Chapter 11

Compressible Flow

IntroductionCompressible flow –variable density, and equation of state is importantIdeal gas equation of state—simple yet representative of actual gases at pressures and temperatures of interest Energy equation is important, due to the significant variation of temperature.

RTP ρ=11.1 Ideal gas relationship

For ideal gas, internal energy u=u(T) ( )u duc

T dTυ υ∂

= =∂

22 1

1

T

Tdu c dT u u c dTυ υ= → − = ∫

constant pressure specific heat:

2 1 2 1( )u u c T Tυ− = −-For moderate changes in temperature:

2 1 2 1( )ph h c T T− = −

Enthalpy h=h(T)

( ) ( )ph u u T RT h Tρ

= + = + =

constant pressure specific heat: ( )p ph dhcT dT∂

= =∂

22 1

1

Tp pT

dh c dT h h c dT= → − = ∫-For moderate changes in temperature

Since h=u+RT, dh=du+RdTor

pdh du R c c RdT dT υ= + → − =

( 1.4 for air) and 1 1

pp

c Rk Rk c cc k kυυ

= ∴ = =− −

Entropy 1st Tds equation

(1/ )Tds du pd ρ= +

1 (1/ )

1

ph u dh du pd dp

Tds dh dp

ρρ ρ

ρ

= + → = + +

∴ = −

Q

---2nd Tds equation

(1 / ) (1 / )(1 / )

(1 / ) p

du p dT Rds d c dT T T

c dTdh Rdp dpT T T p

υρ ρρ

ρ

= + = +

= − = −

2 12 1

1 2

2 2

1 1

ln ln( )

ln ln( )p

Ts s c RTT pc RT p

υρρ

− = +

= −

For constant , :p vc c

For adiabatic and frictionless flow of any fluid

2 1 2 2

1 2 1 1or ln( ) ln( ) ln( ) ln( ) 0p

T T pc R c RT T pυ

ρρ

+ = − =

2 10 or 0 isentropic flowds s s= − = ←

2 2 2 21

1 1 1 1ln( ) ln( ) ( ) ( )

1

kkkT TR R

k T Tρ ρρ ρ

−= → =−

2 2 2 21

1 1 1 1ln( ) ln( ) ( ) ( )

1

kkkT p T pkR R

k T p T p−= → =

−

2 2 21

1 1 1( ) ( ) ( ) const, for isentropic flow

kkk

kT p pT p

ρρ ρ

−∴ = = ⇒ = (11.25)

Comparison of isentropic and isothermal compression

Isentropic process path,

Pvk = const

Pv = const

11.2 Mach number and speed of sound

Ma , --local flow velocity, --speed of soundV V cc

=

Sound generally consists of weak pressure pulses that move through air.Consider 1-D of infinitesimally thin weak pressure pulse moving at the speed of sound through a fluid at rest.

fluid at rest Observer moving with control volume

-conservation of mass

( ) ( )Ac A c Vρ ρ δρ δ= + −

δρρδ cV =

c c V c Vρ ρ ρδ δρ δρδ= − + −

-linear momentum conservation( )( )( ) ( )c cA c V c V A pA p p Aρ δ ρ δρ δ δ− + − + − = − +

( )c cA c V cA pAρ δ ρ δ∴ − + − = −

/V Ac pA V p cδ ρ δ ρδ δ− = − → =

( )( )c V A cAρ δρ δ ρ+ − =Q (continuity)

2 p p pc c ccδ δ δδρ

δρ δρ∴ = → = ⇒ =

V cρδ δρ=Q (continuity)

Isentropic process path

-Alternative derivation using conservation of energyinstead of momentum equation

(loss) 0 for frictionless flow; and (z) 0δ δ2 2( ) 0 =

2 2p c V c pV

cδ δ δρδρ

−∴ + − = →

or p c Vδ δρ

=

(from continuity: )p V c V ccδ ρδ δρ ρδ δρ→ = = =

2 P pc cδ δδρ δρ

∴ = ⇒ =

2(loss) --from (5.103)

2p V g zδ δ δ δρ

⎛ ⎞+ − =⎜ ⎟

⎝ ⎠

Assume the frictionless flow through the control volume is adiabatic, then the flow is isentropic.In the limit 0p pδ → ∂ →

s

pcρ

⎛ ⎞∂= ⎜ ⎟∂⎝ ⎠

- For isentropic flow of idea gas kp cρ=

1 1 k kk

s

p p pck k k RTk c RTkρ ρρ ρρ

− −⎛ ⎞∂= = = = ⇒ =⎜ ⎟∂⎝ ⎠

- More generally, use bulk modulus of elasticity

/vs

dp pEd

ρρ ρ ρ

⎛ ⎞∂= = ⎜ ⎟∂⎝ ⎠

/vc E ρ∴ =

vE c→ ∞ ∴ → ∞- For incompressible flow

(11.36)

(11.34)

11.3 Category of compressible flow- effect of compressibility on CDof a sphere

Why CD increases with Ma?Can you explain physically?

(from S.R. Turns, Thermal-Fluid Sciences)

- Imagine the emission of weak pressure pulses from a point source

- stationary point source

cttr wave )( −=

where t → present time, twave → time wave emitted

- moving point source V< C

- source moving at V=c

- source moving at V>c

- Category of fluid flow1. Incompressible flow Ma 0.3≤

unrestricted, linear symmetrical and instantaneous pressure communication.

