chapter 5: compressible flows · 2 1d compressible flows and euler equations 3 riemann problem...

Chapter 5: Compressible FlowsME5343 CFD

Dr. Stanley Ling

Department of Mechanical EngineeringBaylor University

April 23, 2020

Dr. Stanley Ling (Baylor University) Chapter 5: Compressible Flows April 23, 2020 1 / 60

Outline

1 Governing Equations As Conservation Laws

2 1D Compressible Flows and Euler Equations

3 Riemann Problem & Godunov’s Method

4 Approximate Riemann Solvers

5 Riemann-Solver-Free Methods

6 Reconstruction & Higher-Order Finite-Volume Methods


Basic laws of fluid motion

Assuming a single fluid (not single phase), the three-dimensionalmotion of the fluid is governed by the conservation laws:

∂ρ

∂t+∂ρu

∂x+∂ρv

∂y+∂ρw

∂z= 0 (1)

∂ρu

∂t+∂ρuu

∂x+∂ρuv

∂y+∂ρuw

∂z= −∂p

∂x+∂τxx∂x

+∂τyx∂y

+∂τzx∂z

(2)

∂ρv

∂t+∂ρvu

∂x+∂ρvv

∂y+∂ρvw

∂z= −∂p

∂y+∂τxy∂x

+∂τyy∂y

+∂τzy∂z

(3)

∂ρu

∂t+∂ρwu

∂x+∂ρwv

∂y+∂ρww

∂z= −∂p

∂z+∂τxz∂x

+∂τyz∂y

+∂τzz∂z

(4)

∂ρE

∂t+∂ρEu

∂x+∂ρEv

∂y+∂ρEw

∂z= −

(∂pu

∂x+∂pv

∂y+∂pw

∂z

)+∂(uτxx + vτyx + wτzx)

∂x+∂(uτxy + vτyy + wτzy )

∂y

+∂(uτxz + vτyz + wτzz)

∂z−(∂qx∂x

+∂qy∂y

+∂qz∂z

)(5)


Compact form

Using vector representation, we can rewrite the equations as:

∂ρ

∂t+∇ · ρu = 0 (6)

∂ρu∂t

+∇ · ρuu = −∇p +∇ · τ (7)

∂ρE

∂t+∇ · ρEu = −∇ · (pu) +∇ · (τ · u)−∇ · q (8)

where the total energy E = e + |u|2/2 is the sum of internal andkinetic energy.

The three equations represent the conservation of mass, momentum,and energy, respectively.

Recall momentum is a vector, so Eq. (13) represent three equationsactually.

Recall tensor product between two vectors uu = uiuj gives a tensor.


Conservative form

Rewriting the equations in the following conservative form will notonly make it more compact, but also reveals its conservative nature.

∂

∂t

ρρuρE

+∇ ·

ρuρuu + pI − τρuH − τ · u + q

= 0

where H = E + p/ρ is the total Enthalpy.

An even more compact form can be written as

∂U∂t

+∇ · F = 0 (9)

where

U =

ρρuρE

, and the fluxes F =

ρuρuu + pI − τρuH − τ · u + q


Equation of state

The viscous stress and heat fluxes are given by the constitutiverelations. For Newtonian fluids,

τ = µ(∇u +∇uT ) , q = −κ∇T (10)

The conservation laws for Newtonian fluids are also referred to as theNavier-Stokes Equations.

Fluid properties: µ, κ, which are considered as known.

Primary variables: ρ, u, v ,w , e

Secondary (dependent) variables: p,T , c

There are 8 unknowns and only 5 equations. Three more equationsare required to close the system, which are provided by the thermalrelations, such as the equations of state (EOS).

For calorically perfect gas, the 3 additional equations are given as

e = CvT , e =p

(γ − 1)ρ, c =

√γp/ρ (11)


Governing equations for incompressible flows

All fluid flows are compressible to some extent, yet when the Machnumber M = |u|/c → 0, one can approximate the flow asincompressible.

The governing equations for Incompressible flows is an asymptoticlimit of the general conservation laws listed above:

∇ · u = 0 (12)

∂ρu∂t

+∇ · ρuu = −∇p +∇ · τ (13)

Important differences compared to the general N-S equations:The mass conservation becomes the divergence free condition(∇ · u = 0);The energy equation is decoupled from the fluid motion. The energyequation can be solved for temperature if needed.Pressure p losses its physical meaning as a thermal property, and servesas an approximation of the real pressure and solved by the Poissonequation.


