Solution of the St Venant Equations / Shallow-Water equations of open channel flow

transcript

Dr Andrew SleighSchool of Civil EngineeringUniversity of Leeds, UK

www.efm.leeds.ac.uk/CIVE/UChile

Shock Capturing Methods

Ability to examine extreme flows Changes between sub / super critical

Other techniques have trouble with trans-critical Steep wave front Front speed

Complex Wave interactions Alternative – shock fitting Good, but not as flexible

More recent

Developed from work on Euler equations in the aero-space where shock capturing is very important (and funding available)

1990s onwards Euler equations / Numerical schemes:

Roe, Osher, van Leer, LeVeque, Harten, Toro Shallow water equations

Toro, Garcia-Navarro, Alcrudo, Zhao

E.F. Toro. Riemann Solvers and Numerical Methods for Fluid Dynamics. Springer Verlag (2nd Ed.) 1999.

E.F. Toro. Shock-Capturing Methods for Free-Surface Flows. Wiley (2001)

E.F. Toro. Riemann Problems and the WAF Method for Solving Two-Dimensional Shallow Water Equations. Philosophical Transactions of the Royal Society of London. A338:43-68 1992.

Dam break problem

The dam break problem can be solved

It is in fact integral to the solution technique

Conservative Equations

As earlier, but use U for vector

USUFU xt

fo SSgAgI

I1 and I2

Trapezoidal channel Base width B, Side slope SL= Y/Z

Rectangular, SL = 0

Source term

dBhI L

)3/4(2

nQQS f

Rectangular Prismatic

Easily extendible to 2-d

Integrate over control volume V

fo SSgh

dUUSdtUFdxUVV

2-dimensions

In 2-d have extra term:

friction

USUGUFU yxt

SSghUS

22)3/1(

Normal Form

Consider the control volume V, border Ω

dUUSdUHdVUt VV

UGnUFnUGUFUH 21sincos

Rotation matrix

H(U) can be expressed

UFUGnUFnUH TT 121

cossin0

sincos0

cossin0

sincos0

Finite volume formulation

Consider the homogeneous form i.e. without source terms

And the rectangular control volume in x-t space

dtUFdxU

xi-1/2 xi+1/2

Fi-1/2Fi+1/2

xi-1/2 xi+1/2

Fi-1/2Fi+1/2

Finite Volume Formulation

The volume is bounded by xi+1/2 and xi-1/2 tn+1 and tn

The integral becomes

dttxUFdttxUFdxtxUdxtxU

,,,, 2/12/11

We define the integral averages

And the finite volume formulation becomes

dxtxUx

,1 dxtxU

dttxUFt

2/12/1 dttxUFt

2/12/1

2/12/11

Up to now there has been no approximation

The solution now depends on how we interpret the integral averages

In particular the inter-cell fluxes Fi+1/2 and F1-1/2

Finite Volume in 2-D

The 2-d integral equation is

H(u) is a function normal to the volume

dUUSdUHdVUt VV

dUGnUFndUHFs

Using the integral average

Where |V| is the volume (area) of the volume then

If the nodes and sides are labelled as :

Where Fns1 is normal flux for side 1 etc.

443322111 LFnLFnLFnLFn

tUU ssss

FV 2-D Rectangular Grid

For this grid

Solution reduces to

Fi-1/2Fi+1/2

Gj-1/2

Gj+1/2

i+1/2,j-1/2

i-1/2,j+1/2

i-1/2,j-1/2

i+1/2,j+1/2

i, jFi-1/2

Fi+1/2

Gj-1/2

Gj+1/2

i+1/2,j-1/2

i-1/2,j+1/2

i-1/2,j-1/2

i+1/2,j+1/2

2/1,2/1,,2/1,2/11

jijijijini

Flux Calculation

We need now to define the flux Many flux definitions could be used to that

satisfy the FV formulation

We will use Godunov flux (or Upwind flux)

Uses information from the wave structure of equations.

Godunov method Assume piecewise linear data states

Means that the flux calc is solution of local Riemann problem

i-1 i i-1

Fi+1/2Fi-1/2Cells

Data states

U(0)i+1/2U(0)i-1/2

i-1 i i-1

Fi+1/2Fi-1/2Cells

Data states

U(0)i+1/2U(0)i-1/2

Riemann Problem

The Riemann problem is a initial value problem defined by

Solve this to get the flux (at xi+1/2)

0 xt UFU

nxxifU

xxifUtxU

FV solution

We have now defined the integral averages of the FV formulation

The solution is fully defined First order in space and time

2/12/11

Dam Break Problem

The Riemann problem we have defined is a generalisation of the Dam Break Problem

Dam wall

Deep water at rest

Shallow water at rest

Dam wall

Deep water at rest

Shallow water at rest

Dam Break Solution Evolution of solution

Wave structure

Water levels at time t=t*

Velocity at time t=t*

Rarefaction

Exact Solution

Toro (1992) demonstrated an exact solution

Considering all possible wave structures a single non-linear algebraic equation gives solution.

