Fluid 11

1

FUNDAMENTALS OFFUNDAMENTALS OFFLUID MECHANICSFLUID MECHANICS

Chapter 11 Analysis of Chapter 11 Analysis of Compressible FlowCompressible Flow

JyhJyh--CherngCherng ShiehShiehDepartment of BioDepartment of Bio--Industrial Industrial MechatronicsMechatronics Engineering Engineering

National Taiwan UniversityNational Taiwan University

2

MAIN TOPICSMAIN TOPICS

Ideal Gas RelationshipsIdeal Gas RelationshipsMach Number and Speed of SoundMach Number and Speed of SoundCategories of Compressible FlowCategories of Compressible FlowIsentropic Flow of an Ideal GasIsentropic Flow of an Ideal GasNonisentropicNonisentropic Flow of an Ideal GasFlow of an Ideal GasTwoTwo--Dimensional Compressible FlowDimensional Compressible Flow

3

Ideal Gas RelationshipsIdeal Gas Relationships

4

IntroductionIntroduction

Fluid compressibility is a very important consideration in numerFluid compressibility is a very important consideration in numerous ous engineering applications of fluid mechanics. For example,engineering applications of fluid mechanics. For example,►►The measurement of highThe measurement of high--speed flow velocities requires speed flow velocities requires

compressible flow theory.compressible flow theory.►►The flows in gas turbine engine components are generally The flows in gas turbine engine components are generally

compressible.compressible.►►Many aircraft fly fast enough to involve a compressible flow Many aircraft fly fast enough to involve a compressible flow

field.field.In this study of compressibility effects, we mainly consider theIn this study of compressibility effects, we mainly consider thesteady, onesteady, one--dimensional, constant (including zero) viscosity, dimensional, constant (including zero) viscosity, compressible flow of an ideal gas.compressible flow of an ideal gas.

5

Ideal Gas relationships Ideal Gas relationships 1/21/2

Before to develop compressible flow equation, we need to become Before to develop compressible flow equation, we need to become more familiar with the fluid.more familiar with the fluid.The equation of state for an ideal gasThe equation of state for an ideal gas

The gas constant, R, represents a constant for each distinctThe gas constant, R, represents a constant for each distinct ideal gas ideal gas or mixture of ideal gasesor mixture of ideal gases

gasMR λ=

RTp ρ= R is gas constant

λλis the universal gas constant.is the universal gas constant.MMgasgas is the molecular weight of the ideal gas or gas mixture.is the molecular weight of the ideal gas or gas mixture.

(1)(1)

(2)(2)

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Ideal Gas relationships Ideal Gas relationships 2/22/2

For an ideal gas, For an ideal gas, internal energy is a function of temperature internal energy is a function of temperature onlyonly. Thus, the ideal gas specific heat at constant volume. Thus, the ideal gas specific heat at constant volume

Where the subscript v on the partial derivative refers to Where the subscript v on the partial derivative refers to differentiation at constant specific volume, v=1/differentiation at constant specific volume, v=1/ρρ..

dTud

Tuc

vv

((=⎟

⎠⎞

⎜⎝⎛∂∂

= is function of temperature onlyis function of temperature only

( )12v12v TTcuudTcud −=−⇒= (((

∫=−2

1

T

Tv12 dTcuu ((

(3)(3)

(4)(4)

(5)(5)

7

Enthalpy Enthalpy 1/21/2

The fluid property enthalpyThe fluid property enthalpy

For ideal gasFor ideal gas

ρ+=

puh ((

dTchh P12 =−((

)(uu Ｔ(( =

From the equation of stateFrom the equation of state RTp=

ρ

)(hh Ｔ((

=

The ideal gas specific heat at constant pressure

dThd

Thc

pP

((

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=

RTuh += ((

(6)(6)

(7)(7)

)TT(chh 12P12 −=−((

(8)(8)

(9)(9)

8

Enthalpy Enthalpy 2/22/2

RdT

uddT

hdRdTudhd

RTuh

+=⇒+=⇒

+=((

((

((

Rcc vp =−

The specific heat ratio, k, is defined as

dThd

Thc

pP

((

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

= dTud

Tuc

vv

((=⎟

⎠⎞

⎜⎝⎛∂∂

=

v

P

cck =

1kkRcP −

=

(10)(10)

(11)(11)

(12)(12)

(13)(13)

1k1RcV −

=(14)(14) (15)(15)(12) + (13)(12) + (13)

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Example 11.1 Internal Energy, Enthalpy, Example 11.1 Internal Energy, Enthalpy, and Density for an Ideal Gasand Density for an Ideal Gas

Air flows steadily between two sections in a long straight portiAir flows steadily between two sections in a long straight portion of on of 44--in.in.--diameter pipe as is indicated in Figure E11.1. The uniformly diameter pipe as is indicated in Figure E11.1. The uniformly distributed temperature and pressure at each section are Tdistributed temperature and pressure at each section are T11=540=540°°R, R, pp11=100 =100 psiapsia, and T, and T22=453=453°°R, pR, p22=18.4 =18.4 psiapsia. Calculate the (a) change . Calculate the (a) change in internal energy between sections (1) and (2), (b) change in in internal energy between sections (1) and (2), (b) change in enthalpy between sections (1) and (2), and (c) change in densityenthalpy between sections (1) and (2), and (c) change in densitybetween sections (1) and (2).between sections (1) and (2).

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Example 11.1 Example 11.1 SolutionSolution1/21/2

)TT(cuu 12v12 −=−((

The change in internal energy between sections (1) and (2)The change in internal energy between sections (1) and (2)

1k1RcV −

= (15)(15)

From Table 1.7From Table 1.7 )Rlbm/()lbft(3.53)Rslug/()lbft(1716R4.1k °⋅⋅=°⋅⋅==

( )lbm/lbft600,11)R540R453()Rlbm/()lbft(133

TTcuu 12v12

⋅−=°−°×°⋅⋅=−=− ((

)TT(chh 12P12 −=−((

(9)(9)v

P

cck = (13)(13)

11


)Rlbm/()lbft(186...kcc vP °⋅⋅===

lbm/lbft200,16...)TT(chh 12P12 ⋅−==−=−((

RTp ρ= (1)(1)

3

1

1

2

2

1

1

2

212

ft/lbm389.0...

Tp

Tp

R1

RTp

RTp

−==

⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=ρ−ρ⇒

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Entropy Entropy 1/31/3

For any pure substance including ideal gases, the For any pure substance including ideal gases, the ““first T first T dsdsequationequation”” isis

T is absolute temperatureT is absolute temperature⎟⎟⎠

⎞⎜⎝

⎛ρ

+=1pdudTds (

dp1hdTds ⎟⎟⎠

⎞⎜⎝

⎛ρ

−=(

dp11pdudhd ⎟⎟⎠

⎞⎜⎝

⎛ρ

+⎟⎟⎠

⎞⎜⎝

⎛ρ

+= ((

““second T second T dsds equationequation””

⎟⎟⎠

⎞⎜⎝

⎛ρ

+=ρ

1dyR

TdTcds v

PdpR

TdTcds P −=

(16)(16)

(17)(17)

(16)+(17)(16)+(17) (18)(18)

(1)+(3)+(16)(1)+(3)+(16) (19)(19)

(20)(20)(1)+(7)+(18)(1)+(7)+(18)

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If cIf cpp and and ccvv are assumed to be constant for a given gasare assumed to be constant for a given gas

2

1

1

2v12 lnR

TTlncss

ρρ

+=−

1

2

1

2P12 p

plnRTTlncss −=−

(19)(19)

(20)(20)

(21)(21)

(22)(22)

14


0pplnR

TTlnclnR

TTlnc

ss,0ds

1

2

1

2P

2

1

1

2v

21

=−=ρρ

+⇒

==

k

1

2

1

21k/k

1

2pp

TT

⎟⎟⎠

⎞⎜⎜⎝

⎛ρρ

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

ttanconspk =ρ

For the adiabatic and frictionless flowFor the adiabatic and frictionless flow

Constant entropy flow Constant entropy flow or isentropic flowor isentropic flow(23)(23)

(24)(24)

(25)(25)

15

Example 11.2 Entropy for an Ideal GasExample 11.2 Entropy for an Ideal Gas

For the air flow of Example 11.1, calculate the change in entropFor the air flow of Example 11.1, calculate the change in entropy, y, ss22--ss11, between sections (1) and (2)., between sections (1) and (2).

16

Example 11.2 Example 11.2 SolutionSolution

2

1

1

2v12 lnR

TTlncss

ρρ

+=− (21)(21)

56.4psia4.18

R453R540

psia100pT

Tp

2

2

1

1

2

1 =°

°==

ρρ

)Rlbm/()lbft(5.57...pT

TplnR

TTlncss

2

2

1

1

1

2v12 °⋅⋅==⎟⎟

⎠

⎞⎜⎜⎝

⎛+=−

1

2

1

2P12 p

plnRTTlncss −=− (22)(22)

)Rlbm/()lbft(5.57...pplnR

TTlncss

1

2

1

2P12 °⋅⋅==−=−

17

Mach Number and Mach Number and Sonic SpeedSonic Speed

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Mach NumberMach Number

The The Mach numberMach number, Ma, was a dimensionless measure of , Ma, was a dimensionless measure of compressibility in a fluid flow.compressibility in a fluid flow.The Mach number is defined as the ratio of the value of the locaThe Mach number is defined as the ratio of the value of the local l flow velocity, V, to the local flow velocity, V, to the local speed of soundspeed of sound, c., c.

cVMa =

What we perceive as sound generally consists of weak pressure puWhat we perceive as sound generally consists of weak pressure pulse lse that move through air. When our ear drums respond to a successiothat move through air. When our ear drums respond to a succession n of moving pulse, we hear sounds.of moving pulse, we hear sounds.

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Speed of Sound Speed of Sound 1/61/6

To better understand the notion of speed of sound, we analyze thTo better understand the notion of speed of sound, we analyze the e oneone--dimensional fluid mechanics of an infinitesimally thin, weak dimensional fluid mechanics of an infinitesimally thin, weak pressure pulse moving at the speed of sound through a fluid at rpressure pulse moving at the speed of sound through a fluid at rest. est. Select an infinitesimally thin control volume that move with theSelect an infinitesimally thin control volume that move with thepressure pulse.pressure pulse.

Ahead of the Ahead of the pressure pulsepressure pulse

Behind the Behind the pressure pulsepressure pulse

The speed of the weak The speed of the weak pressure pulse is pressure pulse is considered constant and considered constant and in one direction only; in one direction only; thus, out control volume thus, out control volume is inertial.is inertial.

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For an observer moving with this control volume, it appears as iFor an observer moving with this control volume, it appears as if f fluid is entering the control volume through surface area A withfluid is entering the control volume through surface area A withspeed c at pressure p and densityspeed c at pressure p and densityρρand leaving the control volume and leaving the control volume with speedwith speed……..

Entering CVEntering CV

Leaving CVLeaving CV

21


Apply the Apply the continuity equationcontinuity equation to the flow through this control to the flow through this control volumevolume

)Vc(A)(Ac δ−δρ+ρ=ρ

V))((Vcc δδρ−ρδ−ρ=ρ

δρ=ρδ cV

(26)(26)

(27)(27)

(28)(28)

small

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Apply the Apply the linear momentum equationlinear momentum equation to the flow through this to the flow through this control volumecontrol volume

A)pp(AA)Vc)()(Vc(cAc δ+−ρ=δ−δρ+ρδ−+ρ−

cpV

pAAc)Vc(cAcδ

=ρδ

δ−=ρδ−+ρ−

δρδ

=δρδ

=pcpc2

This expression for the speed of sound results from application This expression for the speed of sound results from application of the of the conservation of mass and conservation of linear conservation of mass and conservation of linear momentum principlesmomentum principles to the flow through the control volume.to the flow through the control volume.

(29)(29)

(30)(30)

(31)(31)

(26)+(29)(26)+(29)

(28)+(30)(28)+(30)

23


Apply the Apply the conservation of energy principleconservation of energy principle to the flow through to the flow through this control volumethis control volume

δρδ

=δρδ

=pcpc2

This expression for the speed of sound results from application This expression for the speed of sound results from application of the of the conservation of mass and conservation of energy conservation of mass and conservation of energy principlesprinciples to the flow through the control volume.to the flow through the control volume.

)loss(zg2

Vp 2

δ=δ+⎟⎟⎠

⎞⎜⎜⎝

⎛δ+

ρδ

02c

2)Vc(p 22

=−δ−

+ρδ

( ) VcV 2 δ<<δ

cpV δ

=ρδgδz≒0 δ(loss)= 0

(32)(32)

(33)(33)

(28)+(33)(28)+(33) (31)(31)

24


δρδ

=pc

Further assume that the frictionless flow Further assume that the frictionless flow through the control volume is adiabatic, through the control volume is adiabatic, then the flow is isentropic.then the flow is isentropic.

0pp →∂→δ s

pc ⎟⎟⎠

⎞⎜⎜⎝

⎛ρ∂∂

=

For isentropic flow of an ideal gasFor isentropic flow of an ideal gas

RTkkpkpk)ttancons(p

))(ttancons(p

1kk

1k

s

k

=ρ

=ρρ

=ρ=⎟⎟⎠

⎞⎜⎜⎝

⎛ρ∂∂

⇒

ρ=

−−

RTkc =

sv

p/d

dpE ⎟⎟⎠

⎞⎜⎜⎝

⎛ρ∂∂

ρ=ρρ

=Bulk modulus of elasticity for any fluidBulk modulus of elasticity for any fluid

(34)(34)

(35)(35)

ρ== vERTkc(36)(36)

(37)(37)

(38)(38)

Used to designate that partial differentiation Used to designate that partial differentiation occurs at constant entropy.occurs at constant entropy.(25)(25)

25

Example 11.3 Speed of Sound Example 11.3 Speed of Sound // SolutionSolution

Verify the speed of sound for air at 0Verify the speed of sound for air at 0℃℃ listed in Table B.4.listed in Table B.4.

In Table B.4, we find the speed of sound of air at 0In Table B.4, we find the speed of sound of air at 0℃℃ given as 331.4 m/s.

RTkc =

)Kkg/(J9.286R4.1k ⋅==

s/m4.331)kg/J(4.331)401.1)(K15.273)](Kkg/(J9.286[RTkc 2/1

=

=⋅==

26

Categories of Categories of Compressible FlowCompressible Flow

27

Compressibility and Mach Number Compressibility and Mach Number 1/21/2

The effects of compressibility become more significant as the MaThe effects of compressibility become more significant as the Mach ch number increase. The incompressible flows can only occur at low number increase. The incompressible flows can only occur at low Mach number.Mach number.The compressibility has a large influence on other important floThe compressibility has a large influence on other important flow w variables.variables.

28

Compressibility and Mach Number Compressibility and Mach Number 2/22/2

Experience has demonstrated Experience has demonstrated that compressibility can have a that compressibility can have a large influence on other large influence on other important flow variables.important flow variables.For example. The variation of For example. The variation of the drag coefficient of a sphere the drag coefficient of a sphere with Reynolds number and with Reynolds number and Mach number. Mach number. Compressibility effects can be Compressibility effects can be of considerable importance.of considerable importance.