Compressible flows: inviscid vs viscous

For compressible flows, the general N-S equations are to be solved

All three conservation laws are coupled;Pressure p remains as a thermal property, and its relation with internalenergy is given by the equation of state.

Inviscid and viscous regions in a flow field. Viscous effects areimportant only in the boundary layer.

Important flow compressible flow features, such as shock andexpansion waves are located in the inviscid region.

The inviscid limit of the N-S equation is the Euler Equations

The inviscid solution outside the boundary layer serves as theboundary conditions for the boundary-layer equations.

Even with the computer power today, it is still often too expensive tosimulate the whole flow domain as a viscous flow.


Euler equations for inviscid compressible flows

The conservative form is the same as the general N-S equation

∂U∂t

+∇ · F = 0

where

U =

ρρuρE

but the fluxes change to

F =

ρuρuu + pIρuH


1D Euler equations

For one-dimensional compressible flows, the Euler equations can bewritten as

∂U∂t

+∂F∂x

= 0 (14)

where

U =

ρρuρe

but the fluxes change to

F =

ρuρu2 + pρuH


Quasi-linear form of 1D Euler equations

The Euler equations can be written in quasi-linear form as

∂U∂t

+ A(U)∂U∂x

= 0 (15)

where A is the Jacobian matrix, defined as

A =∂F∂U

=

0 1 0−1

2 (γ − 3)u2 (3− γ)u γ − 1u[ 1

2 (γ − 1)u2 − H]2 H − (γ − 1)u2 γu

We will perform the eigendecomposition of the Jacobian matrix A as

A = EΛE−1, or Λ = E−1AE (16)

where E and Λ are the eigenvector and diagonal eigenvalue matrices,respectively.


Characteristic form of 1D Euler equations

The eigenvalues and eigenvectors for A can be solved, then thediagonal eigenvalue matrix can be written as

Λ =

u − c 0 00 u 00 0 u + c

namely λ1 = u − c , λ2 = u, and λ3 = u + c.

The eigenvector matrix can be written as

E =

↑ ↑ ↑e1 e2 e3

↓ ↓ ↓

=

1 1 1u − c u u + cH − uc 1

2u2 H + uc


Characteristic form of 1D Euler equations (continued)

Then we use the eigenvector matrix to transform for the solution tothe eigensapce,

U = EW (17)

then

E∂W∂t

+ AE∂W∂x

= 0

∂W∂t

+ Λ∂W∂x

= 0 (18)

which is the characteristic form of the Euler Equations

The key feature of the characteristic form is that, the three PDE’s aredecoupled

∂Wj

∂t+ λj

∂Wj

∂x= 0 (19)

where j = 1, 2, 3 and take the form of the advection equation!

In other words, the eigen-transformed solution W are advectedindependently as three different characteristic waves


Characteristic waves for 1D Euler equations

A schematic of the characteristic waves (or sometimes simply referredto as characteristics) is given as below

x

t

1u + c

1u

1u − c

x

1

t

u + c1

u1u − c

(a) Subsonic flows (b) Supersonic flows


Characteristic method for solving 1D Euler equations

After we obtained the solution on the eigenspace, namely W , we cantransform it back to U ,

Ui = EjiWj (20)

or

U =3∑

j=1

Wje j (21)

Similarly, for the flux

F =3∑i

Wiλje j (22)

And finally, the Euler equation as

3∑i

(∂

∂tWje j +

∂

∂xWiλje j

)= 0 (23)


Finite-Volume Method

The basic idea of the finite-volume method (FVM) is to solve thegoverning equations in the integral form

Two important features: it enforces conservation and requires lesssmoothness of the solution, which makes it appealing to compressibleflows, especially for those with discontinuities.

FVM for 1D Euler equation

∆xduidt

+ Fi+ 12− Fi− 1

2= 0 (24)

where Fi− 12, Fi+ 1

2are the fluxes from the left and right cell surfaces.

ui represent the mean of the solution within the i cell. Note that ui isdifferent from the solution at location xi , though sometimes it issimply denoted by ui . This is a fundamental difference between thefinite-difference and finite-volume methods.


Flux Calculation

First-order FVM simply considers the solution in the cell is constant,while higher-order method (order n) will assume the solution varies asa polynomial of order n − 1.In FVM, reconstructing the solution within the cell based on themean value is referred to as reconstructionThe reconstructed solutions on the left and right sides of a cell surfaceis different in general, which ends up to be the Riemann problem.This is a challenge but also an opportunity.

x

u

i i + 1i − 1i + 1

2i − 12

ui−1 uiui+1

ui−1

ui−1ui+1

Fi−1/2 Fi+1/21st order2nd order

Exact


The Riemann Problem

The Riemann problem for the Euler equation can be stated as

∂U∂t

+∂F (U)

∂x= 0 (25)

where

U(x , 0) =

{UL, if x < 0

UR , ifx > 0

It is a generalization of the 1D shock-tube problem.


Shock Tube

The shock tube is an instrument used to replicate of shock waves.

The shock consists of a driver (high pressure) gas and a driven (lowpressure) gas, separated by a diaphragm. When the diaphragm isremoved, the pressure difference drives fluid to move and alsogenerates the expansion fan, contact, and shock.

x

t

Driver Gas Driven Gas

ρ4, p4 ρ0, p0 u4 = u0 = 0Initial Conditions:Diaphragm

o

ShockContactExpansion Fan

ρ ρ4

ρ0

Tail

HeadTail Contact Shock

(4)(3) (2) (1) (0)

(4) (3) (2)(1) (0)

t = t1

t = t1

t = 0


Characteristics and Riemann’s Invariants

Recall the characteristic form of the Euler equations

∂Wj

∂t+ λj

∂Wj

∂x= 0 (26)

Wj remains unchanged as it is advected along the correspondingcharacteristics, and thus are referred to as Riemann invariants. If weuse a simply expression for entropy s = γp/ρ, we obtains

W1 = W+ = u +2c

γ − 1λ1 = λ+ = u + c (27)

W2 = W− = u − 2c

γ − 1λ2 = λ− = u − c (28)

W3 = s λ3 = u (29)

The first two characteristics represent sound waves that propagate tothe left and right; while the third is fluid particle trajectory.


Characteristics and Riemann’s Invariants (continued)

Another way to derive the Riemann’s invariants is to write the Eulerequations in terms of ρ, u, s as

∂ρ

∂t+ u

∂ρ

∂x+ ρ

∂u

∂x= 0 (30)

ρ∂u

∂t+ ρu

∂u

∂x+∂p

∂x= 0 (31)

∂s

∂t+ u

∂s

∂x= 0 (32)

Multiply Eq. (30) with c2 and Eq. (31) with ±c , we can get[∂

∂t+ (u ± c)

∂

∂x

]p ± ρc

[∂

∂t+ (u ± c)

∂

∂x

]u = 0 (33)

introducing the derivative operator along the positive and negativecharacteristics as D±,

D±p ± ρcD±u = 0 (34)


Characteristics and Riemann’s Invariants (continued)

The differential equation can be integrated to get the Riemanninvariants,

W± =

∫ (D±p

ρc± D±u

)=

∫cD±ρ

ρ± u (35)

The isentropic relation between ρ and c is given asp/ργ = c2/(γργ−1) = const, the derivative of which gives

2

γ − 1dc =

c

ρdρ (36)

Therefore

W± =

∫2

γ − 1D±c ± u =

2c

γ − 1± u (37)


Characteristics and Solution for the Shock-Tube Problem

x

t

Driver Gas Driven Gas

ρ4, p4 ρ0, p0 u4 = u0 = 0Initial Conditions:Diaphragm

o


ρ ρ4

ρ0

Tail


λ− λ+

u

u4 = 0 u0 = 0


p p4

p0


x

(4)(3) (2) (1) (0)

(4) (3) (2)(1) (0)

u2 = u1

p2 = p1 = 0


Solution for the Expansion Fan

Isentropic relations:

p4

ργ4=

p3

ργ3=

p2

ργ2(38)

Riemann invariants (dx/dt = λ+):

u4 +2c4

γ − 1= u3 +

2c3

γ − 1= u2 +

2c2

γ − 1

Combine them we can obtain

u2 = u4 −2c4

γ − 1

[(p2

p4

) γ−12γ

− 1

](39)


Solution within the Expansion Fan

The waves in region (3) are negative characteristic (λ−), so

dx/dt = x/t = u − c , (40)

Furthermore, the Riemann invariant W+ remains constant along thepositive characteristics λ+ crossing the expansion fan, so

u4 +2c4

γ − 1= u +

2c

γ − 1(41)

