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Numerical Hydraulics Block 4 Numerical solution of open channel flow Markus Holzner 1

Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

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Page 1: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Numerical Hydraulics

Block 4 – Numerical solution

of open channel flow

Markus Holzner

1

Page 2: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Contents of the course

Block 1 – The equations

Block 2 – Computation of pressure surges

Block 3 – Open channel flow (flow in rivers)

Block 4 – Numerical solution of open channel flow

Block 5 – Transport of solutes in rivers

Block 6 – Heat transport in rivers

2

Page 3: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

- Finite Volume discretization

- Finite differences method

- Characteristics method

3

Numerical solution of

open channel flow

Page 4: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Basic equations of open channel flow in

variables h and v for rectangular channel

Continuity („Flux-conservative form“)

Momentum equation

4

0)v(

x

h

t

h

2 4/3

v (v)v

v v

2

S E

e hy

st hy

hg gI gI

t x x

hbI r

k r h b

Page 5: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Basic equations of open channel flow in

variables h and q for rectangular channel

Continuity

Momentum equation

5

0h q

t x

2

2 4/3 2

( / )( )

2

S E

e hy

st hy

q q h hgh gh I I

t x x

q q hbI r

k r h h b

Page 6: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Basic equations of open channel flow for

general cross-section in variables A and Q

Continuity

Momentum equation

6

Q0

A

t x

2

0

2 4/3 2

( / )1 1 (Q / )

Q Q

wsp

e

e hy

st hy u

A bQ Ag gI gI

A t A x x

AI r

k r A L

Page 7: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Boundary conditions

• At inflow boundary usually the inflow

hydrograph should be given

• At the outflow boundary we can use

– water level (also time variable e.g. for tide)

– water level-flow rate relation (e.g. weir formula)

– slope of water level or energy

• In supercritical flow two boundary conditions are

necessary for one boundary (for both v and h)

7

Page 8: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Boundary conditions

• Number of boundary conditions from number

of characteristics

In 1D: subcritical flow: IB: 1, OB: 1

superciritical flow: IB: 2, OB: 0

IB = Inflow boundary, OB = Outflow boundary

t t

8

Page 9: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Discretized basic equations in variables h

and v for rectangular channels

Momentum equation

Explicit method

i = 2,…, Nx

i = 2,…, Nx

Which differences to chose in

order to be able to build in lower

and upper boundary

conditions?

Continuity („Flux-conservative form“)

9

1 1( v v ) /new old old old old old

i i i i i ih h t h h x

2 2

1 10

1

2 4/3

1

v v ( )v v

2

v v

2

old old old oldnew old i i i ii i e

old old oldi i i

e hy old

st hy i

h ht g t t gI t gI

x x

h bI r

k r h b

Page 10: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Discretized basic equations in variables h

and v for rectangular channels

Boundary conditions (i = Nx+1) example

Explicit method

Boundary conditions (i = 1) example

10

1( ) . . v ( )q f h i e f h

h or weir formula

1 1 2 2 1 1 1 1( v v ) / v /( )new old old old old old new new new

inh h t h h x q b h

1

1

( )

v v

new

Nx 0

new new

Nx i

h = h weir

computation as i 1,...,Nx

Page 11: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Discretized basic equations in variables h

and v for rectangular channels

Explicit method

11

Explicit method requires stability condition

Courant-Friedrichs-Levy (CFL) criterium must be fulfilled:

c is the relative wave velocity with respect to average flow

/( v )t x c

( ) / ( )c gh gA h b h

Page 12: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Non-conservative form:

vv 0

h hh

t x x

v vv S E

hg I I g

t x x

v0

hh

t x

21

2S E

q qg I I g h

t h x h x

0h q

t x

2 2

2S E

q q ghgh I I

t x h

q = vh

12

Conservative form:

Page 13: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Matrix formulation of the last form of

the equations

13

Page 14: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Assignment

Determine the wave propagation (water surface profile,

maximum water depth, outflow hydrograph) for a rectangular

channel with the following data:

width b = 10 m, kstr = 20 m-1/3/s

length L = 10‘000 m, bottom slope IS=0.002

Inflow before wave, base flow Q0 = 20 m3/s

Boundary condition downstream: Weir with water depth 2.2 m

Boundary condition upstream: Inflow hydrograph

Inflow hydrograph Q (is added to base flow Q0):

Time (h) 0 0.5 1.0 1.5 2.0 2.5

Q (m3/s) 0 50 37.5 25 12.5 0

14

Page 15: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Inflow/Outflow hydrographs

time steps in 10s

Q (m3/s)

about 4 h

15

Page 16: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

1D Shallow water equations

- The total differential for v=v(x,t)

and h=h(x,t) is:

