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Velocity of efflux out of a water tank
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[QUOTE="Chestermiller, post: 5456985, member: 345636"] [b][Mentor Note -- see post #19 for a small typo correction in the math in this post][/b] MODEL SOLUTION Equation Eqn. 10 of the previous post tells us that the velocity of the effluent stream from the tank v[SUB]x[/SUB] will be a function of the area ratio a/A, the initial depth of the (inviscid) fluid in the tank h[SUB]0[/SUB], and the fluid depth at any arbitrary time t, h(t). The general analytic solution to this equation, subject to the initial condition ##v_x=0## at h=h[SUB]0[/SUB] is given by:$$v_x=\sqrt{2gh\frac{\left[1-(h/h_0)^{\frac{1-2r}{r}}\right]}{1-2r}}\tag{11}$$where ##r=(a/A)^2##. For the special limiting cases in which ##r=1/\sqrt{2}## and ##r=1##, this solution reduces to: $$v_x=\sqrt{-4gh\ln(h/h_0)}\tag{for r=1/√2}$$ $$v_x=\sqrt{2g(h_0-h)}\tag{for r = 1}$$ For the case of r =1 (i.e., the case in which the exit hole area is equal to the tank area), the above equation for the efflux velocity v[SUB]x[/SUB] is, as expected, just that predicted for free fall. MODEL RESULTS The results calculated from Eqn. 11 for the efflux v_x (normalized by ##\sqrt{2gh_0}## as a function of the (dimensionless) fluid depth ratio h/h[SUB]0[/SUB] and the (dimensionless) area ratio a/A are shown in the figure below. [ATTACH=full]99836[/ATTACH] In all cases, the efflux velocity is equal to zero initially (i.e., when h = h[SUB]0[/SUB]), and then rises rapidly as the fluid, both inside the tank and in the efflux, accelerates. However, as the depth of fluid in the tank continues to decrease, the efflux velocity passes through a maximum and then begins to decrease. Eventually, as the fluid depth approaches zero, the efflux velocity, of course, also drops to zero. (In the case of a/A = 1, the maximum velocity is attained just as the tank reaches empty.) The results in the above figure are fully consistent with what Haruspex has been saying all along, to wit, the efflux velocity v[SUB]x[/SUB]is always less than the Torricelli velocity based on the initial depth of fluid in the tank ##\sqrt{2gh_0}##. However, it is also of interest to compare v[SUB]x[/SUB] with the Torricelli velocity calculated on the basis of the instantaneous depth of fluid in the tank h(t), rather than on the depth at time zero. The figure below shows the efflux velocity ##v_x## normalized in terms of the Torricelli velocity calculated on the basis of the instantaneous depth h(t) plotted as a function of dimensionless depth ##h/h_0## at a selection of values of the area ratio a/A. [ATTACH=full]99855[/ATTACH] According to the results in the figure, for (small) values of the area ratio a/A less than 0.5, the dimensionless efflux velocity ##v_x/\sqrt{2gh(t)}## levels off to a constant value as the depth of fluid in the tank decreases. The smaller the value of a/A, the more rapidly the dimensionless velocity levels off. From our analytic solution (Eqn. 11), the value to which the dimensionless velocity levels off is given by: $$(v_x)_{level}=\sqrt{\frac{2gh(t)}{[1-2(a/A)^2]}}\tag{12}$$ For the case of a/A = 0.1 in the OP of this thread, we see from the figure that, once the depth h(t) has decreased to about 90% of the initial depth h[SUB]0[/SUB], the velocity has already leveled off. For small values of a/A, Eqn. 12 for ##(v_x)_{level}## can be expressed, to linear terms in ##(a/A)^2## by: $$(v_x)_{level}~\approxeq \frac{\sqrt{2gh}}{[1-(a/A)^2]}$$ By an strange coincidence, this is the same relationship for the efflux velocity that our high schooler OP had written in his second post of this thread. He was able to obtain this result by the fortuitous cancellation of two errors: 1. incorrectly assuming that the steady state Bernoulli equation could be applied this transient problem and 2. failing to take the square root of the denominator in obtaining ##v_x## from his equation for ##v_x^2##. This pretty much completes what haruspex and I wanted to cover. Chet [/QUOTE]
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Velocity of efflux out of a water tank
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