I've rencently been thinking about several systems related to fluid mechanics, and I've asked myself the following question: If I have a cylidrical tank with a small hole in the bottom (the surface area of the hole is small in comparison with that from the tank) and I'm filling the tank with water at a constant flow, how can I determine the speed of the water both coming from the whole and rising up the tank: (A) with respect to the height of the water in the tank, (B) with respect to time. (This is not homework, although it sounds like so) I know how to determine the maximum height of a system like this one with a specific flow Q. If the surface area of the cylinder is A and that from the hole is S, we have: [tex]Av_A=Q=Sv_S[/tex] thus [tex]v_S=\frac{Q}{S}[/tex] And, using Bernouilli's equation: [tex]P_{at}+\rho g h=p_{at}+ \frac{1}{2}\rho v_S^2[/tex] [tex]h=\frac{v_S^2}{2g}[/tex] But that is considering the end point. Where the system is 'at rest' (there's the same amount of fluid in the tank and we can treat it like a pipe), but I don't know how to study the intermediate steps. Thanks for your help :)
You need to consider the viscosity of the fluid and the diameter and length of the hole, then solve for the back pressure equaling the static fluid pressure at the depth of the hole.
You need to do a transient mass balance on the volume of liquid in the tank. If V is the volume of liquid in the tank, then V = hA. The time rate of change of V is equal to the flow rate in minus the flow rate out. Write a differential equation for this.
Not quite right. You need to take a look at the Navier-Stokes equations for incompressible fluids. Or you can just look up the equations for flow thru an orifice.
If the pressure head in the tank is much higher than the frictional pressure drop for flow through the oriface, the latter can be neglected (which is what, at the OPs level of experience, the problem statement intended). The mass balance on the tank is [tex]A\frac{dh}{dt}=Q-Sv_s[/tex] with [itex]v_s=\sqrt{2gh}[/itex] So, [tex]A\frac{dh}{dt}=Q-S\sqrt{2gh}[/tex] Just solve this differential equation subject to some initial height h_{0}.
Why is the flow through the hole given by:? [tex]Q-S\sqrt{2gh}[/tex] I understand it is the flow that comes in minus the flow that goes out. But when I think of it with the continuity equation, the flow is supposed to be constant. When does the continuity equation apply?
This actually is the continuity equation (mass balance equation) on the macroscopic scale (input minus output = accumulation). On the microscopic scale, if the fluid is incompressible, the divergence of the velocity vector is still equal to zero everywhere in this situation. You know this because, at any instant of time, the downward velocity of the fluid in the tank is not varying with depth. Chet