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Velocity of efflux out of a water tank
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[QUOTE="haruspex, post: 5450649, member: 334404"] As I wrote, the velocity will not be constant. It will rise to a peak, then decline as the water level drops. It can never exceed, or even quite reach, √(2gh) as that would violate energy conservation (which is the basis of Bernoulli). Here's an analysis of what happens over time: A tank of horizontal area A contains liquid density ##\rho## to an initial depth y[sub]0[/sub]. At one side near the base, there is a small hole area a. Rate of conversion from GPE to KE: ##\rho Ay\dot y\ddot y-\frac 12{\dot y}^3\rho A \left(\frac{A^2}{a^2}-1\right)=-\rho A y g \dot y## Writing ##k=(\frac Aa)^2-1##: ##y\ddot y-\frac 12k{\dot y}^2=-gy## Writing ##v=\dot y## and ##v'=\frac{dv}{dy}##, we have ##\ddot y = \frac{d}{dt}v =\frac{dy}{dt}\frac{d}{dy}v= vv'##. ##yvv'-\frac 12kv^2=-gy## Writing ##w=v^2## ##yw'=kw-2gy## General solution: ##w=By^{k}##. Particular Integral: ##w=cy## ##cy=kcy-2gy## ##c=\frac{2g}{k-1}## Using initial condition: ##0=By_0^{k}+y_0\frac{2g}{k-1}## ##B=-\frac{2g}{(k-1)y_0^{k-1}}## ##w=\frac{2gy}{k-1}(1-\left(\frac{y}{y_0}\right)^{k-1})## Peak velocity is at w'=0: ##w=\frac{2gy}k## ##\frac{k-1}{k}=(1-\left(\frac{y}{y_0}\right)^{k-1})## ##ky^{k-1}=y_0^{k-1}## ##y=y_0(k)^{-\frac 1{k-1}}## ##w_{max}=y_0(k)^{-\frac 1{k-1}}\frac {2g}k## ##v_{max}=\sqrt{y_0(k)^{-\frac 1{k-1}}\frac {2g}k}## Edit: in the original version of the above I left out a factor of 1/2 in the first equation. This led to 2k appearing subsequently everywhere that should have been just k. [/QUOTE]
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Velocity of efflux out of a water tank
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