2. Compressible subsonic flow 0.3 Ma 1.0< <unrestricted, but noticeably asymmetrical pressure communication

3. Compressible supersonic flow Ma 1≥

formation of Mach wave, pressure communicationrestricted to zone of action

4. transonic flow 0.9 Ma 1.2≤ ≤ (modern aircraft)

5. hypersonic flow Ma 5≥ (space shuttle)

Example 11.4 Mach cone

Video: Mach cone of an airplane

(Note the condensed cloud across the shock wave.)

(Can you estimate the airplane speed?)

Mach cone from a rifle bullet

from M. Van Dyke, An Album of Fluid Motion

Ma=0.978

from Gas Dynamics Lab, The Penn. State University, 2004

11.4 Isentropic flow of an idea gas- no heat transfer and frictionless

11.4.1 Effect of variation in flow cross-section area

- conservation of mass

const.m AVρ= =&

- Conservation of momentum for a inviscid and steady flow0

21 ( ) 02

dp d V dzρ γ+ + =

2dp dV

VVρ= −

Since , ln ln lnm AV c A V cρ ρ ′= = ∴ + + =&

2

differentiation 0

( )

d dA dVA V

dV d dA dpV A V

ρρ

ρρ ρ

→ + + =

→ − = + = (11.44)

2

2 2

2

2

(1 )

(1 )/

dA dp d dp d VA dpV V

dp Vdp dV

ρ ρ ρρ ρρ ρ

ρρ

= − = − ⋅

= −

Since , and Mas

p Vccρ

⎛ ⎞∂= =⎜ ⎟∂⎝ ⎠

22 (1 Ma )dp dA

AVρ∴ − =

2 21

1 Madp dA dV

A VVρ− = − =

−

(11.45)

(11.47)

(11.48)

converging duct

diverging duct

2 21

1 Madp dA dV

A VVρ− = − =

−

Combining (11.44) and (11.48):2

2 21 Ma

1 Ma 1 Mad dA dA d dA

A A Aρ ρρ ρ

+ = ⇒ =− −

(11.49): For subsonic flow, density and area changes are in the same direction; for supersonic flow, density and area changes are in the opposite direction.

2From (11.48): (1 Ma )dA AdV V

= − −

When Ma 1 0 dAdV

= → = ⇒ The area associated with Ma=1 is either a minimum or a maximum.

(11.49)

impossible

Therefore the sonic conduction Ma=1 can be obtained in a converging-diverging duct at the minimum area location.

For subsonic flow → converging diverging nozzleFor supersonic flow →converging diverging diffuser

throat

supersonic subsonicsubsonic

Converging-diverging nozzle Converging-diverging diffuser

throat

What is the physical reason that supersonic flow develops in the diverging duct as long as sonic flow is reached at the throat?

Why can V only be accelerated to c in the converging duct, no matter how low the back pressure Pe is?

throat

VavgVavg

P0

Pe

ΔP-driven flowΔP-induced flow

11.4.2 Converging–Diverging Duct Flow- For an isentropic flow

0

0

constantk k

ppρ ρ

= =

- streamwise equation of motion for steady, frictionless flow2

( ) 0, neglected2

dp Vd dzγρ+ =

1/ 1/00 0

0( / )k k

k kpp p pρ ρ

ρ ρ⎛ ⎞

= → =⎜ ⎟⎝ ⎠Q

1 / 20

1 /0

( ) 02

k

kp dp Vd

pρ+ =

1 11/ 20

00

[ ] 01 2

k kkk kpk Vp p

k ρ

− −

− − =−

20

0

[ ] 01 2

pk p Vk ρ ρ

− − =−

2 2

0 0[ ] 0 or ( ) 0 ( )1 2 2 1p p

kR V V kRT T c T T ck k

∴ − − = − − = =− −

Q

2

0 ( ) 02

Vh h⇒ − + = where h0 is the stagnation enthalpy

20

1 1 2kRT kRT Vk k

= +− −

20

12

kkRT kRT V−= +

-For an idea gas

00

0

, p pRT RTρ ρ

= =

22 20 1 11 1 Ma ( )

2 2T k V k kRT cT kRT

− −= + = + =

20

11 [( 1) 2]Ma

TT k

=+ − (11.56)

With p RTρ=

10 0 0 0

0 0 0

( ) kk k

p pp T pp T p

ρ ρρ ρ ρ ρ

⎛ ⎞= = ⇒ =⎜ ⎟

⎝ ⎠Q

1 11

0 0 0 0 0 0 0

( ) ( ) ( )k k

k k kp p T p T p Tp p T p T p T

−−

−= → = ⇒ =

12

0

1[ ]1 [( 1) 2]Ma

kkp

p k−∴ =

+ −

11

20

1[ ]1 [( 1) 2]Ma

k

kρρ

−=+ −

using isentropic relation

20

11 [( 1) 2]Ma

TT k

=+ − (11.56)

(11.60)

(11.59)

12

0

1[ ]1 [( 1) 2]Ma

kkp

p k−=

+ −

11

20

1[ ]1 [( 1) 2]Ma

k

kρρ

−=+ −

20

11 [( 1) 2]Ma

TT k

=+ −

Figure D1 (p. 718)Isentropic flow of an ideal gas with k = 1.4.

