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Math Help - conical tank

  1. #1
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    conical tank

    Use Torricelli's principle to find the time it takes to empty a conical tank of circular cross section standing on its apex whose angle is 45 and has an outlet of cross sectional area 1.0cm. The tank is initially full of water and at time t = 0 the outlet is opened and the water flows out. The initial depth of the water in the tank is 2m.

    Torricelli's principle: √2gh
    Volume of a Cone: V= (1/3) π r h

    I think you set up a differential equation here
    dV/dt = -a√2gh I'm not sure if this is right place to start.
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    Quote Originally Posted by Automaton View Post
    Use Torricelli's principle to find the time it takes to empty a conical tank of circular cross section standing on its apex whose angle is 45 and has an outlet of cross sectional area 1.0cm. The tank is initially full of water and at time t = 0 the outlet is opened and the water flows out. The initial depth of the water in the tank is 2m.

    Torricelli's principle: √2gh
    Volume of a Cone: V= (1/3) π r h

    I think you set up a differential equation here
    dV/dt = -a√2gh I'm not sure if this is right place to start.
    for the cone ...

    r = h\tan(22.5)

    V = \frac{\pi}{3} [h\tan(22.5)]^2 h

    V = \frac{\pi}{3}\tan^2(22.5) h^3

    \frac{dV}{dt} = \pi \tan^2(22.5) h^2 \cdot \frac{dh}{dt}

    let b = \pi \tan^2(22.5)

    \frac{dV}{dt} = bh^2 \cdot \frac{dh}{dt}

    \frac{dV}{dt} = -a\sqrt{2gh} = -a\sqrt{2g} \cdot \sqrt{h}

    let d = a\sqrt{2g}

    \frac{dV}{dt} = -d\sqrt{h}


    bh^2 \cdot \frac{dh}{dt} = -d\sqrt{h}

    h^{\frac{3}{2}} \, dh = -\frac{d}{b} \, dt

    integrate and find h as a function of t.

    I get the time to empty to be about 46 minutes.
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  3. #3
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    is the height not equal to 2m, if the tank is full and has a depth of 2m?
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    Quote Originally Posted by Automaton View Post
    The tank is initially full of water and at time t = 0 the outlet is opened and the water flows out. The initial depth of the water in the tank is 2m.
    does that answer your question?
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    thats where i got my question from in the first place. I wasn't sure if h was known from that information or if h still needed to be found. If i integrate the left side I get
    but i'm still not on how to integrate:
    -d/b dt

    i understand b is just being used to represent the pi*tan(22.5),
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  6. #6
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    Quote Originally Posted by Automaton View Post
    thats where i got my question from in the first place. I wasn't sure if h was known from that information or if h still needed to be found. If i integrate the left side I get
    but i'm still not on how to integrate:
    -d/b dt

    i understand b is just being used to represent the pi*tan(22.5),
    I can't see your image for your integration of the left side. As far as the right side, -\frac{d}{b} is just a constant ...

     <br />
\int -\frac{d}{b} \, dt = -\frac{d}{b}t + C<br />
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  7. #7
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    left side is [2h^(5/2)]/5
    ok thats what i was getting but i wasn't sure because i didn't know what 'd' was.
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  8. #8
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    Quote Originally Posted by Automaton View Post
    left side is [2h^(5/2)]/5
    ok thats what i was getting but i wasn't sure because i didn't know what 'd' was.
    look at my solution again ... d is defined there.
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  9. #9
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    ok now I see it. Sorry about that. I must have over look it. Thanks so much for your help. I was completely lost on this question
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