Velocity above a bump in open water channel using Bernoulli

[mh1985]

Homework Statement

If in an open water channel, the approach speed of an incompressible ideal fluid, of constant depth h1, is not too large, a small bump in height H (H<< h1) in the base of the channel will cause a dip Δh (Δh << H) in the water level directly above it, such that h1 = Δh + H + h2, where h2 is the height of the fluid above the bump.

Under these conditions, such an arrangement can be used for measuring the flow velocity

Q. Assuming no losses, show that the velocity v2, directly above the bump is given by

$$v_2 = [\frac{2gΔh} {(\frac{h1}{h2})^2 - 1}]^\frac{1}{2}$$

Homework Equations

$$0.5*u^2 + gh_1 = 0.5* v_2^2 + g(h_2 + H)<br /> <br /> g (h_1 - (H + h_2) = 1/2 (v_2^2 - u^2) = g \delta h$$

The Attempt at a Solution

We want the expression for $$v_2$$

$$v_1h_1 = v_2h_2$$------->$$v_1 = \frac {v_2h_2}{h_1}$$

Thanks for the help!

 

[mh1985]

can anyone help?