29

To illustrate some curious To illustrate some curious features of compressible flowfeatures of compressible flow

Emission of weak pressure pluses from Emission of weak pressure pluses from a point sourcea point source

30

Emission of Pressure Pulse Emission of Pressure Pulse 1/51/5

Image the emission of weak pressure pulse from a point source.Image the emission of weak pressure pulse from a point source.These pressure wave are spherical and expand These pressure wave are spherical and expand radiallyradially outward from outward from the point source at the speed of sound, c.the point source at the speed of sound, c.For a stationary point source, the wave pattern is symmetrical.For a stationary point source, the wave pattern is symmetrical.When the point source moves to the left with a constant velocityWhen the point source moves to the left with a constant velocity, V, , V, the wave pattern is no longer symmetrical.the wave pattern is no longer symmetrical.If instead of moving the point source to the left, we held the pIf instead of moving the point source to the left, we held the point oint source stationary and source stationary and moved the fluid to the right with velocity V.moved the fluid to the right with velocity V.

31

Pressure Wave Pattern Pressure Wave Pattern 1/21/2

(a) Pressure waves at t = 3s, V = 0; (b) pressure waves at t = 3(a) Pressure waves at t = 3s, V = 0; (b) pressure waves at t = 3s, V< c.s, V< c.

32

Pressure Wave Pattern Pressure Wave Pattern 2/22/2

(c) pressure waves at t = 3s, V = c.(c) pressure waves at t = 3s, V = c. (d) pressure waves at t = 3s, V > c.(d) pressure waves at t = 3s, V > c.

33


When When the point source and the fluid are stationarythe point source and the fluid are stationary, the pressure , the pressure wave pattern is symmetrical and an observer anywhere in the wave pattern is symmetrical and an observer anywhere in the pressure field would hear the same sound frequency from the poinpressure field would hear the same sound frequency from the point t source.source.When When the velocity of the point source (or the fluid) is very smallthe velocity of the point source (or the fluid) is very small in in comparison with the speed of sound, the pressure wave pattern wicomparison with the speed of sound, the pressure wave pattern will ll still be nearly symmetrical.still be nearly symmetrical.When the point source moves in fluid at rest (or when fluid moveWhen the point source moves in fluid at rest (or when fluid moves s past a stationary point source), the pressure wave patterns varypast a stationary point source), the pressure wave patterns vary in in asymmetry, with asymmetry, with the extent of asymmetry depending on V/c.the extent of asymmetry depending on V/c.

34


When V/c < 1, the flow is When V/c < 1, the flow is considered considered subsonic subsonic and and compressible. A stationary compressible. A stationary observer will hear a different observer will hear a different sound frequency coming from the sound frequency coming from the point source depending on where point source depending on where the observer is relative to the the observer is relative to the source because the wave pattern is source because the wave pattern is asymmetrical. asymmetrical. This This phenomenon is called the phenomenon is called the Doppler effectDoppler effect..

The pressure information can still travel The pressure information can still travel unrestricted throughout the flow field, but unrestricted throughout the flow field, but not symmetrically or instantaneously.not symmetrically or instantaneously.

35


When V/c =1, pressure wave When V/c =1, pressure wave are not present ahead of the are not present ahead of the moving point source. The moving point source. The flow is flow is sonicsonic. If you were . If you were positioned to the left of the positioned to the left of the moving point source, you moving point source, you would not hear the point would not hear the point source until it was coincident source until it was coincident with your location.with your location. Mach waveMach wave

The pressure waves are all tangent to a plane that is perpendicuThe pressure waves are all tangent to a plane that is perpendicular to the flow lar to the flow and that passes through the point source. This plane is called aand that passes through the point source. This plane is called a Mach waveMach wave. . The communication of pressure information is restricted to the rThe communication of pressure information is restricted to the region of flow egion of flow downstream of the downstream of the Mach waveMach wave..

36


When V/c >1, the flow is When V/c >1, the flow is supersonicsupersonic. A cone (. A cone (Mach Mach conecone) that is tangent to the ) that is tangent to the pressure wave can be pressure wave can be constructed to represent the constructed to represent the Mach wave that Mach wave that separates the separates the zone of silence from the zone zone of silence from the zone of action.of action.

Ma1

Vcsin ==αThe angle of Mach cone (39)(39)

37

Compressible Flow VisualizationCompressible Flow Visualization

An abrupt density change can be visualized in a flow field by usAn abrupt density change can be visualized in a flow field by using ing special optics. special optics. Flow visualization methods include the Flow visualization methods include the schlierenschlieren, shadowgraph , , shadowgraph , and interferometer techniques.and interferometer techniques.

The The schlierenschlieren visualization of visualization of flow (supersonic to subsonic) flow (supersonic to subsonic) through a row of compressor through a row of compressor airfoils.airfoils.

38

Categories of Fluid FlowCategories of Fluid Flow

Incompressible flow: Ma Incompressible flow: Ma ≤≤ 0.3. Unrestricted, nearly symmetrical 0.3. Unrestricted, nearly symmetrical and instantaneous pressure communication.and instantaneous pressure communication.Compressible subsonic flow: 0.3 < Ma < 1.0. Unrestricted but Compressible subsonic flow: 0.3 < Ma < 1.0. Unrestricted but noticeably asymmetrical pressure communication.noticeably asymmetrical pressure communication.Compressible supersonic flow: Ma Compressible supersonic flow: Ma ≥≥ 1.0. Formulation of Mach 1.0. Formulation of Mach wave; pressure communication restricted to zone of action.wave; pressure communication restricted to zone of action.

Transonic flow: 0.9 Transonic flow: 0.9 ≤≤ Ma Ma ≤≤ 1.2.1.2.Hypersonic flow: Ma > 5.Hypersonic flow: Ma > 5.

39

Example 11.4 Mach ConeExample 11.4 Mach Cone

An aircraft cruising at 1000An aircraft cruising at 1000--m elevation, z, above you moves past in m elevation, z, above you moves past in a flyby. How many seconds after the plane passes overhead do youa flyby. How many seconds after the plane passes overhead do youexpect to wait before you hear the aircraft if it is moving withexpect to wait before you hear the aircraft if it is moving with a a Mach number equal to 1.5 and the ambient temperature is 20Mach number equal to 1.5 and the ambient temperature is 20℃℃??

40


The angle The angle αα is related to the elevation of the plane, z, and the is related to the elevation of the plane, z, and the ground distance, xground distance, x

Vt1000tan

xztan 11 −− ==α

Ma1sin =α

)]Vt/1000(sin[tan1aM 1−= caMV ⋅=

From Table B.4 (20From Table B.4 (20℃℃), c=343.3m/s. Using Ma=1.5), c=343.3m/s. Using Ma=1.5

s17.2t

t)s/m3.343)(5.1(1000tansin

11.51

=⇒

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡=

−

The speed of the aircraftThe speed of the aircraft

41

Isentropic Flow ThroughIsentropic Flow ThroughConvergingConverging--Diverging Duct Diverging Duct

42

Isentropic Flow of an Ideal GasIsentropic Flow of an Ideal Gas

Consider the steady, oneConsider the steady, one--dimensional, isentropic flow of an ideal dimensional, isentropic flow of an ideal gas with constant specific heat values (cgas with constant specific heat values (cpp and and ccvv). ). Shaft work cannot be involved.Shaft work cannot be involved.Consider flows through finite control volume with Consider flows through finite control volume with uniformly uniformly distributed velocitiesdistributed velocities and fluid properties at each section of flow. and fluid properties at each section of flow. An isentropic flow is not achievable with actual fluids because An isentropic flow is not achievable with actual fluids because of of friction. Nonetheless, the study of isentropic flow trends is usfriction. Nonetheless, the study of isentropic flow trends is useful eful because because it helps us to gain an understanding of actual it helps us to gain an understanding of actual compressible flow phenomenacompressible flow phenomena..

43

Effect of CrossEffect of Cross--Sectional Area Sectional Area 1/101/10

When fluid flows steadily through a conduit that has a flow crosWhen fluid flows steadily through a conduit that has a flow crosss--sectional area that varies with axial distance, the conservationsectional area that varies with axial distance, the conservation of of mass (continuity) equationmass (continuity) equation

In In chapter 3chapter 3, Newton, Newton’’s second law was applied to the s second law was applied to the inviscidinviscid(frictionless) and steady flow of a fluid particle. For(frictionless) and steady flow of a fluid particle. For either either compressible or incompressible flow along the compressible or incompressible flow along the streamwisestreamwise direction,direction,

ttanconsAVm =ρ=&

0dz)V(d21dp 2 =γ+ρ+

(40)(40)

(41)(41)

Can be dropped because of its small size in comparison to the other terms

44


An appropriate equation of motion in the An appropriate equation of motion in the streamwisestreamwise direction for direction for the steady, onethe steady, one--dimensional, and isentropic (adiabatic and dimensional, and isentropic (adiabatic and frictionless) flow of an ideal gas is obtained from frictionless) flow of an ideal gas is obtained from EqsEqs. (41).. (41).

VdV

Vdp

2 −=ρ

ttanconsVlnAlnlnttanconsAVm =++ρ⇒=ρ=&

AdAd

VdV 0

VdV

AdAd

+ρρ

=−

=++ρρ

(42)(42)

(43)(43)

(44)(44)

(42)+(44)(42)+(44)A

dAd/dp

V1Vdp 2

2 =⎟⎟⎠

⎞⎜⎜⎝

⎛ρ

−ρ

(45)(45)

(43)(43)

0dz)V(d21dp 2 =γ+ρ+ 2Vρ÷

45


AdA)aM1(

Vdp 2

2 =−ρ

cVMa =

)aM1(1

AdA

VdV

2−−=

(34)+(45)+(46)(34)+(45)+(46)

Mach number definitionMach number definition (46)(46)

(47)(47)

)aM1(aM

AdAd

2

2

−−=

ρρ

(42)+(47)(42)+(47) (48)(48)

(44)+(48)(44)+(48) (49)(49)

s

pc ⎟⎟⎠

⎞⎜⎜⎝

⎛ρ∂∂

= (34)(34)

46


)aM1(1

AdA

VdV

2−−=

((aa) A diverging duct. () A diverging duct. (bb) A converging duct.) A converging duct.

When the flow is subsonic When the flow is subsonic (Ma<1), velocity and section (Ma<1), velocity and section area changes are area changes are in oppositein oppositedirections.directions.When the flow is supersonic When the flow is supersonic (Ma>1), velocity and area (Ma>1), velocity and area changes changes in the samein the same direction.direction.

(48)(48)

47


A diverging duct will accelerate a A diverging duct will accelerate a supersonic flow.supersonic flow.A converging duct will decelerate a A converging duct will decelerate a supersonic flow.supersonic flow.Subsonic flow through a converging Subsonic flow through a converging duct involves an increase of velocity.duct involves an increase of velocity.Subsonic flow through a diverging Subsonic flow through a diverging duct is accompanied by a velocity duct is accompanied by a velocity decrease.decrease.

((aa) A diverging duct.) A diverging duct.((bb) A converging duct.) A converging duct.

48


For subsonic flow (Ma<1), density and area changes are For subsonic flow (Ma<1), density and area changes are in the in the samesame direction.direction.For supersonic flow (Ma>1), density and area changes are For supersonic flow (Ma>1), density and area changes are in the in the oppositeopposite direction.direction.

)aM1(aM

AdAd

2

2

−−=

ρρ

(49)(49)

49


)aM1(VA

dVdA

)aM1(1

AdA

VdV 2

2 −−=⇒−

−=

For Ma =1, this equation requires that For Ma =1, this equation requires that dA/dVdA/dV=0.=0.This means that the area associated with Ma=1 is either a minimuThis means that the area associated with Ma=1 is either a minimum m or a maximum amount.or a maximum amount.

If the flow enter convergingIf the flow enter converging--divergingdiverging duct duct were were subsonic,subsonic, the fluid velocity would the fluid velocity would increase in the converging portion of the increase in the converging portion of the duct, and achievement of a sonic condition duct, and achievement of a sonic condition (Ma=1) at the (Ma=1) at the minimum area locationminimum area locationappears possible.appears possible.

(50)(50)

50


If the flow enter convergingIf the flow enter converging--divergingdiverging ductductwere were supersonic,supersonic, the fluid velocity would the fluid velocity would decrease in the converging portion of the decrease in the converging portion of the duct and the sonic condition at the minimum duct and the sonic condition at the minimum area is possible.area is possible.

If the flow enter divergingIf the flow enter diverging--convergingconverging ductductwere were subsonicsubsonic, the fluid velocity would , the fluid velocity would decreasedecrease in the diverging portion of the duct in the diverging portion of the duct and the sonic condition and the sonic condition could not be could not be attainedattained at the maximum area.at the maximum area.

51


If the flow enter divergingIf the flow enter diverging--convergingconverging ductductwerewere supersonicsupersonic, the fluid velocity would , the fluid velocity would increaseincrease in the diverging portion of the duct in the diverging portion of the duct and the sonic condition and the sonic condition could not be could not be attainedattained at the maximum area.at the maximum area.

52

Conclusion Conclusion 10/1010/10

1.1. Sonic condition (Ma=1) can be attained in a convergingSonic condition (Ma=1) can be attained in a converging--diverging diverging duct at the minimum area location.duct at the minimum area location.

2.2. This minimum area location is often called the This minimum area location is often called the throatthroat of the of the convergingconverging--diverging duct. diverging duct.

3.3. To achieve To achieve supersonic flow from a subsonic statesupersonic flow from a subsonic state in a duct, a in a duct, a convergingconverging--diverging area variations is necessarydiverging area variations is necessary..

4.4. A convergingA converging--diverging duct is referred as a convergingdiverging duct is referred as a converging--diverging diverging nozzle.nozzle.

5.5. A convergingA converging--diverging duct can also decrease a supersonic flow to diverging duct can also decrease a supersonic flow to subsonic conditions.subsonic conditions.

6.6. A convergingA converging--diverging duct can be a nozzle or a diffuser depending diverging duct can be a nozzle or a diffuser depending on whether the flow in the converging portion of the duct is on whether the flow in the converging portion of the duct is subsonic or supersonic.subsonic or supersonic.

53

ConvergingConverging--Diverging Duct Flow Diverging Duct Flow 1/111/11

To To develop equationsdevelop equations that help us determine how other important that help us determine how other important flow properties vary in the steady isentropic flow of an ideal gflow properties vary in the steady isentropic flow of an ideal gas as through a variable area duct.through a variable area duct.

((aa) A converging) A converging--diverging duct. (diverging duct. (bb) A diverging) A diverging--converging duct.converging duct.

The atmospheric pressure The atmospheric pressure and temperature would and temperature would represent the stagnation state represent the stagnation state of the flowing fluid.of the flowing fluid.

54


For the isentropic flow of an ideal gasFor the isentropic flow of an ideal gas

The The streamwisestreamwise equation of motion for steady and frictionless flow equation of motion for steady and frictionless flow can be expressed for an ideal gascan be expressed for an ideal gas

k0

0k

pttanconspρ

==ρ

02

Vddp 2=⎟

⎟⎠

⎞⎜⎜⎝

⎛+

ρ

02

Vdpdpp 2

k/1o

k/10 =⎟⎟

⎠

⎞⎜⎜⎝

⎛+

ρ

It is convenient to use the stagnation state of the fluid as a referenceas a reference state for compressible flow calculations.

Stagnation stateStagnation state

Note: The stagnation state is can Note: The stagnation state is can also be achieved by also be achieved by isentropicallyisentropicallydecelerating a flow to zero velocity.decelerating a flow to zero velocity.