The above two equations can be used to solve for u and c , and thenp and ρ can be determined

u

c4=

2

γ + 1

(1 +

γ − 1

2

u4

c4+

x

c4t

)(42)

c

c4=γ − 1

γ + 1

(u4

c4− x

c4t

)+

2

γ + 1(43)

ρ

ρ4=

[2

γ + 1+γ − 1

γ + 1

(u4

c4− x

c4t

)] 2γ−1

,p

p4=

(ρ

ρ4

)γ(44)


Solution for the Shock

Transfer reference frame to the shock, namelyu0 = u0 − us , u1 = u1 − usThen the flow becomes steady and the Euler equations can beintegrated to get the Rankine-Hugoinot relations

ρ0u0 = ρ1u1 (45)

ρ0u20 + p0 = ρ1u

21 + p1 (46)

ρ0H0u0 = ρ1H1u01 (47)

where H = e + 1/2u2 + p/ρ.From Eq. (46)

u1 = u0 −p1 − p0

m(48)

where the mass flow rate

m = ρ0u0 = ρ1u1 , (49)

according to Eq. (45).


Solution for the Shock (continued)

The mass flow rate can also be expressed in terms of p and ρ fromEq. (46):

m2 =p1 − p01ρ1− 1

ρ0

(50)

The density ratio across the shock can be expressed in terms for p as

ρ1

ρ0=

γ−1γ+1 + p1

p0

γ−1γ+1

p1p0

+ 1(51)

Combine the two relations above, we obtain

m =

(p1 + γ−1

γ+1p0

2γ+1

1ρ0

)1/2

(52)


Solution for the Shock (continued)

Since u2 (Eq. (39)) is equal to u1 (Eqs. (48) and (52)), and p1 = p2

at the contact, we finally obtain

u4 −2c4

γ − 1

[(p1

p4

) γ−12γ

− 1

]= u0 −

p1 − p0(p1+ γ−1

γ+1p0

2γ+1

1ρ0

)1/2(53)

where p4, u4, p0, u0, ρ0 are given by the initial conditions. Forshocktube, u4 = u0 = 0. Eq. (53) can be solved for the only unknownp1. Then use p1 to obtain other unknown propertiesp2, u2 = u1 = uc , ρ2, ρ1, us .

A useful numerical solution for the shocktube problem:http://silver.neep.wisc.edu/˜shock/tools/gdcalc.html


Full solution for the Riemann Problem

x

t

o


(a) Left expansion fan & right shockx

t

o

Shock Contact Expansion Fan

(b) Left shock & right expansion fan

x

t

o

Expansion Fan

(c) Left expansion fan & right expansion fan

Expansion Fan

x

t

o

ShockContact

ShockContact

(d) Left shock & right shock

ρL, uL, pL ρR, uR, pR ρL, uL, pL ρR, uR, pR

ρL, uL, pL ρR, uR, pR ρL, uL, pL ρR, uR, pR


Full solution for the Riemann Problem (continued)

Left expansion fan:

u∗ = uL −2cLγ − 1

[(p∗pL

) γ−12γ

− 1

](54)

Right expansion fan:

u∗ = uR +2cRγ − 1

[(p∗pR

) γ−12γ

− 1

](55)

Left shock:

u∗ = uL − (p ∗ −pL)

(p ∗+γ−1

γ+1pL2

γ+11ρL

)−1/2

(56)

Right shock:

u∗ = uR + (p ∗ −pR)

(p ∗+γ−1

γ+1pR2

γ+11ρR

)−1/2

(57)


Full solution for the Riemann Problem (continued)

Case (a): Left expansion fan and right shock: Eqs. (54) and (57)

Case (b): Left shock and right expansion fan: Eqs. (55) and (56)

Case (c): Left expansion fan and right expansion fan: Eqs. (54) and(55)

Case (d): Left shock and right shock: Eqs. (56) and (57)


The Godunov’s Method

Godonov’s method is the foundation for FVM for compressible flows.The basic idea is to use the exact solution of the Riemann problem tosolve the Euler equations numerically.

x

t

i i + 1i − 1i + 1

2i − 12

ui ui+1ui−1

Fi−1/2 Fi+1/2

tn

tn+1


The Godunov’s First Method

Godonov’s first method:

Un+1i =

1

∆x

∫ xi+1/2

xi−1/2

U(x , tn+1)dx (58)

where U(x , tn+1) is the exact solution of the Riemann problem usingUn

i as the initial conditions at each cell surfaces.

