Water leaking from box and the relationship of volume and height.












0












$begingroup$


Suppose we have a container that has a base of area $b$ and we fill it up with water.



Volume of water = $b cdot h$, where $h$ is height.



Hence, $mathrm{d}v/mathrm{d}t = b cdot mathrm{d}h/mathrm{d}t$.



The container has a small hole of area $a$ at the bottom corner and so water is constantly leaking.



From my research:



1) I found that $mathrm{d}v/mathrm{d}t = -a cdot $ velocity of water.



2) Velocity = $sqrt{2gh}$ where $g$ is the gravitational constant.



And so $mathrm{d}v/mathrm{d}t = -a cdot sqrt{2gh}$.




Initially I did this:



$mathrm{d}v/mathrm{d}t = -a cdot sqrt{2gh} = B * mathrm{d}h/mathrm{d}t$



$1/sqrt{h}mathrm{d}h = -a sqrt{2g}/b mathrm{d}t$



Integrate both sides



$2sqrt{h} = -a sqrt{2g}/b cdot t$



$h = (a^2 cdot g)/(2b^2) cdot t^2$



But this doesn't make sense.




I realise though that the $h$ from $mathrm{d}v/mathrm{d}t = -a cdot sqrt{2gh}$ is also constantly changing but I'm still not sure what to do. I tried deriving the equation again but it did not get me anywhere.



I want to find the function how much height or volume has decreased after time t. Can someone help?






calculus integration functions derivatives physics







share|cite|improve this question











$endgroup$












  • $begingroup$
    You'll also have to use the equation of continuity.
    $endgroup$
    – Gokul
    Feb 10 '15 at 15:48










  • $begingroup$
    equation of continuity? can you please elaborate?
    $endgroup$
    – Unknown7
    Feb 10 '15 at 15:54










  • $begingroup$
    Do you know how Torricelli's law is derived? If no, how do you know the velocity of efflux is $sqrt{2gh}$? The law (for water flowing out of a container) is derived by using the equation of continuity. Read here
    $endgroup$
    – Gokul
    Feb 10 '15 at 15:58












  • $begingroup$
    Sweet! Thank you so much
    $endgroup$
    – Unknown7
    Feb 10 '15 at 16:28
















0












$begingroup$


Suppose we have a container that has a base of area $b$ and we fill it up with water.



Volume of water = $b cdot h$, where $h$ is height.



Hence, $mathrm{d}v/mathrm{d}t = b cdot mathrm{d}h/mathrm{d}t$.



The container has a small hole of area $a$ at the bottom corner and so water is constantly leaking.



From my research:



1) I found that $mathrm{d}v/mathrm{d}t = -a cdot $ velocity of water.



2) Velocity = $sqrt{2gh}$ where $g$ is the gravitational constant.



And so $mathrm{d}v/mathrm{d}t = -a cdot sqrt{2gh}$.




Initially I did this:



$mathrm{d}v/mathrm{d}t = -a cdot sqrt{2gh} = B * mathrm{d}h/mathrm{d}t$



$1/sqrt{h}mathrm{d}h = -a sqrt{2g}/b mathrm{d}t$



Integrate both sides



$2sqrt{h} = -a sqrt{2g}/b cdot t$



$h = (a^2 cdot g)/(2b^2) cdot t^2$



But this doesn't make sense.




I realise though that the $h$ from $mathrm{d}v/mathrm{d}t = -a cdot sqrt{2gh}$ is also constantly changing but I'm still not sure what to do. I tried deriving the equation again but it did not get me anywhere.



I want to find the function how much height or volume has decreased after time t. Can someone help?






calculus integration functions derivatives physics







share|cite|improve this question











$endgroup$












  • $begingroup$
    You'll also have to use the equation of continuity.
    $endgroup$
    – Gokul
    Feb 10 '15 at 15:48










  • $begingroup$
    equation of continuity? can you please elaborate?
    $endgroup$
    – Unknown7
    Feb 10 '15 at 15:54










  • $begingroup$
    Do you know how Torricelli's law is derived? If no, how do you know the velocity of efflux is $sqrt{2gh}$? The law (for water flowing out of a container) is derived by using the equation of continuity. Read here
    $endgroup$
    – Gokul
    Feb 10 '15 at 15:58












  • $begingroup$
    Sweet! Thank you so much
    $endgroup$
    – Unknown7
    Feb 10 '15 at 16:28














0












0








0





$begingroup$


Suppose we have a container that has a base of area $b$ and we fill it up with water.



