up x the euler equations uep (/)gtryggva/cfd-course/2013-lecture-14.pdfcomputational fluid dynamics!...

Computational Fluid Dynamics

Grétar Tryggvason !Spring 2013!

http://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!


�

p = ρRT; e = e T( ); cv = de /dT

�

h = h T( ); cp = dh /dT

R = cp − cv; γ = cp /cv; p = γ −1( )ρe�

H = h + u2 /2; h = e + p /ρDefine!


The Euler equations for 1D flow:!

0)/(

2 =⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

++

∂∂+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

∂∂

ρρρρ

ρρρ

pEupu

u

xEu

t

�

E = e + u2 /2

Ideal Gas:!

where!


�

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!



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


�

∂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:!


Find!

Computational Fluid Dynamics Exact 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 Problem!Exact Solution!

Computational Fluid Dynamics The shock tube problem!

L! R!

Shock!Contact!Expansion Fan!

uL, pL, ρL! uR, pR, ρR!

L! R!3! 2!5!


L! R!

Shock!Contact!Expansion Fan!

Pressure!

Density!


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

given:!

�

pL , ρL , uL ,

�

pR , ρR , uR

L! R!2!3!4!

�

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! R!2!3!4!


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! R!2!3!4!

�

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! R!2!3!4!

�

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! R!2!3!4!

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

subsonic!supersonic !


0 100 200 300 400 5000

2

4

6

8

10x 104 pressure

Exact Solution:!

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!


Solve using second order Lax-Wendroff !


For the fluid-dynamic system of equations (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!Computational Fluid Dynamics

�

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!

L-W with artificial viscosity!

Lax Fredrich!

Leap Frog!

j-1 j j+1

Computational Fluid Dynamics L-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;


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



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


for istep=1:2000!for 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)= …..!end!time=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);end!for 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)) );!end!time=time+dt,istep!plot(p,'b','linewidth',2),title('pressure'); pause!if(time > 0.005)break,end!end!


Solve using 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!

Flux Splitting!


Here we solve the 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 upwind!Computational 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 2D flow the 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!


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


�

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

0 2 4 6 8 100

2

4

6

8

10x 104

PressurePressure

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

Final time: 0.005!

First order upwind!


Numerical solutions of the one-dimensional Euler equations! Upwind/flux splitting! Lax-Wendroff/artificial viscosity !

Summary!

up x the euler equations uep (/)gtryggva/cfd-course/2013-lecture-14.pdfcomputational fluid dynamics!...

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