(25)(25)

(41)(41) (51)(51)

Incorporating (25) into (51)Incorporating (25) into (51) (52)(52)

55


Integrating (52) (52) between the common stagnation statestagnation state of the flowing fluid to the state of gas at any location in the converging-diverging duct

02

Vpp1k

k 2

0

0 =−⎟⎟⎠

⎞⎜⎜⎝

⎛ρ

−ρ−

( ) 02

VTTc2

oP =−− 02

Vhh2

o =⎟⎟⎠

⎞⎜⎜⎝

⎛+−

((

Stagnation enthalpyStagnation enthalpy

( ) 02

VTT1k

kR 2

o =−−−

(53)(53)

(54)(54)(53)+(1)(53)+(1)

(14)+(54)(14)+(54) (55)(55)

( )12P12 TTchh −=−((

1kkRcP −

= (14)(14)

56


With this equation we can calculate the temperature of an With this equation we can calculate the temperature of an ideal gas anywhere in the convergingideal gas anywhere in the converging--diverging duct if the diverging duct if the flow is steady, oneflow is steady, one--dimensional, and isentropic, dimensional, and isentropic, provided provided we know the value of the local Mach number and the we know the value of the local Mach number and the stagnation temperature.stagnation temperature.

2a2

1ko M11

TT

−+= Pressure variation ?Pressure variation ?

(46)+(36)+(54)(46)+(36)+(54)

(56)(56)

RTkc = (36)(36)

cVMa = (46)(46)

( ) 02

VTT1k

kR 2

o =−−−

(54)(54)

57


Develop an equation for Develop an equation for pressure and density variationpressure and density variation

RTp=

ρ o

o

o TT

pp

=ρρ

1k/k

oo TT

pp

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

1k/k

22

1ko Ma11

pp

−

− ⎥⎥⎦

⎤

⎢⎢⎣

⎡

+=

1k/1

22

1ko Ma11

−

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+=

ρρ

−

(57)(57)

(57)+(25)(57)+(25) (58)(58)

(58)+(56)(58)+(56) (59)(59)

(56)+(57)+(59)(56)+(57)+(59) (60)(60)

k0

0k

pttanconspρ

==ρ (25)(25)

58

TT--s Diagram For Isentropic Flow s Diagram For Isentropic Flow

A useful means of keeping track of the state of an isentropic flA useful means of keeping track of the state of an isentropic flow of ow of an ideal gas involves a an ideal gas involves a temperaturetemperature--entropy (Tentropy (T--s) diagram.s) diagram.An isentropic flow is confined An isentropic flow is confined to a vertical line on a Tto a vertical line on a T--s s diagram. The vertical line is diagram. The vertical line is representative of flow between representative of flow between the stagnation state and any the stagnation state and any state within the convergingstate within the converging--diverging nozzle.diverging nozzle.

59


22

1ko Ma11

TT

−+=

1k/k

22

1ko Ma11

pp

−

− ⎥⎥⎦

⎤

⎢⎢⎣

⎡

+=

Fluid temperature decreases with Fluid temperature decreases with an increase in Mach number.an increase in Mach number.

Fluid pressure decreases with Fluid pressure decreases with an increase in Mach number.an increase in Mach number.

Lower fluid temperature and pressures are Lower fluid temperature and pressures are associated with higher Mach number in an associated with higher Mach number in an isentropic convergingisentropic converging--diverging duct.diverging duct.

TT Ma Ma

pp Ma Ma

(56)(56)

(59)(59)

60


One way to produce flow through a convergingOne way to produce flow through a converging--diverging duct is to diverging duct is to connect the downstream end of the convergingconnect the downstream end of the converging--diverging duct to a diverging duct to a vacuum pump.vacuum pump.When the pressure at the downstream end of the duct (the back When the pressure at the downstream end of the duct (the back pressure) is decreased slightly, air will flow from the atmosphepressure) is decreased slightly, air will flow from the atmosphere re through the duct and vacuum pump.through the duct and vacuum pump.

22

1k0 Ma11

TT

−+=

1k/k

22

1k0 Ma1

1pp

−

− ⎥⎦

⎤⎢⎣

⎡+

=

1k/1

22

1k0 Ma1

1−

⎥⎦

⎤⎢⎣

⎡+

=ρρ

−

Used to describe steady Used to describe steady flow through the flow through the convergingconverging--diverging duct.diverging duct.

+ T+ T--s diagrams diagram

61


If the pressure in the duct is only slightly less than atmospherIf the pressure in the duct is only slightly less than atmospheric ic pressure, the Mach number levels in the duct will be low pressure, the Mach number levels in the duct will be low EqsEqs. (59). (59)and the variation of density in the duct is also small and the variation of density in the duct is also small EqsEqs. (60). (60)..

There is a small amount of There is a small amount of fluid flow acceleration in the fluid flow acceleration in the converging portion of the duct converging portion of the duct followed by flow deceleration followed by flow deceleration in the diverging portion of the in the diverging portion of the duct. duct. EqsEqs. (40). (40)

The TThe T--s diagram for s diagram for VenturiVenturi meter flowmeter flow

62


When the When the back pressure is lowerback pressure is lower further, the fluid flow from the further, the fluid flow from the rest upstream of the converging portion may be accelerated to arest upstream of the converging portion may be accelerated to amaximum maximum Mach number 1 at the nozzle throatMach number 1 at the nozzle throat..When Ma=1 at the throat of the convergingWhen Ma=1 at the throat of the converging--diverging duct, we have diverging duct, we have a condition called a condition called choked flowchoked flow..Use the stagnation state for which Ma=0 as a reference conditionUse the stagnation state for which Ma=0 as a reference condition..Use the state associated with Ma=1 as Use the state associated with Ma=1 as another reference conditionanother reference conditioncalled critical state, denoted ( )called critical state, denoted ( )**..

1k/k

22

1ko Ma11

pp

−

− ⎥⎥⎦

⎤

⎢⎢⎣

⎡

+= Ma=1Ma=1 1k/k

o

*

1k2

pp −

⎥⎦⎤

⎢⎣⎡

+= (61)(61)(59)(59)

Critical pressure ratioCritical pressure ratio

63


Ma=1Ma=11k

2TT

o

*

+=

22

1ko Ma11

TT

−+=

1k1

*o

o

*

o

*

1k2

TT

Pp −

⎟⎠⎞

⎜⎝⎛

+=⎟

⎠

⎞⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

ρρ

The relationship between the The relationship between the stagnation and critical states.stagnation and critical states.

(63)(63)

(65)(65)

64


For k=1.4, the nominal value of k for airFor k=1.4, the nominal value of k for air

atm*

4.1k

atm*

4.1k

atm*

4.1k

634.0

T833.0T

p528.0p

ρ=ρ

=

=

=

=

= (62)(62)

(64)(64)

(66)(66)

65

Example 11.5 Isentropic Flow in a Example 11.5 Isentropic Flow in a Converging DuctConverging Duct

A converging duct passes air steadily from standard atmospheric A converging duct passes air steadily from standard atmospheric conditions to a receiver pipe as illustrated in Figure E11.5a. Tconditions to a receiver pipe as illustrated in Figure E11.5a. The he throat (minimum) flow crossthroat (minimum) flow cross--sectional area of the converging ducts sectional area of the converging ducts is 1is 1××1010--44 mm22. Determine the mass . Determine the mass flowrateflowrate through the duct if the through the duct if the receiver pressure is (a) 80 receiver pressure is (a) 80 kPakPa (abs), (b) 40 (abs), (b) 40 kPakPa (abs). Sketch (abs). Sketch temperaturetemperature--entropy diagrams for situations (a) and (b)entropy diagrams for situations (a) and (b)

66


ttanconsAVm =ρ=& (40)(40)

To determine the mass flowrate through the converging duct we use (40)

ththth VAm ρ=&1k/1

2th2

1ko

th

Ma11

−

⎥⎦

⎤⎢⎣

⎡+

=ρρ

−(60)(60)

1k/k

2th2

1ko

th

Ma11

pp

−

− ⎥⎦

⎤⎢⎣

⎡+

=(59)(59)

4.1km/kg23.1 30 ==ρ

(62)(62) )abs(kPa3.53p528.0p528.0p atm0*

4.1k ====

Stagnation densityStagnation density

(11.5.1)(11.5.1)

(11.5.2)(11.5.2)

(11.5.3)(11.5.3)

67


If the receiver pressure, pIf the receiver pressure, prere is greater than or equal to p*, then is greater than or equal to p*, then ppthth=p=prere. If p. If prere<p*, the <p*, the ppthth=p* and the flow is choked.=p* and the flow is choked.With With ppthth, p, p00, and k known, Ma, and k known, Mathth can be obtained from can be obtained from EqsEqs. (11.5.3. (11.5.3))and and ppthth can be obtained from can be obtained from EqsEqs. (11.5.2).. (11.5.2).

kRTMacMaV ththththth ==

The flow velocity at the throatThe flow velocity at the throat

(11.5.4)(11.5.4)(36)+(46)(36)+(46)

The value of temperature at the throat The value of temperature at the throat TTthth

(56)(56) 2th2

1ko

th

Ma11

TT

−+= (11.5.5)(11.5.5)

68


For pFor prere=80 =80 kPakPa (abs) > 53.3 (abs) > 53.3 kPakPa (abs) = p*, we have (abs) = p*, we have ppthth=80 =80 kPakPa..Then from Then from EqEq. (11.5.3). (11.5.3)

1k/k

2th2

1ko

th

Ma11

pp

−

− ⎥⎦

⎤⎢⎣

⎡+

= (11.5.3)(11.5.3) 587.0Math =

From From EqEq. (11.5.2). (11.5.2)1k/1

2th2

1ko

th

Ma11

−

⎥⎦

⎤⎢⎣

⎡+

=ρρ

−(11.5.2)(11.5.2) 3

th m/kg04.1=ρ

From From EqEq. (11.5.5). (11.5.5)

(11.5.5)(11.5.5) K269Tth =2th2

1ko

th

Ma11

TT

−+=

69


From From EqEq. (11.5.4). (11.5.4)

(11.5.4)(11.5.4) s/m193)kg/J(193V 2/1th ==

From From EqEq. (11.5.1). (11.5.1)

(11.5.1)(11.5.1) s/kg0201.0m =&

kRTMacMaV ththththth ==

ththth VAm ρ=&

70


For pFor prere=40 =40 kPakPa (abs) < 53.3 (abs) < 53.3 kPakPa (abs) = p*, we have (abs) = p*, we have ppthth=p*=53.3kPa (abs) and Ma=p*=53.3kPa (abs) and Mathth=1. The converging duct is chocked. =1. The converging duct is chocked. Then from Then from EqEq. (11.5.2). (11.5.2)

1k/1

2th2

1ko

th

Ma11

−

⎥⎦

⎤⎢⎣

⎡+

=ρρ

−(11.5.2)(11.5.2) 3

th m/kg780.0=ρ

From From EqEq. (11.5.5). (11.5.5)

(11.5.5)(11.5.5) K240Tth =2th2

1ko

th

Ma11

TT

−+=

s/m310)kg/J(310V 2/1th == s/kg0242.0m =&

71


72

Example 11.6 Use of Compressible Flow Example 11.6 Use of Compressible Flow Graphs in Solving ProblemsGraphs in Solving Problems

Solve Example 11.5 using Solve Example 11.5 using Figure D.1 of Appendix DFigure D.1 of Appendix D..

73


Need the density and velocity of the air at the converging duct Need the density and velocity of the air at the converging duct throat to solve for mass throat to solve for mass flowrateflowrate fromfrom

ththth VAm ρ=& (11.6.1)(11.6.1)

Since the receiver pressure, pSince the receiver pressure, prere=80 =80 kPakPa (abs) > 53.3 (abs) > 53.3 kPakPa (abs) = p*, (abs) = p*, we have we have ppthth= p= prere..

792.0kPa101kPa80

pp

o

th ==

From Figure D.1, From Figure D.1, for p/pfor p/p00=0.792=0.792, we get from the graph, we get from the graph

85.094.0TT59.0Ma th

0

thth =

ρρ

== (11.6.2)(11.6.2) (11.6.3)(11.6.3)

74


From (11.6.2) (11.6.3)

s/kg0202.0)...m/kg04.1(VAm 3ththth ==ρ=&

33thth m/kg04.1)m/kg23.1)(85.0(K271)K288)(94.0(T ==ρ==

(36)+(46) s/m194...kRTMacMaV ththththth ====

For the receiver pressure, pFor the receiver pressure, prere=40 =40 kPakPa (abs) < 53.3 (abs) < 53.3 kPakPa (abs) = p*, (abs) = p*, we have we have ppthth= 53.3 = 53.3 kPakPa (abs)..(abs)..

From Figure D.1, From Figure D.1, for Ma=1for Ma=1, we get from the graph, we get from the graph

64.083.0TT th

0

th =ρρ

= (11.6.4)(11.6.4) (11.6.5)(11.6.5)

75


From (11.6.4) (11.6.5)From (11.6.4) (11.6.5)

s/kg024.0)...m/kg04.1(VAm 3ththth ==ρ=&

33thth m/kg79.0)m/kg23.1)(64.0(K240)K288)(83.0(T ==ρ==

(36)+(46)(36)+(46) s/m310...kRTMacMaV ththththth ====

76

Example 11.7 Static to Stagnation Example 11.7 Static to Stagnation Pressure RatioPressure Ratio

The static pressure to stagnation pressure ratio at a point in aThe static pressure to stagnation pressure ratio at a point in a flow flow stream is measured with a stream is measured with a PitotPitot--static tube (Figure 3.6) as being static tube (Figure 3.6) as being equal to 0.82. The stagnation temperature of the fluid is 68equal to 0.82. The stagnation temperature of the fluid is 68ººFF. . Determine the flow velocity if the fluid is (a) air, (b) helium.Determine the flow velocity if the fluid is (a) air, (b) helium.

Figure 3.6Figure 3.6The The PitotPitot--static tubestatic tube

77

Example 11.7 Example 11.7 Solution Solution 1/21/2

(36)+(46)(36)+(46) RTkMaV =

To determine the flow velocityTo determine the flow velocity

For air, For air, p/pp/p00=0.82=0.82; thus from Figure D.1; thus from Figure D.1

0.94TT0.54Ma0==

R496R]460)(0.94)[(68T °=°+=

s/ft590)lbm/lbft(104)...54.0(RTkMaV 2/1 =⋅===

1.4k =

78

Example 11.7 Example 11.7 Solution Solution 2/22/2

For helium, For helium, p/pp/p00=0.82=0.82 and and

R488R]460)[(68)499.0](2/)166.1[(1

1T 2 °=°+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−+=

s/ft1580)lbm/lbft(279)...499.0(RTkMaV 2/1 =⋅===

1.66k =

1k/k

22

1ko Ma11

pp

−

− ⎥⎥⎦

⎤

⎢⎢⎣

⎡

+= (59)(59) 0.499Ma =

22

1ko Ma11

TT

−+= (56)(56)

79

Choked Flow Choked Flow 1/21/2

When Ma=1 at the throat of the convergingWhen Ma=1 at the throat of the converging--diverging duct, we have diverging duct, we have a condition called a condition called choked flowchoked flow..For choked flow through the convergingFor choked flow through the converging--diverging duct, the diverging duct, the conservation of mass equation yieldconservation of mass equation yield

0

0*

o

o

*

* T/TT/T

Ma1

AA

⎟⎟⎠

⎞⎜⎜⎝

⎛ρρ

⎟⎟⎠

⎞⎜⎜⎝

⎛

ρρ

=

*** VAAV ρ=ρ ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

ρρ

=VV

AA **

*

kRTV ** =

(40)(40) (67)(67)

(36)+(46)(36)+(46) RTkMaV =(68)(68) (69)(69)

(67)+(68)+(69)(67)+(68)+(69) (70)(70)

80

Choked Flow Choked Flow 2/22/2

The variation of area ratio with Mach number The variation of area ratio with Mach number

For isentropic flow For isentropic flow of an ideal gas (k=1.4)of an ideal gas (k=1.4)

)1k(2/)1k(2

* 2/)1k(1Ma]2/)1k[(1

Ma1

AA

−+

⎭⎬⎫

⎩⎨⎧

−+−+

=(56)+(60)+(56)+(60)+(63)+(65)+(70)(63)+(65)+(70) (71)(71)

The variation of area ratio with Mach The variation of area ratio with Mach number for isentropic flow of an ideal number for isentropic flow of an ideal gas (gas (kk = 1.4, linear coordinate scales).= 1.4, linear coordinate scales).

