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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.

enter image description here

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)?

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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.

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  • $\begingroup$ Hm, I see what you're saying. So essentially you are taking into consideration the cross sectional area at the top and velocity of the water going down from there. However, doesn't Torricelli's Law say that the velocity of the water going down from the top is negligible, which is what made crossing out V1 on the left side of Bernoulli's equation possible in the first place? $\endgroup$ – Horse Oct 8 '16 at 18:02
  • $\begingroup$ yup ... exactly, but we are talking of very slight changes here which doesn't rule out Toricelli's law from working $\endgroup$ – Knight Oct 8 '16 at 18:04
  • $\begingroup$ If the velocity V1 from the top, then, is negligible, how is your statement "but the velocity with which the water moves down (V1) increases, emptying the tank quickly as the flow rate is high" valid? $\endgroup$ – Horse Oct 8 '16 at 18:06
  • $\begingroup$ Ah, so then that change in velocity V1 is also only ever so slightly increased? $\endgroup$ – Horse Oct 8 '16 at 18:09
  • $\begingroup$ Yeah but , we shouldn't forget the fact that Toricelli's law involves a simplification and moreover negligible doesn't mean zero right. $\endgroup$ – Knight Oct 8 '16 at 18:12
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If you change the cross section of the hole, the velocity v2 according to Torricelli will stay the same. Bernoullis equation will also hold! There is no contradiction because when the water flow A2·v2 at the hole increases due to the larger hole diameter the velocity v1 of the water surface in the vessel will increase so that A1·v1=A2·v2 still holds.

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  • $\begingroup$ So when we are looking at the continuity equation in this context, we do not make A1 and A2 the two different areas of the exit hole, but rather look at A1 as the cross sectional area at the top and A2 as the area of the exit hole - is that correct? $\endgroup$ – Horse Oct 8 '16 at 18:21
  • $\begingroup$ You are correct! The continuity equation of Bernoulli holds for the flow in the cylindrical vessel with cross sectional area A1 with velocity v1 und the flow through the hole with area A2 and velocity v2. (You have one hole only in this set up.) So if you increase A2 and v2 stays the same, then v1 of the water surface in the vessel has to increase because the cross sectional area of the vessel A1 is the same. $\endgroup$ – freecharly Oct 8 '16 at 18:31
  • $\begingroup$ Ah, I see. But how do we know that the v2 stays the same? Is it just because of Torricelli's law where v = sqrt(2gh)? $\endgroup$ – Horse Oct 8 '16 at 18:39
  • $\begingroup$ Yes you are right! The velocity v2 at the increased hole stays the same according to Torricelli's law. $\endgroup$ – freecharly Oct 8 '16 at 18:44
  • $\begingroup$ Thus, the change in the area A2 of the exit hole is compensated by the velocity V1 of the water at the top. I get it! Thank you so much, friend :-) $\endgroup$ – Horse Oct 8 '16 at 19:10

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