compressible flows 1 - lth · issues related to compressible flows ... compressible flow. 6. ......

Compressible Flows 1

Compressible flow

What if you have a supersonic wind tunnel and way too much time on your hands..?

3

Issues related to compressible flows

•Governing equations might change character

• Shocks• Preferential direction of propagation•Boundary conditions• Small timesteps (compared to

incompressible)

4

p

v1

Compressible flow?

• Compressibility:

• τ : property of the fluid• Water: 5x10-10 m2/N @1atm• Air: 10-5 m2/N @1atm

v2

dpdv

v1

−=τ Tdpdv

v

−=

1τ

sdpdv

v

−=

1τ

dpd τρρ =

5

Compressible flow

6

Speed of sound

• What is sound?• da, dp,dρ,dT: small

dTTddppdaa

++++

ρρT

pa

ρ

( )( )( )( )22 )( daaddppap

ddaaa

++++=+

++=

ρρρ

ρρρρd

dpa =

For a calorically perfect gas: RTa γ=

7

Limit of compressibility?( )

xu

xu

xu

xu

∂∂

≈∂∂

+∂∂

=∂

∂ ρρρρxu

xu

∂∂

<<∂∂ ρρ

i.e.

Can be written as:

VdVd

<<ρρ

VV =

ρdadp 2=

speed of sound

From Bernoulli: VdVdp ρ−=

11 22

2

22 <<⇔<<⇔<< MaV

Vdp

adp

ρρ

Mach number 3.0≤M

Usually the limit is set to:

8

Flow regimes0.3 < M < 1

Subsonic M < 1 M < 1M > 1

Transonic

M < 1 M > 1

SupersonicWhich is the most difficultto compute?

M<0.3 Incompressible

M>>1 Hypersonic

9

Examples

[www.eng.vt.edu/fluids/msc/gallery/gall.htm]

10

Examples

[www.eng.vt.edu/fluids/msc/gallery/gall.htm]

11

Examples

ExamplesPrismatic airfoil

Inviscid flowGrid refinement

Coarse grid Refinement 1 Refinement 2

Coarse: Cdp=0.0500R1: Cdp=0.0518R2: Cdp=0.0530

ExamplesPrismatic airfoilInviscid vs viscous

InviscidCdp=0.0500

ViscousCdp=0.0500Cd,tot=0.0539

ExamplesPrismatic airfoilWall refinement

coarse refined

Coarse: Cdp=0.0500, Cd,tot=0.0539Refined:Cdp=0.0531, Cd,tot=0.0569

ExamplesPrismatic airfoilWall refinement

Converged: Cdp=0.0500, Cd,tot=0.0539Non-converged:Cdp=0.0495, Cd,tot=0.0533

Residual 10-8 Residual 10-3

NACA0015 Cd=0.14

Prisma Cd=0.058

17

Characteristic parameters

• Characteristic sound speed

• Characteristic Mach number

p, ρ, T, V, M

p*, ρ∗, T*, V*, M=1

V*=V+dVdQ=0

** RTa γ=

*/* aVM =

11lim

)1(12 *

2*

2

−+

=−−

+=

∞→ γγ

γγ M

M

MM

18

Total/stagnation parameters

• Stagnation sound speed

• Total density

p, ρ, T, V, M

p0, ρ0, T0, V=0

V*=V+dVds=0

00 RTa γ=

000 / RTp=ρ

19

Isentropic flow relations

• Why?• Compute parameters

after/before shock waves

• Boundary conditions• …

1211

00

12100

20

211

211

211

−−

−−

−+=

=

−+=

=

−+=

γγ

γ

γγ

γγ

γρρ

γ

γ

MTT

MTT

pp

MTT

20

(Quasi) 1D flows

• Assume• Very thin boundary layers• Small rate of change in

area• Large curvature

x

y( )xh

( )xR

( )yxV ,

1<<dxdh

( ) ( )xRxh <<

x

y( )xh

( )xV

21

Isentropic flow with area changes

Continuity ( ) ( ) ( ) constant== mxAxVx ρ

Differentiate the continuity and momentum equations

22

21

10

0

Vdp

MAdA

VdV

dadp

VdVdpA

dAVdVd

ρρ

ρ

ρρ

−=−

=⇒

=

=+

=++

22

Isentropic flow with area changes

22 11

Vdp

MAdA

VdV

ρ−=

−=

0>dA

0<dA

1<M 1>M

00

<>

dpdV

00

<>

dpdV

00

><

dpdV

00

><

dpdV 0=dA ∞→dV

12 =M

?

23

Nozzle flow

•ConvergentAVm ρ=

maxmm = when 1=M**** VAm ρ=

Further decreasing pb will not change the mass flow since

1max =M

( )( )

( )21

00*

1121

21

21

0*

11

0***

max

12

12

12

RTA

RTAVAm

ργ

γ

γγρρ

γγ

γ

−+

−

+

=

=

+

+

==

24

Nozzle flows

•Convergent-divergent11

22

2

* 211

121 −

+

−+

+=

γ

γ

γγ

MMA

A

1211

00

12100

20

211

211

211

−−

−−

−+=

=

−+=

=

−+=

γγ

γ

γγ

γγ

ρρ

γ

γ

MkTT

MTT

pp

MTT

25

Normal shock relations

1

1

1

1

1

1

1

ˆ

Msh

pAVρ

2

2

2

2

2

2

2

ˆ

Msh

pAVρ

iu

01020102

*1

*2121212

121221 11

TTppAATTVVppssMM

=<>><>>><>

ρρ

26

Governing equations

•Are the equations different in different regimes?

RTp

VeE

ufx

ux

puxTk

xq

xEu

tE

fxx

px

uutu

xu

t

iij

iji

j

j

jjj

j

ij

ij

ij

jii

i

i

ρ

ρτ

ρρρ

ρτρρ

ρρ

=

+=

+∂∂

+∂∂

−

∂∂

∂∂

+=∂

∂+

∂∂

+∂∂

+∂∂

−=∂

∂+

∂∂

=∂∂

+∂∂

2

0

2

27

Euler equations

• Inviscid

RTp

VeE

ufx

puxTk

xq

xEu

tE

fxp

xuu

tu

xu

t

iij

j

jjj

j

iij

jii

i

i

ρ

ρρρρ

ρρρ

ρρ

=

+=

+∂∂

−

∂∂

∂∂

+=∂

∂+

∂∂

+∂∂

−=∂

∂+

∂∂

=∂∂

+∂∂

2

0

2

28

Euler equations – conserved form0=

∂∂

+∂∂

+∂∂

+∂∂

zH

yG

xF

tU

+

=

2

2Ve

wvu

U

ρ

ρρρρ

+

+

+=

puVeu

uwuv

pu

u

F

2

2

2

ρ

ρρρ

ρ

+

+

+=

pvVev

vwpv

uvv

G

2

2

2

ρ

ρρ

ρρ

+

+

+

=

pwVew

pw

vwuww

H

2

2

2

ρ

ρ

ρρρ

29

Character of the Euler equations

Subsonic Sonic Supersonic

Steady Elliptic Parabolic Hyperbolic

Unsteady Hyperbolic Hyperbolic Hyperbolic

Classification of PDEs

Hyperbolic

PDomain of dependence Region of influence

Charateristic lines

y

xExamples: Inviscid supersonic flow, unsteady inviscid flow

Classification of PDEs

Parabolic

PDomain of dependence Region of influence

y

x

Known boundary conditions

Known boundary conditionsExamples: Steady boundary layer flow, unsteady heat conduction

Classification of PDEs

Elliptic

P

y

x

Every point influences all other points

Examples: Steady subsonic inviscid flow, incompressible inviscid flow

Flux vector splitting schemes

•Determining the correct flux at the cell faces

• FVS splits the flux contributions into a positive and a negative part

• Two examples shown here: Steger-Warming and van Leer splitting

33

Flux vector splitting schemes

34

𝜕𝜕𝑈𝑈𝜕𝜕𝑡𝑡

+𝜕𝜕𝜕𝜕𝜕𝜕𝜕𝜕

+𝜕𝜕𝜕𝜕𝑑𝑑𝑑𝑑

= 0

�𝛿𝛿𝑣𝑣

𝜕𝜕𝑈𝑈𝜕𝜕𝑡𝑡

𝑑𝑑𝑑𝑑 + �𝛿𝛿𝑣𝑣

𝜕𝜕𝜕𝜕𝜕𝜕𝜕𝜕 +

𝜕𝜕𝜕𝜕𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 0

�𝛿𝛿𝑣𝑣

𝜕𝜕𝑈𝑈𝜕𝜕𝑡𝑡

𝑑𝑑𝑑𝑑 + �𝑠𝑠

𝜕𝜕𝑑𝑑𝑑𝑑 − 𝜕𝜕𝑑𝑑𝜕𝜕 = 0

𝜕𝜕𝑈𝑈𝜕𝜕𝑡𝑡

𝛿𝛿𝑑𝑑 + �𝑓𝑓𝑎𝑎𝑎𝑎𝑎𝑎𝑠𝑠

𝜕𝜕∆𝑑𝑑 − 𝜕𝜕∆𝜕𝜕 = 0

The Euler equations

Integrate over a control volume δv

Use Green’s theorem

Discretise

Steger-Warming splitting

𝜕𝜕𝑈𝑈𝜕𝜕𝑡𝑡

+𝜕𝜕𝜕𝜕𝜕𝜕𝜕𝜕

= 0

𝜕𝜕𝑈𝑈𝜕𝜕𝑡𝑡

+ 𝐴𝐴𝜕𝜕𝑈𝑈𝜕𝜕𝜕𝜕 = 0

𝐴𝐴 =𝜕𝜕𝜕𝜕𝜕𝜕𝑈𝑈

𝑇𝑇−1𝐴𝐴𝑇𝑇 = Λ

Consider the 1D Euler equation:

Using the Jacobian

it can be written as

The Jacobian can be reformulated using its eigenvalues and eigenvectors. Here T-1 is a matrix whose rows are the left eigenvectors of A and Λ is a diagonal matrix containing the eigenvalues of A 35

Steger-Warming splitting

36

𝜆𝜆1 = 𝑢𝑢

𝜆𝜆2 = 𝑢𝑢 + 𝑎𝑎

𝜆𝜆3 = 𝑢𝑢 − 𝑎𝑎

𝜕𝜕 = 𝐴𝐴𝑈𝑈 = 𝑇𝑇Λ 𝑇𝑇−1

𝜕𝜕 = 𝜕𝜕+ + 𝜕𝜕− = 𝐴𝐴+𝑈𝑈 + 𝐴𝐴−𝑈𝑈

𝜕𝜕𝑈𝑈𝜕𝜕𝑡𝑡

+𝜕𝜕𝜕𝜕+

𝜕𝜕𝜕𝜕 +𝜕𝜕𝜕𝜕−

𝜕𝜕𝜕𝜕 = 0

The flux can be written in terms of the Jacobian and U as

Now the flux is split into a positive and a negative contribution

Hence, we can write the 1D Euler equation as

Λ = Λ+ + Λ−Likewise, the eigenvalues can be split

The eigenvalues for the 1D case are:

Steger-Warming splitting

37

𝜕𝜕− =12𝜌𝜌𝛾𝛾𝑢𝑢 − 𝑎𝑎

1𝑢𝑢 − 𝑎𝑎

12𝑢𝑢 − 𝑎𝑎 2 +

12

3 − 𝛾𝛾𝛾𝛾 − 1

𝜕𝜕+ =12𝜌𝜌𝛾𝛾

2𝛾𝛾 − 1 𝑢𝑢 + 𝑎𝑎2 𝛾𝛾 − 1 𝑢𝑢2 + 𝑢𝑢 + 𝑎𝑎 2

𝛾𝛾 − 1 𝑢𝑢3 +12 𝑢𝑢 + 𝑎𝑎 3 +

12𝑎𝑎

2 3 − 𝛾𝛾𝛾𝛾 − 1 𝑢𝑢 + 𝑎𝑎

Λ+ =𝑢𝑢

𝑢𝑢 + 𝑎𝑎0

Λ− =0

0𝑢𝑢 − 𝑎𝑎

Λ+ = Λ Λ− = 0Assuming the flow is in the positive x-direction, For the supersonic caseAnd for the subsonic case

The fluxes in the subsonic case:

Steger-Warming splitting

38

𝑈𝑈𝑖𝑖𝑛𝑛+1 = 𝑈𝑈𝑖𝑖𝑛𝑛 −Δ𝑡𝑡Δ𝜕𝜕

𝜕𝜕𝑖𝑖+1/2 − 𝜕𝜕𝑖𝑖−1/2

𝜕𝜕𝑖𝑖+1/2 = 𝜕𝜕+ + 𝜕𝜕− 𝑖𝑖+1/2

𝜕𝜕𝑖𝑖+1/2+ = 𝜕𝜕𝑖𝑖+ 𝜕𝜕𝑖𝑖+1/2

− = 𝜕𝜕𝑖𝑖+1−

The discrete equation to solve becomes:

For a first order discretisation:

Note that ’+’ is a backward and ’-’ a forward discretisation

Steger-Warming splitting:• Performs better than central differencing techniques• Captures shocks well but oscillations will occur at sonic

conditions• Causes some problems close to stagnation points• This is due to the split fluxes nor being continously

differentiable

van Leer flux splitting

39

Addresses the problems asociated with Steger-Warming splitting at sonic transition and stagnation points.The following conditions must then be met to esure standard upwinding differences at supersonic flows:1. 𝜕𝜕 = 𝜕𝜕+ + 𝜕𝜕−

2. 𝜕𝜕𝐸𝐸+

𝜕𝜕𝑈𝑈must have all eigenvalues ≥ 0

3. 𝜕𝜕𝐸𝐸−

𝜕𝜕𝑈𝑈must have all eigenvalues ≤ 0

Restrictions in order to eliminate problems with the Steger-Warming scheme (1-3) and achieve uniqueness of the splitting (4):

1. 𝜕𝜕± must be continuous and 𝜕𝜕+ = 𝜕𝜕 𝑓𝑓𝑓𝑓𝑓𝑓 𝑀𝑀 ≥ 1

𝜕𝜕− = 𝜕𝜕 𝑓𝑓𝑓𝑓𝑓𝑓 𝑀𝑀 ≤ −12. The components must exhibit the same symmetry as the original

flux vector in terms of Mach number3. The Jacobians 𝜕𝜕𝐸𝐸

±

𝜕𝜕𝑈𝑈must be continuous and have one eigenvalue

vanish for 𝑀𝑀 < 14. 𝜕𝜕± must be a polynomial in Mach number of lowest possible order

Flux difference splittingThe Roe scheme

40

𝜕𝜕𝑈𝑈𝜕𝜕𝑡𝑡

+𝜕𝜕𝜕𝜕𝜕𝜕𝜕𝜕

= 0

𝑈𝑈 𝜕𝜕, 0 = �𝑈𝑈𝐿𝐿 𝜕𝜕 < 0𝑈𝑈𝑅𝑅 𝜕𝜕 > 0

𝜕𝜕𝑈𝑈𝜕𝜕𝑡𝑡

+ �̂�𝐴𝜕𝜕𝑈𝑈𝜕𝜕𝜕𝜕 = 0

�̂�𝐴 = �̂�𝐴 𝑈𝑈𝐿𝐿,𝑈𝑈𝑅𝑅�̂�𝐴 𝑈𝑈𝐿𝐿,𝑈𝑈𝑅𝑅 → 𝐴𝐴

𝜕𝜕𝑅𝑅 − �𝜕𝜕𝑖𝑖+12

+ �𝜕𝜕𝑖𝑖+12

− 𝜕𝜕𝐿𝐿 = �̂�𝐴+ + �̂�𝐴− 𝑈𝑈𝑅𝑅 − 𝑈𝑈𝐿𝐿

The Roe scheme is a method to determine the flux by solving a linear problem

The Riemann problem for the fluxes in the 1D Euler equation:

Roe’s linear approximation:

Note that the Jacobian is replaced by the matrixAway from the disturbance this matrix should goto the Jacobian, i.e.

The flux difference may be written as:

Flux difference splittingThe Roe scheme

41

𝜕𝜕𝑅𝑅 − �𝜕𝜕𝑖𝑖+12

+ �𝜕𝜕𝑖𝑖+12

− 𝜕𝜕𝐿𝐿 = �̂�𝐴+ + �̂�𝐴− 𝑈𝑈𝑅𝑅 − 𝑈𝑈𝐿𝐿

�𝜕𝜕𝑖𝑖+12

= 𝜕𝜕𝑅𝑅 − �̂�𝐴+ 𝑈𝑈𝑅𝑅 − 𝑈𝑈𝐿𝐿

�𝜕𝜕𝑖𝑖+12

= 𝜕𝜕𝐿𝐿 + �̂�𝐴− 𝑈𝑈𝑅𝑅 − 𝑈𝑈𝐿𝐿

�𝜕𝜕𝑖𝑖+12

=12

𝜕𝜕𝑅𝑅 + 𝜕𝜕𝐿𝐿 − �̂�𝐴 𝑈𝑈𝑅𝑅 − 𝑈𝑈𝐿𝐿

�𝜕𝜕𝑖𝑖+12

=12 𝜕𝜕𝑖𝑖 + 𝜕𝜕𝑖𝑖+1 − �𝑇𝑇𝑖𝑖+1/2 �Λ𝑖𝑖+1/2 �𝑇𝑇𝑖𝑖+1/2

−1 𝑈𝑈𝑅𝑅 − 𝑈𝑈𝐿𝐿

We have now two expressions for the cell face flux

Assuming symmetry one may use the average

A first order formulation of the flux for the Roe scheme. Note that one as before utilises the eigenvalues and eigenvectors to determine the ’Jacobian’:

AUSMAdvection Upstream Splitting Method

42

This technique is similar to van Leer splitting. However, while the convective terms are transported by a convection velocity, the pressure is transported by acoustic waves. Hence, it is reasonable to treat these terms separately.

