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8/10/2019 Unit I Basic Concepts
1/14
Introduction to CompressibleFlow
0 Dt D
The density of a gas changes significantly along a streamline
Compressible Flow
Definition of Compressibility: the fractional change involume of the fluid element per unit change in pressure
p
p
p
pv
dp p +
dp p +dp p +
dp p +
dvv
Compressible Flow
1. Mach Number:
2. Compressibility becomes important for High SpeedFlows 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
soundof speed velocitylocal==
cV
M
8/10/2019 Unit I Basic Concepts
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Compressible Flow
3. Significant changes in velocity and pressure resultin density variations throughout a flow field
4. Large Temperature variations result in densityvariations.
As a result we now have two new variables we must solve for:T &
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 bythe sonic condition
2. Sound Wave/Pressure Waves rise and fall of pressure during the passage of an acoustic/soundwave. The magnitude of the pressure change isvery small.
3. Shock Waves nearly discontinuous propertychanges in supersonic flow. (Explosions, highspeed flight, gun firing, nuclear explosion)
4. A pressure ratio of 2:1 will cause sonic flow
Applications
1. Nozzles and Diffusers and convergingdiverging nozzles
2. Turbines, fans & pumps3. Throttles flow regulators, an obstruction
in a duct that controls pressure drop.4. One Dimensional Isentropic Flow
compressible pipe flow.
8/10/2019 Unit I Basic Concepts
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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
RT p =
Temperature is absolute and the specific volume is(volume per unit mass):
1=v
K)J/(kg287.97kg/kmol28
K)J/(kmol8314air of massMolecular
ConstantGasUniversal ====m
u
M R
R
Thermodynamics Internal Energy &Enthalpy
Internal Energy individual particle kinetic energy.Summation of molecular vibrational and rotational energy.
For an ideal gas
Recall from our integral form of the Energy Equation forEnthalpy of an ideal gas:
dvvu
dT T u
ud T v
! " #$
% &
+!
" #$
% &
=
~~~
( )T vuu ,~~ =
dT cud v=~
( )T uu ~~ =
)(T hh =
dT cdh p=
pvuh += ~
8/10/2019 Unit I Basic Concepts
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Thermodynamics Internal Energy &Enthalpy
dT cdh p=
RdT ud dh
RT uh
pvuh
+=
+=
+=
~
~
~
RT p =
dT cud v=~
Substituting:
const Rcc
Rcc
RdT dT cdT c
RdT ud dh
v p
v p
v p
==
+=
+=
+= ~
Thermodynamics Internal Energy &Enthalpy
Define the ratio of specific heats: const c
ck
v
p =
Then,
1
1
=
=
k R
c
k kR
c
v
p
For Air:c p = 1004 J/kg-K k = 1.4
The 2 nd Law of Thermodynamics & IsentropicProcesses
Combining the 1 st and 2 nd Laws gives us Gibbs Equation
revT Q
ds ! " #$
% &
=
We define entropy by:
''' =
=
=
2
1
2
1
2
1 pdp
RT
dT cds
dpdT cTds
dpdhTds
p
p
dT cdh p=
p R
T =
1
8/10/2019 Unit I Basic Concepts
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The 2 nd Law of Thermodynamics & IsentropicProcesses
For an Isentropic process: adiabatic and reversibleWe get the following power law relationship
1
2
1
212 lnln p
p R
T T
c s s p =
k k
k
T T
p p
!! "
#$$%
& =!! "
#$$%
& =
1
21
1
2
1
2
Control Volume Analysis of a Finite StrengthPressure Wave
c
0=V
T
p
T T
p p
+
+
+
V
Moving Wave of Frontal Area A
The Speed of sound ( c) is the rate of propagation of a press ure wave of infinitesimalstrength through a still fluid.
cV
T
p
=
T T
p p
+
+
+
V cV =
Stationary WaveReference frame moving with wave
( ) ( )( )
( )( )( ) ( )
(A)
021
+
=
++=
+=
++== '''
cV
V cc
AV ccA
dAV cdAcdAnV CS
!Steady State Continuity Equation (Solve for the induced velocity V):
1 2
Control Volume Analysis of a Finite StrengthPressure Wave
Small Amplitude moderate frequency waves areisentropic and
cV
T
p
=
T T
p p
+
+
+
V cV =( ) ( )
( ) ( )(B)
12
V c p
cV ccA A p p pA
V V mdAnV V F CS
x x
=
=+
== '(
"!
