Lecture 3

• Overview of relevant concepts in compressible flow)

– The 1D Euler equation for compressible flow

• Primitive/conservative variables

– Entropy and second law of thermodynamics

– Speed of sound (�)– Relations across a discontinuous interface

• Contact discontinuity

• Normal Shock

– Quasi-1D Nozzle flow

• Some conveniently defined quantiles

1

Reference book chapterfor lectures of week 2

• Book reference– Modern compressible flow: with historical perspective, Third editions, Johan, D.

Anderson• Chapter 2:

• Chapter 3, one-Dimensional flow

• Chapter 5, Quasi-One-Dimensional flow

– Computational fluid mechanics and heat transfer, by J.C. Tannehill, D.

A. Anderson and R. H. Pletcher

• Chapter 6: Numerical methods for inviscid flow equations.

2

3

p

v1

Compressible flow?

• Compressibility:

• t : property of the fluid

– Water: 5x10-10 m2/N @1atm

– Air: 10-5 m2/N @1atm

v2

dp

dv

v

1t T

dp

dv

v

1

t

sdp

dv

v

1

t

dpd t Such a process can be realized through very slow compression causing negligible velocity (|u| ≈ 0) in the fluid, this “ideal” thermodynamic process will maintain a perfectly uniform distribution of all state variables over the entire fluid.

4

Compressible flow

The three-dimensional Euler equations

5

##$ % + #

#' (⃗ % + ##* +⃗ % + #

#, - % = 0, where %, /, (⃗, +⃗ are vectors of 5-component

##$

00102030/

+ ##'

01014 + 5

012013016

+ ##*

02012

024 + 5023026

+ ##,

03013023

034 + 5036

=

00000

where 6 ≡ / + 89 , : = / − <, ℎ = 6 − < , < ≡ >?@A?@B?

4:: internal energy; ℎ: enthalpy; /: Total energy; 6: Total enthalpy; <: kinetic energy;0: density; 1, 2, 3: velocity component in ', *, ,; 5 : pressure.

Thermodynamical relations for calorific pefect gas :Equation of state : 5 = 0IJ : = �AJ , ℎ = �8J, K = LM

NO, �8 − �A = I

where J is temperature, I is universial gas constant; K = 1.4. �A , �8 are specific heat capacity at constant volume/pressure, in later discussion we assume the value of �A , �8 are constant (i.e. their value do not depends on J)

Conservation lawMass: x-Momentum:Y-Momentum:Z-momentum:Energy:

The 1D Euler equations

6

3D Euler 1111D Euler ($, ', *, ,)

##$

00102030/

+ ##'

01014 + 5

012013016

+ ##*

02012

024 + 5023026

+ ##,

03013023

034 + 5036

=

00000

< = 14 + 24 + 34

2

##$

0010/

+ ##'

01014 + 5

016 =

000

##$ % + #

#' (⃗ % = 0,

with % =0

010/

and the flux (⃗ =01

014 + 5016

.

Conservation lawMass: x-Momentum:Energy:

Different ways of expressing the Euler equationsThe primitive variable vs. conservative variable

7

The same Euler equation can be expressed differently by choosing a “core-set” of three symbols.

Case 1: a core-set chosen as three conservativeconservativeconservativeconservative variablesvariablesvariablesvariables: \ = 0, 01, 0/ ]

Case 2: a three-variable core-set chosen from the set of primitive symbols [0, 1, 5, J, :, ℎ, /, 6, _]:a = 0, 1, 6 ]

The transformation between UUUU and VVVV

\(a) ≡%d%4%e

=0

010/

=

fdfdf4

fdfe − K − 1K fd fe − f4

4

2 ; a(\) =

fdf4fe

=016

=

%d%4/%d

K %e%d

− K − 12

%44

%d4)

For any chosen core-set, every variables inside Euler equations can be regarded as a function

of the core-set variables:0 gh,g?,gi = %d

1 gh,g?,gi = %4/%d

6 gh,g?,gi = K %e%d

− K − 12

%44

%d4)

J gh,g?,gi = ⋯5 gh,g?,gi = ⋯: gh,g?,gi = ⋯ℎ gh,g?,gi = ⋯/ gh,g?,gi = ⋯

…

2_.

0 lh,l?,li = fd1 lh,l?,li = f46 lh,l?,li = fe

J lh,l?,li = (fe−f44/2)/m8

5 lh,l?,li = 1 − 1K fd fe − f4

4

2: lh,l?,li = ⋯ℎ lh,l?,li = ⋯/ lh,l?,li = ⋯

…

Different ways of expressing the Euler equationswhy should we bother about this issue?

