DIVh = - (∆w/∆z)

Next, expand out the right hand side of the expression.

- (∆w/∆z) = - (w2 - w1)/∆z = (w1 - w2)/∆z

and put that back into the "master" expression above.

DIVh = (w1 - w2)/∆z

Dine's Compensation

In this case, level 2 is the 200 mb level, level 1 is the 500 mb level. You want to solve for the wind at the 500 mb level, which is at level 1.

w1 = DIVh ∆z + w2

but w2 = 0, and ∆z = 11200 m - 5600 m = 5600 m

so

w1 = 1. 5 X 10-5 s-1 X 5600 m = 8.4 cm s-1

By the way, this is a typical value for strong synoptic scale vertical motion. Note that it is about two orders of magnitude smaller than the typical values for horizontal motion (m s-1).

If you redid this problem assuming that you knew the 500 mb vertical velocity, but were given the task to find the surface divergence or convergence, you could use the same diagram, but its lower half. Then Level 2 would be 500 mb and Level 1 would be the ground or the 1000 mb level. If you solved Dine's Compensation for the divergence in the lower half of the troposphere, you would get

-1. 5 X 10-5 s-1