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anonymous
 one year ago
hey srry guys about last link , I didn't realise my attachment wasn't uploaded so : it's a theoretical question and would love to know ur opinion :)
anonymous
 one year ago
hey srry guys about last link , I didn't realise my attachment wasn't uploaded so : it's a theoretical question and would love to know ur opinion :)

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phi
 one year ago
Best ResponseYou've already chosen the best response.1you have left out background material. Is the liquid going in at a fixed rate? if so, you are putting in (or emptying) a fixed volume per unit time Volume = Area of base * height and h= Volume/Area if volume changes linearly with time and the base area is fixed, then h changes linearly with time.

phi
 one year ago
Best ResponseYou've already chosen the best response.1in that case, a cylinder such as a "rain meter" is an example.

phi
 one year ago
Best ResponseYou've already chosen the best response.1do you have to empty it? what about filling the container?

phi
 one year ago
Best ResponseYou've already chosen the best response.1this sounds more like physics than math. Offhand I don't know how fast fluid will drain from a hole. Does the rate depend on the water pressure (hence the height) ? It does sound like if you make a really tall container and measure the change in height only over the "top part", the level will be linear or very close to linear.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1next for a dw:1442852608375:dw excusing the crap art, what we have is a vessel that has an as yet undefined contour but which is symmetrical in that it revolves around the h axis, so this is just obe class of solution. fluid leaves at the bottom of the vessel at h = 0 through a pipe of xsectional area a, moving at velocity v. thus for a given height of water we can say \( \dot V = av \) and as \(v = \sqrt{2gh} \) we have \( \dot V = a\sqrt{2gh} \)\) if we say that the radius of the vessel ir related to h by \(r = f(h)\), then an elemental volume of this shape is given by \(dV = \pi f^2(h) dh \implies \dot V = \pi f^2(h) \dot h\) you seem to want \(\dot h = const. \) if i understand correctly so we can say that \(\pi f^2(h) \dot h = \pi f^2(h)k = a\sqrt{2gh}\) meaning \[f(h) = \sqrt[4]{\frac {2ga^2}{\pi^2 k^2}}\times h^{\frac{1}{4}} \\= const. \times h^{\frac{1}{4}}\] that's a bit of a stream of consciousness that may have some merit but needs verification, devil will be in detail specifically an easy way to check is to do it in reverse. imagine a vessel with contour given by \(f(h) = \sqrt[4]{\frac {2ga^2}{\pi^2 k^2}}\times h^{\frac{1}{4}}\) or if you like \( f(h) = const. \times h^{\frac{1}{4}}\)] with a leak cross section area a at it base. what is \(\dot h\) for that vessel given that flow velocity through the hole is governed by \(v = \sqrt{2gh}\)?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1and now for a more compacted approach(!!) we don't need the calculus, we just use continuity for incompressible fluids if the xsect area of the vessel at height h is \(A = A(h)\) we can say that \(A \, \dot h = a \sqrt{2gh} \) and if we go again with the symmetry of a volume of revolution, we can say \(\pi R^2 \dot h = a \sqrt{2gh} \) where \(R = R(h)\) that gives the same solution: \(R^2 = \frac{a}{\pi \, k}\sqrt{2gh}\), again where \(R = R(h)\), so summary upshot is that \( R \propto h^{\frac{1}{4}}\) same answer, less bs, ergo better and i suspect that the \(A = A(h)\) formulation might allow you to explore prisms/ non symmetrical xsections more easily.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 hey thank you so much for taking out ur time and answering the question i figure it out how to get a linear h vs t graph from Torricelli equation

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0looks you have typed my name by mistake :) @ayeshaafzal221
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