CEE 262A H YDRODYNAMICS 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 L x 1 =0 x 1 =L H x 3 =0 x 3 =H (A turbulent stress) “Rigid lid”

Mass: x 1 -Momentum: x 3 -Momentum: Simplify (a) (b) Scale 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 Since we are looking to make We find that ~ ~ ~

Mass: x 1 -Mom: since If we choose which reduces to: From prev. slide:

Likewise the x 3 -momentum eqn. becomes To end up with PC flow we must require two things: If these conditions are both satisfied and we get rid of all of the small stuff, what is left is: 2 Pressure-friction Hydrostatic

Assumption 1: H/L << 1 "The Long Box" Assumption 2: H L Vertical diffusion time scale << Horizontal advection time scale Horizontal acceleration << Vertical friction

The second equation implies that p * = p * (x 1 * ). However, since the boundary conditions are independent of x 1 *, u 1 * should only be a function of x 3 *. This implies that the pressure gradient must be constant, i.e. not a function of x 1 *. Thus, we are back to the PC equations we solved before. But, how do we find ? We obtain an extra condition on u 1 * If we integrate continuity from x 3 * =0 to x 3 * = 1, we find that 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 that results from integration of the x 1 momentum equation No slip on bottom Specified stress on top No net flow =2a=3/2

Thus putting it all together, we find that So that the velocity we find at the top is (to compare to our PC soln.) 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:

Why the 3/2 ? Look at stress distribution and balance of forces 1 -1/2 Net force per unit length = +3/2  0 = 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

What happens when we blow a 7 m/s wind over a 2 km long channel that is 10 m deep?   /  0 =  a C D U 10 2 /  0 =1.0 kg/m 3 * * 49 m 2 /s 2 /  0 = m 2 /s 2 L = 2 km H = 10 m e ~0.005 m 2 /s  H 2 / e = 334 min No stress No slip model (SUNTANS)

What provides the pressure gradient ? In nature, a sloping free surface From hydrostatics: where Water depth without applied stress Super elevation relative to x 1 =0 x 1 =L x 3 =0 x 3 =H Condition required so that domain is still rectangular and that x 1 gradients are small

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.

Since generally all of the water that starts in the lake stays there If we integrate the slope equation wrt x 1 Thus, the maximum change up or down is

Previous example: HF oscillation: T = 2L/c 0 = 2L/(gH) 1/2 =6.7 min (time it takes for a shallow water wave to propagate from one end to the other and back) Predicted Computed: m Less bottom stress! zoomed-in view

Relative to the depth this change is: Normal example Lake Okeechobee during a hurricane