A cylindrical tank of 22.1m high with radius 12.2m is filled to 10m high with water. How much work must be done to pump all the water out of the tank?

Question

(density of water is 1000\ kg/m^3)

I used similar triangles somewhere in my calculations to find the answer, but I'm not sure I'm doing it right...
I got some big number like 1483087pi.

Answer 1

What similar triangles? The tank has a circular cross-section of πr^2

Work = force (weight)*distance, so figure the weight of a thin slice of water (thickness dy).

Then the work done to lift it over the top from depth y is its weight times y. Add up all those work pieces by integrating from 11.1 to 22.1, the range of depths (distance from the top of the tank).

Answer 2

LOL, consider it frozen :)

Well, it would be easier to siphon it out but anyway:

How much work must you do to lift that amount of water from the height of its center of mass in the tank to the top of the tank?

Volume of water = pi r^2(10)
= 10 pi (12.2^2) m^3

center of gravity of water above ground 5 m

top of tank = 22.1 m
so raise the water 22.1 - 5 = 17.1 meters
work done = m g h
= 1000kg/m^3 * 10 pi (12.2^2) * 9.81 * 17.1 Joules

Now you could have done that layer by layer integrating, but really no need.

Answer 3

I suppose you posted that question just to find out who here was a mathematician and who an engineer :)

Answer 4

Nice work, Damon. I like how the engineers always cut to the heart of the problem.

Reminds me of the engineer's way to measure the area of an irregular shape. He traced it onto paper, cut it out, and weighed it. Then he weighed 1 cm^2 of paper, and voila! The area was easy.

Answer 5

Just as a check,

∫[12.1,22.1] 1000 π (12.2^2)*9.8 x dx = 7.83595*10^8

which agrees with Damon's number, as expected.