16

0h h v

v ht x x

S E

v v hv g I I g

t x x

Dv v v x

Dt t x t

Dv h h x

Dt t x t

Page 17: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Characteristic equations

• We multiply the first of the original equations

with a multiplier l and add the two equations up:

• To obtain total differentials in the brackets we

have to choose

17

S E

v v h g hv h v g I I

t x t xl l

l

1,2

g gB

h Al

Page 18: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

• Thus we obtain the characteristic equations:

along

18

along

S E

Dv g Dhg I I

Dt c Dt

S E

Dv g Dhg I I

Dt c Dt

dxv c

dt

dxv c

dt

Page 19: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

• Positive and negative characteristics for

sub-critical, critical and super-critical flow:

Types of characteristics

t t t P P P

C+ C+ C+ C-

C- C-

E E E W W W x x x

Terminology: P, W (West) und E (East) instead of i, i-1, i+1

19

Page 20: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Integration of the characteristic

equations

• Multiplication with dt and integration

– along characteristic line

– and along characteristic line

– yields: 20

P

W

ES

P

W

P

W

dtIIgdhc

gdv

P

E

ES

P

E

P

E

dtIIgdhc

gdv

dxv c

dt

dxv c

dt

Page 21: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

WPWESWP

W

WP ttIIghhc

gvv

EPEESEP

E

EP ttIIghhc

gvv

PWpP hCCv

P n E Pv C C h

WPWESW

W

Wp ttIIghc

gvC

EPEESE

E

En ttIIghc

gvC

( / ) ( / )E E W WC g c C g c

or

This implies a linearisation. The wave

velocity becomes constant in the element. 21

Integration of the characteristic

equations

Page 22: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Grid for subcritical flow (1)

Zeit

x

j

j+1 P

W E

Characteristics start on grid points

Terminology C (Center), W (West) und E (East) instead of i, i-1, i+1

C

Problem: Characteristics

intersect between grid points

in points P at time levels

which do not coincide with

the time levels of the grid.

Results have to be

interpolated.

22

Page 23: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Grid for subcritical flow (2)

Zeit

x

j

j+1

P

W E

Characteristic lines end at point P, starting points do not coincide with

grid points. Values at starting points are obtained by interpolation

from grid point values

C

We choose this

variant!

23

Terminology C (Center), W (West) und E (East) instead of i, i-1, i+1

Page 24: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Interpolation (left)

24

x

tcv

xx

xx

xx

xx

vv

vv LL

WC

LP

WC

LC

WC

LC

L LC L

C W

v c tc c

c c x

Page 25: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Interpolation (right)

25

R RC R C R P R

C E C E C E

v c tv v x x x x

v v x x x x x

R RC R

C E

v c tc c

c c x

Page 26: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Starting point L

• Solution for vL and cL yields:

Inrerpolating analogously for h:

26

WCWC

WCCWC

L

ccvvx

t

vcvcx

tv

v

1

WC

WCLC

L

ccx

t

ccx

tvc

c

1

WCLLCL hhcvx

thh

Page 27: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Starting point R for subcritical flow

• In analogy to point L, variables for point R

vR and cR

The method is an explicit method. The CFL-criterium is automatically fulfilled. 27

1

C E C C E

R

C E C E

tv c v c v

xvt

v v c cx

1

C R C E

R

C E

tc v c c

xct

c cx

R C R R C E

th h v c h h

x

Page 28: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Starting point for supercritical flow

• Starting point of characteristic between W

and C

• Using velocity v-c we obtain

28

WCWC

WCCWC

R

ccvvx

t

vcvcx

tv

v

1

CW

WCRC

R

ccx

t

ccx

tvc

c

1

WCRRCR hhcvx

thh

Page 29: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Final explicit working equations

tIIghc

gvC

LESL

L

Lp

tIIghc

gvC

RESR

R

Rn

P p L Pv C C h P n R Pv C C h

with

and

( / ) ( / )L L R RC g c C g c

Integration from L to P and from R to P

2 equations with

2 unknowns

Boundary conditions

are required as discussed

in FD method

29

Page 30: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Classical dam break problem:

Solution with method of

characteristics

Propagation velocity of fronts slightly too high

30

Page 31: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Matrix form of the St. Venant

equations (1D)

31

Page 32: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Finite volume method

• For simplicity we consider a system without source term. Integrating in space we obtain:

cell boundaries

32

Page 33: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

• Additionally, we integrate in time between tn and tn+1 :

This is the integral form of the equations.

