Compressible Fluid Flow v1 Nh

7/27/2019 Compressible Fluid Flow v1 Nh

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Compressible Fluid Flow

.

MZA@UTPChemEFluidMech


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OutlineOutline

OutlineOutline Concept of compressible fluid

Mach number a criteria for compressible fluid flowcharacterization.

Processes of compressible fluid flow:

Adiabatic flow with friction

Isothermal flow with friction



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Compressibility, Z

A measure of the change in density that will beproduced in the fluid by a specified change in

.

Gases highly compressible.

Liquid very low compressibility.



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In fluid flow there are usuall occur chan esin pressure associated with changes of otherparameters of the flow

For example, changes in the velocity in the.



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ese pressure c anges w , n genera ,

cause density changes which will

the fluid involved will have an

influence on the flow.



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en ens y c anges are mpor an ,

temperature change in the flow that may arise

also influence on the flow



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In other words

,

the temperature changes in the flow

.



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Although the density changes in the flow fieldcan be very important, there exist many situations

of great practical importance in which the effects

of these density and temperature changesare negligible.

Example: Flow of incompressible fluid



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Incompressible fluid flow

The pressure and kinetic energy changes are so small

temperature changes in the fluid flow



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There are however a number of flows that are of reat

practical importance in which this assumption is notadequate.

The density and temperature changes being so large .

In such cases it is necessar to stud thethermodynamics of the flow simultaneously

with its dynamics.



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The study of these flows in which

e c anges n ens y an empera ure are mpor an

is known as

compress e u ow or gas ynam cs.

s c ap er w ocus on gas ows w erecompressibility effects are important.



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ApplicationsApplicationsApplicationsApplications

Although most obvious applications ofcompressible fluid flow theory are

in the design of high speed aircraft,

a knowledge of compressible fluid flow theoryis required in the design and operation

of many devices commonly encountered in

engineering practice.



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Gas and steam turbines e ow n e a ng an nozz es s rea e as compress e.

Reciprocating engines the flow of the gases through the valves and in the intake and

exhaust systems.

Natural gas transmission lines compressibility effects are important in calculating the flow through

such problems.



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Processes of Compressible FlowProcesses of Compressible FlowProcesses of Compressible FlowProcesses of Compressible Flow

Conver ent Diver ent

FlowReservoir Receiver

Thermal insulation

Isentropic flow


, ,


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Isentropic Friction section

Flow

Thermal insulation

Adiabatic flow with friction



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IsentropicFriction section

ow

Isothermal flow with friction



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Speed of Sound, cSpeed of Sound, cSpeed of Sound, cSpeed of Sound, c

Small ressure disturbance that move throu h a

continuous medium

wave (sound wave) propagates through a fluid)



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Sound

A series of small air-pressure disturbances oscillation insinusoidal fashion in the frequency range from

, cyc es per secon .

More rigid material, speed of sound greater.

Note: Other text use the notation a for speed of sound.



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A property of a material / compound.

..(1)kRTkPPc22

S

= =

=

Where:k = specific heat ratio, Cp/CvP = absolute pressure of the fluid (kPa, psi or equivalent)

= 3

R = specific gas constant (kJ/kgK or equivalent)

T = absolute temperature of the fluid (K oroR)



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Mach Number, MaMach Number, MaMach Number, MaMach Number, Ma

For incompressible fluid, Reynolds number, Re is.

In compressible fluid, Mach number, Ma is useful in.

Mach number, Ma a o o u ve oc y an spee o soun .

Dimensionless.

c

VMa =



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BornFebruary 18, 1838

Brno, Austrian Empire

Februar 19, 1916 1916-02-19 a ed 78

Munich, German Empire



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Characteristic of compressible fluid flow:

a < : u son c ow

Ma = 1 : Sonic flow



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Concorde, Ma = 2

(supersonic aircraft)Boeing 747, Ma = 0.85 0.95

(high speed, subsonic aircraft)



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.Sukhoi 30 MKM Ma 2.3

MZA@UTPChemEFluidMechF 18 Ma 1.8 Hawk Ma 0.84


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Analysis of compressible fluid flowAnalysis of compressible fluid flowAnalysis of compressible fluid flowAnalysis of compressible fluid flow

The gas (compressible fluid) is transferred from a very

temperature, TR

pressure, PR

velocity, VR = 0

roper es o u a reservo r are

called reservoir (stagnation)

conditions.