81

Figure D.1Figure D.1

)1k(2/)1k(2

* 2/)1k(1Ma]2/)1k[(1

Ma1

AA

−+

⎭⎬⎫

⎩⎨⎧

−+−+

=

Value of p/pValue of p/p00, T/T, T/T00, , ρρ//ρρ00, and A/A* are graphed in Figure D.1 as a , and A/A* are graphed in Figure D.1 as a function of Mach number for air (k=1.4).function of Mach number for air (k=1.4).

22

1ko Ma11

TT

−+=

1k/k

22

1ko Ma11

pp

−

− ⎥⎥⎦

⎤

⎢⎢⎣

⎡

+=

(56)(56)

(59)(59)

1k/1

22

1ko Ma11

−

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+=

ρρ

−(60)(60)

(71)(71)

82

Example 11.8 Isentropic Choked Flow in a Example 11.8 Isentropic Choked Flow in a ConvergingConverging--Diverging Duct with Subsonic EntryDiverging Duct with Subsonic Entry

Air enters Air enters subsonicallysubsonically from standard atmosphere and flows from standard atmosphere and flows isentropicallyisentropically through a choked convergingthrough a choked converging--diverging duct having a diverging duct having a circular crosscircular cross--sectional area, A, that varies with axial distance from sectional area, A, that varies with axial distance from the throat, x, according to the formulathe throat, x, according to the formula

A=0.1+xA=0.1+x22

where A is in square meters and x is in meters. The duct extwhere A is in square meters and x is in meters. The duct extends ends from x = from x = --0.5 m to x = +0.5 m. For this flow situation, sketch the 0.5 m to x = +0.5 m. For this flow situation, sketch the side view of the duct and graph the variation of Mach number, stside view of the duct and graph the variation of Mach number, static atic temperature to stagnation temperature ratio, T/Ttemperature to stagnation temperature ratio, T/T00, and static pressure , and static pressure to stagnation pressure ratio, p/pto stagnation pressure ratio, p/p00, through the duct from x = , through the duct from x = --0.5 m 0.5 m to x = +0.5 m. Also show the possible fluid states at x = to x = +0.5 m. Also show the possible fluid states at x = --0.5 m, 0 m, 0.5 m, 0 m, and +0.5 m using temperature and +0.5 m using temperature –– entropy coordinates.entropy coordinates.

83


22 x1.0AwhererA +=π=

The side view of the convergingThe side view of the converging--diverging duct is a graph of radius r diverging duct is a graph of radius r from the duct axis as a function of axial distance. For a circulfrom the duct axis as a function of axial distance. For a circular flow ar flow cross section we havecross section we have

1/22x0.1r ⎟⎟⎠

⎞⎜⎜⎝

⎛

π+

= (11.8.3)(11.8.3)

84


1.0x1.0

*AAm1.0A*

22 +

=⇒=

Since the convergingSince the converging--diverging duct is chocked, the diverging duct is chocked, the throat areathroat area is is the critical area, A*the critical area, A*

)1k(2/)1k(2

* 2/)1k(1Ma]2/)1k[(1

Ma1

AA

−+

⎭⎬⎫

⎩⎨⎧

−+−+

= (71)(71)

Values of A/A* can be used in Values of A/A* can be used in EqEq. (71). (71) to calculate corresponding to calculate corresponding values of Mach number, Ma.values of Mach number, Ma.

(11.8.5)(11.8.5)

85


For air with k=1.4, we could enter For air with k=1.4, we could enter Figure D.1Figure D.1 with values of A/A* with values of A/A* and read off values of Mach number.and read off values of Mach number.

With values of Mach number ascertained, we could use With values of Mach number ascertained, we could use EqsEqs. . 56 and 5956 and 59 to calculate related value of T/Tto calculate related value of T/T00 and p/pand p/p00..

22

1ko Ma11

TT

−+=

1k/k

22

1ko Ma11

pp

−

− ⎥⎥⎦

⎤

⎢⎢⎣

⎡

+=

(56)(56)

(59)(59)

For k=1.4 and value For k=1.4 and value of A/A*; thus from of A/A*; thus from Figure D.1 Figure D.1 MaMa

T/TT/T00p/pp/p00

86


1/22x0.1r ⎟⎟⎠

⎞⎜⎜⎝

⎛

π+

=

1.0x1.0

*AA 2+

=

Figure D.1

87


With the air entering the chocked convergingWith the air entering the chocked converging--diverging duct diverging duct subsonicallysubsonically, only one isentropic solution exits for the , only one isentropic solution exits for the converging converging portionportion of the duct.of the duct.Two isentropic flow solutions are possible for the Two isentropic flow solutions are possible for the diverging diverging portionportion of the duct of the duct –– one subsonic, the other supersonic.one subsonic, the other supersonic.If the pressure ratio, p/pIf the pressure ratio, p/p00, is set at 0.98 at x = +0.5 m, the subsonic , is set at 0.98 at x = +0.5 m, the subsonic flow will occur.flow will occur.Alternatively. if p/pAlternatively. if p/p00 is set at 0.04 at x = +0.5 m, the supersonic flow is set at 0.04 at x = +0.5 m, the supersonic flow field will exist.field will exist.

88


The solution values for the entire ductThe solution values for the entire duct

89

Example 11.9 Example 11.9 Isentropic Choked Flow in a Isentropic Choked Flow in a ConvergingConverging--Diverging Duct with Supersonic EntryDiverging Duct with Supersonic Entry

Air enters supersonically with TAir enters supersonically with T00 and pand p00 equal to standard equal to standard atmosphere values and flows atmosphere values and flows isentropicallyisentropically through the chocked through the chocked convergingconverging--diverging duct described in Example 11.8. Graph the diverging duct described in Example 11.8. Graph the variation of Mach number, Ma, static temperature to stagnation variation of Mach number, Ma, static temperature to stagnation temperature ratio, T/Ttemperature ratio, T/T00, and static pressure to stagnation pressure , and static pressure to stagnation pressure ratio, p/pratio, p/p00, through the duct from x = , through the duct from x = --0.5 m to x = +0.5 m. Also 0.5 m to x = +0.5 m. Also show the possible fluid states at x = show the possible fluid states at x = --0.5 m, 0 m, and +0.5 m using 0.5 m, 0 m, and +0.5 m using temperature temperature –– entropy coordinates.entropy coordinates.

90


With the air entering the convergingWith the air entering the converging--diverging duct of Example 11.8 diverging duct of Example 11.8 supersonically instead of supersonically instead of subsonicallysubsonically, a unique isentropic flow , a unique isentropic flow solution is obtained for the converging portion of the duct.solution is obtained for the converging portion of the duct.Now, however, the flow decelerates to the sonic condition at theNow, however, the flow decelerates to the sonic condition at thethroat.throat.The two solutions obtained previously in Example 11.8 for the The two solutions obtained previously in Example 11.8 for the diverging portion are still valid.diverging portion are still valid.Since Since the area variation in the duct is symmetrical with respect to the area variation in the duct is symmetrical with respect to the duct throat, we can use the supersonic flow values obtained the duct throat, we can use the supersonic flow values obtained from from Example 11.8 for the supersonic flow in the converging portion oExample 11.8 for the supersonic flow in the converging portion of f the duct.the duct.

91


The supersonic flow solution for the converging passageThe supersonic flow solution for the converging passage

92


93

Example 11.10 Example 11.10 Isentropic Isentropic UnchokedUnchoked Flow Flow in a Convergingin a Converging--Diverging DuctDiverging Duct

Air enters Air enters subsonicallysubsonically and and isentropicallyisentropically through the convergingthrough the converging--diverging duct described in Example 11.8. Graph the variation ofdiverging duct described in Example 11.8. Graph the variation ofMach number, Ma, static temperature to stagnation temperature raMach number, Ma, static temperature to stagnation temperature ratio, tio, T/TT/T00, and static pressure to stagnation pressure ratio, p/p, and static pressure to stagnation pressure ratio, p/p00, through , through the duct from x = the duct from x = --0.5 m to x = +0.5 m for Ma=0.48. Show the 0.5 m to x = +0.5 m for Ma=0.48. Show the corresponding temperature corresponding temperature –– entropy coordinates.entropy coordinates.

94


Since for this example, Ma=0.48 at x=0m, the isentropic low Since for this example, Ma=0.48 at x=0m, the isentropic low through the convergingthrough the converging--diverging duct will be entirely subsonic and diverging duct will be entirely subsonic and not chocked.not chocked.For air (k=1.4) flowing For air (k=1.4) flowing isentropicallyisentropically through the duct, we can use through the duct, we can use Figure D.1 for flow field quantities. Entering Figure D.1 with Figure D.1 for flow field quantities. Entering Figure D.1 with Ma=0.48 we read off p/pMa=0.48 we read off p/p00=0.85, T/T=0.85, T/T00=0.96, and A/A*=1.4.=0.96, and A/A*=1.4.Even though the duct flow is not chocked in this example and A* Even though the duct flow is not chocked in this example and A* does not therefore exist physically, it still represents a validdoes not therefore exist physically, it still represents a validreference.reference.For given isentropic flow, pFor given isentropic flow, p00, T, T00, and A* are constants., and A* are constants.

95


At x=0 mAt x=0 m 2m07.0*)A/A(

AA* ==22 m10.0x1.0A =+=

With known values of duct With known values of duct area at different axial locations, area at different axial locations, we can calculate we can calculate corresponding area ratios, corresponding area ratios, A/A*, knowing A*=0.07mA/A*, knowing A*=0.07m22..With values of area ratio A/A*, With values of area ratio A/A*, we can we can use Figure D.1use Figure D.1 and and obtain related values of Ma, obtain related values of Ma, T/TT/T00, and p/p, and p/p00..

96


97

Area Ratio vs. Mach Number Area Ratio vs. Mach Number 1/31/3

The isentropic flow behavior for the convergingThe isentropic flow behavior for the converging--diverging duct is diverging duct is summarized in the area ratio summarized in the area ratio –– Mach number graphs.Mach number graphs.The points a, b, and c represent states at axial distance x=The points a, b, and c represent states at axial distance x=--0.5m, 0m, 0.5m, 0m, and +0.5m.and +0.5m.

((aa) Subsonic to subsonic isentropic flow (not choked). ) Subsonic to subsonic isentropic flow (not choked). ((bb) Subsonic to subsonic isentropic flow (Choked).) Subsonic to subsonic isentropic flow (Choked).

98


((cc) Subsonic to supersonic isentropic flow (choked), ) Subsonic to supersonic isentropic flow (choked), ((dd)) SSuperuperssoonniicc ttoo ssuupersonic ipersonic isseennttrroopicpic flofloww (c(chhookkeedd))..

99


((ee)) SSuuperperssoonniic to subsonic isc to subsonic iseennttrrooppiicc flflooww (ch(chookkeded))..((ff) ) SSupupererssoonniicc ttoo ssububssoonniicc iisseennttrropic flow (not choked).opic flow (not choked).

100

Solution Solution 1/31/3

For a given For a given stagnation state (i.e., Tstagnation state (i.e., T00 and pand p00 fixed), ideal gas fixed), ideal gas (k=constant), and converging(k=constant), and converging--diverging duct geometry,diverging duct geometry, an an infinite numberinfinite number of isentropic subsonic to subsonic (not choked) and of isentropic subsonic to subsonic (not choked) and supersonic to supersonic (not choked) flow solutions exist.supersonic to supersonic (not choked) flow solutions exist.In contrast, the isentropic In contrast, the isentropic subsonic to supersonic (choked), subsonic subsonic to supersonic (choked), subsonic to subsonic (choked), supersonic to subsonic (choked), and to subsonic (choked), supersonic to subsonic (choked), and supersonic to supersonic (choked) from solutionssupersonic to supersonic (choked) from solutions are each uniqueare each unique..

Given stagnation state (i.e., TGiven stagnation state (i.e., T00 and pand p00 fixed), ideal gas fixed), ideal gas (k=constant), and converging(k=constant), and converging--diverging duct geometry.diverging duct geometry.Solution?Solution?

101


When the pressure at x=+0.5 (exit) is greater than or equal to When the pressure at x=+0.5 (exit) is greater than or equal to ppII, , isentropic flows are possible.isentropic flows are possible.When the pressure at x=+0.5 (exit) is less than or equal to When the pressure at x=+0.5 (exit) is less than or equal to ppIIII, , isentropic flows are possible.isentropic flows are possible.

When the exit pressure is When the exit pressure is less than less than ppII and greater than and greater than ppIIII, , isentropic flows are no longer possible.isentropic flows are no longer possible.

102


(a) The variation of duct radius with axial distance.

(b) The variation of Mach number with axial distance.

(c) The variation of temperature with axial distance.

(d) The variation of pressure with axial distance.

103

Normal Shock Wave Normal Shock Wave 1/21/2

When the exit pressure is less than When the exit pressure is less than ppII and greater than and greater than ppIIII, , nonisentropicnonisentropic chocked flowschocked flows are possible.are possible.Each abrupt pressure rise within and at the exit of the flow pasEach abrupt pressure rise within and at the exit of the flow passage sage occurs across a very thin discontinuity in the flow called a occurs across a very thin discontinuity in the flow called a normal normal shock wave. shock wave. Except for flow across the normal shock wave, the Except for flow across the normal shock wave, the flow is isentropic. flow is isentropic.

104


The less abrupt pressure rise or drop that occurs after the flowThe less abrupt pressure rise or drop that occurs after the flowleaves the duct is leaves the duct is nonisentropicnonisentropic and attributable to threeand attributable to three--dimensional dimensional oblique shock wavesoblique shock waves or or expansion waveexpansion wave..If the pressure rises downstream of the duct exit, the flow is If the pressure rises downstream of the duct exit, the flow is considered considered overexpandedoverexpanded..If the pressure drops downstream of the duct exit, the flow is cIf the pressure drops downstream of the duct exit, the flow is called alled underexpandedunderexpanded..

105

NonisentropicNonisentropic FlowFlow

106

NonisentropicNonisentropic Flow of an Ideal GasFlow of an Ideal Gas

Actual fluid flows are generally Actual fluid flows are generally nonisentropicnonisentropic..Examples of Examples of nonisentropicnonisentropic flow:flow:

FannoFanno flow: adiabatic flow with friction.flow: adiabatic flow with friction.Rayleigh flow: Flows with heat transfer Rayleigh flow: Flows with heat transfer ((diabaticdiabatic flows) without friction.flows) without friction.

107

FannoFanno FlowFlow

Adiabatic flow with frictionAdiabatic flow with friction

108

FannoFanno Flow Flow Adiabatic flow with friction 1/6Adiabatic flow with friction 1/6

Consider the steady, oneConsider the steady, one--dimensional, and adiabatic flow of an ideal dimensional, and adiabatic flow of an ideal gas through the constant area duct.gas through the constant area duct.For the indicated control volume, the energy equationFor the indicated control volume, the energy equation

the stagnation enthalpythe stagnation enthalpy

( ) innetShaftinnet12

21

22

1L WQzzg2

VVhhm &&& +=⎥⎦

⎤⎢⎣

⎡−+

−+−

ttanconsh2

Vh o

2

==+((

Adiabatic constant area flow.Adiabatic constant area flow.