33

Page 34: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

• Defining:

and

we can write:

34

Page 35: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

• Different schemes can be devised according to the apporach to express fluxes at the cell boundaries

• We distinguish:

- Centered schemes, which give equal weight to neighboring cells and do not need information on direction of wave propagation. They are easy to implement but lead to strong numerical diffusion;

- Upwind schemes, which use wave propagation to express fluxes. More difficult to implement but much more accurate.

Numerical schemes:

35

Page 36: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

• Fluxes are expressed as

• Method is first order accurate and very simple to implement. However, it leads to numerical diffusion that is too strong for practical applications.

• CFL criterion:

Centered scheme: Lax-Friedrichs

and

/( v )t x c 36

Page 37: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

• i+1/2 becomes i and i-1/2 becomes i-1 if the wave propagates in positive x-direction

• Method is more robust in the presence of shocks but numerical diffusion is still considerable.

• CFL criterion:

Simplest upwind scheme: explicit

upwinding

and

/( v )t x c

37

Page 38: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

38

Simplest upwind scheme: explicit

upwinding (2)

Page 39: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

State of the art upwind scheme:

Godunov

• Method requires the solution of the Riemann problem at every cell

boundary and on each time level.

• This amounts to calculating the solution in the regions that form

behind the non-linear waves developing in the Riemann problem as

well as the wave speeds necessary for deriving the complete wave

structure of the solution.

Idea: solve local Riemann

problems forward in time

39

Page 40: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Riemann problem

• Waves are either shocks

(solution is discontinuous)

or rarefactions (solution is

continuous)

• To find the exact solution

in the “star” region, we

need to establish

appropriate jump

conditions

40

Page 41: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Example: HLLE(*) solver

(*) Method is an approximate Riemann solver developed by Harten, Lax and van Leer (1983) and improved by

Einfeldt (1988)

It is assumed that two waves propagate in

opposite directions with velocities SL and SR,

generating a single state in between them:

How to compute UHLLE?

We start from the integral form of the equations (slide 33).

41

Page 42: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

HLLE (2)

Control volume for calculation

of HLLE flux

Integral form of equations:

which gives on the right hand side:

where and

The left hand side can be split into 3 integrals:

(*)

(**)

42

T

L

T

R

x

x

x

x

txdttx

dtdxxdxTxR

L

R

L

00),(()),((

))0,(),(

ufuf

uu

)(),( RLLLRR

x

xTxxdxTx

R

L

ffuuu

dxTxdxTxdxTxdxTxR

R

R

L

L

L

R

L

x

TS

x

x

TS

x

TS

TS ),(),(),(),( uuuu uuu

RRL

TS

TSLL TSxxTSdxTx

R

L

),(

Page 43: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

HLLE (3)

Combining (*) and (**) gives:

Dividing by the length we finally obtain:

Using this expression with the Rankine Hugoniot conditions (see lecture notes)

we get the expression for the HHLE flux to be used in the Godunov scheme:

43

R

L

TS

TSRLLLRR SSTdxTx ffuuu

),(

R

L

TS

TSLR

RLLLRR

LR

HLLE

SS

SSdxTx

SST

ffuuuu

),(1

R

HLLE

RR

HLLE

L

HLLE

LL

HLLE

S

S

uuff

uuff

LR

LRRLRLLRHLLE

SS

SSSS

)( uufff

Page 44: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

2D Shallow water equations

The 2D shallow water equations can be written

in matrix forms as:

44

Page 45: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Difference to 1D

• Additional variable (spec. flow in y-direction) and additional momentum equation.

• Fluxes in x-direction are formulated equal to the fluxes in 1D.

• Fluxes in y-direction are formulated in analgoy to fluxes in x-direction. In addition to e(est) and w(west) the indices n(north) und s(south) are introduced. The direction of upwinding is determined independently from the upwind direction in the x-coordinate.

• The conservation is over the whole element. I.e. source terms and fluxes over east/west and north/south boundaries enter the same balance. 45

Page 46: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Specialties of 2D modelling

• The 2D computational grid is always a projection on the

horizontal plane. Different position of nodes in z-direction

influence the source term (gravity).

• 2D elements are more general, i.e. only in the case of

rectangular elements the 2D problem can be divided into two 1D

problems.

• In the case of general elements (e.g. triangular elements) the

fluxes orthogonal to the element sides must be used. They are

decomposed into components along the orthogonal x/y-

directions.

• Approximately rectangular elements improve computational

accuracy.

• If element sizes vary strongly the numerical error increases fast.

