Asked by Isabella
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?
(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.
(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.
Answers
Answered by
Steve
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).
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).
Answered by
Damon
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.
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.
Answered by
Damon
I suppose you posted that question just to find out who here was a mathematician and who an engineer :)
Answered by
Steve
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.
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.
Answered by
Steve
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.
∫[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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.