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AA214B: NUMERICAL METHODS FOR COMPRESSIBLE FLOWS

AA214B: NUMERICAL METHODS FOR

COMPRESSIBLE FLOWS

Representative Model Problems

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Outline

1 Convection-Diffusion Equation

2 Burgers Equation

3 Inviscid Burgers Equation

4

Scalar Conservation LawsExpansion WavesCompression and Shock WavesContact Discontinuities

5 Riemann Problems1D Riemann Problems for the Euler EquationsRiemann Problems for the Linearized Euler Equations

6 Roes Approximate Riemann Solver for the Euler EquationsSecant ApproximationsRoe AveragesAlgorithm and Performance

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Convection-Diffusion Equation

Combines the diffusion and convection (or advection) equations todescribes physical phenomena where physical quantities aretransferred inside a physical system due to two processes, namely,diffusion and convection

Also referred to by different communities as the advection-diffusion

equation, the drift-diffusion equation, the Smoluchowski equation, orthe scalar transport equation

c

t+ (ac) = (Dc) + S

where c is the variable of interest (species concentration for masstransfer, temperature for heat transfer, ), D is the diffusivity (ordiffusion coefficient), a is the average velocity of the quantity that ismoving, and S describes sources or sinks of the quantity c

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Common simplifications

the diffusion coefficient is constant, there are no sources or sinks, andthe velocity field describes an incompressible flow ( a = v = 0)

ct

+ a c = D2c

in this form, the convection-diffusion equation combines bothparabolic and hyperbolic partial differential equationsstationary convection-diffusion equation

(Dc) (ac) + S = 0

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Why is it a good representative model problem?

for an incompressible flow, the Navier-Stokes equations can bewritten as

(v)

t + v

(v) =

2 (v)

+ (

f

p) (1)

where

2(v) =

2(vx), 2(vy),

2(vz)T

= (vx),

(vy),

(vz)

Tand f is a body force such as gravity

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Why is it a good representative model problem? (continue)

for an incompressible flow, the Navier-Stokes equations can bewritten as

(v)

t +v

(v

) =

2 (

v)

+ (

fp)

compare with the convection-diffusion equation when D is constantand the velocity field describes an incompressible flow ( v = 0)

c

t

+ a c = 2(Dc) + S

= the convection-diffusion equation mimics the incompressibleNavier-Stokes equations


AA21 N CA O S O CO SS OWS
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Burgers Equation

Dropping the pressure term from the incompressible Navier-Stokesequations (1) leads to

(v)

t+ v (v) = 2

(v)

+ f

In one-dimension and assuming that is constant, the aboveequation simplifies to Burgers equation (proposed in 1939 by thedutch scientist Johannes Martinus Burgers)

vx

t + vx

vx

x = 2vx

x2 + gx

where =

, gx =fx


AA214B NUMERICAL METHODS FOR COMPRESSIBLE FLOWS
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Burgers Equation

vx

t+ vx

vx

x=

2vx

x2+ gx

The above equation can be transformed into a linear parabolicequation using the Hopf-Cole transformation (vx = 2x) thensolved exactly

This allows one to compare numerically obtained solutions of thisnonlinear equation with the exact one

For all these reasons, the Burgers equation is often used toinvestigate the quality of a proposed CFD scheme


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Inviscid Burgers Equation

For = 0 and gx = 0, the Burgers equation simplifies to

vx

t+ vx

vx

x= 0

which is known as the inviscid Burgers equationIt is a prototype for equations whose solution can developdiscontinuities (shock waves)

It can be solved by the method of characteristics

It can be written in strong conservation law form as follows

vx

t+

(v2x

2 )

x= 0



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Consider the following inviscid Burgers problem

vx

t+

(v2x

2 )

x= 0

vx(x, 0) =

vxL if x < 0vxR if x > 0

(2)

consider now scaling x and t by a constant > 0

x = x, t = t, > 0

since

t = t, and x = xthe inviscid Burgers equation is not affected by this scalingfurthermore, since the initial condition depends only on the sign of x,it is not affected by the above scaling

