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