(72)(72)

A=constant

Flow is adiabaticFlow is adiabatic

Flow is steadyFlow is steadySmallSmall

109


ttanconsTc2

)V(T 02P

2

==ρ

ρ+

( )0P0 TTchh −=−((

For an ideal gas

ttanconsTc2

VT 0P

2

==+

ttanconsVttanconsAV =ρ→=ρ

Substituting (1) into (74)

ttanconsT)R/p(c2

T)V(T 022P

22==

ρ+

Used to calculate values of fluid temperature corresponding to value of pressure in the Fannoflow.

(9)(9) (73)(73)

(72)+(73)(72)+(73) (74)(74)

ttanconsT)R/p(c2

T)V(T 022P

22

==ρ

+

For particular Fanno flow, the stagnation temperature T0 is fixed

(75)(75)

Determined until laterDetermined until later

(75)(75)

110

TT--s Diagram for s Diagram for FannoFanno FlowFlow1/71/7

From the second Tds relationship, an expression for entropy variation was already derived.

1

2

1

2P12 p

plnRTTlncss −=− (22)(22)

If the temperature, T1, and entropy s1, at the entrance of the Fannoflow duct as reference values

(22)(22)11

P1 pplnR

TTlncss −=− (76)(76)

111

TT--s Diagram for s Diagram for FannoFanno FlowFlow2/72/7

11P1 p

plnRTTlncss −=−

ttanconsT)R/p(c2

T)V(T 022P

22==

ρ+

This curve involves This curve involves a given gas (ca given gas (cppand R) with fixed values of and R) with fixed values of stagnation temperature, densitystagnation temperature, density--velocity product, and inlet velocity product, and inlet temperature, pressure, and entropy.temperature, pressure, and entropy.

TT--s diagram for s diagram for FannoFanno flowflow

(76)(76)

(75)(75)

Taken together result in a curve with T-s coordinates

112

Example 11.11 Compressible Flow with Example 11.11 Compressible Flow with Friction (Friction (FannoFanno Flow)Flow)

Air (k=1.4) enters [section (1)] an insulated, constant crossAir (k=1.4) enters [section (1)] an insulated, constant cross--sectional sectional area duct with the following properties:area duct with the following properties:TT00=518.67=518.67°°RRTT11=514.55=514.55°°RRpp11=14.3 =14.3 psiapsiaFor For FannoFanno flow, determine corresponding value of fluid temperature flow, determine corresponding value of fluid temperature and entropy change for various values of downstream pressures anand entropy change for various values of downstream pressures and d plot the related plot the related FannoFanno line.line.

113


ttanconsT)R/p(c2

T)V(T 022P

22==

ρ+

To plot the Fanno line we use Eq. (75) and (76)

(11.11.1)(11.11.1)

11P1 p

plnRTTlncss −=− (11.11.2)(11.11.2)

)Rlbm/()lbft(187...1k

Rkcp °⋅⋅==−

=From From EqEq. (14). (14)

)Rlbm/()lbft(3.53)Rslug/()lbft(1716R4.1k °⋅⋅=°⋅⋅==

(11.11.3)(11.11.3)

(1)+(69)(1)+(69) kRTMaRTpVRTkMa

RTpV 11

1

111 =ρ==ρ (11.11.4)(11.11.4)

114


2.002.0/1992.01aM 1 =⎟

⎠⎞

⎜⎝⎛ −=From From EqEq. (56). (56)

(11.11.4)(11.11.4) )sft/(lbm7.16)R44.514)](Rlbm/()lbft(3.53{)s/ft1112(2.0)ft/.in144)(psia3.14(V 2

22

⋅=°°⋅⋅

=ρ

992.0R67.518R55.514

TT

o

1 =°°

=

s/ft1112...kRT1 ==

For p= 7psiaFor p= 7psia

(11.11.1)(11.11.1) R3.502TR67.518...)R/p(c2

T)V(T 22P

22

°=⇒°==ρ

+

(11.11.2)(11.11.2) )Rlbm/()lbft(6.33...pplnR

TTlncss

11P1 °⋅⋅==−=−

22

1ko Ma11

TT

−+= (56)(56)

115


For p=6 For p=6 psiapsia T=496.8T=496.8ººR sR s--ss11=39.8 (=39.8 (ftft··lb)/(lbmlb)/(lbm··ººRR))For p=5 For p=5 psiapsia T=488.3T=488.3ººR sR s--ss11=46.3 (=46.3 (ftft··lb)/(lbmlb)/(lbm··ººRR))

For p=4 For p=4 psiapsia T=474.0T=474.0ººR sR s--ss11=52.6 (=52.6 (ftft··lb)/(lbmlb)/(lbm··ººRR))

116


To learn more about To learn more about FannoFanno lineslines

dp1hdTds ⎟⎟⎠

⎞⎜⎝

⎛ρ

−=(

dTchd P=(For ideal gas RTp ρ=

TdTddp

+ρρ

=ρ

⎟⎟⎠

⎞⎜⎝

⎛+

ρρ

−=TdTdRTdTcTds P

VdVdttanconsV −=

ρρ

→=ρFrom continuity equation

⎟⎠⎞

⎜⎝⎛ +−−=

⎟⎠⎞

⎜⎝⎛ +−−=

T1

dTdV

V1R

Tc

dTds

TdT

VdVRTdTcdsT

P

P

(18)(18)

(7)(7) (1)(1) (77)(77)

(1)+(7)+(18)+(77)(1)+(7)+(18)+(77) (78)(78)

(79)(79)

Substituting (79) into (78)Substituting (79) into (78) (80)(80)

Second Tds equation

117


Differentiating (74)Vc

dTdV P−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

T1

Vc

RTc

dTds

2pP

0dTds

=

kRTV aa = RTkc =vs.

The Mach number at state a is 1.

0dTds

=1k

Rkcp −= kRTV aa =

(81)(81)

Substituting (81) into (80) (82)(82)

(83)(83)

At state aAt state a

118


The temperature The temperature at point aat point a is the critical temperature, T*, for the is the critical temperature, T*, for the entire entire FannoFanno line.line.FannoFanno flow corresponding to the portion of the flow corresponding to the portion of the FannoFanno line above the line above the critical temperature must be subsonic, and critical temperature must be subsonic, and FannoFanno flow on the line flow on the line below T* must be supersonic.below T* must be supersonic.The second law of thermodynamics states that, based on all past The second law of thermodynamics states that, based on all past experience, entropy can only remain constant or increase for experience, entropy can only remain constant or increase for adibaticadibatic flowsflows..For For FannoFanno flow to be consistent with the second law of flow to be consistent with the second law of Thermodynamics, flow can only proceed along the Thermodynamics, flow can only proceed along the FannoFanno line line toward state a, the critical state.toward state a, the critical state.

119


The critical state may or may not reached by the flow. If The critical state may or may not reached by the flow. If titi is, the is, the FannoFanno flow is flow is chokedchoked..

((aa) Subsonic ) Subsonic FannoFanno flow. (flow. (bb) Supersonic ) Supersonic FannoFanno flow. (flow. (cc) Normal ) Normal shock occurrence in shock occurrence in FannoFanno flow (an flow (an abruotabruot change from change from supersonic to subsonic flow in the supersonic to subsonic flow in the FannoFanno duct).duct).

subsonicsubsonic

SupersonicSupersonic

120


Subsonic Subsonic FannoFanno flow is accelerated by friction to a higher Mach flow is accelerated by friction to a higher Mach number without chocking.number without chocking.Supersonic Supersonic FannoFanno flow is decelerated by friction to a lower Mach flow is decelerated by friction to a lower Mach number without chocking.number without chocking.Sudden deceleration across a standing normal shock wave.Sudden deceleration across a standing normal shock wave.

121

Qualitative Aspects of Qualitative Aspects of FannoFanno FlowFlow

122

Quantify Quantify FannoFanno Flow Behavior Flow Behavior 1/121/12

To quantify To quantify FannoFanno flow behavior we need to combine a relationship flow behavior we need to combine a relationship that represents the linear momentum law with the set of equationthat represents the linear momentum law with the set of equations s already derived in this chapter.already derived in this chapter.Apply the linear momentum equation to the Apply the linear momentum equation to the FannoFanno flow through the flow through the control volumecontrol volume

( )12X2211 VVmRApAp −=−− &Rx is the frictional force exerted by the Rx is the frictional force exerted by the inner pipe wall on the fluidinner pipe wall on the fluid

ttanconsAVm,AAA 21 =ρ=== &

)VV(VA

Rpp 12X

21 −ρ==− (84)(84)

123


Differential form of (84) , which is valid for Fanno flow through the semi-infinitesimal control volume

VdVADdxdp w ρ=πτ

−−

The wall shear stress is related to the wall friction factor

2w

V8fρ

τ=

(85)(85)

(86)(86)

124


(1)+(36)+(46)+(88)(1)+(36)+(46)+(88) 0V

)V(d2

MakDdxMa

2fk

pdp

2

222 =++ (89)(89)

RTkMaVRTkMacMaV 22 ==⋅=TdT

Ma)Ma(d

V)V(d

2

2

2

2+= (90)(90)

Substituting (86) and A=Substituting (86) and A=ππDD22/4 into /4 into (85)(85)

02

)V(dpD

dx2V

pf

pdp

VdVDdx

2Vfdp

22

2

=ρ

+ρ

+

ρ=ρ−− (87)(87)

(88)(88)

125


(74) is differentiated & divided by T(74) is differentiated & divided by T 0Tc2

)V(dTdT

P

2=+ (91)(91)

Apply energy equation to Apply energy equation to FannoFanno flowflow

ttanconsTc2

)V(T 02P

2

==ρ

ρ+ (74)(74)

0V

)V(dMa2

1kTdT

2

22 =

−+

[ ] 2

22

2

2

Ma2/)1k(1Ma/)Ma(d

V)V(d

−+=

Substituting (14) (36) (46) into (91)Substituting (14) (36) (46) into (91) (92)(92)

(92)+(90)(92)+(90) (93)(93)

126


2

2

2

2

Ma)Ma(d

V)V(d

21

pdp

−=(77)+(79)+(90)(77)+(79)+(90) (94)(94)

((94)+(89)94)+(89) 0DdxMa

2fk

Ma)Ma(d

V)V(d)kMa1(

21 2

2

2

2

22 =+−+ (95)(95)

Incorporating (93) into (95)Incorporating (93) into (95) (96)(96)Ddxf

kMa}Ma]2/)1k[(1{)Ma(d)Ma1(

42

22=

−+

−

(96) can be integrated from one section to another in (96) can be integrated from one section to another in FannoFanno flow flow duct. We elect to use the critical (*) state as a reference and duct. We elect to use the critical (*) state as a reference and to to integrate (96) from an upstream state to the critical state.integrate (96) from an upstream state to the critical state.

127


(97)(97)

Friction factor is constant at an average value over the integraFriction factor is constant at an average value over the integration tion length length ll**-- ll. K is also constant.. K is also constant.

(98)(98)

∫ ∫=−+

−*a

a

M

M

*

42

22

Ddxf

kMa}Ma]2/)1k[(1{)Ma(d)Ma1( l

l

ll is length measured from an arbitrary but fixed upstream is length measured from an arbitrary but fixed upstream reference location to a section on the reference location to a section on the FannoFanno flow.flow.

( )D

*fMa]2/)1k[(1

Ma]2/)1k[(lnk21k

Ma)Ma1(

k1

2

2

2

2 ll −=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−+

+++

−

128


For a given gas, value of f(For a given gas, value of f(ll**-- ll)/D can be tabulated as a function of )/D can be tabulated as a function of Mach number for Mach number for FannoFanno flow. flow. For example, values of fFor example, values of f((ll**-- ll)/)/DD for air Fanno flow are graphed as a function of Mach number in Figure D.2.D.2.Note that the critical state does not have to be exist in the acNote that the critical state does not have to be exist in the actual tual FannoFanno flow being considered, since for any two sections in a given flow being considered, since for any two sections in a given FannoFanno flowflow

)(Df

D)*(f

D)*(f

2112 llllll

−=−

−−

(99)(99)

The physical meaning of (99) The physical meaning of (99) NEXT PAGENEXT PAGE

129


((aa) ) UnchokedUnchoked FannoFanno flow.flow.((bb) Choked ) Choked FannoFanno flow.flow.

130


For a given For a given FannoFanno flow (constant specific heat ratio, duct diameter, flow (constant specific heat ratio, duct diameter, and friction factor) the length of duct required to change the Mand friction factor) the length of duct required to change the Mach ach number from Manumber from Ma11 to Mato Ma22 can be determined from can be determined from EqsEqs. (98) and . (98) and (99) or a graph such as Figure D.2.(99) or a graph such as Figure D.2. To get the values of other fluid To get the values of other fluid properties in the properties in the FannoFanno flow field we need to develop more flow field we need to develop more equations.equations.

131


(90)+(92)(90)+(92) (100)(100)

Integrating Integrating EqsEqs. (100) from any state upstream in a . (100) from any state upstream in a FannoFanno flow to flow to the critical (*) state the critical (*) state

(101)(101)

( ) )Ma(d}Ma]2/)1k[(1{2

1kT

dT 22−+

−−=

2Ma]2/)1k[(12/)1k(

*TT

−+

+=

(68)+(69)(68)+(69) (102)(102)

Substituting (101) into (102)Substituting (101) into (102) (103)(103)

*** TTMa

kRT

RTkMaVV

==

2/1

2

2

Ma]2/)1k[(1Ma]2/)1k[(

*VV

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−+

+=

132


Continuity equation (40)Continuity equation (40) (104)(104)

(104)+(103)(104)+(103) (105)(105)

Ideal gas (1)Ideal gas (1)

(107)(107)

*V*V ρ=ρ

2/1

2

2

Ma]2/)1k[(Ma]2/)1k[(1

V*V

* ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+

−+==

ρρ

*TT

**pp

ρρ

= (106)(106)

(106)+(105)+(101)(106)+(105)+(101)2/1

2Ma]2/)1k[(12/)1k(

Ma1

*pp

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−+

+=

133


The stagnation pressure ratioThe stagnation pressure ratio

(108)(108)

(109)(109)

*o

o*o

o

p*p

*pp

pp

pp

⋅⋅=

By using of By using of EqsEqs. (59) (107). (59) (107)

)]1k(2/)1k[(2

*o

o Ma2

1k11k

2Ma1

pp

−+

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛

+=

134


Value of fValue of f((ll**-- ll)/)/D, T/T*, V/V*, D, T/T*, V/V*, p/pp/p*, and p*, and p00/p/p00* for * for FannoFanno flow of flow of air (k=1.4) are graphed as a function of Mach number in Figure Dair (k=1.4) are graphed as a function of Mach number in Figure D.2..2.