2

0 2p pVc T c T= +

Figure 11.7The (T − s) diagram relating stagnation and static states.

Figure 11.8The T – s diagram for Venturimeter flow.

-Any further decrease of the back pressure will not effect the flow in the converging portion of the duct.

-At Ma=1 the information about pressure can not moveupstream

supersoinc

subsoinc

0.528

0

pp

-Consider the choked flow where at the throat Ma=1, the state is called critical state

Ma =1, critical state (choked flow)

*1

0

2( )1

kkp

p k−=

+

For k=1.4

*

0 1.4

0.528k

pp

=

⎛ ⎞=⎜ ⎟

⎝ ⎠*

1.4 atm0.528kp p= =* *

*1.4 0 atm

0 0 1.4

2 0.833 or 0.833 0.8331 k

k

T T T T TT k T =

=

⎛ ⎞= → = = =⎜ ⎟+ ⎝ ⎠

1* * *0 1 1

*0 0 0 1.4

2 1 2( ) ( ) ( ) 0.6431 2 1

kk k

k

Tp kT p k k

ρ ρρ ρ

− −

=

⎛ ⎞+= = = → =⎜ ⎟+ + ⎝ ⎠

Critical State:Set Ma=1 in (11.56), (11.59), (11.60)

Example 11.50

30

0

101 kPa

1.23 kg/m288 K

Find (a)80 kPa, (b)40 kPa.

p

Tm

ρ

=

=

==&

0Critical pressure * 0.528 =53.3 kPap p=

(a) pa > p* ∴the throat is not choked

1th2

0

80 1[ ] Ma 0.587101 1 [( 1) 2]Ma

kkp

p k−= = → =

+ −1

3 3102

0

1[ ] 1.23 / 1.04 /1 [( 1) 2]Ma

k kg m kg mk

ρ ρ ρρ

−= = ⇒ =+ −

20

1 269 K1 [( 1) 2]Ma

Ma * 193 m/s0.0201 kg/s

T TT k

V kRT Vm VAρ

= ⇒ =+ −

= ⇒ == =&

(b) pb=40 kPa < p*=53.3 the flow is choked at the throat Ma = 1 1

3102

0

1[ ] 0.634 0.78 kg/m1 [( 1) 2]Ma

k

kρ ρ ρρ

−= ⇒ = =+ −

20

1 2401 [( 1) 2]Ma

Ma * 310 m/s0.0242 kg/s

T T KT k

V kRT Vm VAρ

= ⇒ =+ −

= ⇒ == =&

Figure D1 (p. 718)Isentropic flow of an ideal gas with k = 1.4. (Graph provided by Dr. Bruce A. Reichert.)

* ** * *

*

* *

or

with and Ma

A VAV A VA V

V kRT V kRT

ρρ ρρ

= =

= =

** * *0

*0 0

1 1 121 1 2

2

/1Ma /Ma

1 2 1 2 1 ( ) [1 Ma ] [ ]Ma 1 2 1 1 [( 1) 2]Ma

k k

T TA kRTA T TkRT

k kk k

ρ ρρ ρ

− −

→ = =

+ += +

+ + −

Ratio A/A*

122( 1)

*

1 1 [( 1) 2]Ma[ ]Ma 1 [( 1) 2]

kkA k

A k

+−+ −

⇒ =+ −

Figure 11.10 (p. 639)The variation of area ratio with Mach number for isentropic flow of an ideal gas (k = 1.4, linear coordinate scales).

Example 11.8 Air entering subsonically at standard atmospheric condition, A=0.1+x2

1 122 2 20.1( ) ( )A xA r rπ

π π+

= → = =

For the flow to be chocked

*

2

*0 0

At throat, 0 0.10.1 , using (11.71) Ma ,

0.1

x AA x p TA p T

= → =

+= ⇒ ⇒

c

d

a

c

0.98

d

0.04

b

Example 11.9 Air entering supersonically at standard atmospheric condition, A=0.1+x2

For the flow to be chocked

*

2

*0 0

At throat, 0 0.10.1 , using (11.71) Ma ,

0.1

x AA x p TA p T

= → =

+= ⇒ ⇒

d

0.04

c

0.98

b

c

d

a

Example 11.10 Ma=0.48 at throat, A=0.1+x2

The Ma at throat is 0.48

*0 0

** *

Ma 0.48 , ,

0.1 1.4 0.07

p T Ap T A

A AA A

= ⇒

= = ⇒ =

c

b

subsonic-subsonic subsonic-subsonic(choked)

subsonic-supersonic(choked) supersonic-supersonic(choked)

Supersonic-subsonic (choked) Supersonic-supersonic (not-choked)

1. For a given (T0, p0), k and converging-diverging duct geometry, infinite number of isentropic subsonic to subsonic (not choked) and isentropic supersonic to supersonic (not choked) flow solutions exist.

