solution of the st venant equations / shallow-water equations of open channel flow dr andrew sleigh...
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Solution of the St Venant Equations / Shallow-Water equations of open channel flow
Dr Andrew SleighSchool of Civil EngineeringUniversity of Leeds, UK
www.efm.leeds.ac.uk/CIVE/UChile
St. Venant Equations
fo SSgt
u
x
du
x
dg
0
t
db
x
uA
x
Au
St Venant Assumptions of 1-D Flow Flow is one-dimensional i.e. the velocity is uniform over the cross
section and the water level across the section is horizontal.
The streamline curvature is small and vertical accelerations are negligible, hence pressure is hydrostatic.
The effects of boundary friction and turbulence can be accounted for through simple resistance laws analogous to those for steady flow.
The average channel bed slope is small so that the cosine of the angle it makes with the horizontal is approximately 1.
Cunge J A : Practical Aspect of Computational River Hydraulics
Control Volume
h1
z1 z2
h2
bed
free surface
y
s
xx1 x2
h1
z1z1 z2z2
h2
bed
free surface
y
s
xx1x1 x2x2
A1
V1
A2
V2
A1
V1
A2
V2
Continuity
CV in x,t plane between cross sections x=x1 and x=x2 between times t=t1 and t=t2
conservation of mass
02
1 12
2
1 12 dtQQdxAAt
t xx
x
x tt
Momentum
Conservation of momentum
dtdxSSAg
dtdxIg
dtIIg
dtAuAudxuAuA
t
t
x
x fo
t
t
x
x
t
t xx
t
t xx
x
x tt
2
1
2
1
2
1
2
1 2
2
1 2111
2
1 22
122
1 12
Geometric Change Terms
vertical change in cross-section
change in width along the length of the channel.
xhdxxhI
01 ,
xh
hoh
dx
xhI02
d
h-
b
h
()
Integral / Differential Forms Integral form do not require that any flow
variable is continuous We will see later finite difference methods
based on this integral form. Can derive differential form … but
Must assume variables are continuous and diferentable
Replace variable with Taylor’s series
...2
2
2
2
12
t
t
At
t
AAA tt
Differential form
-
0
x
Q
t
A
21
2
gISSgAgIA
Q
xt
Qfo
21 gI
x
hxgA
x
gI
0
fo gASSx
hgAuQ
xt
Q
In terms of Q(x,t) and h(x,t):
Where b = b(h), A=A(h)
t
hb
t
h
h
A
t
hA
01
x
Q
bt
h
02
of SSgAx
hgA
A
Q
xt
Q
remember b = b(h), A=A(h)
Each of these forms are a set of non-linear differential equations which do not have any analytical solution. The only way to solve them is by numerical integration.
In term of u(x,t) and h(x,t):
using
tconshx
A
x
h
h
Au
x
uA
x
Au
x
uA
x
Q
tan
0tan
tconshx
A
b
u
x
hu
x
u
b
A
t
h
0
of SSgx
hg
x
uu
t
u
Characteristic Form
The St Venant equations may be written in a quite different form know as the Charateristic Form.
Writing the equations in this form enables some properties and behaviour of the St Venant equations become clearer.
It will also help identify some stability criteria for numerical integration
will help with the definition of boundary conditions.
Characteristic form
Consider a prismatic channel
0
x
hu
x
u
b
A
t
h
0
of SSgx
hg
x
uu
t
u
Wave speed
Consider the speed, c, of a wave travelling in the fluid.
with respect to x and t gives:
b
Agc
x
hg
x
cc
2t
hg
t
cc
2
022
x
uc
x
cu
t
c 02
of SSgx
uu
x
cc
t
u
Combining
Adding equations (3) and (4) gives
Subtracting equations (3) and (4) gives
fo SSgx
ccv
t
c
x
vcv
t
v
22
fo SSgx
ccv
t
c
x
vcv
t
v
22
Characteristics
Equations (5) and (6) can be rearranged to give respectively
fo SSgt
cv
x
cvcv
22
fo SSgt
cv
x
cvcv
22
Total differential
For some function of x and t In a physical sense.
If the variable is some property of the flow e.g. surface level,
if an observer is moving at velocity v, the observer will see the surface level change
only with time relative to the observers' position.
tdt
dx
xdt
d
The characteristic form of the St Venant equations If we take = (v + 2c)
Total differential is
Compare with
Clearly
and
t
cv
dt
dx
x
cv
dt
cvd
222
fo SSgt
cv
x
cvcv
22
cvdt
dx
fo SSgdt
cvd
2
Characteristic form of the St Venant Equations
These pairs are known as the Characteristic form of the St Venant Equations
fo SSgdt
cvd
2for cv
dt
dx
fo SSgdt
cvd
2
for for
for cvdt
dx
fo SSgdt
cvd
2 cvdt
dxfor
Meaning of the characteristics
1
2
3
Zone of qu iet
x
t
c5
c4
c2
c1
c0
4
5
6
8
7
The paths of these observers that we have talked about can be represented by lines on this graph.
cvdx
dt
1 cvdx
dt
1
Information paths
it takes time for information to travel E.g. a flood wave at u/s end The channel downstream will not receive the
flood for some time. For how long?