8

The 1D Euler equation is often conveniently expressed using the primitive variables mixed with a “un-explicitly” chosen core-set, for instance let a = 0, 1, 6 ] , the equation is

expressed as :

nno \ + n

np q = r, with \(a) =0

010/(9,>,s)

, q(a) =01

014 + 5(9,>,s)016

, where a ≡fdf4fe

=016

.

As a demonstration the same Euler equation can be re-expressed only using another core-set U U U U as:

nno \ + n

np q = r with, \ ≡%d%4%e

=0

010/

uvwxov 0yz

, and q(\) =

y{?

9 + 5(9,{,|) {9 z + 5 9,{,|

where 5(9,{,|) = 0IJ(9,{,|) = 0 }NO

|9 − {?

49? = (K − 1)(z − {?

49 ) , then we expand flux term as:

→ q(\) =

ye��

4{?

9 + K − 1 z

K |{9 − ��d

4{i

9?

������ ����������

%4e��

4g??

gh+ K − 1 %e

K gig?gh

− ��d4

g?i

gi

.

It is very common during numerical computation to calculate certain variable of interest by re-expressing it as a function certain chosen core-set.

Different ways of expressing the Euler equationswhy should we bother about this issue?

9

One benefit of expressing Euler equation only using the core-set symbols is to prevent us from “forget” relations such as the equation of state, it is an essential step to find the Jacobian matrix:

� e×e = nqn\ =

n�hngh

n�hng?

n�hngi

n�?ngh

n�?ng?

n�?ngi

n�ingh

n�ing?

n�ingi

for q(\) =

ye��

4{?

9 + K − 1 z

K |{9 − ��d

4{i

9?

which looks like the following for 1D Euler equation

� =

0 1 0− e��

4{?

9? 3 − K {9 K − 1

−K |{9? + � − 1 {i

9i K |9 − e ��d

4{?

9? K {9

for \ =0

010/

=0yz

The “core-set” is necessary for obtaining [�], however, it is quite difficult to read and tomanipulate (for instance, to find its eigenvalues) if [�] is still expressed by the core-set \.... Wecan then rewrite [�] back to the primitive form which give a more “friendly” look as:

� =

0 1 0K − 3

2 14 3 − K 1 K − 1

−16 + K − 12 1e 6 − (K − 1)14 K1

1_: z0 = �AJ + 14

2 = �8J cA�8

+ 14

2 = 1K ℎ + 14

2 + K − 1K

14

2 = 6 + K − 1K

14

2 .

Conservative / Non-conservative form of 1D Euler equations

10

NoncNoncNoncNonconservative form of 1D Euler equation.Continuity eq �

�o 0 + 0#p1 = 0Momentum eq. 0 �

�o 1 + #p5 = 0Energy eq. 0 �

�o (�AJ + >?

4 ) + #p 51 = 0

��$ � ≡ #

#$ � + 1 ##' �

CCCConservative form of 1D Euler eq.

nno

0010/

+ nnp

01011 + 5

016=

000

→→ nno

0 ⋅ 10 ⋅ 1

0 ⋅ (�AJ + >?

4 )+ n

np

01 ⋅ 101 ⋅ 1

01 ⋅ (�AJ + >?

4 ) + n

np

05

51= 0

� = 1 (continuity eq.) : 0 ��o 1 + �

�o 0 + 0 n>np ⋅ 1 = 0

��o 0 + 0 n>

np =0

= 0 ##$ � + 1 #

#' � + � ##$ 0 + 1 #0

#' + 0� #1#'

= 0 ��$ � + � �

�$ 0 + 0� #1#'

= 0 ��$ � + �

�$ 0 + 0 #1#' ⋅ �

##$ (0 ⋅ �) + #

#' (01 ⋅ �)

What about entropy _ ? Where is the second law of thermodynamics?

11

CCCConservative form of 1D Euler eq.##$

001

0(�AJ + 14

2 )+ #

#'

01011 + 5

01(�AJ + 14

2 ) + 51=

000

NoncNoncNoncNonconservative form.Continuity eq �

�o 0 + 0#p1 = 0Momentum eq. 0 �

�o 1 + #p5 = 0Energy eq. 0 �

�o (�AJ + >?

