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Homework Help: Difference of pressure of fluids in containers.

  1. Aug 1, 2004 #1
    Hello everyone,

    I have a question that pertains to the physics of fluids and how the pressure of which changes in different given containers, which contains water. Basically, here is what I’m trying to do:

    The shape of the container is that defined by a the function:
    [tex]f(x) = \frac{5}{8} x[/tex]
    Which is rotated about the x-axis from the limits 1 to 8, with a circular hole in the smaller part of the cone. So, essentially it is a cut-off cone with a height of 7 and large radius of 5 (from top). The cone container is flipped so that the cone is “upside-down” (or that the larger radius is upward), and water will then flow out at an unknown rate. The following link gives a diagram I drew out to better illustrate what is happening:

    (Copy and paste in new browser window, since angelfire doesn’t allow external linking)

    The essential problem is that I need to find a function in time of the water’s rate of the change of the container’s volume with pressure taken into account, so that I could simply find it’s instantaneous change of the Volume, height, or radius. My thought process has gone as follows:

    First, I needed to write a formula for the volume of the cut-off cone. And since the cone is defined by a function, I used calculus to get:
    [tex]\int_{1}^{x} \pi ( \frac{5}{8} x )^2 dx [/tex]
    [tex]\int_{1}^{x} \pi \frac{25}{64} x^2 dx [/tex]
    With [itex]x[/itex] being the cone’s height from “0” (although the cone does not start until x=1).

    My next step was to find out the container’s rate of flow, or pressure by calculating it initially at the point of which the tank is full, then writing a formula representing the rate of flow in terms of time, which is what I am ultimately looking for. I started by using the Pressure Force relation:
    [tex]P = \frac{F}{A}[/tex]
    [tex]P = \frac{ma}{A}[/tex]
    Where F is force which equals mass*acceleration, and A is the area, in meters squared. I had then calculated the volume of the container when it is full (purely as a test), and got 209.03 meters cubed, which equals 209,030 Kg in mass. And because we are dealing with water, the volume density relation is 1cm cubed = 1 gram.

    I was just about to calculate the pressure, but it suddenly hit me that the shape of the container might possibly affect the pressure of the fluid. In others words, there might be more pressure from fluids in a container that is a cylinder as opposed to a cone, even if the volume of the fluids are the same. If there is such a relation, it is very important to take that into account for calculating a rate of flow from the container, but unfortunately, I do not know where to start to find this relation, if any exists at all. Essentially, all I need to know is how pressures of fluids are affected by the shape of the container of which those fluids are in. But, if anyone would like to double check all my previous work and concepts, I’d gladly appreciate it. Of course, any help is appreciated.

  2. jcsd
  3. Aug 3, 2004 #2
    Hi Josh

    First could you please tell me what you mean by rotating the container that has the equation
    [tex]f(x) = \frac{5}{8}x[/tex]
    about the x-axis? [Analytically, is just a straight line in the x-y or x-z plane (depending on which coordinate axis you choose f(x) to be plotted against)]. Does the problem define the container this way or are you trying to express the cutoff this way?

    The question otherwise seems interesting...

  4. Aug 4, 2004 #3
    to:Izytang, I think p=Dgh, where D=density, g gravity, and so on... it's more applicable than p=F/A, and the pressure in a liquid behaves according to P= p_0 + Dgh due the earth's gravitational field. If for intance, your thinking because in this case the pressure at the bottom of this cone is different in comparison to a cylinder with the same area at the bottom, just think about the action-reaction forces on the side-wall of the cone that reduces the net force at the bottom. In order to solve this problem you need to use bernoulli's equation and the equation of continuity --->
    gh+(1/2)(v_1)^2 =(1/2)(v_2)^2 and (A_1)(v_1)=(A_2)(v_2), but for this you need the area of an opening at the bottom (maybe?).
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