RESERVOIR Receiver



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Stagnation conditions are those that would exist if the

brought to rest (velocity = 0)



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Assumptions:

Flow is one-dimensional

Velocity gradients within a cross section areneglected

Friction is restricted to wall shear

Shaft work is zero

Gravitational effects are negligible



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Steady, frictionless, adiabaticSteady, frictionless, adiabaticSteady, frictionless, adiabaticSteady, frictionless, adiabatic

flowflowflowflow



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Convergent Divergent


Thermal insulationThroat

Provided the reservoir conditions (TR and PR) how

interest?

Need for relation between reservoir condition and


point of interest.


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From THERMODYNAMICS, open-system energy balancefrom reservoir (R) to point/state of interest (1):

22 VV

1R2

gz2

gz

++=

++

, R ,

.. 1RP1R1 ==



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For ideal gas: CRC vP +=

Ck

v

P=

( )1kCP

=

Then Eq. (2) will become;

Rk2

( ) ( )TT2Chh2V 1RP1R2

1 ==

( )= 1k 1R1


( )

= T1k

1

11


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Taking the term RkT1 to the left hand side:

( )1

TT

1k2

RkTV

1

R

1

2

1

=

From Eq. (1), for condition at point 1 11 kRTc =21

21

11c =

2

kRTcS

=

=

=

Then; .(3)( ) 1T1kc 1R

2

1

1 =



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V

= From Eq (3), substitute Ma and rearrange:

c

..(4)( ) 121kMa

TT

2

1R +=

,temperature and Mach numberat point of interestfor a given fluid.



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For compressible fluid, pressure and density change

The isentropic (frictionless, adiabatic) relation is givenb :

(a)T

P

1k

k

RR

=11

1

(b)TT 1k

1

R

1

R

=



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Substituting Eq. (4) into (a) and (b):

..(5)( )1

1kMa

P

1kk

2

1R

+

=2P1

12 ..

12

1

1

R

+

=



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Steady, frictionless, adiabatic flow can be achieved iffluid flow in a variable cross sectional area nozzle.

Then, there is a need to find relation between areaperpendicular to the flow with respect to the area of


.


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In any flow, mass is conserved.

From continuit e uation: AVAV =

11R VA =


RR1


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RR

11

1

R

VA =

Using equation above is not practical since: A is ver lar e

VR = 0From previous assumptions

Therefore other reference point is needed apart fromthe reservoir



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In any such flow there will be a state where Ma = 1.

s s ca e e cr ca s a e.

Properties related to critical state is called critical* .

(Dont confuse yourself with definition of Pcror Tcr)

Then the relation is written as:

1 VA =


11


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Relation between area perpendicular to theow an area a cr ca s a e:

( )( )1k2

1k2

+

)

1

1k

12

Ma

1

A

A

+=

2

At subsonic

to get the fluid go faster, one must reduce thecross sectional area perpendicular to the flow.

At supersonic to get the fluid go faster, one must increase


the cross sectional area perpendicular to the flow.


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Similarly, mass flow rate of fluid can be determinedfrom continuity equation with respect to the critical

2

RRT

kPm

( )( )( )1k2

1k

1

1kA

+

+

=

For air with k = 1.4: RT6847.0m R

R

=



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When the gas is blown down, there will be a decrease.

=,

Mass flow rate can also be determined usin :

VAVAm ==



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Example 1Example 1Example 1Example 1

Air flowing through an insulated, frictionless nozzle is, .

Determine: The Mach number Ma

The temperature, T

The density, The air velocity, V

at a location in the duct where the pressure is 430 kPa.



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SolutionSolutionSolutionSolution

Assumption:

r c on ess, nsu a e uc a a a c

Air is an ideal gas

. , p . ,R = 0.287 kJ/kgK

Given: TR = 400 K, PR = 500 kPa



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For isentropic gas flow:k

( ) 121kMa

PP

1k

k2

1

1

R

+=

T

T

P

P 1k

1

R

1

R

=

1P2

Mak

1k

R

1

=

P

PTT

k

R

1R1

=

5002 4.114.1

( )

500

430400

4.1

14.1

=

( )

469.0

43014.1

=

=K383 =



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Density, 11

RT

P= The speed of sound

( )( )383287.0430

=K

mkPa383287.04.1

213

=

3m

kg3.91 =

mkPa

Kkg

2

13

Fluid velocity, V1m

kg.

