Download Hydraulik II

Numerical Hydraulics
Open channel flow 1
Wolfgang Kinzelbach with
Marc Wolf and
Cornel Beffa
1
Saint Venant equations in 1D
b(h)
lb
A(h)
z
h
• continuity
• momentum equation
2
Saint Venant equations in 1D
• continuity (for section without inflow)
Q A

0
x t
• Momentum equation from integration of
Navier-Stokes/Reynolds equations over
the channel cross-section:
0
hP
v
v
  v   g

t
x
x Rhy 
3
Saint Venant equations in 1D
• In the following we use:
  1
• and
0 

8
v
2
    (Re, k / R ) 
hy
4
Saint Venant equations in 1D
The friction can be expressed as energy loss
per flow distance:
0
E / V

  gI R
Rhy
x
Using friction slope and channel slope
IR 
 1 v2
8 Rhy g
z
IS 
x
Alternative: Strickler/Manning equation for IR
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Saint Venant equations in 1D
we finally obtain
v
v
h
v
 g  IS  IR   g
t
x
x
 (vA(h)) A(h)

0
x
t
For a rectangular channel : A  bh
h
h
v
v h 0
t
x
x
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Approximations and solutions
•
•
•
•
Steady state solution
Kinematic wave
Diffusive wave
Full equations
7
Steady state solution
(rectangular channel)
dv
dh
v
 g  IS  IR   g
dx
dx
dh
dv
v
h
0
dx
dx
Solution: 1) approximately, 2) full
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Steady state solution
(rectangular channel)
Approximation: Neglect advective
acceleration
 IS  IR   0
Normal flow
Full solution (insert second equation into first):
2
I

I
dh
v
2
S
R

 0 with Fr 
2
dx 1  Fr
gh
yields water surface profiles
9
Classification of profiles
hgr = water depth at critical flow
hN = water depth at uniform flow
Is = slope of channel bottom
Igr = critical slope
Horizontal channel bottom
Is = 0
H2: h > hgr
H3: h < hgr
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Classification of profiles
Mild slope:
hN > hgr
Is < Igr
M1: hN <h > hgr
M2: hN > h > hgr
M3: hN > h < hgr
Steep slope:
hN < hgr
Is > Igr
S1: hN <h > hgr
S2: hN < h < hgr
S3: hN > h < hgr
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Classification of profiles
Critical slope
hN = hgr
IS = Igr
C1: hN < h
C3: hN > h
Negative slope
IS < 0
N2: h > hgr
N3: h < hgr
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Numerical solution
(explicit FD method)
Subcritical flow: Computation in upstream direction
h( x  x)  h( x) I S  I E ( x  x)

2
x
1  Fr ( x  x)
Solve for h(x)
Supercritical flow: Computation in downstream direction
h( x  x)  h( x) I S  I E ( x)

2
x
1  Fr ( x)
Solve for h(x+x)
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