4 ) + #p(51) = 0

0 ��$ (�AJ) + 5#p1 = 0

05

��$ (�AJ) − 5

0 ��$ 0 = 0

“Reversible” energyExchange of kinetic energy

with mechanic (pressure) work

0 ��$

14

2 + 1#p5 = 0

�AI

1J

��$ J − 1

0 ��$ 0 = 0�

�$ (�A ln J − I ln 0) = 0

Introduce Entropy: s ≡�A ln J − I ln 0 + ���_$.

��$ _ = 0

multiple by 1

��$ � ≡ #

#$ � + 1 ##' �No discontinuity

is allowed !!!Only for smooth flow!

Continuity eq.

Eq. of state

Reminder from thermodynamics:

J�_ = �: + 5� (10)

This _-equation can replace energy

equation in smooth flow!

Also note, Entropy is not an external

quantity, it is implied by the Euler equations

Isentropic/ homentropic flow

12

(1)This equation holds for 3D-Euler (zero viscosity, non conducting, adiabatic)

equations using (���o ≡ n�

no + 1 n�np + 2 n�

n� + 3 n�n  )

(2)Before encounting a discontinuity, the entropy will maintain constant along the

trajectory of a material point moving with speed 1(', $).

(3)If the entropy _ remain constant along a streamline in a steady flow, it is called

isentropic flow

(4)If the entropy _ holds constant over the entire flow domain, a.k.a. homentropic

flow.

It is indeed true for irrotaitional, potential flow

In any smooth (containing no shock) (quasi) 1D flow, a material point will

move across all domain

��$ _ ≡ #_

#$ + 1 #_#' = 0

Alternative/equivalent entropy definitions:s = �8 ln J − I ln 5 + ���_$

_ = ln 50�

NO+ ���_$

s = �A ln 5 − �8 ln 0 + ���_$

5 = 0IJln 5 = ¡�0 + ¡�I + ¡�J

�(ln 5) = �(¡�0) + �(¡�J)

�(_)

I = �8 − �AK = �8/�A

Isentropic flow: 0 = � _ = �A ⋅ � ln5 − �8 ⋅ � ln 0 ⇒ u £¤8u £¤ 9 = K ⇒ u 8

u 9 = K 89 = KIJ

Entropy definitions ≡ �A ln J − I ln 0 + ���_$

Does it remind you the simple advection equation?

Isentropic/ homentropic flow

13

��$ _ ≡ #_

#$ + 1 #_#' = 0

14

The speed of sound

• What is sound?

– ”Wave” propagation of small disturbance

• �1, �5, �0, �J, …• Continuous and differentiable (sound is NOT a discontinuous shock)

– Chain rule can be applied: u

up �§ = � uup § + § u

up �– Isentropic process

» �_ = 0

1 = 050J

0 + �15 + �50 + �0J + �J

�

15

Speed of sound(a) Derivation using steady Euler equation in conservative form

1 + �15 + �50 + �0J + �J

150J

Mass conservation: ( 01 = ���_$. across ')

0 + �0 1 + �1 = 01

Choose Choose Choose Choose a Control volume a Control volume a Control volume a Control volume traveling with (i.e. relatively stationarystationarystationarystationary to) the wave front of sound:

Momentum conservation (014 + 5 = ���_$. across ')

0 + �0 1 + �1 4 + 5 + �5 = 014 + 5

� = 1 = �5�0

©ª«

= KIJ ©

##$

0010/

+ ##'

01014 + 5

016=

000

Steady Euler equation in conservativeconservativeconservativeconservative form

� = 1

Mass conservation:Momentum conservation:Energy conservation: Constant entropy : �_ = 0

(_ =const. across x)

16

Speed of sound(b) Another derivation using steady Euler eq.

� 01 = 0� 014 + 5 = 0

0�1 + 1�0 = 014�0 + 201d1 + �5 = 0

� ≡ 1 = �5�0

©ª«

= KIJ ©

Mass Eq.Momentum Eq.Energy eq.

##$

0010/

+ ##'

01014 + 5

016=

000

Mass Eq. Momentum Eq.

_ =const. across x

chain rules

17

Speed of sound(c) Derivation from the eigenvalues of the Jacobian matrix of unsteady Euler eq.

¬d,e = 1 ± �5�0

©ª«

= 1 ± �

¬4 = 1

UnsteadyUnsteadyUnsteadyUnsteadyconservation eq.