2. For choked condition, the flow solutions are each unique.

or , isentropic flow is possible, isentropic flow is not possible

ext ext

ext

p p p pp p p

Ι ΙΙ

Ι ΙΙ

≥ ≤

≤ ≤

isentropic isentropic

Entropy generation

Oblique shock wave (3-D)

Non-isentropic choked flow

underexpanded

overexpanded

V11.6 Supersonic nozzle flow

V11.5 Rocket engine start-up

From Zucrow and Hoffman, Gas Dynamics

11.4.3 Constant Area Duct FlowFor constant area isentropic duct flow, the flow velocity, thus the fluid enthalpy and temperature are constant.

11.5 Nonisentropic flow of an ideal gas

Fanno flow – adiabatic flow with friction. Rayleigh flow – constant area duct flow with heat transfer but without friction

pbp0

11.5.1 Adiabatic constant area duct flow with friction (Fanno flow)

Consider steady 1-D ideal gas constant area duct flow with friction

(11.75)

Energy equation:2 2

2 12 1 2 1

2

0 0 0

2

0

2 2 2

0 02 2 2

[ ( )]2

, ( )2

const2

( ) ( ) , where const2 2 /

p

p

p p

V Vm h h g z z Q W

Vh h h h c T T

VT Tc

V V TT T T T Vc c p Rρ ρ ρρ

−− + + − = +

+ = − = −

+ = =

⇒ + = ⇒ + = =

& &&

Continuity: const constm VA Vρ ρ= = ⇒ =&

(stagnation temp.= const)

2 22 1

1 1

11 1

1 1 1

1( )

ln ln

ln ln (11.76)

, , are considered reference values from the entrance.

p

p

p

Tds dh dp

dT dpds c RT p

T ps s c RT p

T ps s c RT p

p T s

ρ= −

= −

− = −

− = −

Ex 11.11

Eq. 11.75 allows us to calculate T for p in the Fanno flow.Tds equation (2nd law):

From (11.75) and (11.76), the Fanno line for variation of p-T-scan be obtained.

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=284K=284KTT11=286K=286Kpp11=99kPa(abs)=99kPa(abs)For For FannoFanno flow, determine corresponding value of fluid temperature and flow, determine corresponding value of fluid temperature and entropy change for various values of downstream pressures and plentropy change for various values of downstream pressures and plot the ot the related related FannoFanno line.line.

Example 11.11Example 11.11

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)

KkgJkRkcp ⋅==−

= /1004...1

From From EqEq. (14). (14)

KkgJRk ⋅== /286.94.1

(11.11.3)(11.11.3)

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

RTpV 11

1

111 =ρ==ρ (11.11.4)(11.11.4)

Example 11.11Example 11.11

2.002.0/1399.0

1M 1 =⎟⎠⎞

⎜⎝⎛ −=aFrom From EqEq. (56). (56)

(11.11.4)(11.11.4) )/(8.81)286)(/6.289()/339(2.01099V 2

3

smkgKKkgJsmPa

⋅=⋅

×=ρ

993.02882861 ==

KK

TT

o

smkRT /339...1 ==

For p= 48 For p= 48 kPakPa

(11.11.1)(11.11.1) KKRpc

TVTP

7.278T288...)/(2

)(22

22

=⇒==+ρ

(11.11.2)(11.11.2) )/(7.181...lnln11

1 KkgJppR

TTcss P ⋅==−=−

22

1ko Ma11

TT

−+= (56)(56)

Example 11.11Example 11.11

For p=48kPa T=278.7KFor p=48kPa T=278.7K ss--ss11=181.7J/(kg=181.7J/(kg‧‧K)K)For p=41kPa T=275.6KFor p=41kPa T=275.6K ss--ss11=215.7J/(kg=215.7J/(kg‧‧K)K)

For p=34kPa T=270.6KFor p=34kPa T=270.6K ss--ss11=251.0J/(kg=251.0J/(kg‧‧K)K)

Analysis of Fanno lineTds equation

dp dpTds dh dh RTpρ

= − = −

For an ideal gas,

;

or

Pdh c dTdp d dTp RTp T

ρρρ

=

= = +

Q

( )

P

P

dpTds c dT RTpd dTc dT RT

Tρρ

= −

= − +

const, or d dVVV

ρρρ

= = −Continuity:

( ) ( + )

1 1- (- )

P P

P

d dT dV dTTds c dT RT c dT RTT V T

cds dVRdT T V dT T

ρρ

→ = − + = − −

→ = +

2

0 const 2

P

p p

cV VdV dVT T dTc c dT V

+ = = → = − ⇒ = −

21- ( )p pc cds R

dT T TV∴ = + (11.82)

Energy eq.:

2 21For 0 ( )p p p

pc c cds R c R c RT

dT T TV Vυ= → = + → − = =

( / )p a aV c c RT kRTυ= =

So, the Mach number at state a is 1.

0, subsonic

0, supersonic

a

a

ds V kRTdTds V kRTdT

< < →

> > → supersonic

Since T0 is constant on the Fanno line, the temperature at point a is the critical temperature T*.

subsonic

Ma=1

supersonic

-2nd law ds>0

subsonic flow (acceleration)

Normal shock supersonic flow (deceleration)

Summary of Fanno flow behavior

Friction drags acceleration by pressure drop Friction helps deceleration by pressure rise

-To quantify the Fanno flow behavior, we need to combine relationship that represents the linear momentum law with the set of equations already derived.

