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CEE 262A
HYDRODYNAMICS
Lecture 13
Wind-driven flow in a lake
Wind-Driven Flow in a Lake
Assumption (i) Steady Forcing
(ii) Two-dimensional
(iii) H/L<<1
Lx1=0 x1=L
Hx3=0
x3=H0
2 20 0 * 10 a Du C U (A turbulent stress)“Rigid lid”
Mass:
x1-Momentum:
x3-Momentum:
03
3
1
1
x
u
x
u
2 21 1 1 1 1
1 3 2 21 3 1 1 3
1 '
e
u u u u upu u v
t x x x x x2 2
3 3 3 3 31 3 2 2
1 3 3 1 3
1 '
e
u u u u upu u v
t x x x x x
Simplify
(a)
(b) Scale
0)(
t
* * *1 1 1 1
* *3 3 3 3
x Lx u Uu p p
x Hx u Wu
Effective viscosity – assumed constant
The governing equations with hydrostatic pressures removed are:
To find U and we use an assumed force balance:
pressure ~ friction
The free surface condition gives
21*
3
eu
ux
Since we are looking to make
We find that
22 *
*or converted to scaling ~
ee
u HUu U
H~
2 2* *
2 2
2*
e ee
u H uU
L H H H
Lu
H
~
~
212
0 1 3
1
eup
x x
Mass: 0*3
*3
*1
*1
x
u
H
W
x
u
L
U
L
HUW
x1-Mom: *3
*1*
3
2
*1
*1*
1
2
x
uu
L
U
x
uu
L
U
2 2 * 2 *** 1 1
* 2 *2 2 *21 1 3
e
u u up U Uv
H x L x H x
since*
3*
33 uL
HUWuu
**31
* *1 3
0
uu
x xIf we choose
which reduces to:
From prev. slide:
2 22 2 * * 2 * 2 *** ** 1 1 1 1
1 32 * * * *2 *21 3 1 3 1
e
u H u u u uH p Hu u
L x x x x L x
2 2 2* * 2 * 2 *2 2 ** *3 3 3 3*
1 32 * * * *2 *21 3 3 3 1
e
u u u uu H H p H Hu u
L x x x L x L x
Likewise the x3-momentum eqn. becomes
To end up with PC flow we must require two things:
22 2*
2*
1
1
e
e
H
L
u H H H
L L u H
If these conditions are both satisfied and we get rid of all of the small stuff, what is left is:
2 ** *1
* *2 *1 3 3
0 0
up p
x x x
2
Pressure-friction Hydrostatic
Assumption 1: H/L << 1 "The Long Box"
Assumption 2:
HL
Vertical diffusion
time scale
<< Horizontal advection
time scale
Horizontal acceleration << Vertical friction
The second equation implies that p* = p*(x1*). However, since the
boundary conditions are independent of x1*, u1
* should only be a function of x3
*. This implies that the pressure gradient must be constant, i.e. not a function of x1
*. Thus, we are back to the PC equations we solved before.
But, how do we find ? We obtain an extra condition on u1*
* *1 p x
If we integrate continuity from x3*=0 to x3
* = 1, we find that
1 1** * *1
3 1 3* *1 10 0
1* * * *
1 3 1 1
0
0
constant=0 since 0 0
udx u dx
x x
u dx u x
2 ** *1
* *2 *1 3 3
0 0
up p
x x x
The pressure gradient we need is the one that imposes no net flow
So to proceed, we now use three conditions to constrain the quadratic velocity profile
* *2 *1 3 3 u ax bx c
that results from integration of the x1 momentum equation
* *1 3
**1
3*3
1* *
1 3
0
(a) 0 at 0 0
(b) 1 at 1
2 1 1 2
(c) 0 03 2
3 1 and
4 2
u x c
ux
x
a b b a
a bu dx
a b
No slip on bottom
Specified stress on top
No net flow
=2a=3/2
* *2 *1 3 3
3 1
4 2 u x xThus putting it all together, we find that
So that the velocity we find at the top is (to compare to our PC soln.)
* *1 3
2*
1 3 0
11
4or in dimensional terms
4
e
u x
u Hu x H U
-0.1 -0.05 0 0.05 0.1 0.15 0.2 0.250
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
u1*
x 1*
Since P = 3 (no net flow for PC flow), we can now compute the dimensional pressure gradient using our computed surface velocity and the definition:
0 02
1
2 3
2eU Pp
x H H
Why the 3/2 ? Look at stress distribution and balance of forces
1
-1/2Net force per unit length = +3/20 = Pressure force/unit length = H
Shear stress Pressure
Note: For turbulent flows it has been found that the stress on the bottom is nearly zero, so that the 3/2 should really be 1 for real flows
-0.2 -0.1 0 0.1 0.2 0.3 0.4 0.50
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
u1*
x 3*
What happens when we blow a 7 m/s wind over a 2 kmlong channel that is 10 m deep?
/0=a CD U102/0=1.0 kg/m3 * 0.002 * 49 m2/s2/0=10-4 m2/s2
L = 2 km
H = 10 me~0.005 m2/s H2/e = 334 min
No stress No slip
model (SUNTANS)
What provides the pressure gradient ?
In nature, a sloping free surface
From hydrostatics: )( 3' xHgp
)(' HHH where
H Water depth without applied stress
Super elevation relative to )f(xHx 13
x1=0 x1=L
x3=0
x3=H
Condition required so that domain is still rectangular and that x1 gradients are small
'3 3
1 1 1 1
2*
1
2*
1
2*
1
( ) ( )
3but from =3:
2
3 or...
2
3
2
pg H x g H x g
x x x x
upP
x H
ugx H
u
x gH
Now if we compute the horizontal pressure gradient, we can solve for the surface slope
This enables us to now integrate to find the water surface change due to winds.
0 02
1
2 3
2eU Pp
x H H
Since generally all of the water that starts in the lake stays there
L
dxx0 11 0)(
2*
1 1
3
2
Ux D
gH
If we integrate the slope equation wrt x1
2 2 22 2* * *
1 1 1 100
2 22* *
2*
1
3 3 3
2 4 4
3 30
4 4
3
2 2
L
L u u ux D dx x Dx L DL
gH gH gH
u u LL DL D
gH gH
u Lx
gH
2*3
4
u L
gH
Thus, the maximum change up or down is
0 5 10 15 20 25 30 35 40 450
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2x 10
-3
c0 t/2L
h at
x=
L (m
)Previous example:
2*3
4
u L
gH
HF oscillation:T = 2L/c0 = 2L/(gH)1/2
=6.7 min(time it takes fora shallow water waveto propagate from oneend to the other and back)
Predicted
Computed:0.001 mLess bottom stress!
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2x 10
-3
c0 t/2L
h at
x=
L (m
)
zoomed-in view
2 4 2 2*
4
~ 10 m s
~ 10 m
~ 10m
u
L
H
31
4000
H
2*
2
3
4
u L
H gHRelative to the depth this change is:
Normal example
Lake Okeechobee during a hurricane
22 30 10
2 3 2 2*
4
3 4
2.6 10 40 4.2 Pa
~ 4 10 m s
~ 4 10 m
~ 3m
4 10 4 1031.2 m
4 9.8
a DC U
u
L
H