=

=

12.410.469

ca1

=

=s

.


s82.5 =


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Adiabatic Flow with FrictionAdiabatic Flow with FrictionAdiabatic Flow with FrictionAdiabatic Flow with Friction


Thermal insulation

Occurs when a gas flows through a length of pipe at

high velocity.

If pipe is insulated or flow is fast, heat transfer is


considered negligible adiabatic.


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Adiabatic Flow with FrictionAdiabatic Flow with FrictionAdiabatic Flow with FrictionAdiabatic Flow with Friction


Thermal insulation

Effect of friction due to the flow will cause the entropy

o ow ng gas o ncrease en ropy s no cons an Therefore isentropic relation cannot be applied in the


analysis.


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Ffriction

T

V

+T + dT

V + dV

dx

Applying the momentum balance:

Net pressure force Force due to wall shear stress= Mass flow rate x (Velocity out Velocity in)

( )[ ] ( )[ ] frictionFPdPPAVdVVm ++=+



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Ffriction

T

V

+T + dT

V + dV

dx

Applying the continuity equation:

AV = constantV = constant (since A is constant) V = ( + d)(V + dV)


F


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Ffriction

T

V

+T + dT

V + dV

dx

Applying the energy balance:

( )dVVV22

+= 22

P + dP = ( + d) R (T + dT)


F


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Ffriction

T

V

+T + dT

V + dV

dx

Also from Mach numberdefinition:

VV

2

2

( )dVVkRTc

22 +=

dTTkR +


F


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Ffriction

T

V

+T + dT

V + dV

dx

The equations represents a set of equations withunknown dP, dT, d, dV and dMa

Have to be solved accordingly to obtain appropriateex ressions.


F


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Ffriction

T

V

+T + dT

V + dV

dx

In momentum balance, there exist the term wallshear stress,

wall.

In pipeline system, this is expressed as dimensionlessvalue friction factor, f

Most compressible gas flows in duct involve turbulentflow



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Solving from the equation (for circular pipe):

( )

+++ =

2

2

2

21

22

Ma1k2

1Maln1k111x4 f

( )

+ 21221 Ma1k2

1

The equation describe the change of Ma over a givenlength.

,condition Ma 1.



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When friction is involved, flows tend to reach soniccondition Ma 1. (Ma2 approaching 1)

By setting Ma2 = 1, the length of duct required to givee va ue o a1 s o a ne as max mum eng , max

(or critical length, L*)

++ 22 M11M1*

( ) +

+

=2

1

21 Ma1k

2112

n2kMakD


ExampleExample

ExampleExample


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ExampleExampleExampleExample

Air flows in a 5 cm diameter pipe. The air entersa a = . an s o eave a a = . . e erm ne

the length of pipe required. What would be the

Assume f = 0.002 and adiabatic flow.


SolutionSolution

SolutionSolution


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Flow is adiabatic.

( )Ma1k2

1

1Maln1k111x4

2

2

2

21

22

f

+++ =

( )Ma1k2

1 21221

+

( )( )

2.111

1

.2

1.5

2.5ln

2(1.4)

141

1.5

1

2.5

1

1.4

1

050

L00204

22

2

22

.

..

.

.

+

+

++

=

m1850L2

.=



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Maximum pipe length, L*:

( )1Ma1kln

2k1k

MkMa1

DL4

2

12

2

1*f +++ =a

2

22

1

*

+

( ) ( ) ( )( )2.51412

112

.ln1.42

.2.541.

050

22

.

.

...

+

+

=

m72L .* =



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Pressure relation: ( )

+ 22Ma

2

1k1

MaP

( )

+

=2112 Ma2

1k1MaP

Temperature relation:

( )

+ 22Ma

1k1

( ) +=

2

12 Ma

21k1T



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Density relation:

( ) 2Ma1k1 +

( ) 22

1

2

1

2

2

1

2

1

Ma

1k

1

Ma

a

TP

+

==


Isothermal Flow with FrictionIsothermal Flow with Friction



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Isothermal Flow with FrictionIsothermal Flow with FrictionIsothermal Flow with FrictionIsothermal Flow with Friction

IsentropicFriction section


Occurs in long, small, uninsulated pipe in contact withenvironment transmit sufficient heat to kee the flow

isothermal. E.g.: flow of natural gas through long distance pipelines.


FfrictiondQ


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V

+V + dV

+ d

dx

Applying the continuity equation:

AV = constantV = constant (since A is constant)V = ( + d)(V + dV)

=

ddV


FfrictiondQ


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V

+V + dV

+ d

dx

Applying the momentum balance:

Net pressure force Force due to wall shear stress= Mass flow rate x (Velocity out Velocity in)

( )[ ] ( )[ ] frictionFPdPPAVdVVm ++=+


FfrictiondQ


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V

+V + dV

+ d

dx

Applying the energy balance:

+

VdVV

22

= 22 PP

=


FfrictiondQ


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V

+V + dV

+ d

dx

From ideal gas EOS (P = RT)P + dP = ( + d) RT

=

ddP

P


FfrictiondQ


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V

+V + dV

+ d

dx

Also from Mach numberdefinition:

V

Ma=(since T constant, c constant)dVV

dMaMa

c

+=+

dV

M

dMa =


FfrictiondQ


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V

+V + dV

+ d

dx

The equations represents a set of equations withunknown dP, dT, d, dV and dMa

expressions.



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Q

1

V1

1

2

V2

2

x

1 2

1

2

2

1

2

1

1

2

Ma

Ma

P

P

V

V==

=

= 2

Mal

111L2f


121 MaMaMa2kD

Q


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V11Ma

V22Ma

x

For isothermal flow with friction, Ma tend to reachk

1

1 When Ma ~ , dq ~ infinity.

An infinite amount of heat must be transferred or removed to kee

k

the temperature of the gas constant.

Limiting value for Ma =

k1





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By setting Ma2 = , maximum length Lmax or L*1

2kMa1L4 *12

1

ankMaD

+

=


ExampleExample

ExampleExample


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pppp

Air flows through a 5 cm diameter pipeline. The flow, .

pressure of 900 kPa, and exit at Ma = 0.5. Determine: Len th of the i e.

Maximum pipe length and the correspondingpressure.

Mass flow rate of the air.

e spec c ea ra o or a r an e mean r c onfactor may be taken as 1.4 and 0.004 respectively.


SolutionSolution

SolutionSolution


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Length of pipe:

( )( ) 0.1

0.5

ln0.5

1

0.1

1

1.42

1

050

L0040222.

.

=

m2204L .=

( ) ( )( )0.11.41L00404 22

*.

( )( )m43233L

..

0.11.4050

2

.*.

=



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Pressure at the maximum length:

MaPFrom 12 =

1Maand*PPL*,at 2 ==

kPa51069000.1

P*

.

.==

1.4



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At T = 293 K,c = = 10.85 m/s

V1 = Ma1 c = (0.1)(10.85) = 1.85 m/skRT

( ) kg710kPa900P 3311 ===.

K293Kkg

2870

.

851050710

m

2

11

=

=

...

kg0.0389 =


ExampleExample

ExampleExample


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Natural gas flows through a 0.075 m diameter pipeline.

assumed to be isothermal with a temperature of 15

o

C.The Mach number and pressure at the inlet are 0.09and 900 kPa, respectively. If the mean friction factorfor the flow is 0.002, determine the Mach number at

.of the pipe and the exit pressure with this length of pipe.

Assume the flow is steady and the specific heat ratiofor the natural gas is 1.3.


SolutionSolution

SolutionSolution


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Assumption: Isothermal, steady flow, ideal gas

= Maln111L22

22f

( )( )

=Ma

ln11175000202 2

121

.

=Ma

l111

4

...

2

2.

0.09Ma0.00812.6

2


Unknowns Ma2 solve through trial-and-error

Ma2, uess LHS RHS Error


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0.1 40 8.916 -31.084

0.2 40 37.068 -2.932

0.25 40 40.3 0.300

o grap a vs rror

y = 64.76x - 15.88

0

0.5

,Ma, x = 0.245

-1.5

-1

-0.5 0.2 0.25

Ma

-3.5

-3-2.5

-2


Error


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MaP 21

1

2 =

kPa6330900

2450

P2 .

.

.==

Maximum possible length:

2kMaln

kMaD

2

12

1

1 +

=

( )( ) ( )( )( )0.091.3ln0.091.30.091.31

0750

L00204 22 +

=.

*.


m838.2L =*


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MaPFrom 12 =

1Maand*PPL*,at 2

21

==

kPa4299000.09

P*

.

.==

1.3



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Compressible Fluid Flow v1 Nh

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