In that case higher order schemes have to be used. 46

Page 47: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Flows with free water surface (Navier-Stokes approach, vertically 2D or 3D)

• For the solution of the partial differential equations the

domain has to be discretized. This is not immediately

possible as the position of the surface is not a priori known.

An iterative procedures is necessary.

• There are different ways to tackle the problem:

– Surface Tracking: the grid follows the free surface.

– Solution of an additional advection equation i.e. on a fixed

grid information is transported with the advective flow. The

information is the water contents of a cell (Fraction of volume

- FOV) or the distance from a datum to the surface (LS).

– On a fixed grid particles are moved convectively with the flow

(Marker in cell - MAC).

47

Page 48: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

48

Page 49: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Comments on the

Navier-Stokes approach

• Additional non-linearity means additional computational effort and the danger of non-convergence.

• The discretisation is considerably more difficult and requires efficient grid generators. Inappropriate discretisation may lead to wrong solutions.

• Numerical diffusion can smooth out the position of the surface.

• The solution of the flow is considerable better if vertically curved streamlines exist.

• The Navier-Stokes approach is more for local phenomena, shallow equation approach for global flow phenomena.

49

Page 50: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

HEC-RAS

50

Page 51: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Conceptual model of HEC

channel

floodplain

Storage

(without flow)

1D but still taking into account channel and floodplains

51

Page 52: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

How to stay 1D in energy, momentum,

and piezometric head?

z = elevation of water surface

is the same for channel and

flood plains

= hp in our nomenclature

52

Page 53: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Some definitions

Conveyance K for each subdivision:

• If there is no foodplain, the model describes what we did so far

• If there is a floodplain, the channel is subdivided into several

sections with the same water level but possibly different friction

coefficients

Total flow:

subdivision

53

3/2

,

1ihyi

i

iEii rAn

KwithIKQ

N

i

iQQ1

Page 54: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Equations for channel and floodplain

lc

f

f

f

fc

c

c

qqt

S

t

A

x

Q

qt

A

x

Q

Indices: f = floodplain, c = channel

cfE

f

f

f

fff

fcE

c

c

c

ccc

MIx

zgA

x

Qv

t

Q

MIx

zgA

x

Qv

t

Q

,

,

)(

)(

Continuity equations:

Momentum equations:

z = elevation of the water surface, ql=lateral inflow, S area of non-conveying cross-sec.

qf = flow from floodplain to channel (per length),

qc = flow from channel to floodplain (per length) ,

Mf, Mc corresponding momentum fluxes (per length)

The right hand sides are eliminated by adding the equations for channel

and floodplain 54

Page 55: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Equations combined in 1D

Final 1D equations

The unknowns are Q and z (= hp)

Ac, Af and S are known functions of z

IE,c and IE,f are known functions of z and Q

55

0)/)1(()/(

0))1(()(

,,

2222

fE

f

fcE

c

c

f

f

c

c

fc

Ix

zgAI

x

zgA

x

AQ

x

AQ

t

Q

x

Q

x

Q

t

A

)/(

)1(

fcc

fc

fc

KKK

withQQQQQ

SAAA

)( workpreviousourinx

hI

x

z

hzz

S

bottom

Page 56: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Equations in difference form

Momentum equation

with

Continuity equation

velocity coefficient

equivalent x-coordinate

56

Qv

QvQvAAA

xIAxIAxAI

ffcc

fc

ffEfccEcEE

,,

0.

fE

EEE

ffccI

x

zgA

x

vQ

xt

xQxQ

0

lff

f

cc Qx

t

Sx

t

Ax

t

AQ

)()()1(

5.0

11

11,

1

1

1

1

,

j

i

j

i

j

i

j

ispacei

j

i

j

i

j

i

j

itimei

FFFFF

FFFFF

Page 57: Block 4 Numerical solution of open channel flow · 2017-11-26 · Contents of the course Block 1 – The equations Block 2 – Computation of pressure surges Block 3 – Open channel

Solution method

Implicit, linearized Finite Difference scheme

Linearization method: Example: term v2, a more complicated term

would be Q2/Ac

122221

1

22

j

i

j

ii

j

i

j

iii

j

i

j

i

j

i

i

j

i

j

i

vvvvvvvvvv

vvv

Implicit scheme:

All spatial derivatives are taken as a weighted

average between old and new time, weight θ

All time derivatives are taken in the middle of

The spatial discretization interval

Solution of big equation system for 2Nx unknowns per time step,

where Nx is the number of nodes

As the equation system is sparse, techniques for sparse

matrices are used

x

t

j i+1

j+1

i x

t θ

0.5

57