The line C0 represents the velocity of flood wave
Everything below C0 is zone of quite
Zones of Dependence and Influence The idea that characteristics carry information
at a certain speed gives two important concepts
c1c2
c1c2
Q x
t
Dom ain of dependence of P
P Zone of influence of Q
Stability
These zones imply a significance to numerical methods and stability of any solver
The numerical method must take only information from within the domain of dependence of P
this limits the size of time step
Calculation with characteristics If we know the solution at points 3 and 5 Can determine the solution at point 6.
For C5-6 then
For C3-6 then
jigdt
cvd
2 jigtcvcv 5566 22
33
1
cvdx
dt
55
1cvdx
dt
jigdt
cvd
2 jigtcvcv 3366 22
Characteristics solution
Adding equations
Subtracting
jigtccvv
v
3535
6 2
243535
6
ccvvc
MOC on Rectangular Grid
x
t
t
t+ tP
L RO EW
x
t
x
P
W O ERL
x
xx L R
t
t+ t
t
Midstream discretisation
Away from boundaries
x
P
W O ERL
x
xx L R
t
t+ t
t
LL cvdt
dx PP cv
dt
dx
OOL cvt
x
OOR cvt
x
WOL
OL vvx
xvv
EOR
OR vvx
xvv
Solution
LLLPP SfSogtcvcv 22
RRRPP SfSogtcvcv 22
SfSogdt
cvd
2
SfSogdt
cvd
2
RLRLRL
P SfSoSfSogt
ccvv
v
22
RLRLRL
P SfSoSfSogtccvv
c
424
Stability
Considering characteristics
t
x
t+ tP
L RM
C 1C 2
t1= v+c
dtdx
1= v-cdtdx
cvdt
dx
cv
xt
max
9.0cv
xt
Boundary conditions
Upstream boundary: backward characteristic
Downstream boundary: forward characteristic
x
P
W OL
x L
t
t+ t
t
bo
und
ary
P
O ER
x
x R
t
t+ t
t
bo
und
ary
Boundary Conditions
The second equation is a boundary condition equation
Upstream depth boundary hP = H(t) pp ghc
RRRPP SfSogtcvcv 22
RPRRP SfSogtccvv 2
Spuer-critical - mid
Right characteristic moves to left
solution method is exactly the same x
P
W O ERL
x
x
x L
R
t
t+ t
t
Super-critical - upstream
No characteristics
P
O ERt
bo
und
ary
L
Super-critical downstream
2 characteristics No boundary condition
bo
und
ary
x
P
W ORLx
x L
R
t
t+ t
t
Finite Difference Schemes
Two basics classes Implicit Explicit
Commercial packages use implicit Explict
for high accuracy (sometimes!) Testing / understanding behaviour Class examples!
Which Equations ?
Not always clear what equations a being used! What are the shallow water equation?
We will look at schemes based on the integral equations:
02
1 12
2
1 12 dtQQdxAAt
t xx
x
x tt
dtdxSSAg
dtdxIg
dtIIg
dtAuAudxuAuA
t
t
x
x fo
t
t
x
x
t
t xx
t
t xx
x
x tt
2
1
2
1
2
1
2
1 2
2
1 2111
2
1 22
122
1 12
Homogeneous Integral Equations Without the gravity / frictions terms
02
1 12
2
1 12 dtQQdxAAt
t xx
x
x tt
02
1 112
2122
1 12 dtIAugIAudxQQt
t xx
x
x tt
Grid based
Consider the grid …
i i+1i-1
n
n+1
t
x
B
A D
C
t
x
i i+1i-1
n
n+1
t
x
B
A D
C
t
x
Integrate around the cell
Considering the cell ABCD, Integral can be written in the general vector form :
ABCD
dtfGfdx 0
1
2
gIA
fG
A
Qf
Gridpoints / Variables
Variables are all known or will be calculated at the grid points xi represents the value of x at position i tn represents the value of t at position n
Derivation approximates values:
ni
nii GGtxG 1, 1 n
inin fftxf 1, 1
10 10
Substitute in approximations
And the equation become
011
11
1
1
11
11
111
1
dtGGGG
dxffff
n
n
i
i
t
t
ni
ni
ni
ni
x
x
ni
ni
ni
ni
011
111
111
111
1
tGGGG
xffffni
ni
ni
ni
ni
ni
ni
ni
Difference equation
Divide through byΔx Δt
And can see that it is an approximation of
i.e. starting with integral form discretisation is also valid for diferential form
0
1111 11
111
111
x
GGGG
t
ffff ni
ni
ni
ni
ni
ni
ni
ni
0
x
fG
t
f
Several schemes
This is a general discretisation scheme Vary the parameters ψ and θ Get a family of different finite difference schemes
Features are They are implicit for values of ψ > 0. else explicit. They link together only adjacent nodes.