To avoid wave interaction, ∆t ≤ ∆x/2amax

, where amax is the maximumwave speed.

x

t

i i + 1i − 1i + 1

2i − 12

ui ui+1ui−1

Fi−1/2 Fi+1/2

tn

tn+1


The Godunov’s Second Method

Let’s first integrate the Euler equations over a grid cell and one timestep: ∫ tn+1

tn

∫ xi+1/2

xi−1/2

∂U∂t

dxdt +

∫ tn+1

tn

∫ xi+1/2

xi−1/2

∂F (U)

∂xdxdt = 0

(59)∫ xi+1/2

xi−1/2

(Un+1 −Un)dx +

∫ tn+1

tn

[F (U i+1/2)− F (U i−1/2)

]dt = 0

(60)

Then we employ the cell average solution and the time average flux as

U i =

∫ xi+1/2

xi−1/2

Udx (61)

F (U i±1/2) =

∫ tn+1

tnF (U i±1/2)dt (62)


The Godunov’s Second Method (continued)

Then we obtain

U in+1

= U in − ∆t

∆x

[F (U i+1/2)− F (U i−1/2)

](63)

The key of the Godunov’s second method is to use the exact solutionof the Riemann problem at x/t = 0 to evaluate F (U i±1/2).

As shown in figure above, F (U i±1/2) are constant in time, so

U in+1

= U in − ∆t

∆x

[F (U i+1/2)− F (U i−1/2)

](64)

Since we only need the solution at xi+1/2, as long as it is notinfluenced by the waves from the neighboring cell surfaces xi−1/2 orxi+3/2, it will be fine. Therefore, the time constraint will be

∆t ≤ ∆x

amax(65)


The Godunov’s First and Second Methods

The second method is better than the first method, because

The second method is less expensive, since it does not requireintegration of the exact solution as in the first method. Actually in thesecond method we don’t even need to determine the entire structure ofthe solution;The second method allows a bigger time step.

Therefore, the second method is sometimes simply referred to as theGodunov’s method.

These methods were introduced in Godunov’s PhD Dissertation“Difference Methods for Shock Waves” in 1954, which was arevolutionary breakthrough in CFD research for compressible flows.


Exact vs Approximate Riemann Solvers

Exact Riemann solvers: Godunov’s methods. Its application is limiteddue to the computational complexity & expense

Most of the information obtained by solving the Riemann problemexactly is never used in the flux computation

Approximate Riemann solvers: linearized Riemann solvers

∂U∂t

+ A∂U∂x

= 0 (66)

Linearization is fulfilled by replacing the original Jacobian matrix A(U)that depends on U , by the matrix A(UL,UR) that depends only onthe left and right states.Constructing the linearized Jacobian matrix A is the key.Several important requirements must be satisfied.


Essential Features of Approximate Jacobian Marix

Similar eigenstructures. The matrix A must have three realeigenvalues which are similar to A, retaining the hyperbolic nature ofthe Euler equation and the wave pattern of Riemann problem.

Consistency. A reduces to A when the difference between the leftand right states is small, namely

A(UL = U ,UR = U) = A(U) (67)

Correct speeds for discontinuities. Roe (1981) suggested that, toretain the correct speed of a discontinuity,

A(UL,UR)(UR −UL) = FR − F L , (68)

which ensures that the approximate Jacobian matrix leads to aconservative method. Note that F L = F (UL) and FR = F (UR)


Propagation speed of an isolated discontinuity

Let’s take the shock as an example to show why Eq. (68) canguarantee the correct speed.

x

u

xL

FL FR

i xRxs

us

Consider the Euler equation a cell contains a shock, using FVM,

d

dt

∫ xR

xL

Udx + FR − F L = 0 (69)

Since the solution jumps across the shock location xs(t), we dividethe integration into two parts

d

dt

∫ xs(t)

xL

Udx +d

dt

∫ xR

xs(t)Udt + FR − F L = 0 (70)


Propagation speed of an isolated discontinuity (continued)

Since the integration limits are time dependent, we use the Lebniz’stheorem,∫ xs(t)

xL

∂U∂t

dt + U(xs,L(t), t)dxsdt

+∫ xR

xs(t)

∂U∂t

dt −U(xs,R(t), t)dxsdt

+ FR − F L = 0 (71)

ULus −URus + FR − F L = 0 , so us(UR −UL) = FR − F L (72)

Similar derivations can be done on the linearized equation,

d

dt

∫ xR

xL

Udt + A(UR −UL) = 0 , so us(UR −UL) = A(UR −UL)

(73)

To ensure the same us , Eq. (68) is reuqired


Approximate Riemann Solver of Roe

Two naive constructions of A(UL,UR)

A = A(

1

2(UL + UR)

)

A =1

2

(A(UL) + A(UR)

)do not satisfy the essential requirements listed above.