Steady State Momentum Equation:(Find p and c)
1 2
Now combine A & B and solve for the speed of sound:
0of limitin the
1
2
2
=
!! " #
$$% &
+
=
+
=
pc
p pc
const p
k =
8/10/2019 Unit I Basic Concepts
6/14
Control Volume Analysis of a Finite StrengthPressure Wave
Calculating the Speed of Sound for an ideal gas:
const p
k =
p
k p
=
kRT pk c ==
kRT c = Typical Speeds of SoundFluid c (m/s)Gases:H2 1,294Air 340
Liquids:Water 1,490Et hyl Alc oh ol 1, 200
Data From White 2003
4.1=v
p
c
ck
For Air:
K)J/(kg287 = R
Example 1: Speed of sound calculation
Determine the speed of sound in Argon (Ar) at 120 oC. MW = 40kg/kmol:
kRT c =
668.1=v
p
c
ck
K)J/(kg9.2070kg/kmol4
K)J/(kmol8314 ===m
u
M R
R
( )( ) 1-ms8.318393J/kgK 9.207668.1 == K c
Movement of a sound sourceand wave propagation
V = 0
V < c V > c
Source moves to the right at a speed V
M cV 1
sin ==
Zone of silence
V t
3 c t
V t V t
Mach cone
8/10/2019 Unit I Basic Concepts
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Example 2: a needle nose projectile traveling at aspeed 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
200 m
M =3
x
Example 2: a needle nose projectile traveling at aspeed 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
( )( )( )
mm
x
xm
M
McV
kRT c
o
5655.19tan
200
200tan
5.1931
sin1
sin
m/s6.10412.3473
m/s2.3473002874.1
11
==
=
=! " #$
% &
=! " #$
% &
=
===
===
Steady Isentropic Flow Control VolumeAnalysis
Applications where the assumptions of steady,uniform, isentropic flow are reasonable:
1. Exhaust gasses passing through the bladesof a turbine.
2. Diffuser near the front of a jet engine3. Nozzles on a rocket engine4. A broken natural gas line
8/10/2019 Unit I Basic Concepts
8/14
Steady Isentropic Flow
p
T
h
V
dp p
dT T
d
dhh
+
+
+
+
dV V +
dx
( )( )( )( )
dAdV d dV Ad dAdV AdV dAVd VAd VdA AV VA
dA AdV V d VA
AV AV dAnV CS
+++++++=
+++=
+== ' 2221110!
Steady State Continuity Equation:
1 2
Steady Isentropic Flow
p
T
h
V
dp p
dT T
d
dhh
+
+
+
+
dV V +
dx
( )( )( )( )
dAdV d dV Ad dAdV AdV dAVd VAd VdA AV VA
dA AdV V d VA
AV AV dAnV CS
+++++++=
+++=
+== ' 2221110!
Steady State Continuity Equation:
Only retain 1 st order differential terms & divideBy VA
V dV d
AdA ++=
0
1 2
~ 0 ~ 0 ~ 0 ~ 0
Steady Isentropic Flow
p
T
h
V
dp p
dT T
d
dhh
+
+
+
+
dV V +
dx
( ) ( ) ( )12122
12
2 ~~2
pvu pvu z z g V V
mW Q s ++++
=
"
""
Steady State Energy Equation with
1 inlet & 1 exit:
Neglecting potential energy and recalling: pvuh += ~
12
21
22
2hh
V V m
W Q s +
=
"
""
( )122
12
2
2T T c
V V m
W Q p
s +
=
"
""Assuming and ideal gas:
1 2
8/10/2019 Unit I Basic Concepts
9/14
Steady Isentropic Flow
p
T
h
V
dp p
dT T
d
dhh
+
+
+
+
dV V +
dxSteady State Energy Equation with 1 inlet& 1 exit, neglecting potential energy &assuming Isentropic duct flow:
1
21
2
22
22h
V h
V +=+
Assuming and ideal gas:
1
21
2
22
22T c
V T c
V p p
+=+
1
21
2
22
1212 RT
k k V
RT k
k V
+=
+
1 2
Stagnation Conditions
Assume the area A2is so big V 2 ~ 0, then
ohhV
h =+= 12
12 2
1
2
Stagnation enthalpy
T c
V T p
o += 2
2
1
21
2
22
22T c
V T c
V p p
+=+
Similarly, as we adiabatically bring a fluid parcel to zero velocitythere is a corresponding increase in temperature
Insolatedwalls
Stagnation Temperature
Stagnation Conditions maximum velocity
T c
V T
po
+=2
2
If the temperature, T is taken taken down to absolute zero,then (+) can be solved for the maximum velocity:
(+)
o pT cV 2max =
No higher velocity is possible unless energy is added to theflow through heat transfer or shaft work.
8/10/2019 Unit I Basic Concepts
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Stagnation Conditions Mach number relations
12
11
21
12
2
22
2
2
2
+
=+
=
+=
+=
M k
cV k
T T
cT V
T T
T cV
T
o
p
o
po For Ideal gases:
11
1 =!
" #$
% &
=
k kRT T
k kR
T c p
pc2c
12
1 2 +
= M k
T T o
Recall, that the Mach number is defined as:cV
M =
Stagnation Conditions Isentropic pressure &density relationships
11
211
121
12
1
12
1
! " #$
% &
+
=! " #$
% &
=
! " #$
% &
+
=! " #$
% &
=
k k oo
k k
k k
oo
M k
T T
M k
T T
p p
12
1 2 +
= M k
T T o
Critical Values: conditions when M = 1
! " #$
% &
+=
12*
k T T
o
1*
1
2 ! "
#$%
&
+
=k
k
o k p
p
11
*
12 !
