where the * denotes a dimensional quantity.

CIS888.11V/EE894R/ME894V A Case Study in Computational Science & Engineering

0AuP2u

)1(RT

xA1

2u

)1(RT

t

x)T(R

xP

x)Au(

A1

t)u(

0x

)Au(A1

t

***2****

**

2****

*

*

**

*

*

*

*2**

**

**

*

***

**

*

We will apply several numerical methods to find a steady state solution of the unsteady quasi-1D equations of frictionless compressible flow.

For adiabatic, frictionless flow, i.e. isentropic flow, the governing equations in dimensional form are:

where the * denotes a dimensional quantity.


Non-dimensionalization of governing equations

Non-dimensionalize using:P = P*/Pref; Pref = 1 Lref = 1 cm.T = T*/Tref; Tref = 1 tref = Lref/uref

/ref; ref = Pref/Rtref = 1/R t = t*/tref

u = u*/uref

RRTu ref2ref

2ref

*

LAA


Non-dimensionalized governing equations

Since tref = Lref/uref, this simply reduces to:

Similarly, the momentum and energy equations reduce to:

0xuA

ALu

tt ref

refref

ref

ref

0xuA

A1

t

x

)T(1x

AuA1

t)u( 2

02

Au)1(uATxA

12u)1(T

t

32


MacCormack’s explicit method

Before proceeding with this numerical scheme, it is necessary to introduce artificial dissipation into each of the governing equations for stability:

2

2a

xxuA

A1

t

2

2a

2

xu

x)T(1

xAu

A1

t)u(

2

2a

32

xT

2Au)1(uAT

xA1

2u)1(T

t


MacCormack’s explicit method

Discretize the domain x [0,L] into N interior points and two boundary points labelled 0 and N+1, each spaced x apart:

Next, the predictor and corrector parts of the algorithm are applied at each time step for each governing equation, at all interior points.

… …

0 1 2 i N N+1


Predictor equations (MacCormack’s explicit method)

n1i

ni

n1i2

na

ini

ni1i

n1i

n1i

i

ni

1ni

2)x(

)t(

AuAu)x(A

)t(

n

1ini

n1i2

na

ni

ni

n1i

n1i

i2n

ini1i

2n1i

n1i

ini

ni

1ni

1ni

uu2u)x(

)t(

TT)x(

)t(

AuAu)x(A

)t(uu


Predictor equations (MacCormack’s explicit method) - contd.

n1i

ni

n1i2

na

i3n

inii

ni

ni

ni

1i3n

1in

1i1in

1in

1in

1ii

2ni

ni

ni

ni

21ni

1ni

1ni

1ni

TT2T)x(

)t(

Au2

)1(ATu

Au2

)1(ATu)x(A

)t(u

2)1(T

u2

)1(T


Corrector equations

1n1i

1ni

1n1i2

1na

1i1n

1i1n

1ii1n

i1n

ii

1ni

ni

1ni

2)x(

)t(

AuAu)x(A

)t(21

1n1i

1ni

1n1i2

1na1n

1i1n

1i1n

i1n

i

1i21n

1i1n

1ii21n

i1n

ii

1ni

1ni

ni

ni

1ni

1ni

uu2u)x(

)t(TT

)x()t(

AuAu)x(A

)t(uu21

u


Corrector equations - contd.

1n1i

1ni

1n1i2

1na

1i31n

1i1n

1i

1i1n

1i1n

1i1n

1i

i31n

i1n

ii1n

i1n

i1n

ii

21ni

1ni

1ni

1ni

2ni

ni

ni

ni

21ni

1ni

1ni

1ni

TT2T)x(

)t(

Au2

)1(

ATu

Au2

)1(ATu)x(A

)t(

u2

)1(Tu2

)1(T21

u2

)1(T


Coefficient of artificial viscosity

One form for the artificial viscosity is as follows:

for the predictor step, and

for the corrector step.)x(uC n

ini

na

)x(uC 1ni

1ni

1na

• C=0.125 is a recommended value for this algorithm, for the non-dimensionalization shown in slide #2 of this lecture.

• Note that there is nothing special about this value of 0.125 - you are encouraged to try different values as well as different expressions for a and explore their effects on the final solution as well as on the convergence to the final steady state solution.


Initial Conditions

• For obtaining supersonic solutions in the case study problem, you may prescribe any linearly decreasing pressure distribution p(x) from the inlet (say starting at a non-dimensional value of 1) to (say 0.015 at) the exit.

• Similarly, a linear profile can be prescribed for T(x) varying from 1 (non-dimensional) at the inlet to 0.2 at the exit.

• Linear for u(x) starting from 0.15 at inlet to 1.2 at exit. (x) can be calculated from P(x)/T(x).• None of these initial profiles are unique, but their

judicious specification can accelerate the solution to the desired steady state solution.


Boundary Conditions• Inlet: Total pressure Po and Total temperature To are specified.– Implicitly extrapolate for u(i=0):

– Determine T(i=0) from total temperature To and extrapolated value of u(i=0):

)2i(u)1i(u2)0i(u

0)2i(u)1i(u2)0i(u0x

u

inlet2

2

2

)0i(u)1(T)0i(TT2

u)1(T2

oo2


Boundary Conditions (contd.)