Exact Solution

Consider the local Riemann problem

Wave structure

0 xt UFU

xifUxU

Right wave, (u+c)Left wave (u-c)

Star region, h*, u*

tShear wave

hL, uL, vL hR, uR, vR

Star region, h*, u*

tShear wave

PossibleWave structures

tRight Shock

Left Rarefaction Shear wave

Left Shock Right RarefactionShear wave

Right RarefacationLeft Rarefaction

Shear wave

Right ShockLeft Shock

Shear wave

tRight Shock

Shear wave

Across left and right wave h, u change v is constant

Across shear wave v changes, h, u constant

Determine which wave

Which wave is present is determined by the change in data states thus:

h* > hL left wave is a shock h* ≤ hL left wave is a rarefaction

h* > hR right wave is a shock h* ≤ hR right wave is a rarefaction

Solution Procedure

Construct this equation

And solve iteratively for h (=h*). The functions may change each iteration

uhhfhhfhf LLLL ,,

f(h) The function f(h) is defined

And u*

uhhfhhfhf LLLL ,,

shockhhifhh

nrarefactiohhifghgh

shockhhifhh

nrarefactiohhifghgh

LR uuu

LLRRRL hhfhhfuuu ,,2

Iterative solution

The function is well behaved and solution by Newton-Raphson is fast (2 or 3 iterations)

One problem – if negative depth calculated! This is a dry-bed problem. Check with depth positivity condition:

RLLR ccuuu 2

Dry–Bed solution

Dry bed on one side of the Riemann problem

Dry bed evolves Wave structure is

different.

Wet bed

Dry bed

Wet bed

Dry bed

Wet bed

Dry bed

Wet bed

Dry bed

Dry-Bed Solution

Solutions are explicit Need to identify which applies – (simple to do)

Dry bed to right

Dry bed to left

Dry bed evolves h* = 0 and u* = 0 Fails depth positivity test

LL cuc 23

1* LL cuu 2

RR cuc 23

1* RR cuu 2

gch /2**

Shear wave

The solution for the shear wave is straight forward. If vL > 0 v* = vL

Else v* = vR

Can now calculate inter-cell flux from h*, u* and v* For any initial conditions

Complete Solution

The h*, u* and v* are sufficient for the Flux But can use solution further to develop exact

solution at any time. i.e. Can provide a set of benchmark solution

Useful for testing numerical solutions.

Choose some difficult problems and test your numerical code again exact solution

Complete Solution

Rarefaction

Difficult Test Problems

Toro suggested 5 testsTest No. hL (m) uL (m/s) hR (m) uR (m/s)

1 1.0 2.5 0.1 0.0

2 1.0 -5.0 1.0 5.0

3 1.0 0.0 0.0 0.0

4 0.0 0.0 1.0 0.0

5 0.1 -3.0 0.1 3.0

Test 1: Left critical rarefaction and right shockTest 2: Two rarefactions and nearly dry bedTest 3: Right dry bed problemTest 4: Left dry bed problemTest 5: Evolution of a dry bed

Exact Solution

Consider the local Riemann problem

Wave structure

0 xt UFU

xifUxU

Star region, h*, u*

tShear wave

Star region, h*, u*

tShear wave

Returning to the Exact Solution We will see some other Riemann solvers that

use the wave speeds necessary for the exact solution.

Return to this to see where there come from

PossibleWave structures

tRight Shock

Shear wave

tRight Shock

Shear wave

Across left and right wave h, u change v is constant

Across shear wave v changes, h, u constant

Conditions across each wave

Left Rarefaction wave

Smooth change as move in x-direction Bounded by two (backward) characteristics Discontinuity at edges

Left bounding characteristic

hL, uL h*, u*

Right bounding characteristic

LL cudtdx /**/ cudtdx

Crossing the rarefaction

We cross on a forward characteristic

States are linked by:

constant2 cu

** 22 cucu LL

** 2 ccuu LL

Solution inside the left rarefaction The backward characteristic equation is For any line in the direction of the rarefaction

Crossing this the following applies:

Solving gives

On the t axis dx/dt = 0

cucu LL 22

dxcuc LL 2

dxcuu LL 2

LL cuc 23

1 LL cuu 2

Right rarefaction

Bounded by forward characteristics Cross it on a backward characteristic

In rarefaction

If Rarefaction crosses axis

RR cuc 23

** 22 cucu RR RR ccuu ** 2

dxcuc RR 2

dxcuu RR 22

RR cuu 23

Shock waves

Two constant data states are separated by a discontinuity or jump

Shock moving at speed Si Using Conservative flux for left shock

Conditions across shock

Rankine-Hugoniot condition

Entropy condition

λ1,2 are equivalent to characteristics. They tend towards being parallel at shock