" #$
% &
+=
k
o k
21
*
12 !
" #$
% &
+=
k cc
o
8/10/2019 Unit I Basic Concepts
11/14
Critical Values: conditions when M = 1
8333.01
2* =! " #$
% &
+=
k T T
o
5283.01
2 1* =! " #$
% &
+=
k k
o k p p
9129.01
2 11
*
=! " #$
% &
+=
k
o k
For Air k = 1.4
9129.01
2 21
*
=! " #$
% &
+=
k cc
o
In all isentropic flow, all critical values are constant.
Critical Values: conditions when M = 1
Critical Velocity: is the speed of sound c*
21
21
***
12
12 !
" #$
% &
+=!
" #$
% &
+===
k kRT
k ckRT cV oo
21
*
12 !
" #$
% &
+=
k cc
o
Example 3: Stagnation ConditionsAir flows adiabatically through a duct. At point 1 the velocity
is 240 m/s, with T 1 = 320K and p1 = 170kPa. Compute(a) T o(b) P o(c) r o(d) M
(e) V max(f) V *
8/10/2019 Unit I Basic Concepts
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Steady Isentropic Duct Flow
p
T
h
V
dp p
dT T d
dhh
+
+
+
+
dV V +
dx
Recall, for Steady isentropic flow Continuity:
V dV d
AdA ++=
0
For compressible, isentropic flow the momentum equation is:
VdV dpdV dp +=+= 2
02
Bernoullis Equation!neglecting gravity
Substitute () into (*)
()
(*)
V dV d
AdA =
!! "
#$$%
& =+=
dpd
V dp
V dpd
AdA
22
1
1 2
Steady Isentropic Duct Flow
p
T
h
V
dp p
dT T d
dhh
+
+
+
+
dV V +
!! " #
$$% &
=dpd
V dp
AdA
21
Recall that the speed of sound is:
= pc 2
!! "
#$$%
& =!
" #$
% &
=2
2
222 111
cV
V dp
cV dp
AdA
Substituting the Mach number:cV
M =
( )22 1 M V dp
AdA =
Describes how the pressure behaves in nozzles and diffusers
under various flow conditions
1 2
Nozzle Flow Characteristics
1. Subsonic Flow : M < 1 and dA < 0, then dP < 0:indicating a decrease in pressure in a c onverging
channel.2. Supersonic Flow: M > 1 and dA < 0, then dP > 0:indicating an increase in pressure in a convergingchannel.
3. Subsonic Flow : M < 1 and dA > 0, then dP > 0 :indicating an increase in pressure in a divergingchannel.
4. Supersonic Flow: M > 1 dA > 0, then dP < 0 :indicating a decrease in pressure in a divergingchannel.
( )22 1 M V dp
AdA =
P P
PP
PP
P P
8/10/2019 Unit I Basic Concepts
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Steady Isentropic Duct Flow NozzlesDiffusers and Converging Diverging Nozzles
( )22 1 M V dp
AdA =
Describes how the pressure behaves in nozzles and diffusersunder various flow conditions
VdV dp +=
0
Recall, the momentum equation here is:
VdV dp =
Now substitute (**) into () :
( )12 = M V dV
AdA
( )12 = M V A
dV dA
Or,
()
(**)
Nozzle Flow Characteristics
1. Subsonic Flow : M < 1 and dA < 0, then dV > 0:indicating an accelerating flow in a convergingchannel.
2. Supersonic Flow: M > 1 and dA < 0, then dV < 0:indicating an decelerating flow in a convergingchannel.
3. Subsonic Flow : M < 1 and dA > 0, then dV < 0 :indicating an decelerating flow in a divergingchannel.
4. Supersonic Flow: M > 1 dA > 0, then dV > 0 :indicating an accelerating flow in a divergingchannel.
( )12 = M V dV
AdA
Converging-Diverging Nozzles
AminSubsonic Supersonic
M = 1
Amax
Subsonic
Supersonic
M < 1Subsonic
Supersonic M > 1
Flow can not be sonic
8/10/2019 Unit I Basic Concepts
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Choked Flow The maximum possible mass flow through a
duct occurs when its throat is at the sonic conditionConsider a converging Nozzle:
VA RT
pVAm == "
o
o
o
T
p
r preceiver
plenum
e
e
V
p
Mass Flow Rate (ideal gas):
kRT V
cV
M ==
MA RT
k p AkRT M
RT p
m =="
MA RT
k pm ="
Choked Flow
MA RT
k pm ="
Mass Flow Rate (ideal gas):
12 12
1 ! " #$
% &
+
=k
k
o M k
p p
Recall, the stagnation pressure and Temperature ratio and substitute:
( )k k
oo M
k MA
RT k
pm
+
! " #$
% &
+=12
1
2
21
1"
12
1 2 +
= M k
T T o
( ) ( )12
12
* 1121
+
!! " #$$%
& ++=
k k
k M k
M A A
The critical area Ratio is:
( )k k
oo
k RT
k A pm
+
! " #$
% & +
=12
1
*
21
"
If the critical area (A*) is where M=1: