > 1 and dA < 0, then dP > 0:
indicating an increase in pressure in a converging
channel.
Subsonic Flow: M < 1 and dA > 0, then dP > 0 :
indicating an increase in pressure in a diverging
channel.
Supersonic Flow: M > 1 dA > 0, then dP < 0 :
indicating a decrease in pressure in a diverging
channel.
P
P
P
P
P
P
P
P
12
Steady Isentropic Duct Flow – Nozzles
Diffusers and Converging Diverging Nozzles
Describes how the pressure
behaves in nozzles and diffusers
under various flow conditions
dA
dp
=
(1 − M 2 )
ρV 2
A
(††)
Recall, the momentum equation here is:
0=
dp
ρ
dp
+ VdV
ρ
= −VdV (**)
Now substitute (**) into (††) :
dA dV
=
(M 2 − 1)
A
V
Or,
(
dA
A
=
M 2 −1
dV V
)
Nozzle Flow Characteristics
(
)
dA dV
=
M 2 −1
A
V
1.
2.
3.
4.
Subsonic Flow: M < 1 and dA < 0, then dV > 0:
indicating an accelerating flow in a converging
channel.
Supersonic Flow: M > 1 and dA < 0, then dV < 0:
indicating an decelerating flow in a converging
channel.
Subsonic Flow: M < 1 and dA > 0, then dV < 0 :
indicating an decelerating flow in a diverging
channel.
Supersonic Flow: M > 1 dA > 0, then dV > 0 :
indicating an accelerating flow in a diverging
channel.
Converging-Diverging Nozzles
Amin
Subsonic
Supersonic
M=1
Amax
Subsonic
Supersonic
M<1
M>1
Subsonic
Supersonic
Flow can not be sonic
13
Choked Flow – The maximum possible mass flow through a
duct occurs when it’s throat is at the sonic condition
Consider a converging Nozzle:
receiver
po
pr
pe
To
Ve
ρo
plenum
Mass Flow Rate (ideal gas):
m& = ρ VA =
m& =
p
M
RT
p
VA
RT
M =
kRT A = p
V
=
c
V
kRT
k
MA
RT
k
MA
RT
m& = p
Choked Flow
Mass Flow Rate (ideal gas):
m& = p
k
MA
RT
Recall, the stagnation pressure and Temperature ratio and substitute:
k
po  k − 1 2
 k −1
=
M + 1
p  2

m& = p o
To k − 1 2
=
M +1
T
2
k +1
k
k − 1 2  2 (1− k )

MA  1 +
M 
RT o
2


If the critical area (A*) is where M=1:
k +1
m& = p o A *
k  k + 1  2 (1− k )


RT o  2 
The critical area Ratio is:
A
1
=
A* M
 2 + (k − 1)M 2

k +1

k +1
 2 ( k −1 )


14