The top of a cylindrical storage tank for gasoline at a service station (gasolin

Question

The top of a cylindrical storage tank for gasoline at a service station (gasoline weight 42 pounds per cubic foot) is 4 feet below ground level. The axis of the tank is horizontal and it diameter and length are 5 feet and 12 feet, respectively. Find the work done in pumping the entire contents of the full tank to a height of 3 feet above ground level.

Explanation / Answer

First, find the volume of a generic slab of gasoline that is h feet from the bottom of the tank and is ?h feet thick. We can split the cylindrical storage tank into generic rectangular prism-like slabs. The volume of each slab is: dV = LWH = 12W?h. We need to find the general width of each slab in terms of h. To do this, we need to find how wide the base of the circle is at a distance of h from the bottom of the tank. The diameter of the tank is 5 feet, so the radius is 5/2 feet. Taking the center of the circle to be at the origin, the equation is: x^2 + y^2 = (5/2)^2 ==> x^2 + y^2 = 25/4. Then, the width of the circle in terms of the distance, h, from the bottom of the circle is: x^2 + (5/2 - h)^2 = 25/4 ==> x = v[25/4 - (5/2 - h)^2] = v(5h - h^2). Then, W = 2x = 2v(5h - h^2). Thus: dV = 12Wh?h = 24v(5h - h^2)?h. (As a sanity check, h varies from 0 to 5, so the volume of the cylinder is: V = ? 24v(5h - h^2) dh (from h=0 to 5) = 75p, which agrees with V = pr^2*h = p(5/2)^2(12) = 75p.) This is the volume of a generic slab of gasoline. The weight can be found by multiplying the volume by the density to yield: dF = Vd = [24v(5h - h^2)?h](42) = 1008v(5h - h^2)?h. This is also the required force to lift this slab. Then, the generic slab of gasoline must travel up 5 - h feet to get to the top of the tank, 4 feet to get to ground level, and another 3 feet to move the gasoline 3 feet above ground level. In total, the gasoline travels (5 - h) + 4 + 3 = 12 - h feet. Thus, the required work to lift this slab up is: dW = Fd = 1008v(5h - h^2)(12 - h)?h. Therefore, since h varies from 0 to 5, the required work is: W = ? 1008v(5h - h^2)(12 - h) dh (from h=0 to 5) = 29925p ft-lbs.

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