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across

  • 3 years ago

Let's do some cool math: A tank in the form of a right-circular cylinder standing on end is leaking water through a circular hole in its bottom. Suppose the tank is \(10\) feet in height and has radius \(2\) feet and the circular hole has radius \(1/2\) inch. If the tank is initially full, how long will it take to empty? Whoever gets this right first gets a cookie.

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  1. across
    • 3 years ago
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    Hint 1: Torricelli's law states that\[v=\sqrt{2gh},\]where \(v\) is the speed of the water leaving at the bottom of a tank of height \(h\). \(g\) stands for the acceleration due to gravity, as usual.

  2. LolWolf
    • 3 years ago
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    I really want to work this out... but, unless I'm missing something, using Bernoulli's, it'd take too long, and I don't think I have the attention span...

  3. Algebraic!
    • 3 years ago
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    ~30 min

  4. across
    • 3 years ago
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    @Algebraic!, how did you get that? ;P

  5. Algebraic!
    • 3 years ago
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    math&stuff

  6. across
    • 3 years ago
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    That's not a valid explanation! ;O

  7. Algebraic!
    • 3 years ago
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    where's my cookie?

  8. LolWolf
    • 3 years ago
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    Roughly 30.275 mins.

  9. sara12345
    • 3 years ago
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    idk differential equations, but i can try setting up one :p

  10. LolWolf
    • 3 years ago
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    Yeah, I forgot about Torricelli's law... I had started trying to derive it from Bernoulli's and was like... nope.

  11. sara12345
    • 3 years ago
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    flowrate outside = \(\pi (1/2)^2\sqrt{2gh} \) cubic inch per sec

  12. hartnn
    • 3 years ago
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    using this differential equation : dh/dt = -Ah/Aw sqrt (2gh) take g= 32 ft/s^2 Ah=Area of hole , Aw is area of water ,so Aw/Ah = 576 so (h^(-1/2))dh = -(8/576)dt so 2 (sqrt h) = -(8/576) t put h= 10 to get t = 30.36 minutes.

  13. across
    • 3 years ago
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    You guys are right. The answer is roughly \(30\) minutes. I guess I'll have to start posting Millennium Prize problems. That way, you guys will take at least five minutes longer to answer correctly. ;P

  14. LolWolf
    • 3 years ago
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    I mean, it's not hard to compute, just a bunch of numbers and converting feet to metres. So, converting everything, we get the very 'elegant' expressions of: \[ f_\text{rate}=(0.0127)^2\pi\sqrt{2\cdot3.048g}\approx0.003918\text{ m}^3/\text{s}\\ vf_\text{rate}\approx30.275\text{ mins.} \]

  15. LolWolf
    • 3 years ago
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    I meant: \[ \frac{v}{f_\text{rate}} \]... oops.

  16. Algebraic!
    • 3 years ago
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    still waiting...

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