Chapter 13, Problem 56P

(II) In Fig. 13–55, take into account the speed of the top surface of the tank and show that the speed of fluid leaving the opening at the bottom is

$v_1=\sqrt{\frac{2gh}{\left(1-A_1^2/A_2^2\right)},}$

where h = y₂ – y₁, and A₁ and A₂ are the areas of the opening and of the top surface, respectively. Assume $A_1\ll A_2$ so that the flow remains nearly steady and laminar.