Getting the rate of drain from a tank [closed]












0














A tank with a top radius of 1m, a bottom radius of 0.5m and a height of 2m is initially filled with water. Water drains through a square hole of side 3cm in the bottom.



How do I get the rate of drain,
begin{equation}
frac{partial V}{partial t}
end{equation}

of the tank?










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closed as off-topic by Shailesh, Jyrki Lahtonen, Leucippus, José Carlos Santos, Mostafa Ayaz Nov 25 at 9:06


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, Jyrki Lahtonen, Leucippus, José Carlos Santos, Mostafa Ayaz

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    0














    A tank with a top radius of 1m, a bottom radius of 0.5m and a height of 2m is initially filled with water. Water drains through a square hole of side 3cm in the bottom.



    How do I get the rate of drain,
    begin{equation}
    frac{partial V}{partial t}
    end{equation}

    of the tank?










    share|cite|improve this question













    closed as off-topic by Shailesh, Jyrki Lahtonen, Leucippus, José Carlos Santos, Mostafa Ayaz Nov 25 at 9:06


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, Jyrki Lahtonen, Leucippus, José Carlos Santos, Mostafa Ayaz

    If this question can be reworded to fit the rules in the help center, please edit the question.
















      0












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      0


      1





      A tank with a top radius of 1m, a bottom radius of 0.5m and a height of 2m is initially filled with water. Water drains through a square hole of side 3cm in the bottom.



      How do I get the rate of drain,
      begin{equation}
      frac{partial V}{partial t}
      end{equation}

      of the tank?










      share|cite|improve this question













      A tank with a top radius of 1m, a bottom radius of 0.5m and a height of 2m is initially filled with water. Water drains through a square hole of side 3cm in the bottom.



      How do I get the rate of drain,
      begin{equation}
      frac{partial V}{partial t}
      end{equation}

      of the tank?







      calculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 25 at 6:24









      Gabriel Henry J. Lopez

      31




      31




      closed as off-topic by Shailesh, Jyrki Lahtonen, Leucippus, José Carlos Santos, Mostafa Ayaz Nov 25 at 9:06


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, Jyrki Lahtonen, Leucippus, José Carlos Santos, Mostafa Ayaz

      If this question can be reworded to fit the rules in the help center, please edit the question.




      closed as off-topic by Shailesh, Jyrki Lahtonen, Leucippus, José Carlos Santos, Mostafa Ayaz Nov 25 at 9:06


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, Jyrki Lahtonen, Leucippus, José Carlos Santos, Mostafa Ayaz

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          1 Answer
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          0














          Conservation of mass and constant density tells us change in volume in the tank equals change in volume out of tank.



          Thinking area times velocity gives you rate of volume change out of tank, by dimensional analysis.



          Bernouli s principle says



          $frac{1}{2}v^2=gh$



          $v=sqrt{2gh}$



          Area=side length of square hole squared.



          Keep in mind h varies with time.



          Volume in tank:
          $V=frac{pi}{3}(h+2)(r_1+frac{h(r_2-r_1)}{2})^2-frac{pi}{3}2r_1^2$



          So :



          $Asqrt{2gh}=frac{dV}{dt}$



          Will give you a differential equation in h as function of t.



          Insert back into expression for V to get it as function of t. Differentiate to find alternative expression for derivative.






          share|cite|improve this answer






























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0














            Conservation of mass and constant density tells us change in volume in the tank equals change in volume out of tank.



            Thinking area times velocity gives you rate of volume change out of tank, by dimensional analysis.



            Bernouli s principle says



            $frac{1}{2}v^2=gh$



            $v=sqrt{2gh}$



            Area=side length of square hole squared.



            Keep in mind h varies with time.



            Volume in tank:
            $V=frac{pi}{3}(h+2)(r_1+frac{h(r_2-r_1)}{2})^2-frac{pi}{3}2r_1^2$



            So :



            $Asqrt{2gh}=frac{dV}{dt}$



            Will give you a differential equation in h as function of t.



            Insert back into expression for V to get it as function of t. Differentiate to find alternative expression for derivative.






            share|cite|improve this answer




























              0














              Conservation of mass and constant density tells us change in volume in the tank equals change in volume out of tank.



              Thinking area times velocity gives you rate of volume change out of tank, by dimensional analysis.



              Bernouli s principle says



              $frac{1}{2}v^2=gh$



              $v=sqrt{2gh}$



              Area=side length of square hole squared.



              Keep in mind h varies with time.



              Volume in tank:
              $V=frac{pi}{3}(h+2)(r_1+frac{h(r_2-r_1)}{2})^2-frac{pi}{3}2r_1^2$



              So :



              $Asqrt{2gh}=frac{dV}{dt}$



              Will give you a differential equation in h as function of t.



              Insert back into expression for V to get it as function of t. Differentiate to find alternative expression for derivative.






              share|cite|improve this answer


























                0












                0








                0






                Conservation of mass and constant density tells us change in volume in the tank equals change in volume out of tank.



                Thinking area times velocity gives you rate of volume change out of tank, by dimensional analysis.



                Bernouli s principle says



                $frac{1}{2}v^2=gh$



                $v=sqrt{2gh}$



                Area=side length of square hole squared.



                Keep in mind h varies with time.



                Volume in tank:
                $V=frac{pi}{3}(h+2)(r_1+frac{h(r_2-r_1)}{2})^2-frac{pi}{3}2r_1^2$



                So :



                $Asqrt{2gh}=frac{dV}{dt}$



                Will give you a differential equation in h as function of t.



                Insert back into expression for V to get it as function of t. Differentiate to find alternative expression for derivative.






                share|cite|improve this answer














                Conservation of mass and constant density tells us change in volume in the tank equals change in volume out of tank.



                Thinking area times velocity gives you rate of volume change out of tank, by dimensional analysis.



                Bernouli s principle says



                $frac{1}{2}v^2=gh$



                $v=sqrt{2gh}$



                Area=side length of square hole squared.



                Keep in mind h varies with time.



                Volume in tank:
                $V=frac{pi}{3}(h+2)(r_1+frac{h(r_2-r_1)}{2})^2-frac{pi}{3}2r_1^2$



                So :



                $Asqrt{2gh}=frac{dV}{dt}$



                Will give you a differential equation in h as function of t.



                Insert back into expression for V to get it as function of t. Differentiate to find alternative expression for derivative.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 25 at 7:31

























                answered Nov 25 at 6:53









                TurlocTheRed

                828311




                828311















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