1 1 2 2 2 1( )xp A p A R m V V− − = −

1 2 2 1 1 2( ), ( and )xRp p V V V A A A m AV CA

ρ ρ→ − − = − = = = =&Q

-Therefore, for the semi-infinitesimal control volumes

w Ddxdp VdVA

τ π ρ− − =

2

2

8with , 4

w Df AVτ πρ

= =

2

2 2

2

( )or 02 2

V dxdp f VdVD

dp f V dx d Vp p D p

ρ ρ

ρ ρ

→ − − =

+ + =

2 2( ) 02 2

dp f V dx d Vp p D p

ρ ρ+ + =

2 22 2

2 21 ( ) (Ma )(1 Ma ) Ma 02 2Ma

d V d fk dxkDV

+ − + =

(11.88)

After some derivation, (11.88) becomes

{ }2 2

2 4

(1 Ma ) (Ma )or 1 [( 1) 2]Ma Ma

d dxfDk k

−=

+ −

-Integrate to the critical state

(11.96)

* *2 2Ma 1

2 4Ma

(1 Ma ) (Ma )1 [( 1) 2]Ma Ma

d dxfDk k

= −=

⎡ ⎤+ −⎣ ⎦∫ ∫

l

l

( ) ( )2 *2

2 2

1 Ma1 1 [( 1) 2]Ma2Ma 1 [( 1) 2]Ma

fk knk k Dk

− −⎡ ⎤+ ++ =⎢ ⎥

+ −⎣ ⎦

l ll (11.98)

-Note that the critical state does not have to exist, since for any two section in the Fanno flow,

( ) ( ) ( ).*21

12*

llllll

−=−

−−

Df

Df

Df

-Other fluid properties in the Fanno flow can also be derived, as summarized below.

Summary of Property Relations for Fanno Flow

(11.98)

(11.101)

(11.103)

(11.107)

(11.109)

(11.105)

Note: These equations correlate ratio of properties associated with different positions (a certain position and the choke position), between which friction loss exists.

( ) ( )2 *2

2 2

2

12 2

2

1

12

2

0 0

0 0

1 Ma1 1 [( 1) 2]MaMa 2 1 [( 1) 2]Ma

( 1) 2 ,* 1 [( 1) 2]Ma

[( 1) 2]Ma , * 1 [( 1) 2]Ma

,* *

1 ( 1) 2 ,* * * Ma 1 [( 1) 2]Ma

* 1 2* * * Ma 1

fk knk k k DT k

kT

V kkV

V

V

p T kkp T

p p p pp p p p k

ρ

ρ

ρ

ρ

−

− −⎡ ⎤+ ++ =⎢ ⎥+ −⎣ ⎦+

=+ −

⎡ ⎤+= ⎢ ⎥+ −⎣ ⎦

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎡ ⎤+= = ⎢ ⎥+ −⎣ ⎦

= =+

l ll

( ) ( )1

2 121 [( 1) 2]Ma

kk

k

+−⎡ ⎤⎛ ⎞ + −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

different positionssame position

Note: For p/p0, p*/p0*, T/T0at the same position, isentropic relations in terms of Ma or Fig. D1 can be used.

Figure D2 (p. 719)Fanno flow of an ideal gas with k = 1.4.

*p

p

0

0 *pp

Note: These curves correlate ratio of properties associated with different positions (a certain position and the choke position), between which friction loss exists.

Example 11.12 Choked Fanno flow0 0Given 101 kPa, 288 Kp T= =

What is the maximum flow rate at the duct?

-For maximum flow rate, the flow must be choked at the exit.( ) ( )

4.01.0

202.0121*

=×

=−

=−

Df

Df llll

0,11 1 11 * * * *

0

Fig. D2 Ma 0.63 1.1, 0.66, 1.7, 1.16pT V p

T V p p→ = → = = = =

1 1 11

0 0,1 0,1

Fig. D1 with Ma 0.63 0.93, 0.76, 0.83T pT p

ρρ

= → = = =

1 1 1.65 kg/sm V Aρ= =&

30,1 0,1 0,1 1 0,1/ 1.23 0.83 1.02 kg/mp RTρ ρ ρ= = → = =

0*

*0 2

0

* * *2 1

Since = =288 K

2 0.8333 (0.8333) 2401

310m/s ( ) = 0.66 205m/s

T C

T T T TT k

V RT k V V V

= = ⇒ = = ⇐+

= = ⇐ → =

*0 0,2 0,1

1 87kPa1.16

p p p= = × =

1 0

1 1 1

Or, perhaps more logical,0.93 0.93 288

268 K

= Ma

0.63 1.4 386.9 268 207m/s

T T

V kRT

= × = ×

=

×

= × × ×=

*1

2 0,11 0,1

1 0.76 101 45kPa1.7

ppp pp p

= = × × =

with friction, use Fig. D2

Example 11.13 Effect of duct length on choked Fanno flow

1 11

0 0,1

Fig. D1 with Ma 0.70 0.72, 0.79pp

ρρ

= → = =

If the flow is choked.

