Question

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 cone with "r" being the bottom radius and "R" being the top radius, and h being height.

Answer 1

Like so.?

Answer 2

Yes, exactly. Can you help me?

Answer 3

ill, try my best.

Answer 4

Thank you

Answer 5

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.

Answer 6

162000 is incorrect. Did we have to include gravity anywhere? I know I had to on some other similar problems.