To derive A, we first consider the parameterization

U(θ) = UL + θ(UR −UL) (74)

where 0 ≤ θ ≤ 1 and dU/dθ = UR −UL


Approximate Riemann Solver of Roe (continued)

The flux difference can be written as

FR − F L =

∫ 1

0

dF (U(θ))

dθdθ =

∫ 1

0

dF (U(θ))

dUdUdθ

dθ

=

(∫ 1

0

dF (U(θ))

dUdθ

)(UR −UL) . (75)

According to Eq. (68)

A =

∫ 1

0

dF (U(θ))

dUdθ . (76)

The above integration is not available in closed form in general.However, Roe found a clever way to integrate and to find the explicitexpressions of A.



A new parameter vector is introduced, z(θ) = zL + θ(zR − zL), and

z =

z1

z2

z3

=√ρ

1uH

The variable U and flux F can be expressed in terms of z as

U =

z21

z1z2z1z3γ + γ−1

2γ z22

, F =

z1z2γ−1γ z1z3 + γ+1

2γ z22

z2z3

Then the variable and flux differences can be expressed as

UR −UL =

(∫ 1

0

dU(z(θ))

dzdθ

)(zR − zL) = B(zR − zL) , (77)

FR − F L =

(∫ 1

0

dF (z(θ))

dzdθ

)(zR − zL) = C (zR − zL) , (78)



Each term in the matrices can be integrated to yield

B =

2z1 0 0z2

2 z1 0z3γ

γ−1γ z2

z1γ

, C =

z2 z1 0γ−1γ z3

γ+1γ z2

γ−1γ z1

0 z3 z2

,where z = (zL + zR)/2.

Then the approximate Jacobian matrix can be identified as

A = C B−1

=

0 1 012 (γ − 3)( z2

z1)2 (3− γ)( z2

z1) γ − 1

− z2z3z1

+ (γ − 1)( z2z1

)3 γ z3z1− 3

2 (γ − 1)( z2z1

)2 γ( z2z1

)


Approximate Jacobian Matrix

Finally the approximate Jacobian matrix can be written as

A =

0 1 012 (γ − 3)u2 (3− γ)u γ − 1

u[ 12 (γ − 1)u2 − H]2 H − (γ − 1)u2 γu

The approximate matrix A looks identical to the exact A, by replacingthe exact u and H by the Roe’s averages variables

u =

√ρLuL +

√ρRuR√

ρL +√ρR

(79)

H =

√ρLHL +

√ρRHR√

ρL +√ρR

(80)


Flux Calculation With the Approximate Jacobian Matrix

Now we are going to use A to calculate the flux at the cell surface.The exact and linear Riemann problems, we define two controlvolumes as follows on the left and right of the t-axes:

x

t

o


(a) Exact Riemann Problemx

t

o

Left Contact Right

(b) Linearized Riemann ProblemUL, F(UL)

F(UR)UL, F(UL) UR, F(UR)

F(UL)UR, F(UR)

F0,LF0,R F(UR)F(UL)

F0,LF0,RΔt

For the left control volume for the exact Riemann problem, we applythe integral form of the Euler equation (Eq. (60)), the fluxes on theleft and right surfaces are

1

∆t

∫ tn+1

tnF (UL)dt = F (UL),

1

∆t

∫ tn+1

tnF (U(0−, t))dt = F 0,L (81)

Recall that for the Gundonov’s second method, F 0,L is average flux atx = 0, and is what we would need eventually.


