Let's say we have a Torricelli's Law apparatus, where, in the picture below, we are concerned about the velocity v coming out of the bottom-most spigot that is a height h below the top of the water.
The law states that $V=\sqrt{2gh}$. Essentially, the speed of the efflux in a Torricelli apparatus is directly proportional to and affected only by the height of the fluid above it.
We also know, however, that in fluid dynamics, volume flow rate is constant, demonstrated quantitatively by the continuity equation $Av = \textrm{constant},$ or $$A_1v_1 = A_2v_2\iff v_1 = \frac{A_2}{A_1}\cdot v_2$$
We can interpret this as: v is inversely proportional to the area of the hole of the container it is flowing through.
My question is now this: if we changed the area of the spigot – ever so slightly making it greater or less, but not so much as to deem the hole too big for Torricelli's Law to work – but kept the height h of the hole the same, would velocity change (as the continuity equation would suggest), or stay the same (as Torricelli's Law would suggest)?
Best Answer
The velocity would stay the same no matter whether the area is slightly increased or decreased provided the hole is at the same height and is quantified by $V=\sqrt{2gh}$
And at the same time continuity principle isn't violated.Where you've gone wrong is comparing the same thing. It should be like ..
Case 1 : Area increased
Velocity stays the same,but the flow rate increases.
Area at exit point (A2) increased and velocity (V2) stays the same. Similarly cross-sectional area of the tank (A1) is the same but the velocity with which the water moves down (V1) increases emptying the tank quickly as the flow rate is high.
So (A1V1)=(A2V2) ------ since if A2 raises then V1 raises.
Case 2 : Area decreased
Velocity stays the same,but the flow rate decreases.
Area at exit point (A2) decreased and velocity (V2) stays the same. Similarly cross-sectional area of the tank (A1) is the same but the velocity with which the water moves down (V1) decreases emptying the tank comparitively slowly as the flow rate is low.
So (A1V1)=(A2V2) ------ since if A2 decrease then V1 too decreases.