Volume of water = $b cdot h$, where $h$ is height.



Hence, $mathrm{d}v/mathrm{d}t = b cdot mathrm{d}h/mathrm{d}t$.



The container has a small hole of area $a$ at the bottom corner and so water is constantly leaking.



From my research:



1) I found that $mathrm{d}v/mathrm{d}t = -a cdot $ velocity of water.



2) Velocity = $sqrt{2gh}$ where $g$ is the gravitational constant.



And so $mathrm{d}v/mathrm{d}t = -a cdot sqrt{2gh}$.




Initially I did this:



$mathrm{d}v/mathrm{d}t = -a cdot sqrt{2gh} = B * mathrm{d}h/mathrm{d}t$



$1/sqrt{h}mathrm{d}h = -a sqrt{2g}/b mathrm{d}t$



Integrate both sides



$2sqrt{h} = -a sqrt{2g}/b cdot t$



$h = (a^2 cdot g)/(2b^2) cdot t^2$



But this doesn't make sense.




I realise though that the $h$ from $mathrm{d}v/mathrm{d}t = -a cdot sqrt{2gh}$ is also constantly changing but I'm still not sure what to do. I tried deriving the equation again but it did not get me anywhere.



I want to find the function how much height or volume has decreased after time t. Can someone help?






calculus integration functions derivatives physics







share|cite|improve this question











$endgroup$




Suppose we have a container that has a base of area $b$ and we fill it up with water.



Volume of water = $b cdot h$, where $h$ is height.



Hence, $mathrm{d}v/mathrm{d}t = b cdot mathrm{d}h/mathrm{d}t$.



The container has a small hole of area $a$ at the bottom corner and so water is constantly leaking.



From my research:



1) I found that $mathrm{d}v/mathrm{d}t = -a cdot $ velocity of water.



2) Velocity = $sqrt{2gh}$ where $g$ is the gravitational constant.



And so $mathrm{d}v/mathrm{d}t = -a cdot sqrt{2gh}$.




Initially I did this:



$mathrm{d}v/mathrm{d}t = -a cdot sqrt{2gh} = B * mathrm{d}h/mathrm{d}t$



$1/sqrt{h}mathrm{d}h = -a sqrt{2g}/b mathrm{d}t$



Integrate both sides



$2sqrt{h} = -a sqrt{2g}/b cdot t$



$h = (a^2 cdot g)/(2b^2) cdot t^2$



But this doesn't make sense.




I realise though that the $h$ from $mathrm{d}v/mathrm{d}t = -a cdot sqrt{2gh}$ is also constantly changing but I'm still not sure what to do. I tried deriving the equation again but it did not get me anywhere.



I want to find the function how much height or volume has decreased after time t. Can someone help?






calculus integration functions derivatives physics




calculus integration functions derivatives physics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 10 '15 at 15:31







Unknown7

















asked Feb 10 '15 at 12:06









Unknown7Unknown7

61




61












  • $begingroup$
    You'll also have to use the equation of continuity.
    $endgroup$
    – Gokul
    Feb 10 '15 at 15:48










  • $begingroup$
    equation of continuity? can you please elaborate?
    $endgroup$
    – Unknown7
    Feb 10 '15 at 15:54










  • $begingroup$
    Do you know how Torricelli's law is derived? If no, how do you know the velocity of efflux is $sqrt{2gh}$? The law (for water flowing out of a container) is derived by using the equation of continuity. Read here
    $endgroup$
    – Gokul
    Feb 10 '15 at 15:58












  • $begingroup$
    Sweet! Thank you so much
    $endgroup$
    – Unknown7
    Feb 10 '15 at 16:28


















  • $begingroup$
    You'll also have to use the equation of continuity.
    $endgroup$
    – Gokul
    Feb 10 '15 at 15:48










  • $begingroup$
    equation of continuity? can you please elaborate?
    $endgroup$
    – Unknown7
    Feb 10 '15 at 15:54










  • $begingroup$
    Do you know how Torricelli's law is derived? If no, how do you know the velocity of efflux is $sqrt{2gh}$? The law (for water flowing out of a container) is derived by using the equation of continuity. Read here
    $endgroup$
    – Gokul
    Feb 10 '15 at 15:58












  • $begingroup$
    Sweet! Thank you so much
    $endgroup$
    – Unknown7
    Feb 10 '15 at 16:28
















$begingroup$
You'll also have to use the equation of continuity.
$endgroup$
– Gokul
Feb 10 '15 at 15:48




$begingroup$
You'll also have to use the equation of continuity.
$endgroup$
– Gokul
Feb 10 '15 at 15:48












$begingroup$
equation of continuity? can you please elaborate?
$endgroup$
– Unknown7
Feb 10 '15 at 15:54




$begingroup$
equation of continuity? can you please elaborate?
$endgroup$
– Unknown7
Feb 10 '15 at 15:54












$begingroup$
Do you know how Torricelli's law is derived? If no, how do you know the velocity of efflux is $sqrt{2gh}$? The law (for water flowing out of a container) is derived by using the equation of continuity. Read here
$endgroup$
– Gokul
Feb 10 '15 at 15:58






$begingroup$
Do you know how Torricelli's law is derived? If no, how do you know the velocity of efflux is $sqrt{2gh}$? The law (for water flowing out of a container) is derived by using the equation of continuity. Read here
$endgroup$
– Gokul
Feb 10 '15 at 15:58














$begingroup$
Sweet! Thank you so much
$endgroup$
– Unknown7
Feb 10 '15 at 16:28




$begingroup$
Sweet! Thank you so much
$endgroup$
– Unknown7
Feb 10 '15 at 16:28










2 Answers
2






active

oldest

votes


















0












$begingroup$

$$huge {rm Toricelli's;Law}\large v=sqrt{frac{2{rm gh}}{1-a^2/b^2}}approxsqrt{2{rm gh}}tag{$all bequiv (a/b)to0$}$$
where $rm a,b,v,g,h$ are cross-section area of hole ,base area of cylinder, velocity of water from hole, Acceleration due to gravity and Height of the liquid above the hole.




This can be derived using equation of continuity:
$$av=bV$$
where $V$ is speed of horizontal surface of water. And Bernoulli's equation:
$$rm P+frac12rho v^2+rho gh= constant$$
where $rm P,rho,v,g,h$ are Pressure,Density,Velocity,Acceleration due to gravity and Height of the point of the liquid where we are calculating the constant.
I believe this can also be derived by Energy Conservation.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    $frac{dV}{dt} = a sqrt{2gy}$



    $frac{asqrt{2g} dt}{A} = -frac{1}{sqrt{y}} dy $



    Upon integration we get $t=frac{Asqrt{2h}}{asqrt{g}}$



    Where $a$ is cross section area of hole and $A$ is cross section of tank $h$ is the height till which water is filled.






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      0












      $begingroup$

      $$huge {rm Toricelli's;Law}\large v=sqrt{frac{2{rm gh}}{1-a^2/b^2}}approxsqrt{2{rm gh}}tag{$all bequiv (a/b)to0$}$$
      where $rm a,b,v,g,h$ are cross-section area of hole ,base area of cylinder, velocity of water from hole, Acceleration due to gravity and Height of the liquid above the hole.




      This can be derived using equation of continuity:
      $$av=bV$$
      where $V$ is speed of horizontal surface of water. And Bernoulli's equation:
      $$rm P+frac12rho v^2+rho gh= constant$$
      where $rm P,rho,v,g,h$ are Pressure,Density,Velocity,Acceleration due to gravity and Height of the point of the liquid where we are calculating the constant.
      I believe this can also be derived by Energy Conservation.






      share|cite|improve this answer











      $endgroup$


















        0












        $begingroup$

        $$huge {rm Toricelli's;Law}\large v=sqrt{frac{2{rm gh}}{1-a^2/b^2}}approxsqrt{2{rm gh}}tag{$all bequiv (a/b)to0$}$$
        where $rm a,b,v,g,h$ are cross-section area of hole ,base area of cylinder, velocity of water from hole, Acceleration due to gravity and Height of the liquid above the hole.