Flux Calculation (Continued)

For the right control volume for the exact Riemann problem, the leftand right fluxes

1

∆t

∫ tn+1

tnF (U(0+, t))dt = F 0,R ,

1

∆t

∫ tn+1

tnF (UR)dt = F (UR),

(82)

Similarly, we can obtain the fluxes for the linearized Riemann problem:For the left control volume, the left and right fluxes

1

∆t

∫ tn+1

tnF (UL)dt = F (UL),

1

∆t

∫ tn+1

tnF (U(0−, t))dt = F 0,L

(83)

For the right control volume, the left and right fluxes

1

∆t

∫ tn+1

tnF (U(0+, t))dt = F 0,R ,

1

∆t

∫ tn+1

tnF (UR)dt = F (UR),

(84)


Flux Calculation (Continued 2)

Now If we want to keep the change the variables, namely∫ 0−

xL(Un+1 −Un)dx and

∫ xR0+ (Un+1 −Un)dx for the left and right

control volumes, the same for both the exact and linearized Riemannproblems, then the flux differences must be the same

F 0,L − F (UL) = F 0,L − F (UL) (85)

F (UR)− F 0,R = F (UR)− F 0,R (86)

To be consistent, we required

F 0,L = F 0,R = F 0, F 0,L = F 0,R = F 0 (87)

Then we have

F 0 = F (UL) + F 0 − F (UL) = F (UL) + A(UL,U(0))(U(0)−UL)(88)

F 0 = F (UR) + F 0 − F (UR) = F (UR) + A(UL,U(0))(U(0)−UR)(89)


Eigenstructure of Approximate Jacobian Matrix of Roe

The beauty of the approximate matrix of Roe, A, is that it almostidentical eigenvalue as the exact Jacobian matrix A

λ1 = u − c , λ2 = u, λ3 = u + c (90)

where

c =

√(γ − 1)

(H − 1

2u2

)(91)

Similarly the right eigenvector matrix is

E =

↑ ↑ ↑e1 e2 e3

↓ ↓ ↓

=

1 1 1u − c u u + c

H − uc 12 u

2 H + uc

Recall what have obtained for A, everything is the same exceptreplacing ( ) by ( )


Eigenstructure of Approximate Jacobian Matrix of Roe

If we apply the eigentransform on solution differences between the leftand right states, leading to an expansion in terms of the eigenvectors:

UR −UL = E α =3∑

i=1

αi e i , (92)

recall that e i are the eigenvectors for the approximate Jacobian A.

x

t

o

Left Contact Right

UL

UR

UR − ULα1e1

α2e2

α3e3

xo

U

U(0) − UL = ∑λi≤0

αiei

UR − U(0) = ∑λi≥0

αiei

U(0)

Left-going waves

Right-going waves


Flux Calculation (Continued 3)

Plug the eigen-expansion back to the flux calculation

A(UL,U(0))(U(0)−UL) = A(UL,U(0))∑λi≤0

αi e i =∑λi≤0

λi αi e i

(93)

A(UL,U(0))(UR −U(0)) = A(UR ,U(0))∑λi≥0

αi e i =∑λi≥0

λi αi e i

(94)

Recall that A∑αie i = EΛE−1Eα = EΛα =

∑λiαie i

Therefore, the fluxes become

F 0 = F (UL) +∑λi≤0

λi αi e i (95)

F 0 = F (UR)−∑λi≥0

λi αi e i (96)


Roe Flux

Since the above two fluxes calculated from the left and right statesmust be the same for consistency, so

F 0 =1

2

(F (UL) + F (UR)

)− 1

2

3∑i

|λi |αi e i (97)

which is known as the Roe’s flux.

The first term represents a central difference approximation of theflux.

The second term provides numerical dissipation based on the linearwaves. Without this term, the central difference approximation of theflux is not stable.

The use of Roe’s flux is easy, since explicit expressions have beengiven for λi and e i . Only αi need to be computed based on thesolutions on the left and right states.


Further Remarks on Approximate Riemann Solvers

Linearized Riemann solvers do not recognize unphysical “expansionshock”, therefore, the entropy-fix method is required.

Another very popular approximate Riemann solver is the HLL(Harten, Lax, and van Leer) family of solvers. The decompositions ofUR −UL are not rigorously derived from the eigenvectors andeigenvalues as Roe. Instead some simpler physical arguments are usedto “guess? these values. That is why HLL solvers have many differentversions, such as HLLE, HLLC, and others

AUSM (Advection Upstream Splitting Method). The AUSMrecognizes that the inviscid flux consist of two physically distinctparts, i.e., convective and pressure fluxes. The former is associatedwith the flow (advection) speed, while the latter with the acousticspeed. Both are formulated using the eigenvalues of the Jacobianmatrix.


