Speed of a fluid in a funnel

[Raptor11122]

1. The problem
A funnel is in the form of a cone of semi-angle alpha and is placed with its vertex downwards. It is filled with water to a depth H and then the water is allowed to flow out of the funnel through a small hole at the vertex. If the stream of water has a cross-section of area A, and the velocity of the fuid in the stream when the depth of the water in the funnel is h and the rate at which the water level is decreasing is equal to U. Show further that:

$U^2 = \frac{A}{ \pi \tan(\alpha)} (\frac{2g}{h^3})^3$ (Equation 1)2. The attempt at a solution
$A = \pi r^2 = \pi h^2 tan(\alpha)^2$

as

$tan(\alpha) =\frac{r}{h}$ for a cone

$uA \delta = UA , u =$velocity of fluid coming out and $\delta << 1$

using Bernoulli means that

$u= \sqrt{2gh}$

from this you can show that

$u^2 = \frac{A}{ \pi \tan(\alpha)} (\frac{2g}{h^3})^3$

The problem is I have a $u^2$ instead of $U^2$ as asked for in the question. What am I doing wrong?

 

[Chestermiller]

In this problem, A is supposed to be the cross sectional area of the hole at the bottom of the funnel. In terms of h and tanα, what is the cross sectional area at the upper surface of the liquid in the funnel? In terms of A, what is the volumetric flow rate out the bottom of the funnel?

Chet