Space interval can vary – no loss of accuracy. They are first order, except for the special case of
ψ=θ=0.5 when they are second order.
Preissmann Scheme
ψ = 0.5 gives Preissmann 4-point scheme Time derivative
Space derivative
t
ffff
t
f ni
ni
ni
ni
21
111
x
GGGG
x
fG ni
ni
ni
ni
11 1
111
Equations become …
0122 1
1111
11
1
ni
ni
ni
ni
ni
ni
ni
ni QQQQtAA
xAA
x
012
1
22
1
122
1
121
2
1
2
1
1
21
1
21
1
1
2
111
1
n
jofn
jof
n
jofn
jof
n
i
n
i
n
i
n
i
ni
ni
ni
ni
ASASIASASI
ASASIASASIxtg
gIA
QgI
A
QgI
A
QgI
A
Qt
QQx
QQx
More common form
The terms I1 and I2 are often difficult (expensive / time consuming) (when originally attempted)
Usual form used in packages
01
x
Q
bt
h
02
of SSgAx
hgA
A
Q
xt
Q
Priessmann Function represented by,
0.5 < θ ≤ 1 Continuity:
nini
ni
ni fffff
1
111 2
1
2
0
1
12
2
1
111
11
111
11
11
ni
ni
ni
ni
ni
ni
ni
ni
ni
ni
ni
ni
bbbb
QQQQ
xt
hh
t
hh
01
x
Q
bt
h
Priessmann
Momentum Equation
02
of SSgAx
hgA
A
Q
xt
Q
02
1
2
1
2
1
111
1
2
1
2121
1
2
11
11
EAAAAg
A
Q
A
Q
xA
Q
A
Q
x
t
t
ni
ni
ni
ni
n
i
n
i
n
i
n
i
ni
ni
ni
ni
Source term
So-Sf
21
221
2
1
22
1
2121111
1111
11
111
1
ni
ni
ni
ni
ni
ni
ni
ni
ni
ni
ni
ni
ni
ni
ni
ni
KKKKQQQQQQQQ
x
hh
x
hhE
Unknowns
There are the 4 unknowns
Plus k, b, A which are functions of h and Q
,,,, 11
111
1
nj
nj
nj
nj QQhh
,,,, 11
111
1
nj
nj
nj
nj QQhh
Need to linearise
“Linearise” equations
A, B, C, D and RHS are funtion of the unknowns. But not strongly
RHSDQCQBhAh nj
nj
nj
nj
1
111
11
Boundary Conditions
In implicit scheme specify h or Q Use characteristics to decide appropriately
Or relation between h and Q
2N unknowns (h, Q at each node) 2N – 2 equations from internal points 2 boundary equations
Junction
At a junction each chanel share the same node hjuntion 1 = hjuntion 2 = hjuntion 3 …..= hjuntion n = h
Continuity Sum of inflow and outflow equal to zero
Iterative solution
Need to iterate updating coeficients
Cunge says …
iterates rapidly one or two iterations
Newton-Raphson methods used in packages
Stability
Formally unconditionally stable for all time steps 0.5 < θ ≤ 1
Further away from Cr = 1 less accurate Cr = 20 is common Because of linearisation may fail for extreme
flows or those that are too far from original assumptions
Explicit Schemes
Not used in simulations of real rivers Time step limitations.
Important features of explicit schemes they are simple to implement allow experiment with weights, time-step and
space-step to understand behaviour of the solution.
There are MANY schemes
Leap-Frog
Earliest scheme aplied to wave equations Spatial derivative Temporal
derivative
gives
t
ff
t
f ni
ni
211
x
GG
x
fG ni
ni
211
ni
ni
nin
i
nin
i GGx
tf
Q
Af 111
11
2
Stability
All explict scheme have time step limit Courant confition (CFL)
Cr < 1
cvdt
dx cv
xt
Lax Explict Scheme
Similar to leap from with weighting, α, in time
For 0 ≤ α ≤ 1 Spatial derivative
t
ffff
t
f
ni
nin
ini
21 11
11
x
GG
x
fG ni
ni
211
Lax Explict Scheme
Leads to function solution
Boundary conditions must be applied using method of characteristics
ni
ni
ni
nin
ini
nin
i GGx
tfff
Q
Af 11
111
11
221
Some useful texts
Rather old- still basis of many commercial programs.
Cunge, J.A., Holley, F.M. and Verwey, A. (1980): Practical Aspects of Computational River Hydraulics
Mahmood and Yevjevich (1975): Unsteady Flow in Open Channels - Fort Collins, Colorado
Liggett & Cunge (1975) Preissmann, A. (1960): Propogation des
Intumescenes dans les Canaue et Rivieres - 1st Congress de l'Assoc. Francaise de Calcul, Grenoble