)(Df

D)*(f

D)*(f

2112 llllll

−=−

−−

(99)(99)

(101)(101)2Ma]2/)1k[(12/)1k(

*TT

−+

+= (103)(103)

2/1

2

2

Ma]2/)1k[(1Ma]2/)1k[(

*VV

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−+

+=

(107)(107)

2/1

2Ma]2/)1k[(12/)1k(

Ma1

*pp

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−+

+=

(109)(109))]1k(2/)1k[(

2*o

o Ma2

1k11k

2Ma1

pp

−+

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛

+=

135

Example 11.12 Choked Example 11.12 Choked FannoFanno Flow Flow 1/21/2

Standard atmospheric air [TStandard atmospheric air [T0 0 = 288K, p= 288K, p00=101 =101 kPakPa (abs)] is drawn (abs)] is drawn steadily through a frictionless, adiabatic diverging nozzle intosteadily through a frictionless, adiabatic diverging nozzle into an an adiabatic, constant area duct as shown in Figure E11.12a. The duadiabatic, constant area duct as shown in Figure E11.12a. The duct ct is 2is 2--m long and has an inside diameter of 0.1 m. The average m long and has an inside diameter of 0.1 m. The average friction factor for the duct is estimated as being equal to 0.02friction factor for the duct is estimated as being equal to 0.02. What . What is the maximum mass is the maximum mass flowrateflowrate through the duct? For this maximum through the duct? For this maximum flowrateflowrate, determine the values of static temperature, static pressure, , determine the values of static temperature, static pressure, stagnation temperature, stagnation pressure, and velocity at thestagnation temperature, stagnation pressure, and velocity at the inlet inlet [section (1)] and exit [section (2)] of the constant area duct. [section (1)] and exit [section (2)] of the constant area duct. Sketch a Sketch a temperaturetemperature--entropy diagram for this flow.entropy diagram for this flow.

136

Example 11.12 Choked Example 11.12 Choked FannoFanno Flow Flow 2/22/2

137


Consider the flow through the converging nozzle to be isentropicConsider the flow through the converging nozzle to be isentropicand the flow through the constant area duct to be and the flow through the constant area duct to be FannoFanno flow.flow.A decrease in the pressure at the exit of the constant area ductA decrease in the pressure at the exit of the constant area duct (back (back pressure) causes the mass pressure) causes the mass flowrateflowrate through the nozzle and the duct through the nozzle and the duct to increase.to increase.The flow throughout is subsonic. The maximum The flow throughout is subsonic. The maximum flowrateflowrate will occur will occur when the back pressure is lowered to the extent that the constanwhen the back pressure is lowered to the extent that the constant t area duct chokes and the Mach number at the duct exit is equal tarea duct chokes and the Mach number at the duct exit is equal to 1.o 1.Any further decrease of back pressure will not affect the Any further decrease of back pressure will not affect the flowrateflowratethrough the nozzle through the nozzle –– duct combination.duct combination.

138


For the maximum For the maximum flowrateflowrate condition, the constant area duct must be condition, the constant area duct must be chocked, andchocked, and

4.0)m1.0(

)m2)(02.0(D

)(fD

)*(f 121 ==−

=− llll

(11.12.1)(11.12.1)

4.0D/)*(f 1 =−llEntering Figure D.2 with Entering Figure D.2 with We readWe read

16.1*p

p7.1

*pp66.0

*VV

1.1*T

T63.0Ma

0

1,011

11

===

==

139


Entering Figure D.1 with MaEntering Figure D.1 with Ma11=0.63 =0.63 We readWe read

83.076.0pp93.0

TT

1,0

1

1,0

1

0

1 =ρρ

==

Since TSince T00=288K=288K

s/m73.312)Kg/J(73.312

)4.1)(K49.243)(Kkg/J9.286(k*RT*V

K49.2431.1/T*TK84.267T93.0T

2/1

101

==

⋅==

====

140


)abs(kPa77)abs(kPa10176.0p76.0pK28893.0T93.0T

1,01

01=⋅=⋅=

=⋅=⋅=

=⎥⎥⎦

⎤

⎢⎢⎣

⎡ π=ρ=

=⋅=ρ=ρ

=⋅==

)s/m40.206(4

)m1.0()m/kg02.1(VAm

m/kg02.1m/kg23.183.083.0

s/m40.206s/m73.31266.0*V66.0V

23

111

331,01

1

&

The stagnation temperature, TThe stagnation temperature, T00, remain constant through this , remain constant through this adiabatic flow at a value ofadiabatic flow at a value of

K288TTT 02,01,0 ===

141


The stagnation pressure, pThe stagnation pressure, p00, at the entrance of the constant area , at the entrance of the constant area duct is the same as the constant value of stagnation pressure duct is the same as the constant value of stagnation pressure through the isentropic nozzle.through the isentropic nozzle.

)abs(kPa101pp 01,0 ==

The duct exit pressureThe duct exit pressure

==== )abs(kPa101)76.0(7.1

1ppp

p*p*pp 1,0

1,0

1

12

The duct exit stagnation pressureThe duct exit stagnation pressure

)abs(kPa84)abs(kPa10116.11p

p*p*pp 1,01,0

002,0 ====

142

Example 11.13 Effect of Duct Length on Example 11.13 Effect of Duct Length on Choked Choked FannoFanno FlowFlow

The duct in Example 11.12 is shortened by 50%, but the duct The duct in Example 11.12 is shortened by 50%, but the duct discharge pressure is maintained at the chocked flow value for discharge pressure is maintained at the chocked flow value for Example 11.12, namely,Example 11.12, namely,

PPdd=45 =45 kPakPa (abs)(abs)Will shortening the duct cause the mass Will shortening the duct cause the mass flowrateflowrate through the duct to through the duct to increase or decrease? Assume that the average friction factor foincrease or decrease? Assume that the average friction factor for the r the duct remains constant at a value of f = 0.02.duct remains constant at a value of f = 0.02.

143


We guess that the shortened duct will still choke and check our We guess that the shortened duct will still choke and check our assumption by comparing passumption by comparing pdd with p*. If pwith p*. If pdd < p*, the flow is chocked. < p*, the flow is chocked. If not, another assumption has to be made.If not, another assumption has to be made.

2.0)m1.0(

)m1)(02.0(D

)(fD

)*(f 121 ==−

=− llll

2.0D/)*(f 1 =−ll

For chocked flowFor chocked flow

Entering Figure D.2 with Entering Figure D.2 with We readWe read

73.0*V

V05.1*T

T5.1*p

p7.0Ma 1111 ====

144


Entering Figure D.1 with MaEntering Figure D.1 with Ma11=0.7=0.7We readWe read

79.072.0pp92.0

TT

1,0

1

1,0

1

0

1 =ρρ

==

The duct exit pressureThe duct exit pressure

)abs(kPa5.48)abs(kPa101)72.0(5.1

1ppp

p*p*pp 1,0

1,0

1

12 ====

ppdd<p*. Our assumption of chocked flow is justified.<p*. Our assumption of chocked flow is justified.

145


331,01 m/kg97.0)m/kg23.1(79.079.0 ==ρ=ρ

73.0*V

V1 =

Since TSince T00=288K=288K

s/m37.318)Kg/J(37.318

)4.1)(K34.252)(Kkg/J9.286(k*RT*V

K34.25305.1/T*TK96.264T92.0T

2/1

101

==

⋅==

====

s/kg77.1VAms/m31.232*V73.0V 1111 =ρ=⇒== &

146


ConclusionConclusion1.1. The mass The mass flowrateflowrate associated with a shortened tube is large associated with a shortened tube is large

than the mass than the mass flowrateflowrate for the longer tube.for the longer tube.2.2. This trend is general for subsonic This trend is general for subsonic FannoFanno flow.flow.3.3. For the sane upstream stagnation state and downstream For the sane upstream stagnation state and downstream

pressure, the mass pressure, the mass flowrateflowrate for the for the FannoFanno flow will decrease flow will decrease with increase in length of duct for subsonic flow.with increase in length of duct for subsonic flow.

4.4. If the length of the duct remains the same but the wall frictionIf the length of the duct remains the same but the wall frictionis increased, the mass is increased, the mass flowrateflowrate will decrease.will decrease.

147

Example 11.14 Example 11.14 UnchokedUnchoked FannoFanno FlowFlow

If the same If the same flowrateflowrate obtained in Example 11.12 is desired through obtained in Example 11.12 is desired through the shortened duct of Example 11.13 (the shortened duct of Example 11.13 (((ll22-- ll11 =1 m=1 m), determine the ), determine the Mach number at the exit of the duct, MaMach number at the exit of the duct, Ma22, and the back pressure, p, and the back pressure, p22, , required. Assume f remain constant at a value of 0.02.required. Assume f remain constant at a value of 0.02.

148


From Example 11.12, MaFrom Example 11.12, Ma11=0.63 and from Figure D.2=0.63 and from Figure D.2

2.0D

)*(fD

)*(fD

)*(fD

)(f4.0D

)*(f

2

21121

=−

⇒

−−

−=

−=

−

ll

llllllll

From Figure D.2, MaFrom Figure D.2, Ma11=0.7=0.7

)abs(kPa68)abs(kPa101)76.0(7.1

1)5.1(ppp

p*p

*ppp

5.1*p

p

1,01,0

1

1

22

2

=⎟⎠⎞

⎜⎝⎛==

= From Example 11.12From Example 11.12

149

Rayleigh flowRayleigh flow

Flows with heat transferFlows with heat transferwithout frictionwithout friction

150

Rayleigh Flow Rayleigh Flow Frictionless,Frictionless, heat transfer 1/6heat transfer 1/6

Consider the steady, oneConsider the steady, one--dimensional, and frictionless flow of an dimensional, and frictionless flow of an ideal gas through the constant area duct with heat transfer. Thiideal gas through the constant area duct with heat transfer. This is s is Rayleigh FlowRayleigh Flow..Apply linear momentum equation to Rayleigh flow through the Apply linear momentum equation to Rayleigh flow through the finite control volumefinite control volume

X222111 RVmApVmAp ++=+ && ttancons)V(p2=

ρρ

+ (110)(110)

FrictionlessFrictionless

151


(111)(111)(1)+(110)(1)+(110) ttanconsp

RT)V(p2

=ρ

+

ttanconsV =ρ

For a given Rayleigh flow, the constant, For a given Rayleigh flow, the constant, ρρV, and R are all fixed.V, and R are all fixed.EqsEqs. (111) can be used to determine values of fluid temperature . (111) can be used to determine values of fluid temperature corresponding to the local pressure in a Rayleigh flow.corresponding to the local pressure in a Rayleigh flow.

152

TT--s Diagram for Rayleigh Flows Diagram for Rayleigh Flow

ttanconsp

RT)V(p2

=ρ

+

11P1 p

plnRTTlncss −=−

Rayleigh line.Rayleigh line.

These two equations can be solved These two equations can be solved simultaneously to obtain the Rayleigh simultaneously to obtain the Rayleigh lines.lines.

Using Using EqsEqs. (76),. (76), which was developed earlier from the second which was developed earlier from the second TdsTdsrelationship. relationship. EqsEqs. (111) and . (111) and EqsEqs. (76). (76) can be solved simultaneously to can be solved simultaneously to obtain temperatureobtain temperature--entropy diagramentropy diagram

(111)(111)

(76)(76)

153


The physical meaning of point a ??The physical meaning of point a ??

(112)(112)Differentiating (110)Differentiating (110) VdVdpVdVdp −=ρ

ρ−=

(113)(113)(112)+(18)(112)+(18) VdVhdTds +=(

Substituting (7) into (113)Substituting (7) into (113)

dThd

Thc

pP

((

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=

dTdV

TV

Tc

dTds

VdVdTcVdVhdTds

p

P

+=

+=+=(

(114)(114)

[ ])R/V()V/T(1

TV

Tc

dTdV

TV

Tc

dTds PP

−+=+=

(114)+(112)+(1)+(77)+(79)(114)+(112)+(1)+(77)+(79)

(115)(115)

At point a, At point a, ds/dTds/dT=0=0

(7)(7)

154


Hence, at point a where Hence, at point a where ds/dTds/dT=0, =0, (115)(115) (116)(116)

1Maa =Comparison of (116) and (36)Comparison of (116) and (36)

(117)(117)

(118)(118)

kRTV aa =

Mach number at state aMach number at state a

At point b where At point b where dT/dsdT/ds=0, From =0, From (115)(115)

1p )]R/V()V/T)[(T/V()T/c(

1dT/ds

1dsdT

−−+==

k1Ma0

dsdT

b =⇒= The flow at point b is subsonic.The flow at point b is subsonic.

155


To learn more about Rayleigh flow, we need to consider the energTo learn more about Rayleigh flow, we need to consider the energy y equation in addition to equation already used. Apply the equation in addition to equation already used. Apply the energy energy equationequation to the Rayleigh flow through the finite control volumeto the Rayleigh flow through the finite control volume

or in differential form for Rayleigh flow through the or in differential form for Rayleigh flow through the semsem--infinitesimal control volumeinfinitesimal control volume

innetshaftinnet12

21

22

12 WQ)zz(g2

VVhhm &&((

+=⎥⎥⎦

⎤

⎢⎢⎣

⎡−+

−+−

qVdVhd δ=+(

(119)(119)

The heat transfer per unit mass of fluid in the The heat transfer per unit mass of fluid in the semisemi--infinitesimal control volumeinfinitesimal control volume

156


(120)(120)

From (121), we see clearly that when the Rayleigh flow is subsonFrom (121), we see clearly that when the Rayleigh flow is subsonic ic (Ma<1), fluid heating ((Ma<1), fluid heating (δδq>0) increases fluid velocity while fluid q>0) increases fluid velocity while fluid cooling decreases fluid velocity.cooling decreases fluid velocity.When the Rayleigh flow is supersonic (Ma>1), fluid heating decreWhen the Rayleigh flow is supersonic (Ma>1), fluid heating decrease ase fluid velocity and fluid cooling increases fluid velocity.fluid velocity and fluid cooling increases fluid velocity.

By using into (119)By using into (119))1k/(RkdTdTchd P −==(

12

P kRT)1k(V

dVdT

TV

Tcq

VdV

−

⎥⎥⎦

⎤

⎢⎢⎣

⎡ −+

δ=

(36)+(46)+(1)+(77)+(79)+(112)+(120)(36)+(46)+(1)+(77)+(79)+(112)+(120))Ma1(

1Tcq

VdV

2P −

δ= (121)(121)

157

Qualitative Aspects of Rayleigh Flow Qualitative Aspects of Rayleigh Flow 1/21/2

(a) Subsonic Rayleigh flow. (b) Supersonic Rayleigh flow. (c) Normal shock in a Rayleigh flow.

158

Qualitative Aspects of Rayleigh Flow Qualitative Aspects of Rayleigh Flow 2/22/2

Along the upper portion of Rayleigh line, which includes point bAlong the upper portion of Rayleigh line, which includes point b, , the flow is subsonic. Heating the fluids results in flow accelerthe flow is subsonic. Heating the fluids results in flow acceleration ation to a maximum Mach number of 1 at point a.to a maximum Mach number of 1 at point a.Between Between points b and a along the Rayleigh linepoints b and a along the Rayleigh line, heating the fluids , heating the fluids results in a temperature decrease and cooling the fluids leads aresults in a temperature decrease and cooling the fluids leads atemperature increase.temperature increase.Along the lower portion of the Rayleigh line the flow is supersoAlong the lower portion of the Rayleigh line the flow is supersonic. nic. Rayleigh flows may or may not be chocked.Rayleigh flows may or may not be chocked.As with As with FannoFanno flow, an abrupt deceleration from supersonic flow to flow, an abrupt deceleration from supersonic flow to subsonic flow across a normal shock wave can also occur in subsonic flow across a normal shock wave can also occur in Rayleigh flows.Rayleigh flows.

159

Quantify Rayleigh Flow Behavior Quantify Rayleigh Flow Behavior 1/41/4

To quantify Rayleigh flow behavior we need to develop appropriatTo quantify Rayleigh flow behavior we need to develop appropriate e forms of the governing equationsforms of the governing equations

We elect to use the state of the Rayleigh flow fluid We elect to use the state of the Rayleigh flow fluid at point a as at point a as the reference statethe reference state..Apply the linear momentum equation to Apply the linear momentum equation to Rayleigh flow between any upstream Rayleigh flow between any upstream section and the section, actual or imaged, section and the section, actual or imaged, where state a is attained.where state a is attained.

2a

a

a

a

2

a

2aaa

2

Vp

1pV

pp

VpVp

ρ+=

ρ+

ρ+=ρ+

(122)(122)

160


(123)(123)Substituting (1) into (122) and Substituting (1) into (122) and making use of (36)+(46)making use of (36)+(46) 2

a kMa1k1

pp

+

+=

From (1) From (1) ρρ

= a

aa pp

TT

(124)(124)

From (40) with constant AFrom (40) with constant Aa

aVV

=ρρ

(125)(125)

(125)+(36)+(46)(125)+(36)+(46)a

aTTMa=

ρρ

(126)(126)

(124)+(126)(124)+(126)2

aaMa

pp

TT

⎟⎟⎠

⎞⎜⎜⎝

⎛= (127)(127)

161


(128)(128)

(125)+(126)+(128)(125)+(126)+(128) (129)(129)

(127)+(123)(127)+(123)2

2a kMa1

Ma)k1(TT

⎥⎦

⎤⎢⎣

⎡

+

+=

⎥⎦

⎤⎢⎣

⎡

+

+==

ρρ

2a

a

kMa1Ma)k1(Ma

VV

The energy equation tells us that because of the heat transfer iThe energy equation tells us that because of the heat transfer involved nvolved in Rayleigh flows, the stagnation temperature varies.in Rayleigh flows, the stagnation temperature varies.

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

a,o

a

a

oa,o

oTT

TT

TT

TT

(130)(130)

162


(130)+(56)+(128)(130)+(56)+(128) (131)(131)22

22

a,oo

)kMa1(

Ma2

1k1Ma)1k(2

TT

+

⎟⎠⎞

⎜⎝⎛ −++

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

a,0

a

a

o

a,0

opp

pp

pp

pp

1k/k2

2a,0

0 Ma2

1k11k

2kMa1

k1pp

−

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛

++

+=

(132)(132)

(132)+(123)+(59)(132)+(123)+(59) (133)(133)

163


Value of p/pValue of p/paa, T/T, T/Taa, , ρρ//ρρaa or V/or V/VVaa, T, T00/T/T0,a0,a, and p, and p00/p/p0,a0,a are graphed in are graphed in Figure D.3 as a function of Mach number for Rayleigh flow of airFigure D.3 as a function of Mach number for Rayleigh flow of air(k=1.4).(k=1.4).

(123)(123)2

a kMa1k1

pp

+

+= (128)(128)

(129)(129)

2

2a kMa1

Ma)k1(TT

⎥⎦

⎤⎢⎣

⎡

+

+=

⎥⎦

⎤⎢⎣

⎡

+

+==

ρρ

2a

a

kMa1Ma)k1(Ma

VV

(131)(131)22

22

a,oo

)kMa1(

Ma2

1k1Ma)1k(2

TT

+

⎟⎠⎞

⎜⎝⎛ −++

=

1k/k2

2a,0

0 Ma2

1k11k

2kMa1

k1pp

−

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛

++

+= (133)(133)

164

Example 11.15 Rayleigh FlowExample 11.15 Rayleigh Flow

Air (k=1.4) enters [section (1)] a frictionless, constant flow Air (k=1.4) enters [section (1)] a frictionless, constant flow cross0sectional area duct with the following properties (the samcross0sectional area duct with the following properties (the same as e as in Example 11.11):in Example 11.11):TT00=518.67=518.67°°RRTT11=514.55=514.55°°RRpp11=14.3 =14.3 psiapsiaFor Rayleigh flow, determine corresponding value of fluid For Rayleigh flow, determine corresponding value of fluid temperature and entropy change for various values of downstream temperature and entropy change for various values of downstream pressures and plot the related Rayleigh line.pressures and plot the related Rayleigh line.

165


To plot the Rayleigh line we use Eq. (111) and (76)

11P1 p

plnRTTlncss −=−

)Rlbm/()lbft(187...1k

Rkcp °⋅⋅==−

=From From EqEq. (14). (14)

)Rlbm/()lbft(3.53)Rslug/()lbft(1716R4.1k °⋅⋅=°⋅⋅==

lbm/ft3.13...p

RT)sft/(lbm7.16V 3

1

12 ==⋅=ρ

ttanconsp

RT)V(p2

=ρ

+

166


)Rlbm/()lbft(121...pplnR

TTlncss

11P1 °⋅⋅==−=−

ttanconspsia10.15p

RT)V(p2

==ρ

+

R979T

psia10.15T]R/).in/lb(1065.1[psia5.13p

RT)V(p 232

°=⇒

=°×+=ρ

+ −

With the downstream pressure of p=13.5 psia

With the downstream pressure of p=12.5 psia……

167


168

Example 11.16 Effect of Mach Number and Example 11.16 Effect of Mach Number and Heating/Cooling for Rayleigh FlowHeating/Cooling for Rayleigh Flow

The information in Table 11.2 shows us that subsonic Rayleigh flThe information in Table 11.2 shows us that subsonic Rayleigh flow ow accelerates when heated and decelerates when cooled. Supersonic accelerates when heated and decelerates when cooled. Supersonic Rayleigh flow behaves just opposite to subsonic Rayleigh flow; iRayleigh flow behaves just opposite to subsonic Rayleigh flow; it t decelerates when heated and accelerates when cooled. Using Figurdecelerates when heated and accelerates when cooled. Using Figure e D.3 for air (k=1.4), state whether velocity, Mach number, staticD.3 for air (k=1.4), state whether velocity, Mach number, statictemperature, static pressure, and stagnation pressure increase otemperature, static pressure, and stagnation pressure increase or r decrease as subsonic and supersonic Rayleigh flow is (a) heated,decrease as subsonic and supersonic Rayleigh flow is (a) heated, (b) (b) cooled.cooled.

169


170


Normal shock wave can occur in supersonic flows through Normal shock wave can occur in supersonic flows through convergingconverging--diverging and constant area ducts.diverging and constant area ducts.Normal shock waves involve deceleration from a supersonic flow tNormal shock waves involve deceleration from a supersonic flow to o a subsonic flow, a pressure rise, and an increase of entropy.a subsonic flow, a pressure rise, and an increase of entropy.

To develop the equations that To develop the equations that verify the observed behavior of verify the observed behavior of flows across a normal shock, we flows across a normal shock, we apply first principle to the flow apply first principle to the flow through a control volume that through a control volume that completely surrounds a normal completely surrounds a normal shock wave.shock wave.

171


For steady flow through the control volume, the conservation of For steady flow through the control volume, the conservation of mass principle yieldsmass principle yields

because the flow crossbecause the flow cross--sectional area remains essentially constant sectional area remains essentially constant within the infinitesimal thickness of the normal shock.. Equatiowithin the infinitesimal thickness of the normal shock.. Equation n (134) is identical to the continuity equation used for (134) is identical to the continuity equation used for FannoFanno and and Rayleigh flows.Rayleigh flows.The linear momentum equation describing steady gas flow through The linear momentum equation describing steady gas flow through the control volume isthe control volume is

ttanconsVp 2 =ρ+

ttanconsV =ρ (134)(134)

(135)(135)ttanconsp

RT)V(p2

=ρ

+For ideal gasFor ideal gas

≣≣ Linear momentum equation (111) of Rayleigh flowLinear momentum equation (111) of Rayleigh flow

172


For the control volume containing the normal shock, no shaft worFor the control volume containing the normal shock, no shaft work k is involved and the heat transfer is assumed negligible. Thus, tis involved and the heat transfer is assumed negligible. Thus, the he energy equation can be applied to steady gas flow through the energy equation can be applied to steady gas flow through the control volume to obtaincontrol volume to obtain

(136)(136)

RTp)TT(chh 0p0 ρ=−=−((

For ideal gasFor ideal gasttanconsh

2Vh 0

2==+

((

ttanconsT)R/p(c2

T)V(T o22P

22

==ρ

+

≣≣ Energy equation (75) of Energy equation (75) of FannoFanno flowflow

173

TT--s Diagram for Normal Shocks Diagram for Normal Shock

The steady flow of an ideal gas across a normal shock is governeThe steady flow of an ideal gas across a normal shock is governed d by some of the same equation used for describing by some of the same equation used for describing FannoFanno and and Rayleigh flows (Rayleigh flows (energy equation for energy equation for FannoFanno flowsflows and and momentum momentum equation for Rayleigh flowsequation for Rayleigh flows).).

For For a given a given ρρV, gas (R,k), and V, gas (R,k), and conditions at the inlet of normal conditions at the inlet of normal shock (shock (TTxx, , ρρxx, and , and ssxx),), the conditions the conditions downstream of the shock (state y) will downstream of the shock (state y) will be on both a be on both a FannoFanno line and a Rayleigh line and a Rayleigh line that pass through the inlet state line that pass through the inlet state (state x).(state x).

174

Equation forEquation forNormal Shock WaveNormal Shock Wave

The energy equation for The energy equation for FannoFanno flow and flow and the momentum equation for Rayleigh flow the momentum equation for Rayleigh flow

are valid for flow across normal shocks.are valid for flow across normal shocks.

175

Normal Shock WaveNormal Shock Wave

The second law of thermodynamics requires that entropy must The second law of thermodynamics requires that entropy must increase across a normal shock wave.increase across a normal shock wave.This law and sketches of the This law and sketches of the FannoFanno line and Rayleigh line line and Rayleigh line intersections persuade us to conclude that flow across a normal intersections persuade us to conclude that flow across a normal shock shock can only proceed from supersonic to subsonic flow.can only proceed from supersonic to subsonic flow.Since the state upstream and downstream of a normal shock wave aSince the state upstream and downstream of a normal shock wave are re represented by the supersonic and subsonic intersections of acturepresented by the supersonic and subsonic intersections of actual al and/or imaged and/or imaged FannoFanno and Rayleigh lines.and Rayleigh lines.

We should be able to use equations developed earlier for We should be able to use equations developed earlier for FannoFanno and Rayleigh flows to quantify normal shock wave.and Rayleigh flows to quantify normal shock wave.

176

Normal Shock Wave Normal Shock Wave ppyy/p/pxx1/21/2

For the Rayleigh line For the Rayleigh line

x

a

a

y

x

y

pp

pp

pp

=(137)(137)

((bb) The normal shock in a Rayleigh flow) The normal shock in a Rayleigh flow

(123)(123)2a kMa1

k1pp

+

+=FromFrom

2ya

y

kMa1k1

pp

+

+=

2xa

x

kMa1k1

pp

+

+=

(138)(138)

(139)(139)

Momentum equation for Rayleigh flowMomentum equation for Rayleigh flow

177

Normal Shock Wave Normal Shock Wave ppyy/p/pxx2/22/2

2y

2x

x

y

kMa1kMa1

pp

+

+=(137)+(138)+(139)(137)+(138)+(139) (140)(140)

Equation (140) can also be derived by Equation (140) can also be derived by ……....

2y

2x

x

y

kMa1kMa1

pp

++

= (140)(140)2/1

2

x

y

x

y

Ma]2/)1k[(12/)1k(

Ma1

*pp

p*p

*pp

pp

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−+

+=

=

(107)(107)

2222

2yyy

2xxx

kMaRTk/kVRT/Vp/V

VpVp

===ρ

ρ+=ρ+From linear momentum equation

178

Normal Shock Wave Normal Shock Wave TTyy/T/Txx1/21/2

For the For the FannoFanno flowflow

x

y

x

y

T*T

*TT

TT

= (141)(141)

(a) The normal shock in a (a) The normal shock in a FannoFanno flowflow

(101)(101)2Ma]2/)1k[(1

2/)1k(*T

T−+

+=FromFrom

2y

y

Ma]2/)1k[(12/)1k(

*TT

−+

+=

2x

x

Ma]2/)1k[(12/)1k(

*TT

−+

+=

(142)(142)

(143)(143)

Energy equation for Energy equation for FannoFanno flowflow

179

Normal Shock Wave Normal Shock Wave TTyy/T/Txx2/22/2

(141)+(142)+(143)(141)+(142)+(143) (144)(144)2y

2x

x

y

Ma]2/)1k[(1Ma]2/)1k[(1

TT

−+

−+=

180

Normal Shock Wave MaNormal Shock Wave Mayy1/21/2

To develop an equation to determine the Mach number downstream To develop an equation to determine the Mach number downstream of the normal shock, Maof the normal shock, Mayy, when the Mach number upstream of the , when the Mach number upstream of the normal shock, Manormal shock, Maxx, is known., is known.

x

y

x

y

x

y

TT

pp

ρ

ρ=(1)(1) (145)(145)

yyxx VV ρ=ρ

y

x

x

y

x

y

VV

TT

pp

= (146)(146)

(146)+(36)+(46)(146)+(36)+(46)y

x2/1

x

y

x

y

MaMa

TT

pp

⎟⎟⎠

⎞⎜⎜⎝

⎛= (147)(147)

(147)+(144)(147)+(144) (148)(148)y

x2/1

2y

2x

x

y

MaMa

Ma]2/)1k[(1Ma]2/)1k[(1

pp

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−+

−+=

181

Normal Shock Wave MaNormal Shock Wave Mayy2/22/2

1Ma)]1k/(k2[)]1k/(2[MaMa 2

x

2x2

y−−

−+= (149)(149)(148)+(140)(148)+(140)

Equation (149) can be used to calculate values of Mach number Equation (149) can be used to calculate values of Mach number downstream of a normal shock from a known Mach number of the downstream of a normal shock from a known Mach number of the shock. shock. MaMaxx>1>1……..Ma..Mayy<1<1

182

Normal Shock Wave Normal Shock Wave ppyy/p/pxx TTyy/T/Txx1/21/2

(149)+(144)(149)+(144)2x

2

2x

2x

x

y

Ma)]}1k(2/[)1k{(}1Ma)]1k/(k2}{[Ma]2/)1k[(1{

TT

−+

−−−+= (151)(151)

(40)(40)y

x

x

y

VV

=ρ

ρ(152)(152)

y

x

x

y

x

y

TT

pp

=ρ

ρ(153)(153)(152)+(1)(152)+(1)

1k1kMa

1kk2

pp 2

xx

y

+−

−+

=(149)+(140)(149)+(140) (150)(150)

183

Normal Shock Wave Normal Shock Wave ppyy/p/pxx TTyy/T/Txx2/22/2

(152)+(153)+(150)+(151)(152)+(153)+(150)+(151) (154)(154)

The stagnation pressure ratio across the shockThe stagnation pressure ratio across the shock

(155)(155)

(156)(156)

(59)+(149)+(150)(59)+(149)+(150)

2Ma)1k(Ma)1k(

VV

2x

2x

y

x

x

y

+−

+==

ρ

ρ

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

x,0

x

x

y

y

y,0

x,0

y,0

pp

pp

pp

pp

)1k/(11k1k2

x1kk2

)k1/(k2x2

1k)1k/(k2x2

1k

x,0

y,0

)Ma(

)Ma1()Ma(pp

−+−

+

−−−+

−

+=

184


Value of downstream Mach numbers, MaValue of downstream Mach numbers, Mayy, pressure ration, , pressure ration, ppyy/p/pxx, , temperature ratio, temperature ratio, TTyy/T/Txx, density ratio, , density ratio, ρρyy//ρρxx or velocity ratio or velocity ratio VVyy/V/Vxx, , and stagnation pressure ratio, pand stagnation pressure ratio, p0,y0,y/p/p0,x0,x are graphed in Figure D.4 as a are graphed in Figure D.4 as a function of upstream Mach number, Mafunction of upstream Mach number, Maxx, for the steady flow across a , for the steady flow across a normal shock wave of an ideal gas having a specific heat ratio knormal shock wave of an ideal gas having a specific heat ratio k=1.4.=1.4.