=

the inviscid Burgers problem defined above is scale invariant


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Scale invariance often implies the risk of multiple solutionsif vx(x, t) is the solution of problem (2), then u(x, t) = vx(x, t) isalso a solution of problem (2) for any > 0hence, desiring uniqueness of the solution of the above problem isdesiring u vx that is

vx(x, t) = vx( xt

)

this implies that the solution vx(x, t) is constant on the rays x = ct(it is said to be self-similar)in a homework, it will be shown that more precisely, the solution of

problem (2) isvx(x, t) = vx(

x

t) =

x

t

this solution is called a rarefaction wave centered at the origin(x = t = 0)


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In many circumstances, the uniqueness of the solution is enforced byimposing the condition that characteristics must impinge on adiscontinuity from both sides, which is known as the Lax EntropyCondition

consider a shock located along the curve x = (t) and traveling atthe speed V = dx

dt= d

dt

let vx(t) and vx+ (t) denote the left and right limits of the solutionvx(x, t) of problem (2), respectivelythe Lax Entropy Condition states that

vx+(t) < V < vx(t)

in particular, the Lax Entropy Condition states that the solutionmust jump downfor problem (2), it can be shown that for > 0, the solution jumpsup at the discontinuity: Thus, the only admissible solution that is,the solution in which any shock satisfies the Lax Entropy Condition is the continuous solution which has no shock


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Scalar Conservation Laws

Scalar conservation laws are simple scalar models of the Eulerequations

They can be written in strong conservation form as

ut

+ f(u)x

= 0 (3)

Their integral form in the space-time domain [x1, x2] [t1, t2] is

x2x1

[u(x, t2

) u(x, t1

)] dx +t

2

t1 [f(u(x2, t)) f(u(x1, t))] dt = 0(4)


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The solutions of the integral form (4) may contain jumpdiscontinuities: In this case, the discontinuous solutions are calledweak solutions of the differential form (3)

Jump discontinuities in the differential form (3) must satisfy a jumpcondition derived from the integral form: For a jump discontinuitytraveling at a speed V, the jump condition is

f(u+) f(u) = f(u)+ = V(u+ u) (5)

and therefore is analogous to the Rankine-Hugoniot relations


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Using chain rule, the non conservation form (or wave speed form) ofa scalar conservation law is

u

t + a(u)u

x = 0

where

a(u) =df

du

a(u) is called the wave speed


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Examples

f(u) =u2

2

Burgers equationf(u) = au linear advection

f(u) = u2

u2+c(1u)2, where c is a constant Bucky-Leverett equation

which is a simple model of two-phase flow in a porous medium


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Expansion Waves

Scalar conservation laws support features analogous to simpleexpansion waves

For scalar conservation laws, an expansion wave (or a rarefactionwave) is any region in which the wave speed a(u) increases from leftto right

a (u(x, t)) a (u(y, t)) , b1(t) x y b2(t)

A centered expansion fan is an expansion wave where allcharacteristics originate at a single point in the x t plane hence,the solution of problem (2) is a centered expansion fan

Centered expansion fans must originate in the initial conditions or atintersections between shocks or contacts (see definitions below)


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Compression and Shock Waves

Scalar conservation laws support features analogous to simple

compression and shock waves

For scalar conservation laws, a compression wave is any region inwhich the wave speed a(u) decreases from left to right

a (u(x, t))

a (u(y, t)) , b1(t)

x

y

b2(t)

A centered compression fan is a compression wave where allcharacteristics converge on a single point in the x t planeThe converging characteristics in a compression wave musteventually intersect, creating a shock wave

A shock wave is a jump discontinuity governed by the jumpcondition (5): From the mean value theorem, it follows that

V =df

du() = a(), u u+


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Compression and Shock Waves

A shock wave may originate in a jump discontinuity in the initialconditions or it may form spontaneously from a smooth compressionwave

In addition to the jump condition (5), shock waves must satisfy

(think of the Lax Entropy Condition)

a(u) V a(u+)

If wave speeds are interpreted as slopes in the x t plane, then theabove equation implies that waves (characteristics) terminate onshocks and never originate in shocks (shocks only absorb waves they never emit waves)


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Contact Discontinuities

Scalar conservation laws support features analogous to the contactdiscontinuities

For scalar conservation laws, a contact discontinuity is a jump

discontinuity from u to u+ such that

a(u) = a(u+)

Like contacts in the Euler equations, contacts in scalar conservationlaws must originate in the initial conditions or at the intersections of

shocks.