HLLC Flux

The HLLC flux (Toro 1993) is expressed as

F 0 =

F (UL), if 0 ≤ SL

F (UL) + SL(U∗L −UL), if SL ≤ 0 ≤ S∗

F (UR) + SR(U∗R −UR), if S∗ ≤ 0 ≤ SR

F (UR), if SR ≤ 0

The wave speed for the contact S∗ is estimated as

S∗ =(pR − pL) + ρLuL(SL − uL)− ρRuR(SR − uR)

ρL(SL − uL)− ρR(SR − uR)(98)

The solution in region * are

U∗K = ρK

(SK − uKSK − S∗

) 1S∗

EKρK

+ (S∗ − uK )[S∗ + pK

ρK (SK−uK )

]


HLLC Flux (continued)

The left and right wave speeds SL and SR are estimated based on thepressure p∗

SL = uL − cLqL , SR = uR − cRqR (99)

where

qK =

1, if p∗ ≤ pK[1 + γ+1

2γ

(p∗pK− 1)]1/2

, if p∗ > pK

where K = L and R, and the pressure p∗ is estimated as

p∗ = max(0, p − ρc(uR − uL)/2) (100)

where () = (()L + ()R)/2

Note that here SL and SR are NOT simply speeds of characteristics!They actually represent speeds of the head of the expansion fan andthe shock waves


Riemann-Solver-Free Methods

There exist also another family of methods, the so-calledRiemann-solver-free methods, which do not require to solve theRiemann solvers, even approximately.

Recall the example or the Roe’s method, the additional term in theflux beyond the central difference approximation is introduced toprovide numerical dissipation to stabilize the method.

Efforts have been made to construct this dissipation term in adifferent way, such as the Lax-Friedrichs method and theKurganov-Tadmor (KT) methods

These methods can handle scalar, vector, and combined problems.Therefore, it is easier to be used in conservation laws other than Eulerequations, such as the shallow-water equations.


Kurganov-Tadmor Central Scheme

The Kurganov-Tadmor (KT) flux is given as

F 0 =1

2

(F (UL) + F (UR)

)− cmax

2(UR −UL) (101)

where cmax is the maximum local wave propagation speeds, which arenothing but the eigenvalues of the Jacobian matrixes on the left andright states, namely

λmax = [|λi ,R |, |λi ,L|] ; cmax = sign(λ)λmax (102)

Compare the KT and Roe fluxes, the first terms are identical, thedifference is on the second term which provides the numericaldissipation.


Reconstruction

For FVM, if we assume the solution to be constant, as the meanwithin a cell, then the method is only first order.

If we want a second order method, then we need to allow the solutionto vary linearly with respect to the mean within the cell, with aconstant gradient.

x

u

i i + 1i − 1i + 1

2i − 12

ui−1 uiui+1

ui−1

ui−1ui+1

Fi−1/2 Fi+1/21st order2nd order

Exact

How to calculate the gradient? Using a central differenceapproximation will lead to an unstable method.


Slope Limiters & 2nd Order TVD Scheme

The gradients computed by central differencing will lead to significantnumerical oscillations which will eventually blow up the simulation.

In order to stabilize the method, we have to modify the gradientscomputed by central differencing, by the slope limiters technique.

The ultimate goal is to use numerical viscosity to damp the numericaloscillations to achieve the total variation diminishing (TVD)condition.

Recall the amplitude (dissipation) and phase (dispersion) errors inspatial discretization. The fundamental idea here is to use dissipationerrors to damp the oscillations induced by dispersion errors.(We let a bad guy to kill the other bad guy!)

The first 2nd order TVD scheme is the MUSCL (MonotonicUpstream-centered Scheme for Conservation Laws) schemeintroduced in the seminal paper of van Leer (1979)

The accuracy of a TVD discretization is only 2nd order for smoothregion, and will degrade to first order near discontinuities .


ENO/WENO Schemes

The essentially non-oscillatory (ENO) weighted ENO (WENO)methods provide a different way to evaluate the gradientThe methods first calculate the slope based on finite differencing

sc,i = (ui+1 − ui+1)/2∆x (103)

ENO method picks the minimum slopes from the three candidates,

si = min(sc,i−1, sc,i , sc,i+1) (104)

WENO method uses the weights that biased toward the small slopes

si = wi−1sc,i−1 + wi sc,i + wi+1sc,i+1 (105)

x

u

i i + 1i − 1i + 1

2i − 12

i − 2 i + 2

Exact

ui

sc,i−1

sc,i sc,i+1 sc,i+2

sc,i−2 ui−1

ui+1


chapter 5: compressible flows · 2 1d compressible flows and euler equations 3 riemann problem...

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