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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 :)

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  1. phi
    • one year ago
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    you 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.

  2. phi
    • one year ago
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    in that case, a cylinder such as a "rain meter" is an example.

  3. phi
    • one year ago
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    do you have to empty it? what about filling the container?

  4. phi
    • one year ago
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    this sounds more like physics than math. Off-hand 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.

  5. IrishBoy123
    • one year ago
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    next 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 x-sectional 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 \(v = \sqrt{2gh}\)?

  6. IrishBoy123
    • one year ago
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    and now for a more compacted approach(!!) we don't need the calculus, we just use continuity for incompressible fluids if the x-sect 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 x-sections more easily.

  7. anonymous
    • one year ago
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    .

  8. anonymous
    • one year ago
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    @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

  9. ganeshie8
    • one year ago
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    looks you have typed my name by mistake :) @ayeshaafzal221

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