##$

0010/

+ ##'

01014 + 5

016=

000 ®

�o _ ≡ nno _ + 1 n

np _ = 0

Find 3 eigenvalues!| ± − ¬I| = 0

¬ − 1 ¬ − 1 4 − #5#0 ª

« = 0

##$ 0 + 1 #

#' 0 + 0 ##' 1 = 0

##$ 1 + 1 #

#' 1 + 10

##' 5 = 0

##' 5 0, _ = #5

#0 ª«

##' 0 + #5

#_ ª9

##' _

NonconservativeNonconservativeNonconservativeNonconservativeform:

##$

01_

+1 0 0

10

#5#0 ª

«1 #5

#_ ª9

0 0 1

##'

01_

=000

1 − ¬ 0 010

#5#0 ª

«1 − ¬ #5

#_ ª9

0 0 1 − ¬= 0

18

Limit of compressibility?

x

u

xu

x

u

x

u

x

u

xu

i.e.

Can be written as:

V

dVd

VV

dcdp 2

speed of sound

From Bernoulli: VdVdp

112

2

2

22 M

c

V

V

dp

a

dp

Mach

number

3.0M

Usually the limit is set to:

The relations across a discontinuous interface

19

0d1d = 04145d + 0d1d

4 = 54 + 04144

0d1d �8Jd + 1d4

2 = 0414 �8J4 + 144

2

(⃗d = (⃗4##$ % + #

#' (⃗ % = 0

##$

001

0(�AJ + 14

2 )+ #

#'

01014 + 5

01(�8J + 14

2 ) =

000

There is a second “trivial” solution of

contact discontinuity (i.e. 04 ≠ 0d,

but 14 = 1d = r, 54 = 54), think

about an interface separating gas of

different temperature but sharing the

same pressure (neglect diffusion).

A control volume chosen relatively

stationary to a discontinuity interface

J4(8?,9?)

Use the conservative form of the

governing equations (mass,

momentum, energy conservations)

A first “trivial” solution:(04= 0d, 14 = 1d, 54 = 5d)

0d1d5d

Jd(5d, 0d)

…

041454

J4(8?,9?)…

Assume given Find three unknowns

0d1d = 04145d + 0d1d

4 = 54 + 04144

�8Jd + 1d4

2 = �8J4(8?,9?) + 144

2This gives a relation for shock!

To find the relation across a normal shock

20

0d1d5d

Jd(54, 04)

…

041454

J4(8?,9?)…

A shock discontinuity

Find three unknowns

Three unknowns (³, ´, � ) live in a “phase” space of 3 dimensions.

Three expressions correspond to three curved surfaces (due to nonlinearity):

/'5�:__µ��d ³, ´, � = 0 /'5�:__µ��4 ³, ´, � = 0/'5�:__µ��e ³, ´, � = 0

³

´�

Surface 1/'5d ³, ´, � = 0

Surface 2/'54 ³, ´, � = 0

0d1d = 04145d + 0d1d

4 = 54 + 04144

�8Jd + 1d4

2 = �8J4(8?,9?) + 144

2 A third surface /'5e ³, ´, � = 0 can lead to more than one intersections points.

This explain why there may exist multiple (>1)

solutions in additional to the trivial solution.

Assume given

Normal shock relationsRankine–Hugoniot jump conditions

21

0d1d = 04145d + 0d1d

4 = 54 + 04144

�8Jd + 1d4

2 = �8J4(8?,9?) + 144

2

K

K − 15404

− 5d0d

− 12

104

+ 10d

54 − 5d = 0

y = 0d1d54 − 5d104

− 10d

= y4

Rayleigh line:

Hugoniot curve:

K + 1K − 1

− K − 1K + 1

5#

1/0#

1

1

(Shock) Hugoniot curve

5#

1/0#

1

1

Rayleigh line

Get rid of 14Two remaining unknowns now live in a 2D Phase space

Rayleigh line: 8#�dh

¹#�d= −º = − {?

8h9h

Hugoniot curve: 5# = (�@d��d − d

9#)/(�@d��d

d9# − 1)

Isentropic curve, 8#

9#» = 1

Normalizing: 5# = 8?8h

, 0# = 9?9h

, 1# = >?¼h

Normal shock relationsRankine–Hugoniot jump conditions (cont’d)

22

K + 1K − 1

− K − 1K + 1

5#

1/0#

Reyleigh line: 8#�dh

¹#�d= −º = − {?

8h9h< 0 (negative slope)

Hugoniot curve: 5# =»¾h»¿h� h

¹#»¾h»¿h

h¹#�d

Which of following is possible?5# = 54

5d= ∞? 0# = 04

0d= ∞?

ds>0

The normal shock relationsexpressed using the ³�ℎ number

23

0d1d = 04145d + 0d1d

4 = 54 + 04144

�8Jd + 1d4

2 = �8J4(8?,9?) + 144

2

1d14 = ³∗4

Â44 = 1 + [(K − 1)/2]Âd

4

KÂd4 − (K − 1)/2

545d

= 1 + 2KK + 1 Âd

4 − 19?9h

= >h>?

