unit i basic concepts

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


2/14

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.


3/14

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 += ~


4/14


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

==

+=

+=

+= ~


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


5/14

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


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 =


6/14


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


7/14

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


8/14

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


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


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


9/14


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.


10/14

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


11/14


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 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 *


12/14

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


13/14

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


14/14

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:

unit i basic concepts

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