## Physics Friday 102

Torricelli’s law, mentioned in the previous Physics Friday, is derived from Bernoulli’s Principle, via a few assumptions. Let us have a small hole at the base of a large reservoir, with fluid depth h. Assuming an incompressible fluid and inviscid flow, Bernoulli’s principle says that  is a constant along all points on a streamline, where v is the flow speed, z is the elevation, P is pressure, and ρ is the fluid density. For our problem, our streamlines all run from the upper surface of the fluid to the hole. Thus, choosing zero elevation at the hole, we have
,
where vs is the flow speed at the surface of the fluid, vh the speed at the hole, Ps the pressure at the surface, and Ph the pressure at the hole. However, we should note that Ps and Ph are both just atmospheric pressure, so that Ps=Ph, and the pressure terms cancel, giving us
.
If the hole is sufficiently small compared to the reservoir, then the flow rate out is small enough that the motion of the fluid surface is negligible; setting vs=0, and solving for vh, one gets Torricelli’s Law: .

Now, suppose we relax the last assumption, and do not neglect the speed at which the fluid drops in the reservoir. Assuming a hole of area a, and that the surface of the fluid in the reservoir has area A, then we see that the volume of fluid lost out through the hole is avh; this must equal the volume from the lowering of the surface, and so . Solving for the surface velocity, and plugging into our result from Bernoulli’s principle,
.
This is larger than the speed given by Torricelli’s law by a factor of . For , this factor is approximated by , and thus it is reasonable to neglect it, sticking to just Torricelli’s law.

### 3 Responses to “Physics Friday 102”

1. Shobeir Says:

Why P(h) is not equal P (atmosphere) + Rho* g* h?

• twistedone151 Says:

Because of the hole. If there is no hole, then that above is the pressure. But, instead, the hole allows the fluid contact with the air, and the pressure of the liquid where it has a surface with the air is just the pressure of that air (P(liquid)=P(atmosphere) at the surface of the liquid in the jet as well as the top of the reservoir.

• Shobeir Says:

Thanks…My mistake 🙂

And BTW allow me to thank you for writing such good posts.