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Work needed to pump water out of spherical tank

  1. Feb 13, 2009 #1
    1. The problem statement, all variables and given/known data

    A spherical tank is full of water. If radius r = 3m and height h = 1.5m, find the work W required to pump the water out of the spout. (Use 9.8 for g and 3.14 [tex]\pi[/tex])

    2. Relevant equations

    Work is the integral from a to b of f(x)dx.

    3. The attempt at a solution

    Volume= area * thickness, so I got V= pi*r2dx for any given slice of that sphere. I could be wrong, but I think the density of water is 1000kg/m3, so I multiplied that by the volume to get mass of the water, and then multiplied the mass by 9.8 to get the weight of water. Then I set up my integral from 0 to 1.5 of the answer I got from multiplying all those numbers out * xdx.
    But I got the wrong answer. So how do I work a problem such as this?
  2. jcsd
  3. Feb 13, 2009 #2
    First of all I'm not sure what you mean by the 'height' of a spherical tank.

    Is the situation that water is essentially being lifted vertically to the top ofa spherical tank?

    Assuming that is the case, the work done is the integral (sum) of the weight of each slice of the sphere.

    Draw a diagram: a circle centred on (0,0) with radius 3. Let x be the y-ordinate of a slice. Then the radius of a slice is given by sqrt(3^2 - x^2).

    You can then find the volume, and hence the mass by multiplying the density, of each slice. I'll let you sort out the units.

    Integrate that over the appropriate interval that x ranges over.

    Show us your work if you get stuck.
  4. Feb 14, 2009 #3


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    Science Advisor

    Your mistake is that you have the weight of all of the water in the sphere being lifted from every height in the sphere: in other words, the same weight is being multiple times.

    Analyse the problem using things that are constant- that's one reason for emphasizing "Riemann sums" as the definition of integral: that's often how we set up integrals in applications.

    As Unco suggeted, draw a circle representing the sphere. Put it in "standard position" in a coordinate system so the center is at (0,0) and the radius is 3: [itex]x^2+ y^2= 9[/itex]. Draw a horizontal line at any level in the circle, at y, representing one "layer" of water, with thickness "dy". All the water in that "layer" is at the same height, y, and so must be lifted the same distance, 3- y (notice that y may be from -3 to 3 so this distance is from 3-(-3)= 6 at the bottom of the tank to 3-3= 0 at the top).

    Now, in 3 dimensions, that "layer" of water is a disk whith radius equal to the x coordinate where the line at y crosses the circle: [math]x^2+ y^2= 9[/math] so "radius squared" is [itex]]x^2= 9- y^2[/itex] and the area of the disk is [itex]\pi(9- y^2)[/itex]. The volume of the thin layer is [itex]\pi(9-y^2)[/itex] and since weight density of water is 1000(9.81)= 9810 Newtons/l3, the weight of that layer is [itex]9810\pi(9- y^2)dy[/itex] and, finally, the work done lifting that weight a distance 3- y is [itex]9810\pi(9- y^2)(3- y)dy[/itex].

    You get work required to lift all of the water to the top of the sphere by "adding" the work done on every "layer" of water in the sphere. That is, you integrate that from y= -3 to y= 3.
  5. Jan 18, 2010 #4
    But, what about the work done to get the water out of the spigget at the top? Isn't (3-y) just the distance to the top of the sphere, but before the spigget?
    Last edited: Jan 18, 2010
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