– Before finding (i=0), first find P(i=0) from assumption that flow is isentropic (i.e. loss-free) from supply to inlet, and from specification of Po and To:

)1(

oo

)1(oo

o

o

TTP)0i(P

TT

PP

TPandPP

– Now determine (i=0) from:

)0i(T)0i(P)0i(


Boundary Conditions (contd.)• Exit:

– To find supersonic solution, implicitly extrapolate for all computed variables:

)1Ni(u)Ni(u2)1Ni(u0)1Ni(u)Ni(u2)1Ni(u0

xu

exit2

2

)1Ni()Ni(2)1Ni(0)1Ni()Ni(2)1Ni(0

x exit2

2

)1Ni(T)Ni(T2)1Ni(T0)1Ni(T)Ni(T2)1Ni(T0

xT

exit2

2


Boundary Conditions (contd.)– Pressure is determined from the equation of state:

P(i=N+1) = (i=N+1)T(i=N+1)• Note that a supersonic steady state solution can also be

obtained by prescribing a pressure at the exit plane that is low enough. What the exact value of this pressure would be cannot be determined a priori for more complicated problems.

• Specification of a Pexit that is higher than this value for supersonic flow results in the formation of standing normal shock waves in the diverging portion of the C-D nozzle.

• You are encouraged to experiment with these various boundary conditions in your homework assignments.


Courant condition• Using Fourier stability analysis, it can be shown

that this scheme is stable provided a(t)/(x) 1.

• The quantity a(t)/(x) is called the Courant number and a(t)/(x) 1 is called the Courant condition, where a is the relevant speed of sound in the problem.


FTBS (Forward in Time, Backward in Space) Scheme

Given equations of the form , we forward difference in time and backward difference in the spatial coordinate:

This is valid for a > 0. For a < 0, a forward difference in space must be used. This FTBS scheme is also called an upwind scheme.

xFa

tG

n1i

ni

ni

1ni

n1i

ni

ni

1ni

FF)x()t(aGG

)x(FF

a)t(GG


FTBS method applied to Quasi-1D flow

•Discretize the domain as before with MacCormack’s method (slide#5).

•Next, apply FTBS to the non-dimensionalized equations, and add artificial dissipation to each equation:

n

1ini

n1i2

na

1in

1in

1iini

ni

ini

1ni

2)x(

)t(

AuAu)x(A

)t(


FTBS method - contd.

n

1ini

n1i2

na

n1i

n1i

ni

ni

i2n

1in

1ii2n

ini

ini

ni

1ni

1ni

uu2u)x(

)t(

TT)x(

)t(

AuAu)x(A

)t(uu


FTBS method - contd.

n1i

ni

n1i2

na

1i3n

1in

1i1in

1in

1in

1i

i3n

inii

ni

ni

ni

i

2ni

ni

ni

ni

21ni

1ni

1ni

1ni

TT2T)x(

)t(

Au2

)1(ATu

Au2

)1(ATu)x(A

)t(u

2)1(T

u2

)1(T


Coefficient of artificial viscosity

A form for the artificial viscosity is as follows:)x(uC n

ini

na

• C ranging from 0.125 to 3 is recommended for this algorithm, for the non-dimensionalization shown in slide #2 of this lecture, and may be required to vary with spatial location as well.

• Note that there is nothing special about this range of values for C - you are encouraged to try different values as well as different expressions for a and explore their effects on the final solution as well as on the convergence to the final steady state solution.


Stability, ICs and BCs

• As with the MacCormack scheme, a Fourier stability analysis shows that the Courant condition a(t)/(x) 1 must be satisfied for stable time-marching using the FTBS or first order upwind explicit scheme.

• Similar stability analysis for the FTCS (Forward in Time, Central differenced in Space) is always unstable.

• The BTCS or Backward in Time, Central differenced in Space is unconditionally stable.

• Initial conditions are same as described in slide #11.• Boundary conditions are same as described in slides #12-#15.


• There are problems involving multiple time scales, where stiffness of the set of equations forces you to take (t) smaller than the maximum (t) allowed by the Courant condition for explicit methods.

• This makes computational time immense, so that it is necessary to use implicit methods for such multi-time scale problems.

• Note that time steps in implicit methods are not strictly governed by the Courant condition.


Detour on StiffnessConsider the initial value problemwith y(0)=1.Suppose we apply a simple explicit method such as Euler’s Forward Method:

The exact solution to this problem is of course

ydtdy

nnn1n

nn

n1n

y)t(1y)t(yy

ydtdy

)t(yy

te)t(y


Let us define the error, , as

.tttwherey)t(y)t(

y)t(y)t(

n1n1n1n1n

solutioncomputed

n

solutionexact

nn

)methodnumericalofstabilityondependent(ErroropagationPr

n

)schemenumericalofaccuracyondependent(ErrorTruncation

n)t(

nnn)t(

1n

n)tt(

1n

)t()t(1)t(y)t(1e)t(y)t()t(1)t(ye)t(or

y)t(1e)t( n


Now suppose =106 and tn=2.763x10-5 Suppose the error tolerance is desired to be

and that

we only have to satisfy

provided we do not amplify the previous error (tn). 1263.27n 10~e)t(y

41n 10)t(

4n 10x5.0~)t(

8n

4)t( 10

)t(y10))t(1(e

100tor10)t(or

10))t(1(e8

8)t(


But this means that• Assuming this yn is only one component of a larger system,

this is unacceptable because the step size for the rest of the equations is determined by the time scale of a component which has decreased by 12 orders of magnitude (from 1 to 10-

12) and is probably of no further physical interest! • Another way to view this is that 1/ is the time constant. The

smaller the time constant, the smaller must be the time step just to maintain stability.

• Thus, for stiff equations, the step size depends almost entirely on the stability of the numerical method and not on the accuracy. This is best resolved by using implicit methods which are unconditionally stable.

!10x1tor210x5.0

10)t(1 64

4

where the * denotes a dimensional quantity.

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Transcript of where the * denotes a dimensional quantity.