LiL UUSUFUF **

*USU iiLi

Shock analysis

Change frame of reference, add Si

Rankine-Hugoniot gives

LLL Suu ˆLSuu **ˆ

ghuhghuh

Shock analysis

Mass flux conserved

From eqn 1

LLL uhuhM ˆˆ**

** /ˆ hMu L LLL hMu /ˆ

LLL hhhhgM **2

LL uuuu ** ˆˆ

Left Shock Equation

Equating gives

LLL hhfuu ,**

LLLL hh

hhghhhhf

LLLL qauS

Right Shock Equation

Similar analysis gives

RRR hhfuu ,**

RRRR hh

hhghhhhf

RRRR qauS

Complete equation

Equating the left and right equations for u*

Which is the iterative of the function of Toro

LLL hhfuu ,** RRR hhfuu ,**

0,, LRLLLR hhfhhfuu

0,, *** uhhfhhfhf LRLL

Approximate Riemann Solvers No need to use exact solution Expensive

Iterative When other equations, exact may not exist

Many solvers Some more popular than others

Toro Two Rarefaction Solver

Assume two rarefactions Take the left and right equations

Solving gives

For critical rarefaction use solution earlier

** 2 ccuu LL RR ccuu ** 2

24*RLRL ccuu

Toro Primitive Variable Solver Writing the equations in primitive variables

Non conservative

Approximate A(W) by a constant matrix

Gives solution

0 xt WWAW

LR WWW 2

RLLRRL

hhuuhhh

RLLRRL

cchhuuu

Toro Two Shock Solver

Assuming the two waves are shocks

Use two rarefaction solver to give h0

RLRRLL

uuhqhqh

LLRRRL qhhqhhuuu *** 2

LoL hh

RoR hh

Roe’s Solver

Originally developed for Euler equations Approximate governing equations with:

Where is obtained by Roe averaging

xtxtxt UAUAUUUFU~

~RLhhh

1~RL ccc

Roe’s Solver

Properties of matrix Eigen values

Right eigen vectors

Wave strengths

Flux given by

cu ~~~1 cu ~~~

R ~~1~ )1(

R ~~1~ )2(

LR hhh

nii FFF R

HLL Solver

Harten, Lax, van Leer Assume wave speed Construct volume Integrate round

Alternative gives flux

FRFL F*

0* LRRLRRLL tUtUUxxUxUx

LRRLLLRR SSFFUSUSU /*

LRLRRLRLLR SSUUSSFSFSF /*

HLL Solver

What wave speeds to use? Free to choose One option:

For dry bed (right)

Simple, but robust

TRTRLLL cucuS ,min TRTRRRR cucuS ,min

LLL cuS RRR cuS 2

Higher Order in Space

Construct Riemann problem using cells further away

Danger of oscillations Piecewise

reconstruction

Limiters

i-1 i i+1 i+2

Limiters

Obtain a gradient for variable in cell i, Δi

Gradient obtained from Limiter functions Provide gradients at cell face

Limiter Δi =G(a,b)

iiL xUU 2

1 iiR xUU

iii xx

Limiters

A general limiter

β=1 give MINMOD β=2 give SUPERBEE

van Leer

Other SUPERBEE expression s = sign

0,max,,max,0min

0,min,,min,0max,

aforbaba

aforbababaG

bababaG

asbasbsbaG 2,min,,2min,0max,

Higher order in time Needs to advance half time step MUSCL-Hancock

Primitive variable Limit variable Evolve the cell face values 1/2t:

Update as normal solving the Riemann problem using evolved WL, WR

0 xt WWAW

tWWWAWW

2/12/11

Alternative time advance

Alternatively, evolve the cell values using

Solve as normal

Procedure is 2nd order in space and time

tUU WW

2/12/11

Boundary Conditions

Set flux on boundary Directly Ghost cell

Wall u, v = 0. Ghost cell un+1=-un

Transmissive Ghost cell hn+1 = hn

un+1 = un

Wet / Dry Fonts

Wet / Dry fronts are difficult Source of error Source of instability

Common near tidal boundaries Flooding - inundation

Dry front speed

We examined earlier dry bed problem

Front is fast Faster than characteristic. Can cause problem with time-step / Courant

Wet bed Dry bed

Solutions

The most popular way is to artificially wet bed

Give a small depth, zero velocity Loose a bit of mass and/or momentum

Can drastically affect front speed E.g. a 1.0 dambeak with 1cm gives 38% error 1mm gives 25% error – try it!

Conservation Errors

The conserved variable are h and hu

Often require u

Need to be very careful about divide by zero

Artificial dry-bed depths could cause this

Source Terms

“Lumped” in to one term and integrated Attempts at “upwinding source” Current time-step

Could use the half step value E.g.

2/1CRK

Main Problem is Slope Term

Flat still water over uneven bed starts to move.

Problem with discretisation ofx

zzgh lr

i-1 i i+ 1

d a tu m le v e l

b e d le v e l

w a te r su r fa c e

iz1izlz1iz

1ihihlh1ih

Discretisation

Discretised momentum eqn

For flat, still water

Require

rlini zzgh

tghghhuhu

lni zh

lil zh

A solution

Assume a “datum” depth, measure down

Momentum eqn: Flat surface

zhg ii

zzhzzhg liirii

2 liiriirli zzhzzhhhx

liil zzhh riir zzhh

Some example solutions

Weighted mesh gives more detail for same number of cells

Dam Break - CADAM Channel with 90° bend

Secondary Shocks

Solution of the St Venant Equations / Shallow-Water equations of open channel flow

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