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Thread: Calculus 2 : Work Required to Pump Tank of Water

  1. #1

    Exclamation Calculus 2 : Work Required to Pump Tank of Water

    A water tank in the shape of a hemispherical bowl of radius 4 m is filled with water to a depth of 2 m. How much work is required to pump all the water over the top of the tank? (The density of the water is 1000 kg/m3. Assume g = 9.8 m/s2.)

    The answer is 352800π but I do not understand how the book came to this answer. Please I would like an explanation

  2. #2
    Full Member MarkFL's Avatar
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    Nov 2012
    St. Augustine, FL.
    I prefer to work problems like this in general terms, and then plug our given data into the resulting formula.

    First, let's let:

    [tex]R[/tex] = the radius of the hemispherical tank.

    [tex]\rho[/tex] = the weight density of the fluid.

    [tex]g[/tex] = the acceleration due to gravity.

    Now, let's imagine slicing the contents of the tank horizontally into circular sheets. The radius of each sheet will be a function of its distance from the base of the tank. So, let's orient a vertical [tex]y[/tex]-axis passing through the axis of symmetry of the tank, with its origin at the base of the tank.

    The circular portion of the cross-section of the tank containing our [tex]y[/tex]-axis will lie on the circle:

    [tex]\displaystyle x^2+y^2=R^2[/tex]

    And so, for any circular sheet, its radius will be the [tex]x[/tex]-coordinate of a horizontal line passing through the circle, and so we may state:

    [tex]\displaystyle r^2=R^2-y^2[/tex]

    Hence, the volume of an arbitrary sheet can be given by:

    [tex]\displaystyle dV=\pi\left(R^2-y^2 \right)\,dy[/tex]

    Next, we want to determine the weight [tex]w[/tex] of this sheet. Using the definition of weight density, we may state:

    [tex]\displaystyle \rho=\frac{w}{dV}\,\therefore\,w=\rho\,dV[/tex]

    Next, we observe that the distance [tex]d[/tex] this sheet must be lifted is:

    [tex]\displaystyle d=R-y[/tex]

    Thus, using the fact that work is the product of the applied force and the distance through which this force is applied, we find the work done to lift the arbitrary sheet is:

    [tex]\displaystyle dW=wd=\pi\rho(R-y)\left(R^2-y^2 \right)\,dy[/tex]

    Distributing, we find:

    [tex]\displaystyle dW=\pi\rho\left(R^3-R^2y-Ry^2+y^3 \right)\,dy[/tex]

    Now, if [tex]y_i[/tex] is the initial depth of fluid in the tank, and [tex]y_f[/tex] is the final depth, where [tex]0\le y_f<y_i\le R[/tex], then the total amount of work required to pump out the required amount of fluid is given by:

    [tex]\displaystyle W=\pi\rho\int_{y_f}^{y_i} R^3-R^2y-Ry^2+y^3\,dy[/tex]

    Applying the FTOC, there results:

    [tex]\displaystyle W=\pi\rho\left[R^3y-\frac{1}{2}R^2y^2-\frac{1}{3}Ry^3+\frac{1}{4}y^4 \right]_{y_f}^{y_i}=\frac{\pi\rho}{12}\left[12R^3y-6R^2y^2-4Ry^3+3y^4 \right]_{y_f}^{y_i}[/tex]

    And so we find:

    [tex]\displaystyle W=\frac{\pi\rho}{12}\left(y_i-y_f \right)\left(12R^3-6R^2\left(y_i-y_f \right)-4R\left(y_i-y_f \right)^2+3\left(y_i-y_f \right)^3 \right)[/tex]

    Can you use this formula now to get the desired answer?
    Living in the pools, They soon forget about the sea... Rush, "Natural Science" (1980)

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