        This can be derived using equation of continuity:
        $$av=bV$$
        where $V$ is speed of horizontal surface of water. And Bernoulli's equation:
        $$rm P+frac12rho v^2+rho gh= constant$$
        where $rm P,rho,v,g,h$ are Pressure,Density,Velocity,Acceleration due to gravity and Height of the point of the liquid where we are calculating the constant.
        I believe this can also be derived by Energy Conservation.






        share|cite|improve this answer











        $endgroup$
















          0












          0








          0





          $begingroup$

          $$huge {rm Toricelli's;Law}\large v=sqrt{frac{2{rm gh}}{1-a^2/b^2}}approxsqrt{2{rm gh}}tag{$all bequiv (a/b)to0$}$$
          where $rm a,b,v,g,h$ are cross-section area of hole ,base area of cylinder, velocity of water from hole, Acceleration due to gravity and Height of the liquid above the hole.




          This can be derived using equation of continuity:
          $$av=bV$$
          where $V$ is speed of horizontal surface of water. And Bernoulli's equation:
          $$rm P+frac12rho v^2+rho gh= constant$$
          where $rm P,rho,v,g,h$ are Pressure,Density,Velocity,Acceleration due to gravity and Height of the point of the liquid where we are calculating the constant.
          I believe this can also be derived by Energy Conservation.






          share|cite|improve this answer











          $endgroup$



          $$huge {rm Toricelli's;Law}\large v=sqrt{frac{2{rm gh}}{1-a^2/b^2}}approxsqrt{2{rm gh}}tag{$all bequiv (a/b)to0$}$$
          where $rm a,b,v,g,h$ are cross-section area of hole ,base area of cylinder, velocity of water from hole, Acceleration due to gravity and Height of the liquid above the hole.




          This can be derived using equation of continuity:
          $$av=bV$$
          where $V$ is speed of horizontal surface of water. And Bernoulli's equation:
          $$rm P+frac12rho v^2+rho gh= constant$$
          where $rm P,rho,v,g,h$ are Pressure,Density,Velocity,Acceleration due to gravity and Height of the point of the liquid where we are calculating the constant.
          I believe this can also be derived by Energy Conservation.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 10 '15 at 20:52

























          answered Feb 10 '15 at 12:14









          RE60KRE60K

          14k22155




          14k22155























              0












              $begingroup$

              $frac{dV}{dt} = a sqrt{2gy}$



              $frac{asqrt{2g} dt}{A} = -frac{1}{sqrt{y}} dy $



              Upon integration we get $t=frac{Asqrt{2h}}{asqrt{g}}$



              Where $a$ is cross section area of hole and $A$ is cross section of tank $h$ is the height till which water is filled.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                $frac{dV}{dt} = a sqrt{2gy}$



                $frac{asqrt{2g} dt}{A} = -frac{1}{sqrt{y}} dy $



                Upon integration we get $t=frac{Asqrt{2h}}{asqrt{g}}$



                Where $a$ is cross section area of hole and $A$ is cross section of tank $h$ is the height till which water is filled.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $frac{dV}{dt} = a sqrt{2gy}$



                  $frac{asqrt{2g} dt}{A} = -frac{1}{sqrt{y}} dy $



                  Upon integration we get $t=frac{Asqrt{2h}}{asqrt{g}}$



                  Where $a$ is cross section area of hole and $A$ is cross section of tank $h$ is the height till which water is filled.






                  share|cite|improve this answer











                  $endgroup$



                  $frac{dV}{dt} = a sqrt{2gy}$



                  $frac{asqrt{2g} dt}{A} = -frac{1}{sqrt{y}} dy $



                  Upon integration we get $t=frac{Asqrt{2h}}{asqrt{g}}$



                  Where $a$ is cross section area of hole and $A$ is cross section of tank $h$ is the height till which water is filled.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 3 '18 at 16:03









                  Math Girl

                  631318




                  631318










                  answered Dec 3 '18 at 15:40









                  user622450user622450

                  1




                  1






























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