= �@d Äh?

4@ ��d Äh?J4Jd

= 545d

0d04

= ⋯….

Across a stationary normal shock: Â4 < 1 < Âd

54 > 5dJ4 > Jd04 > 0d_4 > _d

and for total (stagnation) quantiles:JÅ,4 = JÅ,d

0d1d5d

Jd(54, 04)

…

041454

J4(8?,9?)…

(Quasi) 1D flows

Nozzle flow

24

�(�01) = 0�50 + 1�1 = 0

�(ℎ + 14/2) = 0

##$

�0�01

�0(: + 14/2)+ #

#'�01

�011 + �5�01(ℎ + 14/2)

=0

5 #�#'0

Assumption:

Thin boundary layer, 1(x,y) is constant along y.

Smaller rate of change across the Area: u

up �(') ≪ 1

'*

� '0(')1(')5(')

J(0, 5)…

##$

001

0(: + 14/2)+ #

#'

01011 + 5

01(ℎ + 14/2)=

000

1D Euler Eq. in a tube of constant Area

Quasi 1D Euler Eq. in a tube of changing Area � '

0(')1(')5(')

J(0, 5)…

Steady state relations

� 01 = 0� 011 + 5 = 0� ℎ + 14/2 = 0

Flows configurations

�(�01) = 0�50 + 1�1 = 0

�5�0 = �4

The conservation of energy-

flux can be replaced by

�(_) = 0 for Isentropic flow

(no shock), which yield this

isentropic relations.

'

Handle

discontinuity?

Yes

No

(Quasi) 1D flows

Isentropic flow with (nozzle) area changes

25

�(�01) = 0�50 + 1�1 = 0

�5�0 = �4

u>> = uÇ

Çd

Ä?�d = − u89>?

��� + �0

0 + �11 = 0 × 14

 ≡ 1�

1 increaseing

1 decreasing

³ > 1 ³ < 1

Mach number

�1 > 0�5 < 0

�1 < 0�5 > 0

�1 > 0�5 < 0

�1 < 0�5 > 0

 < 1  > 1

�� > 0

�� < 0

 = 1

?

26

Nozzle flow

• Convergent

Further decreasing pb will not change the

mass flow (”choked” since

1max

M

27

Nozzle flows

• Convergent-divergent

Some conveniently defined flow parameters

28

• Static pressure 5 and temperature J for an fluid element in a actual flow domain

5, J, u ⇒ [0 = 8}] , c = rIJ©

, Ma = ¼N , h = CÈT, e = �AJ, s = ln 8

9»NO , … . ]

• Imagine take this fluid element and adiabatically slow it down (if Ma>1) or speed it up (Ma<1) until

its Mach number is 1, the new temperature at this imagined state is J∗ , and the corresponding

imagined sound speed is �∗ = KIJ∗©.

• Imagine take the same fluid element and isentropically slow it down to zero velocity (bring it to

stagnation), the new temperature and temperature at this imagined state is called total temperature

JÅ and total pressure 5Å respecectivly.

Stagnation

reservoir:

JÅ5Å

1Å = 0�Å = KIJÅ© J∗

1∗ = �∗ = KIJ∗©

Real flow domainstatic 5, J, and velocity u = 1

An hypothetical, imaginary flow process(nozzle

flow) which can either be isentropic (shock free) or

adiabatic (if there exits shock)

Ma>1 Ma<1

Question:1) Can we define _∗?2) Can we define _Å?

Relations between the conveniently defined flow parameters

29

�(�01) = 0�50 + 1�1 = 0

�(ℎ + 14/2) = 0

Steady state relations for nozzle flow

Constant energy flux

Valid across

discontinuity

�8JÅ + 0 = �8J + 12 14 = �8J∗ + 1

2 �∗4

JÅJ = 1 + K − 1

2 Â4

]∗

]Ë= N∗

NË

4= 4

�@d

Â∗ = 1�∗ = K + 1 Â4

2 + K − 1 Â4

Stagnation

reservoir:

JÅ5Å

1Å = 0�Å = KIJÅ

© J∗

1∗ = �∗ = KIJ∗©

• Those ”imaganary” flow

parameters help to:

– Compute parameters

after/before shock waves

– Boundary conditions

– …

5Å5 = 0Å

0�

= JÅJ

���d

………

Isentropic

relation give: