computational fluid dynamics the euler equationsgtryggva/cfd-course2017/lecture-10-2017.pdf ·...

Computational Fluid Dynamics


http://www.nd.edu/~gtryggva/CFD-Course/

Grétar Tryggvason

Lecture 10February 27, 2017

Computational Fluid Dynamicshttp://www.nd.edu/~gtryggva/CFD-Course/

The Euler Equations


The Euler equations for 1D flow:

0)/(

2 =⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

++

∂∂+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

∂∂

ρρρρ

ρρρ

pEupu

u

xEu

t

�

E = e + u2 /2

Ideal Gas:

where

The Euler Equations

�

p = ρRT; e = e T( ); cv = de /dT

h = h T( ); cp = dh / dT;h − e = RTR = h /T − e /T = cp − cv;γ = cp / cv;

p = ρ h − e( ) = h / e −1( )ρe = γ −1( )ρeSpeed of sound

c2 = ∂p∂ρ

⎛⎝⎜

⎞⎠⎟ s

= γ RT = γ pρ

p = γ −1( )ρec2 = γ p / ρ

�

H = h + u2 /2; h = e + p /ρand we define

Pressure and energy are related by


The Euler Equations

0)/(

2 =⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

++

∂∂+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

∂∂

ρρρρ

ρρρ

pEupu

u

xEu

t

Expanding the derivative and rearranging the equations

�

∂∂t

ρup

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

+u ρ 00 u 1/ρ0 ρc 2 u

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ∂∂x

ρup

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

= 0

p = γ −1( )ρec2 = γ p / ρ

The Euler Equations

�

∂f∂t

+A ∂f∂x

= 0

�

det(AT − λI) = 0The Characteristics for the Euler Equations are found by finding the eigenvalues for AT

or


Therefore

�

λ1 = u − c;λ2 = u + c;λ3 = u

�

detu − λ 0 0ρ u − λ ρc 2

0 1/ρ u − λ

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

= 0

�

det(AT − λI) = 0 or

�

u − λ( ) u − λ( )2 − c 2( ) = 0

�

u − λ( ) = 0⇒ λ = u

u − λ( )2 − c 2( ) = 0

⇒ u − λ = ±c⇒ λ = u ± c

or:

The Euler Equations

Find

Computational Fluid DynamicsExact Solution:

The Rankine-Hugoniot conditions

�

∂f∂t

+ ∂F∂x

= 0

For a hyperbolic system

The speed of a discontinuity (shock) is found by moving to a frame where a shock moving with speed s is stationary

�

s f[ ] = F[ ]�

∂f∂t

− s ∂f∂x

+ ∂F∂x

= 0

Integrating across the shock yields

0


For the Euler equations:

0)/(

2 =⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

++

∂∂+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

∂∂

ρρρρ

ρρρ

pEupu

u

xEu

t

�

s ρL − ρR( ) = ρu( )L − ρu( )R( )s ρu( )L − ρu( )R( ) = ρu2 + p( )L − ρu2 + p( )R( )s ρE( )L − ρE( )R( ) = ρu(E + p /ρ)( )L − ρu(E + p /ρ)( )R( )

The Rankine-Hugoniot conditions are:


The Shock-Tube ProblemExact Solution

Computational Fluid DynamicsThe shock tube problem

L R

ShockContactExpansion Fan

uL, pL, ρL uR, pR, ρR

L R3 25


L R

ShockContactExpansion Fan

Pressure

Density


Consider the case pL > pR:Shock separates R and 2Contact discontinuity separates 2 and 3Expansion fan separates 3 and L

given:

�

pL , ρL , uL ,

�

pR , ρR , uR

L R234

�

α = γ +1γ −1

cR = γ pR ρRcL = γ pL ρL


P =pLpR

1−γ −1( ) cR / cL( )(P −1)

2γ 2γ + γ +1( ) P −1( )( )⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

2γ / γ −1( )

�

P = p2pR

The Rankine-Hugoniot conditions give a nonlinear relation for the pressure jump across the shock

which can be solved by iteration

given:

�

pL , ρL , uL ,

�

pR , ρR , uR

L R234


The speed of the shock and velocity behind the shock are found using RH conditions:

�

sshock = uR + cRγ −1+ γ +1( )P

2γ⎛

⎝ ⎜

⎞

⎠ ⎟ 1/ 2

�

scontact = u3 = u2 = uL + 2cLγ −1

1− P pRpL

⎛

⎝ ⎜

⎞

⎠ ⎟

γ −12γ

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

Exact Solution:

The speed of the contact is

The left hand side of the fan moves with speed

The right hand side of the fan moves with speed

�

sfL = −cL

s fR = u2 − c3

�

x fR = x0 + sfR t�

x fL = x0 + sfL t�

xcontact = x0 + scontact t�

xshock = x0 + sshockt

L R234

�

P = p2pR


In the expansion fan (4)

�

p4 = p41+ (s− x)cLα t

⎛

⎝ ⎜

⎞

⎠ ⎟

2γγ −1

Exact Solution:

�

ρ4 = ρ41+ (s− x)cLα t

⎛

⎝ ⎜

⎞

⎠ ⎟

2γγ −1

�

u4 = uL + 2γ +1

s− xt

⎛ ⎝ ⎜

⎞ ⎠ ⎟

L R234

�

pL , ρL , uL ,given:

�

cL = γpL ρL

Left uniform state

Behind the contact (3)

�

p3 = p2

�

u3 = u2

�

ρ3 = ρL P pRpL

⎛

⎝ ⎜

⎞

⎠ ⎟ 1/γ

�

P = p2pR


L R234

Behind the shock (2)

ρ2 =1+αPα + P

⎛⎝⎜

⎞⎠⎟ρR

u2 = uL +2cLγ −1

1− P pRpL

⎛⎝⎜

⎞⎠⎟

γ −12γ

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟�

p2 = P pR

�

pR , ρR , uR

given:

Right uniform state

cR = γ pR ρR

�

α = γ +1γ −1

Behind the contact (3)

�

p3 = p2

ρ3 = ρL P pRpL

⎛⎝⎜

⎞⎠⎟

1/γ

�

u3 = u2

�

P = p2pR


The velocity relative to the speed of sound defines the flow regime

c = γ pρ

�

Ma = uc

The Mach number is defined as the ratio of the local velocity over the speed of sound

�

Ma <1Ma >1

subsonicsupersonic


0 100 200 300 400 5000

2

4

6

8

10x 104 pressure

0 100 200 300 400 5000

0.2

0.4

0.6

0.8

1

density

0 100 200 300 400 5000

100

200

300velocity

Pressure

Density

Velocity

0 100 200 300 400 5000

0.2

0.4

0.6

0.8

1Mach number

t=0.0005

Mach Number

Test case:

�

pL =105; ρL =1.0; uL = 0pR =104; ρR = 0.125; uR = 0

Shocktube problem of G.A. Sod, JCP 27:1, 1978

�

t final = 0.005x0 = 0.5

Subsonic case


0 100 200 300 400 5000

0.2

0.4

0.6

0.8

1

density

0 100 200 300 400 5000

2

4

6

8

10

x 104 pressure

0 100 200 300 400 5000

50

100

150

200

250

300

velocity

0 100 200 300 400 5000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Mach number


Pressure

Density

Velocity

Mach Number

Test case:

�

pL =105; ρL =1.0; uL = 0pR =104; ρR = 0.01; uR = 0


�

t final = 0.0025x0 = 0.5

Supersonic case


0 100 200 300 400 5000

2

4

6

8

10x 104 pressure

0 100 200 300 400 5000

0.2

0.4

0.6

0.8

1

density

0 100 200 300 400 5000

100

200

300velocity

Pressure

Density

Velocity

0 100 200 300 400 5000

0.2

0.4

0.6

0.8

1Mach number

t=0.0005

Mach Number

Pressure

Density

Velocity

Mach Number

Supersonic caseSubsonic case


Centered Schemes


For a system of equations the characteristics often have different slopes and we therefore have different “upwind” directions

The simplest approach to solve the Euler equations is by using centered schemes where it is not necessary to take into account the characteristic direction

We will examine two such schemes:

The first order Lax-Friedrich scheme and

The second order Lax-Wendroff scheme

Flux Splitting


First order Lax-Friedrich

scheme


�

∂f∂t

+ ∂F∂x

= 0

Write the Euler equations as

The error term contains an term includingbut by taking we have a first order scheme

Update the solution by

f jn+1 =

12f j+1n + f j−1

n( )− 12ΔthFj+1n −Fj−1

n( )

h2 Δth ~ Δt

f jn+1 = f j

n −ΔthFj+1/2n −Fj−1/2

n( )The scheme is often written as

where Fj+1/2 =12Fj+1 +Fj( )− 12

hΔtf j+1n − f j

n( )


%Euler equation--one-dimensional Lax-Friedrichnx=256; maxstep=150;gg=1.4;p_left=100000;p_right=10000;r_left=1;r_right=0.125;u_left=0;xl=10.0;h=xl/(nx-1);time=0; for i=1:nx,x(i)=h*(i-1);end

r=zeros(1,nx);ru=zeros(1,nx);rE=zeros(1,nx);rn=zeros(1,nx);run=zeros(1,nx);rEn=zeros(1,nx);c=zeros(1,nx);u=zeros(1,nx);m=zeros(1,nx);

for i=1:nx,r(i)=r_right;ru(i)=0.0;rE(i)=p_right/(gg-1);endfor i=1:nx/2; r(i)=r_left; rE(i)=p_left/(gg-1)+0.5*r_left*u_left^2; ru(i)=r_left*u_left;end

rh=r;ruh=ru;rEh=rE;

cmax=sqrt( max(gg*p_right/r_right,gg*p_left/r_left) ); dt=0.45*h/cmax;

for istep=1:maxstep for i=1:nx,p(i)=(gg-1)*(rE(i)-0.5*(ru(i)*ru(i)/r(i)));end for i=2:nx-1 rh(i)=0.5*(r(i+1)+r(i-1))-(0.5*dt/h)*(ru(i+1)-ru(i-1)); ruh(i)=0.5*(ru(i+1)+ru(i-1))-...

(0.5*dt/h)*((ru(i+1)^2/r(i+1))+p(i+1)-(ru(i-1)^2/r(i-1))-p(i-1)); rEh(i)=0.5*(rE(i+1)+rE(i-1))-... (0.5*dt/h)*((ru(i+1)/r(i+1))*(rE(i+1)+p(i+1))...

-(ru(i-1)/r(i-1))*(rE(i-1)+p(i-1)) );end

r=rh;ru=ruh;rE=rEh;%plot(r,'r','linewidth',2); pausetime=time+dt,istepend


Results for LF

0 2 4 6 8 100

0.2

0.4

0.6

0.8

1Density

Grid size128256512

1024

Final time0.0071

�

pL =105; ρL =1.0; uL = 0pR =104; ρR = 0.125; uR = 0



Second order Lax-Wendroff

scheme


The Euler equations:

0)/(

2 =⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

++

∂∂+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

∂∂

ρρρρ

ρρρ

pEupu

u

xEu

t

epueE ργ )1(;2/2 −=+=Add the artificial viscosity to RHS:

where

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

∂∂

∂∂

∂∂=

xu

xu

uh

x10

2ρα

L-W with artificial viscosity


�

f j+ 12

* = 0.5 f jn + f j+1

n( ) − 0.5Δth F j+1n −F j

n( )f jn+1 = f j

n − ΔthF j+ 1

2

* −F j− 12*( )

Solutions of the 1D Euler equation using Lax-Wendroff

F ' = F −αh2ρ01u

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

∂u∂x

∂u∂x

With an artificial viscosity term added to the corrector step


Lax Fredrich

Leap Frog

j-1 j j+1


for istep=1:2000for i=1:nx,p(i)=…..; end

for i=2:nx-1 %prediction step rh(i)= ….. ruh(i)= ….. rEh(i)= …..end

for i=1:nx,ph(i)=…..; end

for i=2:nx-1 %correction step r(i)= ….. ru(i)= ….. rE(i)= ….. end

for i=1:nx,u(i)=ru(i)/r(i);end

for i=2:nx-1 %artificial viscosity ru(i)= ….. rE(i)= …..endtime=time+dt,istep

end

Outline of L-W program



nx=129; artvisc=0.5;hold off%gg=1.4;p_left=100000;p_right=1000;r_left=1;r_right=0.01;gg=1.4;p_left=100000;p_right=10000;r_left=1;r_right=0.125;xl=10.0;h=xl/(nx-1);time=0;

r=zeros(1,nx);ru=zeros(1,nx);rE=zeros(1,nx);p=zeros(1,nx);rh=zeros(1,nx);ruh=zeros(1,nx);rEh=zeros(1,nx);ph=zeros(1,nx);

for i=1:nx,r(i)=r_right;ru(i)=0.0;rE(i)=p_right/(gg-1);endfor i=1:nx/2; r(i)=r_left; rE(i)=p_left/(gg-1); end

rh=r;ruh=ru;rEh=rE;ph=p;

dt=0.25*h/sqrt(1.4*max([700+p_right/r_right,700+p_left/r_left]) )

for istep=1:2000

for i=1:nx,p(i)=(gg-1)*(rE(i)-0.5*(ru(i)*ru(i)/r(i)));end

for i=2:nx-1 %prediction step rh(i)=0.5*(r(i)+r(i+1))-(0.5*dt/h)*(ru(i+1)-ru(i)); ruh(i)=0.5*(ru(i)+ru(i+1))-(0.5*dt/h)*((ru(i+1)^2/r(i+1))+p(i+1)-(ru(i)^2/r(i))-p(i)); rEh(i)=0.5*(rE(i)+rE(i+1))-...

(0.5*dt/h)*((rE(i+1)*ru(i+1)/r(i+1))+(ru(i+1)*p(i+1)/r(i+1))... -(rE(i)* ru(i) /r(i))- (ru(i) *p(i) /r(i)));

end

for i=1:nx,ph(i)=(gg-1)*(rEh(i)-0.5*(ruh(i)*ruh(i)/rh(i)));end

for i=2:nx-1 %correction step r(i)=r(i)-(dt/h)*(ruh(i)-ruh(i-1)); ru(i)=ru(i)-(dt/h)*((ruh(i)^2/rh(i))-(ruh(i-1)^2/rh(i-1))+ph(i)-ph(i-1)); rE(i)=rE(i)-(dt/h)*((rEh(i)*ruh(i)/rh(i))-(rEh(i-1)*ruh(i-1)/rh(i-1))+...

(ruh(i)*ph(i)/rh(i))-(ruh(i-1)*ph(i-1)/rh(i-1)));end

for i=1:nx,u(i)=ru(i)/r(i);end

for i=2:nx-1 %artificial viscosity ru(i)=ru(i)+artvisc*(0.5*dt/h)*( (r(i)+r(i+1))*(u(i+1)-u(i))*abs(u(i+1)-u(i))... -(r(i)+r(i-1))*(u(i)-u(i-1))*abs(u(i)-u(i-1)) );

rE(i)=rE(i)+artvisc*(0.25*dt/h)*... ( (u(i+1)+u(i))*(r(i)+r(i+1))*(u(i+1)-u(i))*abs(u(i+1)-u(i))... -(u(i)+u(i-1))*(r(i)+r(i-1))*(u(i)-u(i-1))*abs(u(i)-u(i-1)) );endtime=time+dt,istepplot(p,'b','linewidth',2),title('pressure'); pauseif(time > 0.005)break,endend

L-W with artificial viscosityComputational Fluid DynamicsL-W with artificial viscosity

Test case:

�

pL =105; ρL =1.0; uL = 0pR =104; ρR = 0.125; uR = 0


Final time: 0.005

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

density

�

α = 0.5nx=128;

Results for LW


0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

density

�

α = 0.5α =1.0α =1.5α = 2.5

nx=128;

Effect of α

0 0.2 0.4 0.6 0.8 10

2

4

6

8

10

x 104 pressure

�

α = 0.5α =1.0α =1.5α = 2.5

L-W with artificial viscosityComputational Fluid Dynamics

0 0.2 0.4 0.6 0.8 10

2

4

6

8

10

x 104 pressure

nx=64; nx=128; nx=256;time=0.5

�

α =1.5

Effect of resolution

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

density



First order upwinding with flux

splitting


Define

�

F = F + + F−

So that

�

∂f∂t

+ ∂F+

∂x+ ∂F−

∂x= 0

Flux Splitting

�

∂f∂t

+ ∂F∂x

= 0

For conservation law

where

Which can be written as

�

∂f∂t

+ A[ ] ∂f∂x

= 0; A[ ] = ∂F∂f

�

λ[ ] = λ+[ ] + λ−[ ]

Are the positive and negative eigenvalues of A


For nonlinear equation the splitting is not unique, different matrices can have the same eigenvalues

The history of flux splitting is interesting in that complex splitting were discovered first and the simpler one later

The simplest one is that Zha-Bilgen

Other examples are due to van Leer and Steger-Warming

Flux SplittingComputational Fluid DynamicsThe Euler Equations

6.5. VELOCITY-PRESSURE METHOD FOR COMPRESSIBLE FLOW 55

The first part can be written as uf and it can be shown that it representsinformation carried by the flow velocity u, so the upwind direction dependsonly on the sign of u. The second part of the fluxes represents informationcarried by u±c and for supersonic flow, where the Mach number is either largerthan 1 or smaller then �1, the upwind direction again depends on the sign ofthe velocity. For �1 < M < 1, the second term is split in two and allocated tothe upwind and downwind directions as a linear function of the Mach number.Thus, following Laney (1998), the scheme is:

F+ = max(u, 0)

2

4

⇢⇢u⇢e

3

5+

2

4

0p+

(pu)+

3

5 (6.133)

and

F� = min(u, 0)

2

4

⇢⇢u⇢e

3

5+

2

4

0p�

(pu)�

3

5 , (6.134)

where we note that the first part depends only on the sign of the velocity. Thevariables in the second part are found by:

p+ = p

8

<

:

0, M �112 (1 +M),�1 < M < 1,

1, M � 1p� = p

8

<

:

1 M �112 (1�M),�1 < M < 1,

0, M � 1(6.135)

and

(pu)+ = p

8

<

:

0, M �112 (u+ c),�1 < M < 1

u, M � 1(pu)� = p

8

<

:

u M �112 (u� c),�1 < M < 1.

0, M � 1(6.136)

For M � 1, F = F+ and for M 1, F = F�, as we intended.To discretize equation (6.130) we first write

@f

@t+

@F+

@x+

@F�

@x= 0, (6.137)

and then approximate the fluxes using the first order upwind scheme:

fn+1j

= fnj

� �t

�x

⇣

F+(fnj

)� F+(fnj�1) + F�(fn

j+1)� F�(fnj

)⌘

. (6.138)

For a numerical implementation it is possible to rewrite (6.138) as

fn+1j

= fnj

� �t

�x

⇣

F̂n

j+1/2 � F̂n

j�1/2

⌘

, (6.139)

where the fluxes at the half-points are found by

F̂n

j+1/2 = F+(fnj

) + F�(fnj+1). (6.140)



F+ = max(u, 0)

2

4

⇢⇢u⇢e

3

5+

2

4

0p+

(pu)+

3

5 (6.133)

and

F� = min(u, 0)

2

4

⇢⇢u⇢e

3

5+

2

4

0p�

(pu)�

3

5 , (6.134)


p+ = p

8

<

:

0, M �112 (1 +M),�1 < M < 1,

1, M � 1p� = p

8

<

:

1 M �112 (1�M),�1 < M < 1,

0, M � 1(6.135)

and

(pu)+ = p

8

<

:

0, M �112 (u+ c),�1 < M < 1

u, M � 1(pu)� = p

8

<

:

u M �112 (u� c),�1 < M < 1.

0, M � 1(6.136)


@f

@t+

@F+

@x+

@F�

@x= 0, (6.137)


fn+1j

= fnj

� �t

�x

⇣

F+(fnj

)� F+(fnj�1) + F�(fn

j+1)� F�(fnj

)⌘

. (6.138)


fn+1j

= fnj

� �t

�x

⇣

F̂n

j+1/2 � F̂n

j�1/2

⌘

, (6.139)


F̂n

j+1/2 = F+(fnj

) + F�(fnj+1). (6.140)



F+ = max(u, 0)

2

4

⇢⇢u⇢e

3

5+

2

4

0p+

(pu)+

3

5 (6.133)

and

F� = min(u, 0)

2

4

⇢⇢u⇢e

3

5+

2

4

0p�

(pu)�

3

5 , (6.134)


p+ = p

8

<

:

0, M �112 (1 +M),�1 < M < 1,

1, M � 1p� = p

8

<

:

1 M �112 (1�M),�1 < M < 1,

0, M � 1(6.135)

and

(pu)+ = p

8

<

:

0, M �112 (u+ c),�1 < M < 1

u, M � 1(pu)� = p

8

<

:

u M �112 (u� c),�1 < M < 1.

0, M � 1(6.136)


@f

@t+

@F+

@x+

@F�

@x= 0, (6.137)


fn+1j

= fnj

� �t

�x

⇣

F+(fnj

)� F+(fnj�1) + F�(fn

j+1)� F�(fnj

)⌘

. (6.138)


fn+1j

= fnj

� �t

�x

⇣

F̂n

j+1/2 � F̂n

j�1/2

⌘

, (6.139)


F̂n

j+1/2 = F+(fnj

) + F�(fnj+1). (6.140)

Zha-Bilgen flux splitting Separate the advective fluxes and linearly interpolate the pressure terms between M=1 and M=-1.

Computational Fluid Dynamics346

The Euler Equations

56 CHAPTER 6. COMPUTATIONAL FLUID DYNAMICS

Using that max(f, 0) = 0.5(f + |f |) and min(f, 0) = 0.5(f � |f |) for any f wecan write the fluxes for �1 < M < 1 as:

F̂n

j+1/2 = F(fnj+1) + F(fn

j

)� 1

2

|un

j+1|

2

4

⇢nj+1

⇢nj+1u

n

j+1

⇢nj+1e

n

j+1

3

5� |un

j

|

2

4

⇢nj

⇢nj

un

j

⇢nj

enj

3

5

!

�1

2

2

4

0pnj+1M

n

j+1

pnj+1c

n

j+1

3

5�

2

4

0pnj

Mn

j

pnj

cnj

3

5

!

, (6.141)

and we use this form in the code shown below. Once the fluxes have been foundby (6.141), the variables are updated using (6.139).

As methods for the Euler equations are developed, new methods are oftentested using problems used to test earlier methods. Several of these tests havebeen used by so many researchers that they have become de facto standards.One such test was introduced by Sod (1978) and consists of discontinuous initialconditions with high pressure gas occupying the left half of a domain and a lowpressure gas in the right half. The solution consists of the high pressure gas ex-panding into the low pressure gas, sending a shock wave ahead and a rarifactionwave in the other direction. Thus, the solution consists of (moving from rightto left): undisturbed gas at low pressure on the right, compressed low pressuregas (separated by a shock from the undisturbed gas); expanded high pressuregas moving into the right hand side of the domain; a rarifaction fan where theinitially compressed gas expands smoothly; and finally the undisturbed highpressure gas on the left. The exact solution is easily found, as explained inchapter 15, and once the conditions in each region have been found, the timeevolution is simply given by expanding the time axis at a constant rate. A codefor this problem, using (6.141) and (6.139) is given below and sample results aregiven in figure 6.26, where we plot the density, the pressure, the velocity andthe Mach number at time 0.005. In all cases we plot results for three di↵erentresolutions. The time step must be limited by the Courant conditions, whichsay that the signal cannot travel more then one grid space in each time step, or(u+ c)�t < �t, and we take the time step to be 0.45 times the maximum step.The shock (the left most discontinuity) is captured reasonably well for the finerresolutions, but the contact discontinuity (the second density discontinuity fromthe right) is relatively smooth for all three resolutions. It is generally found thatmaintaining sharp contacts is more di�cult than keeping shocks sharp.

%project 2--upwind-Zha-Bilgen flux splitting

nx=40*256; tfinal=0.005; xl=10.0; time=0; gg=1.4;

h=xl/(nx-1);for i=1:nx,x(i)=h*(i-1);end

p_left=100000;p_right=10000;r_left=1;r_right=0.125;u_left=0;

56 CHAPTER 6. COMPUTATIONAL FLUID DYNAMICS

Using that max(f, 0) = 0.5(f + |f |) and min(f, 0) = 0.5(f � |f |) for any f wecan write the fluxes for �1 < M < 1 as:

F̂n

j+1/2 = F(fnj+1) + F(fn

j

)� 1

2

|un

j+1|

2

4

⇢nj+1

⇢nj+1u

n

j+1

⇢nj+1e

n

j+1

3

5� |un

j

|

2

4

⇢nj

⇢nj

un

j

⇢nj

enj

3

5

!

�1

2

2

4

0pnj+1M

n

j+1

pnj+1c

n

j+1

3

5�

2

4

0pnj

Mn

j

pnj

cnj

3

5

!

, (6.141)

and we use this form in the code shown below. Once the fluxes have been foundby (6.141), the variables are updated using (6.139).

As methods for the Euler equations are developed, new methods are oftentested using problems used to test earlier methods. Several of these tests havebeen used by so many researchers that they have become de facto standards.One such test was introduced by Sod (1978) and consists of discontinuous initialconditions with high pressure gas occupying the left half of a domain and a lowpressure gas in the right half. The solution consists of the high pressure gas ex-panding into the low pressure gas, sending a shock wave ahead and a rarifactionwave in the other direction. Thus, the solution consists of (moving from rightto left): undisturbed gas at low pressure on the right, compressed low pressuregas (separated by a shock from the undisturbed gas); expanded high pressuregas moving into the right hand side of the domain; a rarifaction fan where theinitially compressed gas expands smoothly; and finally the undisturbed highpressure gas on the left. The exact solution is easily found, as explained inchapter 15, and once the conditions in each region have been found, the timeevolution is simply given by expanding the time axis at a constant rate. A codefor this problem, using (6.141) and (6.139) is given below and sample results aregiven in figure 6.26, where we plot the density, the pressure, the velocity andthe Mach number at time 0.005. In all cases we plot results for three di↵erentresolutions. The time step must be limited by the Courant conditions, whichsay that the signal cannot travel more then one grid space in each time step, or(u+ c)�t < �t, and we take the time step to be 0.45 times the maximum step.The shock (the left most discontinuity) is captured reasonably well for the finerresolutions, but the contact discontinuity (the second density discontinuity fromthe right) is relatively smooth for all three resolutions. It is generally found thatmaintaining sharp contacts is more di�cult than keeping shocks sharp.

%project 2--upwind-Zha-Bilgen flux splitting

nx=40*256; tfinal=0.005; xl=10.0; time=0; gg=1.4;

h=xl/(nx-1);for i=1:nx,x(i)=h*(i-1);end

p_left=100000;p_right=10000;r_left=1;r_right=0.125;u_left=0;



F+ = max(u, 0)

2

4

⇢⇢u⇢e

3

5+

2

4

0p+

(pu)+

3

5 (6.133)

and

F� = min(u, 0)

2

4

⇢⇢u⇢e

3

5+

2

4

0p�

(pu)�

3

5 , (6.134)


p+ = p

8

<

:

0, M �112 (1 +M),�1 < M < 1,

1, M � 1p� = p

8

<

:

1 M �112 (1�M),�1 < M < 1,

0, M � 1(6.135)

and

(pu)+ = p

8

<

:

0, M �112 (u+ c),�1 < M < 1

u, M � 1(pu)� = p

8

<

:

u M �112 (u� c),�1 < M < 1.

0, M � 1(6.136)


@f

@t+

@F+

@x+

@F�

@x= 0, (6.137)


fn+1j

= fnj

� �t

�x

⇣

F+(fnj

)� F+(fnj�1) + F�(fn

j+1)� F�(fnj

)⌘

. (6.138)


fn+1j

= fnj

� �t

�x

⇣

F̂n

j+1/2 � F̂n

j�1/2

⌘

, (6.139)


F̂n

j+1/2 = F+(fnj

) + F�(fnj+1). (6.140)



F+ = max(u, 0)

2

4

⇢⇢u⇢e

3

5+

2

4

0p+

(pu)+

3

5 (6.133)

and

F� = min(u, 0)

2

4

⇢⇢u⇢e

3

5+

2

4

0p�

(pu)�

3

5 , (6.134)


p+ = p

8

<

:

0, M �112 (1 +M),�1 < M < 1,

1, M � 1p� = p

8

<

:

1 M �112 (1�M),�1 < M < 1,

0, M � 1(6.135)

and

(pu)+ = p

8

<

:

0, M �112 (u+ c),�1 < M < 1

u, M � 1(pu)� = p

8

<

:

u M �112 (u� c),�1 < M < 1.

0, M � 1(6.136)


@f

@t+

@F+

@x+

@F�

@x= 0, (6.137)


fn+1j

= fnj

� �t

�x

⇣

F+(fnj

)� F+(fnj�1) + F�(fn

j+1)� F�(fnj

)⌘

. (6.138)


fn+1j

= fnj

� �t

�x

⇣

F̂n

j+1/2 � F̂n

j�1/2

⌘

, (6.139)


F̂n

j+1/2 = F+(fnj

) + F�(fnj+1). (6.140)



F+ = max(u, 0)

2

4

⇢⇢u⇢e

3

5+

2

4

0p+

(pu)+

3

5 (6.133)

and

F� = min(u, 0)

2

4

⇢⇢u⇢e

3

5+

2

4

0p�

(pu)�

3

5 , (6.134)


p+ = p

8

<

:

0, M �112 (1 +M),�1 < M < 1,

1, M � 1p� = p

8

<

:

1 M �112 (1�M),�1 < M < 1,

0, M � 1(6.135)

and

(pu)+ = p

8

<

:

0, M �112 (u+ c),�1 < M < 1

u, M � 1(pu)� = p

8

<

:

u M �112 (u� c),�1 < M < 1.

0, M � 1(6.136)


@f

@t+

@F+

@x+

@F�

@x= 0, (6.137)


fn+1j

= fnj

� �t

�x

⇣

F+(fnj

)� F+(fnj�1) + F�(fn

j+1)� F�(fnj

)⌘

. (6.138)


fn+1j

= fnj

� �t

�x

⇣

F̂n

j+1/2 � F̂n

j�1/2

⌘

, (6.139)


F̂n

j+1/2 = F+(fnj

) + F�(fnj+1). (6.140)



F+ = max(u, 0)

2

4

⇢⇢u⇢e

3

5+

2

4

0p+

(pu)+

3

5 (6.133)

and

F� = min(u, 0)

2

4

⇢⇢u⇢e

3

5+

2

4

0p�

(pu)�

3

5 , (6.134)


p+ = p

8

<

:

0, M �112 (1 +M),�1 < M < 1,

1, M � 1p� = p

8

<

:

1 M �112 (1�M),�1 < M < 1,

0, M � 1(6.135)

and

(pu)+ = p

8

<

:

0, M �112 (u+ c),�1 < M < 1

u, M � 1(pu)� = p

8

<

:

u M �112 (u� c),�1 < M < 1.

0, M � 1(6.136)


@f

@t+

@F+

@x+

@F�

@x= 0, (6.137)


fn+1j

= fnj

� �t

�x

⇣

F+(fnj

)� F+(fnj�1) + F�(fn

j+1)� F�(fnj

)⌘

. (6.138)


fn+1j

= fnj

� �t

�x

⇣

F̂n

j+1/2 � F̂n

j�1/2

⌘

, (6.139)


F̂n

j+1/2 = F+(fnj

) + F�(fnj+1). (6.140)

Using that max(f,0)=0.5*(f+|f|) and min(f,0)=0.5*(f-|f|)

Computational Fluid DynamicsSummary

%upwind-Zha-Bilgen flux splitting

nx=40*256; tfinal=0.005; xl=10.0; time=0; gg=1.4;h=xl/(nx-1);for i=1:nx,x(i)=h*(i-1);endp_left=100000;p_right=10000;r_left=1;r_right=0.125;u_left=0;

r=zeros(1,nx);ru=zeros(1,nx);rE=zeros(1,nx);p=zeros(1,nx);c=zeros(1,nx);u=zeros(1,nx);m=zeros(1,nx);F1=zeros(1,nx);F2=zeros(1,nx);F3=zeros(1,nx);

for i=1:nx, r(i)=r_right;ru(i)=0.0;rE(i)=p_right/(gg-1);endfor i=1:nx/2; r(i)=r_left; rE(i)=p_left/(gg-1); end

cmax=sqrt( max(gg*p_right/r_right,gg*p_left/r_left) ); dt=0.45*h/cmax; maxstep=tfinal/dt;

for istep=1:maxstep for i=1:nx,p(i)=(gg-1)*(rE(i)-0.5*(ru(i)*ru(i)/r(i)));end for i=1:nx,c(i)=sqrt( gg*p(i)/r(i) );end for i=1:nx,u(i)=ru(i)/r(i);end; for i=1:nx,m(i)=u(i)/c(i);end

for i=1:nx-1% Find fluxes

F1(i)=0.5*(ru(i+1)+ru(i))-0.5*(abs(ru(i+1))-abs(ru(i)));

F2(i)=0.5*(u(i+1)*ru(i+1)+p(i+1)+u(i)*ru(i)+p(i))... -0.5*(abs(u(i+1))*ru(i+1)-abs(u(i))*ru(i))... -0.5*(p(i+1)*m(i+1)-p(i)*m(i));

F3(i)=0.5*(u(i+1)*(rE(i+1)+p(i+1))+u(i)*(rE(i)+p(i)))... -0.5*(abs(u(i+1))*rE(i+1)-abs(u(i))*rE(i))... -0.5*(p(i+1)*c(i+1)-p(i)*c(i));

if m(i) > 1, F2(i)=ru(i)*u(i)+p(i); F3(i)=(rE(i)+p(i))*u(i);end if m(i) < -1, F2(i)=ru(i+1)*u(i+1)+p(i+1); F3(i)=(rE(i+1)+p(i+1))*u(i+1);end end

for i=2:nx-2% Update solution

r(i)=r(i)-(dt/h)*(F1(i)-F1(i-1)); ru(i)=ru(i)-(dt/h)*(F2(i)-F2(i-1)); rE(i)=rE(i)-(dt/h)*(F3(i)-F3(i-1)); end

time=time+dt,istepend


The Euler Equations

0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

DensityDensityDensity

0 2 4 6 8 100

50

100

150

200

250

300

VelocityVelocityVelocity�

pL =105; ρL =1.0; uL = 0pR =104; ρR = 0.125; uR = 0

Shock tube problem of G.A. Sod, JCP 27:1, 1978

Grid size64256

1024

Final time0.005


The Euler Equations

�

pL =105; ρL =1.0; uL = 0pR =104; ρR = 0.125; uR = 0

Shock tube problem of G.A. Sod, JCP 27:1, 1978

0 2 4 6 8 100

2

4

6

8

10x 10

4

PressurePressurePressure

0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

Mach NumberMach NumberMach Number

Grid size64256

1024

Final time0.005


Another example: one-dimensional Euler equation using the van Leer vector flux splitting

F− = −ρ4c

(u − c)2

1

(γ −1)u − 2cγ

2c − (γ −1)u⎡⎣ ⎤⎦2

2(γ 2 −1)

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

F+ =ρ4c

(u + c)2

1

(γ −1)u + 2cγ

(γ −1)u + 2c⎡⎣ ⎤⎦2

2(γ 2 −1)

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

Van Leer

First order upwindComputational Fluid Dynamics

van Leer vector flux splitting

ρu

ρu2 + p

ρu(E +pρ

)

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

=ρ4c

(u + c)2

1

(γ −1)u + 2cγ

(γ −1)u + 2c⎡⎣ ⎤⎦2

2(γ 2 −1)

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

−ρ4c

(u − c)2

1

(γ −1)u − 2cγ

2c − (γ −1)u⎡⎣ ⎤⎦2

2(γ 2 −1)

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

�

ρ4c(u + c)2 − ρ

4c(u − c)2

= ρ4c

u2 + 2uc + c 2 − u2 + 2uc − c 2( ) = ρ4c4uc = ρu

For example, the mass flux:

First order upwind


For 1D flow the fluxes are

F± = ± ρ4c(u ± c)2

1

(γ −1)u ± 2cγ

2c ± (γ −1)u[ ]22(γ 2 −1)

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

= ± ρc4(M ±1)2

1

2cγ

γ −12

M ±1⎛⎝⎜

⎞⎠⎟

2c2

γ 2 −11± γ −1

2M⎛

⎝⎜⎞⎠⎟2

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥


for istep=1:maxstep for i=1:nx,c(i)=sqrt( gg*(gg-1)*(rE(i)-0.5*(ru(i)^2/r(i)))/r(i) );end for i=1:nx,u(i)=ru(i)/r(i);end; for i=1:nx,m(i)=u(i)/c(i);end for i=2:nx-1 %upwind rn(i)=r(i)-(dt/h)*(... (0.25*r(i)*c(i)*(m(i)+1)^2) - (0.25*r(i-1)*c(i-1)*(m(i-1)+1)^2)+... (-0.25*r(i+1)*c(i+1)*(m(i+1)-1)^2) - (-0.25*r(i)*c(i)*(m(i)-1)^2) ); run(i)=ru(i)-(dt/h)*(... ( 0.25*r(i)*c(i) *(m(i)+1)^2) *(( 1+0.5*(gg-1)*m(i)) *2*c(i) /gg) - ... ( 0.25*r(i-1)*c(i-1)*(m(i-1)+1)^2)*(( 1+0.5*(gg-1)*m(i-1))*2*c(i-1)/gg) + ... (-0.25*r(i+1)*c(i+1)*(m(i+1)-1)^2)*((-1+0.5*(gg-1)*m(i+1))*2*c(i+1)/gg) - ... (-0.25*r(i)*c(i) *(m(i)-1)^2) *((-1+0.5*(gg-1)*m(i)) *2*c(i) /gg) ); rEn(i)=rE(i)-(dt/h)*(... ( 0.25*r(i)*c(i) *(m(i)+1)^2) *((1+0.5*(gg-1)*m(i))^2 *2*c(i)^2 /(gg^2-1)) - ... ( 0.25*r(i-1)*c(i-1)*(m(i-1)+1)^2)*((1+0.5*(gg-1)*m(i-1))^2*2*c(i-1)^2/(gg^2-1)) + ... (-0.25*r(i+1)*c(i+1)*(m(i+1)-1)^2)*((1-0.5*(gg-1)*m(i+1))^2*2*c(i+1)^2/(gg^2-1)) - ... (-0.25*r(i)*c(i) *(m(i)-1)^2) *((1-0.5*(gg-1)*m(i))^2 *2*c(i)^2 /(gg^2-1)) ); end

First order upwind


�

pL =105; ρL =1.0; uL = 0pR =104; ρR = 0.125; uR = 0


0 2 4 6 8 100

0.2

0.4

0.6

0.8

1 DensityDensityDensityDensityDensity

Effect of resolution

nx=64; maxstep=32nx=128; maxstep=64nx=256; maxstep=128nx=512; maxstep=256

0 2 4 6 8 100

2

4

6

8

10x 104

PressurePressure

nx=64; maxstep=32nx=256; maxstep=128

Final time: 0.005


For 2D flow the van Leer fluxes are

F± = ± ρ4c(u ± c)2

1(γ −1)u ± 2c

γ

vv2

2+(γ −1)u ± 2c[ ]22(γ 2 −1)

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

; G± = ± ρ4c(v ± c)2

1u

(γ −1)v ± 2cγ

u2

2+(γ −1)v ± 2c[ ]22(γ 2 −1)

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

First order upwind

∂f∂t

+ ∂F+

∂x+ ∂F−

∂x+ ∂G+

∂y+ ∂G−

∂y= 0


F− =12ργ(u − c)

1u − c

12(u − c)2 + 1

2c2 3− γ

γ −1⎛⎝⎜

⎞⎠⎟

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

F+ =ρ2γ

(2γ −1)u + c2(γ −1)u2 + (u + c)2

(γ −1)u3 + 12(u + c)3 + 3− γ

2(γ −1)(u + c)c2

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

Steger-Warming

Several other splitting schemes are possible, such as:


F+ = ρ4c(u + c )2

1

(γ −1)u + 2cγ

(γ −1)u + 2c⎡⎣ ⎤⎦2

2(γ 2 −1)

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟

= ρc4uc+1

⎛⎝⎜

⎞⎠⎟

2

1

2cγ1+ γ −1

2uc

⎛⎝⎜

⎞⎠⎟

2c 2

γ 2 −11+ γ −1

2uc

⎛⎝⎜

⎞⎠⎟

2

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟

= ρc4(M +1)2

1

2cγ1+ γ −1

2M⎛

⎝⎜⎞⎠⎟

2c 2

γ 2 −11+ γ −1

2M⎛

⎝⎜⎞⎠⎟

2

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟

Rewrite the flux terms in terms of Mach number:

First order upwind


∂∂t

ρρuρE

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟+

∂∂x

ρc4

( M +1)2

1

2cγ

1+γ −1

2M

⎛⎝⎜

⎞⎠⎟

2c2

γ 2 −11+

γ −12

M⎛⎝⎜

⎞⎠⎟

2

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟

+∂∂x

−ρc4

( M −1)2

1

2cγ

−1+γ −1

2M

⎛⎝⎜

⎞⎠⎟

2c2

γ 2 −11−

γ −12

M⎛⎝⎜

⎞⎠⎟

2

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟⎟

= 0

�

∂f∂t

+ ∂F+

∂x+ ∂F-

∂x= 0


∂f∂t

+∂F+

∂x+∂F-

∂x= 0

f jn+1 = f j

n −Δth

⎛⎝⎜

⎞⎠⎟Fj+ − Fj -1

+( )n − Δth

⎛⎝⎜

⎞⎠⎟Fj+1- − Fj

-( )n

F+ =ρc4

( M +1)2

1

2cγ

1+γ −1

2M

⎛⎝⎜

⎞⎠⎟

2c2

γ 2 −11+

γ −12

M⎛⎝⎜

⎞⎠⎟

2

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟

; F− = −ρc4

( M −1)2

1

2cγ

−1+γ −1

2M

⎛⎝⎜

⎞⎠⎟

2c2

γ 2 −11−

γ −12

M⎛⎝⎜

⎞⎠⎟

2

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟⎟

Where

Solve by

First order upwind


WENO-3

http://www.nd.edu/~gtryggva/CFD-Course/Computational Fluid Dynamics

df jdt

+1Δx

Fj+1/2 −Fj−1/2( ) = 0

Fj+1/2 =ω1Fj+1/2(1) +ω2Fj+1/2

(2)

!ωl =γ l

ε +βl( )2

ωm =!ωm

!ωll=1

2∑

Fj+1/2(1) = −

12Fj−1 +

32Fj

Fj+1/2(2) =

12Fj +

12Fj+1

γ1 =13; γ2 =

23

β1 = f j − f j−1( )2

β2 = f j+1 − f j( )2

ε =10−6

The semi-discrete equation is

The fluxes are the weighted sum

where

The weights are

2nd order upwind2nd order central

Third order WENO

∂∂t

f + ∂∂x

F = 0


The time integration is done by a third order Runga-Kutta

fj

(1) = fj

n +ΔtL f n, tn( )fj

(2) =34fj

n +14fj

(1) +14ΔtL f (1), tn +Δt( )

fj

n+1 =13fj

n +23fj

(2) +23ΔtL f (2), tn + 1

2Δt

"

#$

%

&'

L f , t( ) = −∂F∂x

where

df jdt

+1Δx

Fj+1/2 −Fj−1/2( ) = 0


% one-dimensional Euler by WENO3-RK% Third order scheme--Lax-Friedrich fluxes---compact scheme-3 times Euler% Nothing is done to account for boundaries%------------------------------------------------------------

nx=256; maxstep=150;

gg=1.4;p_left=100000;p_right=10000;r_left=1;r_right=0.125;u_left=0;xl=10.0;h=xl/(nx-1);time=0; for i=1:nx,x(i)=h*(i-1);end

gam1=1/3;gam2=2/3; eps=0.000001;

r=zeros(1,nx);ru=zeros(1,nx);rE=zeros(1,nx);rL=zeros(1,nx);ruL=zeros(1,nx);rEL=zeros(1,nx);rR=zeros(1,nx);ruR=zeros(1,nx);rER=zeros(1,nx);F1=zeros(1,nx); F2=zeros(1,nx); F3=zeros(1,nx);

for i=1:nx,r(i)=r_right;ru(i)=0.0;rE(i)=p_right/(gg-1);endfor i=1:nx/2; r(i)=r_left; rE(i)=p_left/(gg-1)+0.5*r_left*u_left^2; ru(i)=r_left*u_left;end

% These are set here to provide default values for the endpointsfor j=1:nx, rL(j)=r(j);ruL(j)=ru(j);rEL(j)=rE(j); rR(j)=r(j);ruR(j)=ru(j);rER(j)=rE(j); end

cmax=sqrt( max(gg*p_right/r_right,gg*p_left/r_left) ); dt=0.45*h/cmax;

plot(x,r,'r','linewidth',2); pause, % Plot initial conditions for istep=1:maxstep

rn=r; run=ru;rEn=rE; for substep=1:3 % third order % ****** WENO3 ****** for j=3:nx-2, % LEFT values beta1=(r(j)-r(j-1))^2; beta2=(r(j+1)-r(j))^2; omt1=(gam1)/(eps+beta1)^2; omt2=(gam2)/(eps+beta2)^2; omtsum=omt1+omt2; om1=omt1/omtsum; om2=omt2/omtsum; rL(j)=om1*(-0.5*r(j-1)+1.5*r(j))+om2*(0.5*r(j)+0.5*r(j+1));

beta1=(ru(j)-ru(j-1))^2; beta2=(ru(j+1)-ru(j))^2; omt1=(gam1)/(eps+beta1)^2; omt2=(gam2)/(eps+beta2)^2; omtsum=omt1+omt2; om1=omt1/omtsum; om2=omt2/omtsum; ruL(j)=om1*(-0.5*ru(j-1)+1.5*ru(j))+om2*(0.5*ru(j)+0.5*ru(j+1)); beta1=(rE(j)-rE(j-1))^2; beta2=(rE(j+1)-rE(j))^2; omt1=(gam1)/(eps+beta1)^2; omt2=(gam2)/(eps+beta2)^2; omtsum=omt1+omt2; om1=omt1/omtsum; om2=omt2/omtsum; rEL(j)=om1*(-0.5*rE(j-1)+1.5*rE(j))+om2*(0.5*rE(j)+0.5*rE(j+1));

http://www.nd.edu/~gtryggva/CFD-Course/% RIGHT values beta1=(r(j+1)-r(j+2))^2; beta2=(r(j)-r(j+1))^2; omt1=(gam1)/(eps+beta1)^2; omt2=(gam2)/(eps+beta2)^2; omtsum=omt1+omt2; om1=omt1/omtsum; om2=omt2/omtsum; rR(j)=om1*(-0.5*r(j+2)+1.5*r(j+1))+om2*(0.5*r(j+1)+0.5*r(j));

beta1=(ru(j+1)-ru(j+2))^2; beta2=(ru(j)-ru(j+1))^2; omt1=(gam1)/(eps+beta1)^2; omt2=(gam2)/(eps+beta2)^2; omtsum=omt1+omt2; om1=omt1/omtsum; om2=omt2/omtsum; ruR(j)=om1*(-0.5*ru(j+2)+1.5*ru(j+1))+om2*(0.5*ru(j+1)+0.5*ru(j));

beta1=(rE(j+1)-rE(j+2))^2; beta2=(rE(j)-rE(j+1))^2; omt1=(gam1)/(eps+beta1)^2; omt2=(gam2)/(eps+beta2)^2; omtsum=omt1+omt2; om1=omt1/omtsum; om2=omt2/omtsum; rER(j)=om1*(-0.5*rE(j+2)+1.5*rE(j+1))+om2*(0.5*rE(j+1)+0.5*rE(j)); end

for j=1:nx-1, % find the fluxes pL=(gg-1)*(rEL(j)-0.5*(ruL(j)*ruL(j)/rL(j))); pR=(gg-1)*(rER(j)-0.5*(ruR(j)*ruR(j)/rR(j))); F1(j)=0.5*(ruR(j)+ruL(j))-0.5*(h/dt)*(rR(j)-rL(j)); F2(j)=0.5*((ruR(j)^2/rR(j))+pR+(ruL(j)^2/rL(j))+pL)-... 0.5*(h/dt)*(ruR(j)-ruL(j)); F3(j)=0.5*((ruR(j)/rR(j))*(rER(j)+pR)+(ruL(j)/rL(j))*(rEL(j)+pL))-... 0.5*(h/dt)*(rER(j)-rEL(j)); end; % **** End WENO3 ****

for j=3:nx-2, r(j)=r(j)-(dt/h)*(F1(j)-F1(j-1)); ru(j)=ru(j)-(dt/h)*(F2(j)-F2(j-1)); rE(j)=rE(j)-(dt/h)*(F3(j)-F3(j-1)); end;

if substep==2, r=0.75*rn+0.25*r; ru=0.75*run+0.25*ru; rE=0.75*rEn+0.25*rE; % RK elseif substep==3 r=0.33334*rn+0.66666*r; ru=0.33334*run+0.66666*ru; rE=0.33334*rEn+0.66666*rE; % RK end

end % End of sub-iteration for RK-3 time integration plot(x,r,'r','linewidth',2); pause(0.001) time=time+dt; %istepend

% plot(x,r,'k','linewidth',2);


Density Velocity


0 2 4 6 8 100.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

2565121024

0 2 4 6 8 10-50

0

50

100

150

200

250

3002565121024

Subsonic Case


Density Velocity


0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

256512

1024

0 2 4 6 8 10-50

0

50

100

150

200

250

300

350

400

4502565121024

Supersonic Case


In general we solve:

The Euler Equations

f jn+1 = f j

n −ΔthFj+1/2n −Fj−1/2

n( )�

∂f∂t

+ ∂F∂x

= 0

by

Where the fluxes are found in a variety of ways

Fj+1/2 = F f L( ) j+1/2 , fR( ) j+1/2( )

In principle we can solve this problem, the Riemann problem, exactly by assuming constant states and then integrate the fluxes over the time step.


�

f jn+1/ 2 = f j

n − Δt2h

Fj+1/ 2n − Fj−1/ 2

n( )

�

f jn+1 = f j

n − Δth

Fj+1/ 2n+1/ 2 − Fj−1/ 2

n+1/ 2( )�

Fj+1/ 2n+1/ 2 = F f L( ) j+1/ 2

n+1/ 2, f R( ) j+1/ 2

n+1/ 2( )

Predictor step

Variables

�

f j+1/ 2L = f j

n+1/ 2 + 12ΨL f j

n+1/ 2 − f j−1n+1/ 2( )

�

f j+1/ 2R = f j+1

n+1/ 2 − 12ΨR f j+1

n+1/ 2 − f jn+1/ 2( )

Final step

Predictor-corrector method: Limiting the variables

Find:

The Euler Equations


The Riemann problems for a

linear hyperbolic system


The Riemann problem for a system of linear equations with constant coefficients. We start with:

The eigenvalues and eigenvectors of A are given by

If we normalize the eigenvectors to unity then the eigenvectors form an orthonormal basis and we can write

∂f∂t

+ A ∂f∂x

= 0

f = wp

p=1

3

∑ e p

The Euler Equations

Ae p = λpe p

e p = 1

wp = f ⋅e p


Substituting for f in the original equation, we get

Or, using the definition of the eigenvalue and the fact that the eigenvectors are orthonormal, we have three sets of independent equations

Whose solutions are given by

Representing a simple translation of the initial conditions

∂wp

∂tp=1

3

∑ e p +∂wp

∂xA

p=1

3

∑ e p = 0

The Euler Equations

∂wp

∂t+ λp

∂wp

∂x= 0

wp (x,t) = wp0 (x − λpt)

Ae p = λpe p


For discontinuous initial conditions, where

w = wL , x < 0 & w = wR , x > 0; w = w1,w2 ,w3( )

The Euler Equations

The different components of the solution remain piecewise constant but the discontinuity moves left or right, depending on the sign of the eigenvalues.


Since the solution is piecewise constant the fluxes are easily found. In particular, for each of the linear equations the fluxes are:

The Euler Equations

wpλp =wpLλp if λp > 0

wpRλp if λp < 0

⎧⎨⎪

⎩⎪= 12

λp + λp( )wpL + λp − λp( )wpR{ }= 12

λp wpLi +wp

R( )− λp wpRi −wp

L( ){ }

Fj+1/2 =12

λp wpLi +wp

R( )− λp wpRi −wp

L( ){ }e pp∑

Or, for the system


For the Euler equations the situation is more complex.Start with the primitive form

The Euler Equations

We can find the eigenvectors in standard ways and use them to make the matric diagonal

∂w∂t

+C∂w∂x

= 0

w =ρ

up

!

"

###

$

%

&&&

C =u ρ 00 u 1/ ρ0 ρc2 u

!

"

####

$

%

&&&&


The eigenvector of

r1 =100

!

"

###

$

%

&&&; r2 =

ρ2c12ρc2

!

"

###

$

%

&&&; r3 =

− ρ2c

12

− ρc2

!

"

###

$

%

&&&

The Euler Equations

C =u ρ 00 u 1/ ρ0 ρc2 u

!

"

####

$

%

&&&&

are (using arbitrary scaling

l1 =100

!

"

###

$

%

&&&; l2 =

011

!

"

###

$

%

&&&; l3 =

− 1c2

1ρc

− 1ρc

!

"

###

$

%

&&&

Right eigenvectors Left eigenvectorsCrp = λprp CT l p = λpl p


Form the matrices

QC =

1 ρ2c − ρ

2c

0 12

12

0 ρc2 − ρc

2

"

#

$$$$

%

&

''''

The Euler Equations

∂w∂t

+C∂w∂x

= 0Then

C =u ρ 00 u 1/ ρ0 ρc2 u

!

"

####

$

%

&&&&

QC−1 =

1 0 − 1c2

0 1 1ρc

0 1 − 1ρc

"

#

$$$$

%

&

''''

QC−1CQC =

u 0 00 u+ c 00 0 u− c

"

#

$$$

%

&

'''

w =ρ

up

!

"

###

$

%

&&&


Multiply

The Euler Equations

QC−1 ∂w∂t

+QC−1C∂w

∂x= 0

Which can be written as

Λ =QC−1CQC =

u 0 00 u+ c 00 0 u− c

#

$

%%%

&

'

(((

∂w∂t

+C∂w∂x

= 0

To get the characteristic form

∂v∂t+Λ

∂v∂x

= 0∂v0∂t

+u∂v0∂x

= 0

∂v+∂t

+ u+ c( )∂v+∂x

= 0

∂v−∂t

+ u− c( )∂v−∂x

= 0

or


Although the Riemann problem can be solved for the general case when the fluids are not stationary, the solution is expensive (involving solving a nonlinear equation for the pressure ratio across the shock) and the only information that we need from the solution is the flux across the cell boundary.

Therefore, usually we use approximate Riemann solvers. Different approximations are possible but generally the result is a complex process. Here we will only outline it briefly

The Euler Equations

computational fluid dynamics the euler equationsgtryggva/cfd-course2017/lecture-10-2017.pdf ·...

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