𝜕𝜕𝑈𝑈𝜕𝜕𝑡𝑡

+𝜕𝜕𝜕𝜕𝜕𝜕𝜕𝜕 = 0

𝜕𝜕 =𝜌𝜌𝑢𝑢𝜌𝜌𝑢𝑢2𝜌𝜌𝜌𝜌𝑢𝑢

+0𝑝𝑝0

= 𝜕𝜕 𝑎𝑎 +0𝑝𝑝0

𝜌𝜌 = 𝜕𝜕 +𝑝𝑝𝜌𝜌

𝜕𝜕1/2𝑎𝑎 = 𝑢𝑢1/2

𝜌𝜌𝜌𝜌𝑢𝑢𝜌𝜌𝜌𝜌 𝐿𝐿/𝑅𝑅

= 𝑀𝑀1/2

𝜌𝜌𝑎𝑎𝜌𝜌𝑎𝑎𝑢𝑢𝜌𝜌𝑎𝑎𝜌𝜌 𝐿𝐿/𝑅𝑅

∗ 𝐿𝐿/𝑅𝑅 = �∗ 𝐿𝐿 𝑖𝑖𝑓𝑓 𝑀𝑀1/2 ≥ 0∗ 𝑅𝑅 𝑓𝑓𝑡𝑡𝑜𝑜𝑜𝑓𝑓𝑜𝑜𝑖𝑖𝑜𝑜𝑜𝑜

Again 1D Euler equation

Divide the flux in a convective part and a pressure part

The flux at the cell face is evaluated using values on either side of the shock

AUSM

43

𝑀𝑀1/2 ≈ 𝑀𝑀𝐿𝐿+ + 𝑀𝑀𝑅𝑅

−

𝑀𝑀± =±

14𝑀𝑀 − 1 2 𝑖𝑖𝑓𝑓 𝑀𝑀 ≤ 1

12𝑀𝑀 ± 𝑀𝑀 𝑓𝑓𝑡𝑡𝑜𝑜𝑜𝑜𝑜𝑖𝑖𝑜𝑜𝑜𝑜

𝑝𝑝1/2 = 𝑝𝑝𝐿𝐿+ + 𝑝𝑝𝑅𝑅−

𝑝𝑝± =

𝑝𝑝2 1 ± 𝑀𝑀 𝑖𝑖𝑓𝑓 𝑀𝑀 ≤ 1

𝑝𝑝2𝑀𝑀 ± 𝑀𝑀

𝑀𝑀 𝑓𝑓𝑡𝑡𝑜𝑜𝑜𝑓𝑓𝑜𝑜𝑖𝑖𝑜𝑜𝑜𝑜

The convection Mach number can be calculated using van Leer splitting

The pressure is also found by considering splitting

AUSM

44

𝜌𝜌𝑢𝑢𝜌𝜌𝑢𝑢𝑢𝑢 + 𝑝𝑝𝜌𝜌𝑢𝑢𝜌𝜌 1/2

= 𝑀𝑀1/212

𝜌𝜌𝑎𝑎𝜌𝜌𝑎𝑎𝑢𝑢𝜌𝜌𝑎𝑎𝜌𝜌 𝐿𝐿

+𝜌𝜌𝑎𝑎𝜌𝜌𝑎𝑎𝑢𝑢𝜌𝜌𝑎𝑎𝜌𝜌 𝑅𝑅

−12 𝑀𝑀1/2

𝜌𝜌𝑎𝑎𝜌𝜌𝑎𝑎𝑢𝑢𝜌𝜌𝑎𝑎𝜌𝜌 𝐿𝐿

−𝜌𝜌𝑎𝑎𝜌𝜌𝑎𝑎𝑢𝑢𝜌𝜌𝑎𝑎𝜌𝜌 𝑅𝑅

+0

𝑝𝑝𝐿𝐿+ + 𝑝𝑝𝑅𝑅−0

The final expression for the flux at the cell face:

Numerical dissipation term

Boundary conditions for the Euler equations

45

Boundary conditions for the Euler equations are depenent on the character of the equations. In the hyperbolic case the number of conditions that is needed depends on the direction of information transport over the boundary.