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Riemann Problems

In the theory of hyperbolic equations, a Riemann problem (namedafter Bernhard Riemann) consists of a conservation law equippedwith uniform initial conditions on an infinite spatial domain, exceptfor a single jump discontinuity

In one-dimension (1D), for a hyperbolic problem governing the field

u, the Riemann problem centered on x = x0 and t = t0 has thefollowing initial conditions

u(x, 0) =

uL if x < x0uR if x > x0

For example, problem (2) is a Riemann problem

For convenience, the remainder of this chapter uses x0 = 0 andt0 = 0


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Riemann Problems

In 1D, the Riemann problem has an exact analytical solution for theEuler equations, scalar conservation laws, and any linear system ofequations

Furthermore, the solution is self-similar (or self-preserving): Itstretches uniformly in space as time increases but otherwise retainsits shape, so that u(x, t1) and u(x, t2) are similar to each otherfor any two times t1 and t2 in other words, the solution dependson the single variable x

trather than on x and t separately

The Riemann problem is very useful for the understanding of theEuler equations because shocks and rarefaction waves appear ascharacteristics in the solution

Riemann problems appear in a natural way in finite volume methodsfor the solution of equations of conservation laws due to thediscreteness of the grid: They give rise to the Riemann solvers whichare very popular in CFD


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Riemann Problems

1D Riemann Problems for the Euler Equations

Shock Tube

Consider a 1D tube containing two regions of stagnant fluid atdifferent pressures

Suppose that the two regions are initially separated by a rigiddiaphragm

Suppose that this diaphragm is instantly removed (for example, by asmall explosion)

pressure imbalance 1D unsteady flow containing a steadily moving

shock, a steadily moving simple centered expansion fan, and asteadily moving contact discontinuity separating the shock andexpansionthe shock, expansion, and contact separate regions of uniform flow


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Riemann Problems



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Riemann Problems


Shock Tube

The flow in a shock tube has always zero initial velocity

Removing this restriction, the shock tube problem becomes the

Riemann problem and thus is a special case of the Riemann problem

Major result:

like the shock tube problem, the Riemann problem may give rise to ashock, a simple centered expansion fan, and a contact separating theshock and expansion, and the shock, expansion, and contact

separate regions of uniform flowunlike the shock tube however, one or two of these waves may beabsent


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Riemann Problems


Governing Equations

W

t +

Fx

x = 0W = (, vx, E)

T, Fx =

vx, v2x + p, (E + p)vx

TW(x, 0) =

WL = (L, Lv2xL , EL)

T if x < 0WR = (R, Rv

2xR

, ER)T if x > 0

(6)

with p = ( 1)

E v2x2

, and the speed of sound c given by c2 = p

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Ri P bl

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Riemann Problems


Exact Solution

Next, consider the contact discontinuityby definition

vx2 = vx3 (10)

p2 = p3 (11)

Finally, consider the simple centered expansion fanrecall that a simple wave is a wave where all states lie on the sameintegral curve of one of the characteristic familiesrecall that for the 1D Euler equations, an expansion wave is a wavewhere the speeds vx c increase monotonically from left to right

recall that a simple wave in the entropy characteristic family is awave in which vx = cst and p = cst entropy waves cannot createexpansionsit follows that the simple centered expansion fan here is a simplecentered acoustic fan associated with the characteristic curvedx = (vx c)dt



Riemann Problems
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Riemann Problems


Finally, consider the simple centered expansion fan (continue)along the integral curve of a simple centered expansion fanassociated with the characteristic curve dx = (vx c)dt, the twoRiemann invariants 0 and + are constant

0 = s = cst p = and c = cst

1

2

Zdp

c=

2c

1+ cst

+ = vx +

2c

1 for dx = (vx + c)dt

(and

= vx2c

1for dx = (vx c)dt)

hence, along the integral curve of a simple centered expansion fan

associated with the characteristic curve dx = (vx

c)dt and on thischaracteristic curve

s = cst, vx +2c

1= cst, and vx

2c

1= cst

therefore in this flow region, all flow properties are constant anddx = (vx c)dt becomes the straight line x = (vx c)t + cst



Riemann Problems

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Riemann Problems


Finally, consider the simple centered expansion fan (continue)now, along the integral curve of a simple centered expansion fanassociated with the characteristic curve x = (vx c)t + cst

vx +2c

1= vx4 +

2c4 1

hence along this integral curve and on the characteristic curvex = (vx c)t c = vx

xt

the following holds

vx +2

1

vx

x

t

= vx4 +

2c4 1

=

8>>>:

vx(x, t) = 2+1`xt +

12 u4 + c4

c(x, t) = 2+1

`x

t+ 1

2u4 + c4

x

t

p = p4

c

c4

21

(12)



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Riemann Problems


Combine now the shock, contact, and expansion results to determinep2p1 across the shock in terms of the known ration p

4

p1 = pL

pR

simple wave condition vx +2c

1= cst implies

vx3 +2c3

1= vx4 +

2c4 1

(13)

from the third of equations (12) and (13) it follows that

vx3 = vx4 +2c4

1

"1

p3

p4

12

#(14)

from (10), (11) and (13) it follows that

vx2 = vx4 +2c4

1

"1

p2

p4

12

#= vx4 +

2c4 1

"1

p1

p4

p2

p1

12

#

(15)



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Riemann Problems


Solving equation (15) forp4p1 gives

p4

p1=

p2

p1

1 +

12c4

(vx4 vx2) 2

1

(16)

Finally, combining (8) and (16) delivers the nonlinear equation inp2

p1

p4

p1=

p2

p1

1 + 1

2c4

vx4 vx1 c1

p2p1 1

+12

p2p1

1 + 1

21

(17)which can be solved by a preferred numerical method to obtain p2

p1and therefore p2



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Riemann Problems


Once p2 is found, equation (8) gives vx2 , equation (7) gives c2, andequation (9) gives the speed of the shock V, which completelydetermines the state 2

Then, equations (10) and (11) give vx3 and p3 and equation (13)

gives c3, which completely determines state 3Finally, the first, second, and third of equations (12) deliver vx, c,and p inside the expansion, respectively

In some cases (depending on the values of WL and WR), theRiemann problem may yield only one or two waves, instead of three:

To a large extent, the solution procedure described above handlessuch cases automatically



Riemann Problems
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Riemann Problems

Riemann Problems for the Linearized Euler Equations

The exact solution of the Riemann problem (6) is (relatively)expensive because finding p2 requires solving the nonlinearequation (17)

To this effect, approximate Riemann problems are often constructedas surrogate Riemann problems for the Euler equations

Here, the family of approximate Riemann problems based on alinearization of problem (6) is considered in the general case of Ndimensions



Riemann Problems
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Consider the linear Riemann problem

W

t+ A

W

x= 0

W(x, 0) =

WL if x < 0WR if x > 0

(18)

where A is a constant N N matrix whose construction is discussedin the next section

Assume that A is diagonalizable

A = Q1Q, = diag (1,

, N)

and that ri and li, i = 1, ..., N are its right and left eigenvectors,respectively

Ari = iri, ATli = ili (or l

Ti A = il

Ti )



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In the linear case, the change to characteristic variables d = QdW

simplifies to = QW

and leads to the following characteristic form of problem (18)

t+

x= 0

(x, 0) =

L = QWL if x < 0R = QWR if x > 0

The individual form of the above problem is

i

t+ i

i

x= 0, i = 1, ..., N

i(x, 0) =

Li = l

Ti WL if x < 0

Ri = lTi WR if x > 0

(19)



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Riemann Problems
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Since i is constant, the solution of problem (19) is trivial: For

N = 3, it can be written as

(x, t) = (x

t) =

(L1 , L2 , L3)T if x

t< 3

(L1 , L2 , R3)T if 3 >>:

L = R32 1 if x

t< 3 < 2 < 1

L + 3 = R

21 if 3

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CA Aa214b Ch5