1Ma)]1k/(k2[)]1k/(2[MaMa 2

x

2x2

y−−

−+= (149)(149)

2x

2

2x

2x

x

y

Ma)]}1k(2/[)1k{(}1Ma)]1k/(k2}{[Ma]2/)1k[(1{

TT

−+

−−−+= (151)(151)

1k1kMa

1kk2

pp 2

xx

y

+−

−+

= (150)(150)

(154)(154)2Ma)1k(Ma)1k(

VV

2x

2x

y

x

x

y

+−

+==

ρ

ρ

(156)(156))1k/(11k1k2

x1kk2

)k1/(k2x2

1k)1k/(k2x2

1k

x,0

y,0

)Ma(

)Ma1()Ma(pp

−+−

+

−−−+

−

+=

185

Summary of Normal Shock WaveSummary of Normal Shock Wave

186

Example 11.17 Stagnation Pressure Drop Example 11.17 Stagnation Pressure Drop Across a Normal ShockAcross a Normal Shock

Designers involved with fluid mechanics work hard at minimizing Designers involved with fluid mechanics work hard at minimizing loss of available energy in their designs. Adiabatic, frictionleloss of available energy in their designs. Adiabatic, frictionless ss flows involve no loss in available energy. Entropy remains constflows involve no loss in available energy. Entropy remains constant ant for these idealized flows. Adiabatic flows with friction involvefor these idealized flows. Adiabatic flows with friction involveavailable energy loss and entropy increase. Generally, larger enavailable energy loss and entropy increase. Generally, larger entropy tropy increases imply larger losses. For normal shocks, shows that theincreases imply larger losses. For normal shocks, shows that thestagnation pressure drop (and thus loss) is larger for stagnation pressure drop (and thus loss) is larger for heigherheigher Mach Mach number.number.

187


x,0

y,0x,0

x,0

y,0

ppp

pp

1−

=−

)1k/(11k1k2

x1kk2

)k1/(k2x2

1k)1k/(k2x2

1k

x,0

y,0

)Ma(

)Ma1()Ma(pp

−+−

+

−−−+

−

+= k=1.4k=1.4

1k1kMa

1kk2

pp 2

xx

y

+−

−+

=

188

Example 11.18 Supersonic Flow Example 11.18 Supersonic Flow PitotPitotTubeTube

A total pressure probe is inserted into a supersonic air flow. AA total pressure probe is inserted into a supersonic air flow. A shock shock wave forms just upstream of the impact hole and head as illustrawave forms just upstream of the impact hole and head as illustrated ted in Figure E11.18. The probe measures a total pressure of 60 in Figure E11.18. The probe measures a total pressure of 60 psiapsia. . The stagnation temperature at the probe head is 1000The stagnation temperature at the probe head is 1000ººR. The static R. The static pressure upstream of the shock is measured will a wall tap to bepressure upstream of the shock is measured will a wall tap to be 12 12 psiapsia. From these data determine the Mach number and velocity of . From these data determine the Mach number and velocity of the flow.the flow.

189


We assume that the flow along the stagnation We assume that the flow along the stagnation pathlinepathline is isentropic is isentropic except across the shock. Also, the shock is treated as a normal except across the shock. Also, the shock is treated as a normal shock.shock.

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

x

x,0

x,0

y,0

x

y,0

pp

pp

pp

)1k/(11k1k2

x1kk2

)1k/(k2x2

1k

x

y,0

)Ma(

)Ma(p

p−

+−

+

−+

−=

Where pWhere p0,y0,y is the stagnation pressure measured by the probe, and is the stagnation pressure measured by the probe, and ppxx is is the static pressure measured by the wall trap. The stagnation prthe static pressure measured by the wall trap. The stagnation pressure essure upstream of the shock, pupstream of the shock, p0,x0,x, is not measured., is not measured.

(11.18(11.18--1)1)

(11.18(11.18--1)+(156)+(59)1)+(156)+(59)

Called the Called the Rayleigh Rayleigh PitotPitot--tube formulatube formula

190


For k=1.4For k=1.4 5psia12psia60

pp

x

y,0 ==

kRTMacMaV xxxxx ==

To determine the flow velocity we need to know the static To determine the flow velocity we need to know the static temperature upstream of the shock. temperature upstream of the shock.

MaMaxx=1.9=1.9

The stagnation temperature downstream of the shock was measured The stagnation temperature downstream of the shock was measured and found to beand found to be


R1000TTR1000T y,ox,oy,o °==⇒°=

191


From or Figure D.1From or Figure D.1

=== 9.1kRTMaV xxx

R590T59.0TT

xx,o

x °=⇒=2a2

1ko M11

TT

−+=

192

Example 11.19 Normal Shock in a Example 11.19 Normal Shock in a ConvergingConverging--Diverging DuctDiverging Duct

Determine, for the convergingDetermine, for the converging--diverging duct of Example 11.8, the diverging duct of Example 11.8, the ratio of back pressure to inlet stagnation pressure, ratio of back pressure to inlet stagnation pressure, ppIIIIII/pa, that will /pa, that will result in a standing normal shock at the exit (x=+0.5 m) of the result in a standing normal shock at the exit (x=+0.5 m) of the duct. duct. What value of the ratio of back pressure to inlet stagnation preWhat value of the ratio of back pressure to inlet stagnation pressure ssure would be required to position the shock at x=+0.3 m? Show relatewould be required to position the shock at x=+0.3 m? Show related d temperaturetemperature--entropy diagrams for these flows.entropy diagrams for these flows.

193


For supersonic, isentropic flow through the nozzle to just upstrFor supersonic, isentropic flow through the nozzle to just upstream eam of the standing normal shock at the duct exit, we have from the of the standing normal shock at the duct exit, we have from the table of Example 11.8 at x= +0.5m.table of Example 11.8 at x= +0.5m.

04.0pp8.2aM

x,0

xx ==

From Figure D.4 for Max=2.8 we obtainFrom Figure D.4 for Max=2.8 we obtain

0.9pp

x

y =

x,0

III

x,0

x

x

y

x,0

y

pp36.0)04.0)(0.9(

pp

pp

pp

===⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

194


When the ratio of duct back pressure to inlet stagnation pressurWhen the ratio of duct back pressure to inlet stagnation pressure, e, ppIIIIII/p/p0,x0,x, is set equal to 0.36, the air will accelerate through the , is set equal to 0.36, the air will accelerate through the convergingconverging--diverging duct to a Mach number of 2.8 at the duct exit.diverging duct to a Mach number of 2.8 at the duct exit.The air will subsequently decelerate to a subsonic flow across aThe air will subsequently decelerate to a subsonic flow across anormal shock at the duct exit.normal shock at the duct exit.The stagnation pressure ratio across the normal shock, pThe stagnation pressure ratio across the normal shock, p0,y0,y/p/p0,x0,x, is , is 0.38 (Figure D.4 for Ma0.38 (Figure D.4 for Maxx=2.8)=2.8)A considerable amount of available energy is lost across the shoA considerable amount of available energy is lost across the shock.ck.For a normal shock at x=+0.3 m, we note from the table of ExamplFor a normal shock at x=+0.3 m, we note from the table of Example e 11.8 that Ma11.8 that Maxx=2.14 and=2.14 and

10.0pp

x,0

x =

195


From Figure D.4 for MaFrom Figure D.4 for Maxx=2.14 we obtain =2.14 we obtain ppyy/p/pxx=5.2 and Ma=5.2 and Mayy=0.56 =0.56 and and

66.0pp

x,0

y,0 =

From Figure D.4 for MaFrom Figure D.4 for Mayy=0.56 we get=0.56 we get

24.1*A

A y =

For x = +0.3m, the ratio of duct exit area to local area (AFor x = +0.3m, the ratio of duct exit area to local area (A22/A/Ayy) is) is

28.2AA

*AA

*AA842.1

)3.0(1.0)5.0(1.0

AA

y

2y22

2

y

2 ==⇒=++

=

196


With AWith A22/A*=2.28 we use Figure D.1 and find Ma/A*=2.28 we use Figure D.1 and find Ma22=0.26 and=0.26 and

95.0pp

y,0

2 =

63.0)66.0)(95.0(pp

pp

pp

x,0

y,0

y,0

2

x,0

2 ===

197

Compressible Flow vs. OpenCompressible Flow vs. Open--Channel Channel Flow Flow 1/61/6

The propagation of weak pressure pulse (sound wave) in a The propagation of weak pressure pulse (sound wave) in a compressible flow can be considered to be comparable to the compressible flow can be considered to be comparable to the movement of small amplitude waves on the surface of an openmovement of small amplitude waves on the surface of an open--channel flow.channel flow.The influence of the flow velocity on wave pattern is similar:The influence of the flow velocity on wave pattern is similar:

When the flow velocity is less than the wave speed, wave fronts When the flow velocity is less than the wave speed, wave fronts can move upstream of the wave source and the flow is subsonic can move upstream of the wave source and the flow is subsonic (compressible flow) or (compressible flow) or subcriticalsubcritical (open(open--channel flow).channel flow).When the flow velocity is equal to the wave speed, wave fronts When the flow velocity is equal to the wave speed, wave fronts cannot move upstream of the wave source and the flow is sonic cannot move upstream of the wave source and the flow is sonic (compressible flow) or critical (open(compressible flow) or critical (open--channel flow).channel flow).

198


When the flow velocity is greater than the wave speed, the flow When the flow velocity is greater than the wave speed, the flow is supersonic (compressible flow) or supercritical (openis supersonic (compressible flow) or supercritical (open--channel channel flow).flow).Normal shock can occur in supersonic compressible flows.Normal shock can occur in supersonic compressible flows.Hydraulic jumps can occur in supercritical openHydraulic jumps can occur in supercritical open--channel flows.channel flows.

199


For compressible flows, Mach numberFor compressible flows, Mach number

For openFor open--channel flows, Froude numberchannel flows, Froude number

cVMa =

gyVFr oc=

VVococ is the velocity of the channel flowis the velocity of the channel flowccococ is the velocity of a small amplitude wave on the surface of an is the velocity of a small amplitude wave on the surface of an

openopen--channel flowchannel flow

(157)(157)

oc

occVFr =gycoc = (158)(158) (159)(159)

200


For compressible flows, Continuity equationFor compressible flows, Continuity equation

For openFor open--channel flows, Continuity equationchannel flows, Continuity equation

ttanconsAV =ρ

ttanconsybVoc =

(160)(160)

(161)(161)

y and b are the depth and width of the openy and b are the depth and width of the open--channel flow. channel flow. Comparing Comparing EqsEqs. (160) and (161) we note that if flow velocities are . (160) and (161) we note that if flow velocities are considered similar and flow area, A, and channel width, b, are considered similar and flow area, A, and channel width, b, are considered similar, the compressible flow density, considered similar, the compressible flow density, ρρ, is analogous , is analogous to opento open--channel depth, y.channel depth, y.

201


It should be pointed out that the similarity between Mach numberIt should be pointed out that the similarity between Mach numberand Froude number is generally not exact.and Froude number is generally not exact.If compressible flow and openIf compressible flow and open--channel flow velocities are channel flow velocities are considered to be similar, then it follows that for Mach number aconsidered to be similar, then it follows that for Mach number and nd Froude number similarity the wave speeds c and Froude number similarity the wave speeds c and ccococ must also be must also be similar.similar.From the development of the equation for the speed of sound in aFrom the development of the equation for the speed of sound in an n ideal gasideal gas

1kk)ttancons(c −ρ= (162)(162)

202


From equation (162) and (158), we see that if y is to be similarFrom equation (162) and (158), we see that if y is to be similar to to ρρas suggested by comparing equation (160) and (161), as suggested by comparing equation (160) and (161), then k then k should be equal to 2.should be equal to 2.Typically k=1.4 or 1.67, not 2.Typically k=1.4 or 1.67, not 2.

This limitation to exactness is, however, usually not This limitation to exactness is, however, usually not serious enough to compromise the benefits of the serious enough to compromise the benefits of the analogy between compressible and openanalogy between compressible and open--channel flows.channel flows.

203

TwoTwo--Dimensional Compressible Flow Dimensional Compressible Flow 1/81/8

Consider a supersonic flow over a wall with a small change of Consider a supersonic flow over a wall with a small change of directiondirection

The component of velocity The component of velocity parallel to the Mach wave is parallel to the Mach wave is constant across the Mach constant across the Mach wave. That is, Vwave. That is, Vt1t1=V=Vt2t2..The flow accelerates The flow accelerates because of the change in because of the change in direction of the flow.direction of the flow.

Flow acceleration across a Mach wave.Flow acceleration across a Mach wave.

204


Consider a supersonic flow over a wall with several change of Consider a supersonic flow over a wall with several change of directiondirection

The supersonic flow The supersonic flow accelerates because of the accelerates because of the change in flow direction change in flow direction across the Mach wave (also across the Mach wave (also called expansion waves).called expansion waves).Each Mach wave makes an Each Mach wave makes an appropriately smaller angle appropriately smaller angle with the upstream wall.with the upstream wall.

Flow acceleration across a Mach wave.Flow acceleration across a Mach wave.

205


Consider a supersonic flow over a wall with round cornerConsider a supersonic flow over a wall with round corner

A rounder expansion corner A rounder expansion corner may be considered as a may be considered as a series of infinitesimal series of infinitesimal changes in direction.changes in direction.

Corner expansion fan..Corner expansion fan..

206


Consider a supersonic flow over a wall with a small change of Consider a supersonic flow over a wall with a small change of directiondirection

Flow deceleration across a Mach waveFlow deceleration across a Mach wave

The flow decelerates and The flow decelerates and static pressure increases static pressure increases across the Mach wave.across the Mach wave.

207


Consider a supersonic flow over a wall with several change of Consider a supersonic flow over a wall with several change of directiondirection

For several changes in wall For several changes in wall direction, several Mach direction, several Mach waves occur.waves occur.The supersonic flow The supersonic flow decelerates across the Mach decelerates across the Mach wave.wave.

Corner expansion fan..Corner expansion fan..

208


Consider a supersonic flow over a wall with round cornerConsider a supersonic flow over a wall with round corner

A rounder compression A rounder compression corner may be considered as corner may be considered as a series of infinitesimal a series of infinitesimal changes in direction.changes in direction.

Oblique shock wave.Oblique shock wave.

209


Consider a supersonic flow incident on a wedgeConsider a supersonic flow incident on a wedge--shaped leading shaped leading edge with smaller wedge angle.edge with smaller wedge angle.

Supersonic flow over a wedge: (Supersonic flow over a wedge: (aa) Smaller ) Smaller wedge angle results in attached oblique wedge angle results in attached oblique shock.shock.

An attached oblique shock An attached oblique shock can form.can form.

210


Consider a supersonic flow incident on a wedgeConsider a supersonic flow incident on a wedge--shaped leading shaped leading edge with larger wedge angle.edge with larger wedge angle.

Supersonic flow over a wedge: Supersonic flow over a wedge: ((bb) Large wedge angle results in detached ) Large wedge angle results in detached curve shock.curve shock.

A detached, curved shock A detached, curved shock ahead of a blunt object can ahead of a blunt object can result.result.

Fluid 11

Documents

linear momentum

higher mach

abrupt pressure

pressure wave

normal shock

diverging

flow enter

stagnation