𝜕𝜕𝑈𝑈𝜕𝜕𝑡𝑡

+𝜕𝜕𝜕𝜕𝜕𝜕𝜕𝜕 +

𝜕𝜕𝜕𝜕𝑑𝑑𝑑𝑑 = 0

𝜕𝜕𝑈𝑈𝜕𝜕𝑡𝑡

+ 𝐴𝐴𝜕𝜕𝑈𝑈𝜕𝜕𝜕𝜕 + 𝐵𝐵

𝜕𝜕𝑈𝑈𝑑𝑑𝑑𝑑 = 0

𝜕𝜕𝑈𝑈𝜕𝜕𝑡𝑡 + 𝑇𝑇Λ𝑎𝑎𝑇𝑇−1

𝜕𝜕𝑈𝑈𝜕𝜕𝜕𝜕 + 𝑆𝑆Λ𝑏𝑏𝑆𝑆−1

𝜕𝜕𝑈𝑈𝑑𝑑𝑑𝑑 = 0

The 2D Euler equations can be reformulated using eigenvectors and eigen values.

Boundary conditions for the Euler equations

46

𝜕𝜕𝑈𝑈𝜕𝜕𝑡𝑡

+ 𝑇𝑇Λ𝑎𝑎𝑇𝑇−1𝜕𝜕𝑈𝑈𝜕𝜕𝜕𝜕

+ 𝑆𝑆Λ𝑏𝑏𝑆𝑆−1𝜕𝜕𝑈𝑈𝑑𝑑𝑑𝑑

= 0

𝜕𝜕𝑊𝑊𝜕𝜕𝑡𝑡 + Λ𝑎𝑎𝑇𝑇−1

𝜕𝜕𝑊𝑊𝜕𝜕𝜕𝜕 + 𝑇𝑇−1𝑆𝑆Λ𝑏𝑏𝑆𝑆−1𝑇𝑇

𝜕𝜕𝑊𝑊𝑑𝑑𝑑𝑑 = 0

𝑊𝑊 = 𝑇𝑇−1𝑈𝑈

𝜆𝜆1,2 = 𝑢𝑢,𝑢𝑢𝜆𝜆3,4 = 𝑢𝑢 ± 𝑎𝑎

Assuming T-1 is constant:

W are called characteristic variables or Riemann variables

The eigenvalue for this case are

1 & 2 are the streamline charecteristics and 3 & 4 are the wave fronts

Boundary conditions for the Euler equations

47

𝜆𝜆1,2 = 𝑢𝑢,𝜆𝜆3,4 = 𝑢𝑢 ± 𝑎𝑎

Inflow and outflow boundary conditions

t

x

λ1,2

λ3

t

x

λ4

λ1,2

λ3 λ4

Supersonic inflow Subsonic inflow

Boundary conditions for the Euler equations

48

Inflow and outflow boundary conditions

t

x

λ1,2

λ3

t

x

λ4

λ1,2

λ3 λ4

Supersonic inflow Subsonic inflow

Boundary conditions for all variables

Boundary conditions for 1 variable

Boundary conditions for the Euler equations

49

49

t

x

λ1,2

λ3

t

x

λ4

λ1,2

λ3 λ4

Supersonic outflow Subsonic outflow

All values on the boundary can be extrapolated from the interior

Boundary conditions for no. of variables -1

Boundary conditions for the Euler equations

Wall boundary conditions

𝜕𝜕 𝜕𝜕, 𝑑𝑑 = 𝑑𝑑 − 𝑓𝑓 𝜕𝜕

𝑑𝑑 = 𝑢𝑢𝑑𝑑𝑓𝑓𝑑𝑑𝜕𝜕 = 𝑢𝑢

𝑑𝑑𝑑𝑑𝑑𝑑𝜕𝜕 𝑠𝑠𝑠𝑠𝑠𝑠𝑓𝑓𝑎𝑎𝑎𝑎𝑎𝑎

𝜕𝜕𝑝𝑝𝜕𝜕𝑛𝑛 =

𝜌𝜌𝑑𝑑𝑡𝑡2

𝑅𝑅

𝜕𝜕𝑝𝑝𝜕𝜕𝑛𝑛 = 0

A correct BC at ai impermiable wall is that the material derivative of the wall should vanish

This gives the following BC:

Pressure on the wall may be extrapolated from the interior:

An alternative is to solve the normal momentum equation on the wall. At the wall pressure gradient is balanced by the cetrifugal force