Date Rewarded: April 24th, 2008.

A Chomby was walking down a path one day when Jhudora appeared in front of it. "Answer this question, and you may pass," she said.

Suppose you have a perfectly spherical water tank with an inside diameter of 8.6 metres. If the drain at the bottom of the tank can't handle a hydrostatic pressure of more than 50 kilopascals, what is the maximum volume of water, in litres, that can be contained in the tank? Assume that gravitational acceleration is exactly 9.81 m/s

^{2}. Please round to the nearest 10 litre increment, and please submit only a number for your answer. (For example, if you calculate the answer to be 16277 litres, submit 16280 as your answer)

Prize

__Underwater Tour____Click to show/hide solution__

First off, we need to find the maximum height of the water tank that the drain can handle using hydrostatic pressure equation [1]:

P = rho * g * h <=> h = P / (rho * g) = 5.096839959225281 m

where P = 50k Pa = 50,000 kg/m·s

^{2 }is the maximum hydrostatic pressure, rho = 1000 kg/m

^{3}is density of water, g = 9.81 m/s

^{2}is the gravitational acceleration. Remember that do not round h yet.

Then, we can start to calculate the partial volume of the sphere by the following formula (explanation of the formula is at the end of the solution):

V = 1000 * pi * h

where r = d/2 = 4.3 m, is the radius of the spherical water tank, h is the above maximum height.

We know that: 1 m

If you are interested to see how the partial volume of a sphere is derived, you can continue to read. Otherwise stop here.

We can consider the sphere with center at origin (above picture). We cut the partial sphere into small slices. Each slice is a circle, perpendicular to the vertical direction. Using Pythagorean theorem [2], we can calculate area of each circle at a certain height x:

Then we do integration on the whole vertical direction to add all area of the slices to get the volume. Since we choose the origin at the center of the sphere, the limit of the integration should be from -r to h-r

References:

[1] Hydrostatic pressure

[2] Pythagorean theorem

^{2}* (r - h/3) = 212276.11where r = d/2 = 4.3 m, is the radius of the spherical water tank, h is the above maximum height.

We know that: 1 m

^{3}= 1000 dm

^{3}= 1000 liters. We have a 1000 factor in the above formula because the answer should be in liters while we calculated in cubic meters. Now perform rounding as the requirements and we get V = 212,280 liters.

__Results__: 462 people guessed the correct answer earning themselves

**4330 NP**each.

If you are interested to see how the partial volume of a sphere is derived, you can continue to read. Otherwise stop here.

We can consider the sphere with center at origin (above picture). We cut the partial sphere into small slices. Each slice is a circle, perpendicular to the vertical direction. Using Pythagorean theorem [2], we can calculate area of each circle at a certain height x:

Then we do integration on the whole vertical direction to add all area of the slices to get the volume. Since we choose the origin at the center of the sphere, the limit of the integration should be from -r to h-r

References:

[1] Hydrostatic pressure

[2] Pythagorean theorem

## 14 comments:

Awesome thank you! I couldn't figure out the necessary formulas.

Good luck on the contest this round :)

Whats the answer

thanks...I had figured out the hydrostatic pressure part but got stuck on the volume part..calculus just isn't my thing anymore.

You're welcome. Good luck submitting answer.

Very clear hints. This is much appreciated. I much prefer this way of doing things (hints) than the sites who post anwers. Thank you !

No problem. I prefer not giving out answer because those who at least working on the problem deserve the prize.

I came up with a number that ends in 233 should I round down to 230litres then?

Did anyone else get a number with the same last digits or am I off?

you should round down to xxx230 if you get xxx233. However, I didn't get that number. You may want to double check. If you think that number is right, you can submit.

Much clearer than my original plan, which included atmospheric pressure (meaning that the formula would've been Pa = rho*g*h + P(0)e^(-Mgh/RT), with P(0) being the atmos p @ h=0, M in g/mol, etc.).

Should've known that Jhudora can't be THAT evil. Thanks.

P.S.: btw, I found an easier way to solve this w/o the slice formula; see "http://mathforum.org/dr.math/

faq/formulas/faq.sphere.html" under "Spherical Cap" for full explanation. Basically, you calculate r1^2, which is simply 2hr-h^2, then plug it into the V=(Pi/6)*(3r1^2+h^2)*h equation. Finally, subtract that value from the total potential V=(4 Pi/3)*r^3, & convert from cubic meters to liters.

Thanks for letting me know. I believe the way they calculate V also involve into some sort of integration.

If you use their formula, h should always only smaller than r. In the formula I presented, h can be either small or greater than r and it works well so you can calculate the sphere cap that way (just upside down)

There are always several ways to calculate and pick the way you feel comfortable with.

And I hope you enjoy the puzzle.

Agreed about the variability of h vs. r. If that happens, calculus always comes in handy. Who knows? They might even make us use a shell formula or do vectors next time.

yup. TNT is very tricky with their puzzles. That's why I love to solve their puzzles

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