( )*1 1 1

1 * *

0.02 1 0.2, Fig. D2 Ma 0.7, 1.5, 0.73, 0.1

f p VD p V

− ×= = → = = =

l l

** 1

2 011 01

1 0.72 101 48.5kPa ( 45kPa)1.5 d

ppp p p pp p

= = = × × = > =

1m

45kPadP =

*1 0,1 1

1 1 1

0.79 , 0.731.73 kg/s

V Vm AVρ ρ

ρ

= =

→ = =&

-For the same upstream stagnation state and downstream pressure,

or when is fixed: m f m↑→ ↓ ↑→ ↓& &l l

Example 11.14 Unchoked Fanno flow

2

0 const, although there is no friction, why?2Vp p ρ

= + ≠

11.5.2 Frictionless constant-area duct flow with heat transfer (Rayleigh flow)

or( ) ( )2 2

const constV V RT

p pp

ρ ρρ

+ = → + = ──(11.111)

1 1 1 2 2 2 xp A mV p A mV R+ = + +0 (frictionless flow)

Momentum:

Tds eq.: 11 1

pT ps s c n R nT p

− = −l l ──(11.76)

Continuity: V Cρ =

Eqs. (11.111) and (11.76) can be used to construct a Rayleigh line with reference conditions.

Example 11.15 Construction of a Rayleigh line for various downstream p2 (or T2), with entrance conditions T0, T1, p1

0 1 1 1 1Given , , , , and can be calculated.T T p V V Cρ ρ→ =

-Assume p2 (or T2), then from (11.111) T2 (or p2) can be obtained.

-From (11.76), s2 can be obtained.

( )2

11 1

const (11.111)

(11.76)p

V RTp

pT ps s C n R nT p

ρ+ =

− = −l l

-At point a on the Rayleigh line, 0dsdT

=

-After some derivation, we have

[ ]1 (11.115)

( / ) ( / )pcds V

dT T T T V V R= +

−

Discussion on the Rayleigh line

[ ]

( ) 22

2

For 0,

1 ( / ) ( / )

01

1 0

1 1 1

Ma 1

p

a a a

dsdT

cds VdT T T T V V R

kR T V Vk V R

k VkRT kVk k kV kRT

V kRT

=

= +−

⎛ ⎞− + =⎜ ⎟− ⎝ ⎠−

− + =− − −=

= ⇒ =

dp VdVρ= −

VdVdP−=

ρ

Since Tds equation

p

dpTds dh

c dT VdVρ

= −

= +

or

( )1

/ /

p

p

cds V dVdT T T dT

c VT T T V V R

= +

= +−

.

1 1

V Cd dV

Vp RTdp d dTp T

VdV dV dTRT V T

V dTRT V T dV

T V dTV R dV

ρρρρ

ρρ

ρρ

=⎡⎢⎢ = −⎢⎣=

= +

− = − +

− = − +

− =

Derivation of ds/dT

[ ]

[ ]

[ ]

1

2

At point , / 01

( / ) ( / )1 1 0

( / ) ( / )

1( / ) ( / )

1Ma

p

p

p

bb

b dT dscds V

dT T T T V V RdT

ds c Vds T V V RdT T Tc VT T T V V R

T V V RT V RTV R

V RTc kkRT

−

=

= +−

= = =+ −

+ →∞−

∴ = ⇒ = ⇒ =

= = =

- At point b, the flow is subsonic since k＞1.

( )

( ) ( )

( )

2

1 122

122

, 1

/ 1

11 Ma

1 1 Ma 1

p

p

p p

p

p p

p p

kRdh VdV q dh c dT dTk

c dT VdV q

dT VdV qT c T c T

dV V dT V qV T dV kRT k c T

k VdV q V dT q V T V kV c T T dV kRT c T T V R

q V qk kc T RT c T

δ

δ

δ

δ

δ δ

δ δ

− −

−

+ = = =−

+ =

+ =

⎡ ⎤+ =⎢ ⎥

−⎢ ⎥⎣ ⎦

⎡ ⎤− ⎡ ⎤⎛ ⎞∴ = + = − + −⎢ ⎥ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦⎣ ⎦

⎡ ⎤= − + − = −⎢ ⎥

⎣ ⎦( ) 12 2

2

Ma 1 Ma

1 1 Map

k

qc Tδ

−⎡ ⎤+ −⎣ ⎦

=⎡ ⎤−⎣ ⎦

- Now, consider the energy equation,

(11.121)

( )2 2

2 12 1 2 12 net sheftn et

V Vm h h g z z Q W⎡ ⎤−

− + + − = +⎢ ⎥⎣ ⎦

&

-linear momentum2 2

a a ap V p Vρ ρ+ = + where a is the reference state.

22or 1 a

aa a a

p V Vp p p

ρρ+ = +

22 2 , since a a a

a a a aa a a a

kVV V k V kRTp RT kRTρ ρ

ρ= = = =

221 1a

aa a a

p V V kp p p

ρρ+ = + = +

Derivation for Property Relations for Rayleigh Flow

21 1 , where Ma

a

p V k V kRTp p

ρ⎛ ⎞+ = + =⎜ ⎟

⎝ ⎠

2

11 Maa

p kp k

+=

+ (11.123)

2

11 Maa

p kp k

⎛ ⎞+=⎜ ⎟+⎝ ⎠

Q( ) 22

2

1 MaMa

1 Maa

a a a a

kT p T pT p T p k

ρρ

⎡ ⎤+⎛ ⎞= → = =⎜ ⎟ ⎢ ⎥+⎝ ⎠ ⎣ ⎦

Ma Maa

a a a a a

V T T p TV T T p T

ρρ

= = ⇒ =

( )2

1 MaMa Ma

1 Maa

a a

kV TV T k

ρρ

⎡ ⎤+= = = ⎢ ⎥+⎣ ⎦

- Due to heat transfer, T0 varies

( )

