## A tank is full of water. Find the work required to pump the water out of the spout. Use the fact that water we?

A tank is full of water. Find the work required to pump the water out of the spout. Use the fact that water weighs 62.5 lb/ft3. (Assume r = 4 ft, R = 8 ft, and h = 15 ft.)

The tank is a frustum of a cone where r is the radius of the base and R is the radius of the top of the tank

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Bramble

You leave too much undefined. What shape is the tank? Looks like it might be the frustum of a cone, but why should I guess – you should tell me! What are R and r? Is the tank standing with dimension h (is that height?) vertical or horizontal or something else?

Before you get into any profession you’d better improve your written communication. Lack of clarity wastes peoples’ time and costs cash. I would throw you out for it.

Edit : Following your clarifications here is my answer:

1. Assume the spout is at the level of the top of the tank.

2. Divide the water into elementary horizontal circular layers of thickness dy, y being the vertical coordinate.

3.Consider a typical such layer at height y above the base . Its radius = r = 4.(1+y/15) and

Its volume = dV = π.r².dy = 4.π.(1+y/15)².dy and its weight = dW = 4.π.ϱ.(1+y/15)².dy

where ϱ = density of water = 62.5 lb/ft³

4. Work done to remove this layer through the spout = dW.(h-y) = dW.(15-y)

= 4.π.ϱ.(1+y/15)².(15-y).dy

5. Therefore work done to remove all the water through the spout

= W = 4.π.ϱ.∫(1+y/15)².(15-y).dy = 4.π.ϱ.∫(1+2y/15 +y²/225).(15-y).dy

= 4.π.ϱ.∫[15 +y -y²/15 -y³/225].dx

= 4.π.ϱ.[15y +½.y² -y³/45 –(y^4)/900]. The limits on y are 0 and 15 so inserting these gives:

W = 4.π.ϱ.[225 + 112.5 -75 -56.25] = 162,000. ft.lbs.