A conical tank has height 10 ft and radius 8 ft. The tank is buried vertex-down so that the top is 6 ft underground. The tank is filled with kerosene (of density 51.2lb/ft^3).

Question

a) Calculate the total fluid force on the side of the tank.

b) Calculate the work required to pump the oil to ground level.

Answer 1

For the first part, let's determine the force on the side of the tank. Draw a picture of the tank and then draw an arbitrary layer at a depth that is labeled y (which is from the top of the tank to the layer you drew). Then label the radius of that layer R. Make sure your drawing is more of an upside down triangle than a cone. Label the top radius as 8 and the depth of the entire triangle as 10. The triangle is isosceles because it is like looking at the cone straight on. Since the depth from the top of the tank to the layer is y, the depth from the layer to the bottom of the tank is 10-y. Using similar triangles, 8/10 = R/(10-y), and then solving for R, we get R = (40-4y)/5.

The layer drawn as a volume of pi*R^2*dy (dy is the little depth). If you multiply this value times the density of the liquid you get a weight or a force. So the force exerted on the outside of the tank at that layer would be 51.2*Pi*R^2*dy. Now, sub in the (40-4y)/5 for R. So now the force exerted by that layer is:

51.2*Pi*(40-4)^2/25*dy. The bounds of integration are from 0 to 10 because Y at the top of the tank is 0 and Y at the bottom of the tank is 10. The integral is then solved using the power rule. Take the constants and put them on the outside (51.2*pi/25) and integrate the expansion from 0-10 or (1600-160y+ y^2)dy and you get the entire hydrostatic force exerted by all the layers on the sides of the tank because essentially you are adding up all the tiny forces exerted by each layer.

The work needed to empty the tank begins with a similar drawing. Since each layer must travel from it's position to the top of the tank, we have to multiply the weight of the layer (a force) by the distance it must travel (y) to obtain a bit of work. So the solution then is the same as above multiplied one more time by y because it captures the distance each layer must travel to get to the top of the tank.

w = 51.2*π/25∫1600y-160y²+16y³ dy with bounds from 0-10 gives you the total work needed to empty the tank.