( ) ( )

0 0

0, 0,

22

2 2

22

1 Ma 11 [( 1) / 2]Ma1 Ma 1 ( 1) / 2

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

1 Ma

a

a a a

T T TTT T T T

kk

k k

k k

k

=

⎡ ⎤+ ⎡ ⎤⎡ ⎤= + − ⎢ ⎥ ⎢ ⎥⎣ ⎦ + + −⎣ ⎦⎣ ⎦

+ + −=

⎡ ⎤+⎣ ⎦(11.131)

(11.129)

( )

( )

( )( ) ( )

( ) ( )

2

2

2

2

120 0

20, 0,

2 20 0

220, 0,

11 Ma

1 Ma1 Ma

1 MaMa

1 Ma

1 2 1 [( 1) / 2]Ma11 Ma

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

1 Ma

a

a

a

a

kk

a

a a a

a

a a a

p kp k

kTT k

kVV k

kp p pp kp p p p kk

k kT T TTT T T T k

ρρ

−

+=

+

⎡ ⎤+= ⎢ ⎥+⎣ ⎦

⎡ ⎤+= = ⎢ ⎥+⎣ ⎦

+ ⎡ ⎤⎛ ⎞= = + −⎜ ⎟⎢ ⎥++ ⎝ ⎠⎣ ⎦

+ + −= =

⎡ ⎤+⎣ ⎦

(11.123)

(11.128)

(11.129)

(11.133)

(11.131)

Summary of Property Relations for Rayleigh Flow

Figure D3 (p. 720)Rayleigh flow of an idea gas withk = 1.4. (Graph provided by Dr. Bruce A. Reichert.)

a

a

VV,

ρρ

Effect of Ma and Heating/Cooling for Rayleigh Flow(Example 11.16)

Note: p0 is not constant, although frictionless

11.5.3 Normal Shock Waves-Normal shock waves involves:• deceleration from supersonic to subsonic• a pressure rise• an increase of entropy

V11.7 Blast waves

(from Cengel and Cimbala, Fluid Mechanics, 2006)

FIGURE 12–30Schlieren image of a normal shock in a Laval nozzle. The Mach number in the nozzle just upstream (to the left) ofthe shock wave is about 1.3. Boundary layers distort the shape of the normal shock near the walls and lead to flow separation beneath the shock.

V11.6 Supersonic nozzle flow

infinitesimal thin control volume surrounding the shock wave: friction and heat transfer negligible, A=C

constVρ =─continuity:

( )22 const or const

V RTp V p

pρ

ρ+ = + =

─linear momentum（friction negligible）:─same as Rayleigh line

Derivation for Normal Shock Waves

2

0 const2

Vh h+ = =

with 0, 0z qΔ = Δ =

( )( )

2 2

02 2

For an ideal gas,

const2 /p

V TT T

c p Rρ

+ = =─same as Fanno line

─energy:

─Tds relationship: 11 1

pT ps s c n R nT p

− = −l l

Q: The irreversibility in shock wave is not from friction or heat transfer. What is it from?

Since shock wave flows have the same energy eq. for Fanno flows and same momentum eq. for Rayleigh flows, thus for a given ρV, gas (R, k), and conditions at the inlet of the normal shock (Tx, px, sx), the conditions downstream of the shock (state y) will be on both a Fanno line and a Rayleigh line that pass through the inlet state (state x).- For Rayleigh line,

y y a

x a x

p p pp p p

=

2 21 1,

1 Ma 1 May x

a ay x

p pk kp pk k

+ += =

+ +Q

2

21 Ma (11.140)1 Ma

y x

x y

p kp k

+∴ =

+

1* 2

* * 2

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

y y

x x

p p p p kp p p p k

⎡ ⎤+= = ⎢ ⎥+ −⎣ ⎦

1/ 22

2Ma 1 [( 1) / 2]Ma (11.148)Ma 1 [( 1) / 2]Ma

y x x

x y y

p kp k

⎡ ⎤+ −→ = ⎢ ⎥

+ −⎢ ⎥⎣ ⎦

- For Fanno line

Total energy, mass, Tds eqs.

Mom., mass, Tds eqs.

Total energy, mom., mass, Tds eqs.

2*

* 2

1 [( 1) / 2]Ma1 [( 1) / 2]Ma

y y x

x x y

T T kTT T T k

+ −→ = =

+ −

* 2 * 2

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

y x

y x

T Tk kT k T k

+ += =

+ − + −

1/ 22 2

2 2

Combining (11.140) and (11.148)

1 [( 1) / 2]Ma Ma 1 Ma1 [( 1) / 2]Ma Ma 1 Ma

y x x x

x y y y

p k kp k k

⎡ ⎤+ − +→ = =⎢ ⎥

+ − +⎢ ⎥⎣ ⎦

- For Fanno line, Eq. 11.101

22

2Ma [2 /( 1)]Ma

[2 /( 1)]Ma 1x

yx

kk k

+ −⇒ =

− −(11.149)

22

2

1 Ma 2 1(11.149) into (11.140) Ma1 Ma 1 1

y xx

x y

p k k kp k k k

+ −⇒ = = −

+ + + (11.150)

(11.144)2 2

2 2

[1 [( 1) / 2]Ma ]{[2 /( 1)]Ma 1}(11.149) into (11.144){( 1) /[2( 1)]}Ma

y x x

x x

T k k kT k k

+ − − −⇒ =

+ − (11.151)

2 21 10,

120, 1

1 1[ Ma ] [1 Ma ]2 2

2 1[ Ma ]1 1

k kk k

x xy

x kx

k kpp k k

k k

− −

−

+ −+

=−

−+ +

Summary of Property Relations across Shock Wave

2

2

( 1)Ma( 1)Ma 2

y x x

x y x

V kV k

ρρ

+= =

− +

22

2Ma [2 /( 1)]Ma

[2 /( 1)]Ma 1x

yx

kk k

+ −=

− −

2 2

2 2

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

y x x

x x

T k k kT k k

+ − − −=

+ −

22 1Ma1 1

yx

x

p k kp k k

−= −

+ +

(11.149)

(11.150)

If Max is known, property ratios across the shock can be known:

(11.151)

(11.154)

(11.156)

Figure D4 (p. 721)Normal shock flow of an idea gas with k = 1.4. (Graph provided by Dr. Bruce A. Reichert.)

Example 11.17 Stagnation pressure drop across a normal shock

0,

0,

For Ma and y yx

x x

p pp p

↑⇒ ↑ ↓

0,

0,

y

x

pp

1 2.5 5

0.06

0.5

1

Ma x

y

x

pp

1 3 5

10

29

Ma y

1

Across a normal shock, adverse pressure gradient occurs which can cause flow separation. Therefore, shock-boundary layer interactions are of great concern to designers of high speed flow device.

Example 11.18 Supersonic flow pitot tube

0, 0Given: 4114 kPa, 555K, 82 kPay xp T p= = =

2 10, 0, 0,

120, 1

1[ Ma ]2 Rayleigh Pitot tube formula

2 1[ Ma ]1 1

414kPa 582kPa

kk

xy y x

x x x kx

kp p pp p p k k

k k

−

−

+

= = ⇐−

−+ +

= =

0, 0,0,

Fig. D1 Ma 1.9

Ma Ma

, 0.59 327K

678 m/s

x

x x x x x

xx y x

x

x

V c kRTTT T TT

V

→ =

= =

= = ⇒ =

⇒ =

Note: Incompressible calculation for pitot tube would give the wrong result.

Example 11.19 Normal shock in a converging-diverging duct

0 0

(a) ? for normal shock at exit, (b) =? for shock at 0.3 mIIIp p xp p

= =

x=-0.5 x=0.5

2.80.98

0.04

-shock at x=0.3m

0,

0,

0,

From Ex 11.8: Ma 2.14, 0.1

Fig. D4 with Ma 2.14 across the shock

5.2, Ma 0.56, 0.66

(Conditions downstream the shock at =0.3m)

xx

x

x

y yy

x x

pp

p pp p

x

= =

= →

= = =

Note: for isentropic A*=0.1 (a minimum)

0, 0, 0,

0,

0,

9 0.04 0.36

0.38 considerable energy loss

y y x III

x x x x

y

x

p p p pp p p ppp

= = × = =

= −

-shock at x=0.5m

0,

0,

0,

From Ex 11.8: Ma 2.8 and 0.04 at 0.5m

Fig. D4 Ma 2.8(at exit): 9, 0.38

xx

x

y yx

x x

p xpp pp p

= = =

→ = = =

22

2

2 2

2

Consider the isentropic decelerating flow downstream the shock

Also, Fig. D4 1.24 (Here, * is used as dummy)*

0.1 (0.5) 1.8420.1 (0.3)

1.24 1.842 2.28* *

* 0.152.28

y

y

y

y

AA

AAA

AA AA A A

AA

→ =

+= =

+

= = × =

→ = =

2 22

0,

0,2 2

0, 0, 0,

2.28, Fig. D1 (isentropic) Ma 0.26, 0.95*

0.95 0.66 0.63

y

y

x y x

A pA p

pp pp p p

= → = =

→ = = × =

0, 0,

0, 0,

: 0.66, shock at 0.3m > 0.36, shock at exit 0.5my y

x x

p px x

p p= = = =Note

Figure D1 (p. 718)Isentropic flow of an ideal gas with k = 1.4. (Graph provided by Dr. Bruce A. Reichert.)

11.7 Two-Dimensional Compressible FlowSupersonic- Flow acceleration across a Mach wave

1 2

2 1

2 1

t t

n n

V V

V V

V V

=

>

∴ >

- Supersonic flows decelerate across compression Mach wave

from M. Van Dyke, An Album of Fluid Motion

Small wedge angle-attached shock

Large wedge angle-detached shock

from M. Van Dyke, An Album of Fluid Motion

Figure 11.4 (p. 591)The schlieren visualization of flow (supersonic to subsonic) through a row ofcompressor airfoils. (Photograph provided by Dr. Hans Starken, Germany.)

(from Cengel and Cimbala, Fluid Mechanics, 2006)

3-D shock wave around a model space shuttle

from Gas Dynamics Lab, The Penn. State University, 2004