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Time needed to half a volume of liquid

  1. Jul 6, 2015 #1
    1. The problem statement, all variables and given/known data

    A barrel, open on the top, full of water, has a radium ##R=0.25\text{ m}## and a height ##h=0.75\text{ m}##. At its base there is a hole where a pipe is attached. The pipe's length is ##\ell=1.0\text{ m}## and its radius ##r=0.0019\text{ m}##. I would like to find the time necessary for the barrel to lose half its content, if we know that the density of the liquid in the barrel is ##\rho=1.00\cdot 10^3\text{ kg/m}^3## and its viscosity ##\eta=2.5\cdot10^{-3}\text{ N sm}^{-2}##.

    2. Relevant equations

    I know the dependence of the pressure of a static fluid from its height ##h##:
    ##p=p_{\text{atm}}+\rho g h##,​
    Bernoulli's equation, which I do not think to be useful in such a context where the liquid is viscous, and the formula for the volume flow rate in a pipe having radius ##r## and length ##\ell## with the difference of pressure ##\Delta p## at its ends
    ##Q=\frac{\pi r^4\Delta p}{8\eta \ell}##​

    3. The attempt at a solution

    I think that the pressure at the bottom of the barrel is ##p=p_{\text{atm}}+\rho g h##, where ##p_{\text{atm}}## is the external pressure, and the pressure at the exit of the pipe is ##p_{\text{atm}}##, therefore ##\Delta p=\rho g h##
    ##\frac{dV}{dt}=:Q=\frac{\pi r^4 \rho g h}{8\eta \ell}=\frac{\pi r^4 \rho g \frac{V}{\pi R^2}}{8\eta \ell}##​
    The derivative of the inverse of ##t\mapsto V(t)## is
    ##\frac{dt}{dV}=\frac{1}{Q}=\frac{8\eta \ell R^2}{ r^4 \rho g V}##​
    and the fundamental theorem of calculus would give
    ##t\big(\frac{V}{2}\big)-t(0)=\int_0^{V/2} \frac{dt(v)}{dv}dv=\int_0^{V/2}\frac{8\eta \ell R^2}{ r^4 \rho g v}dv##​
    but this last integral diverges, which means that something went wrong, although I do not see where my error is...
    I heartily thank you for any answer!
     
  2. jcsd
  3. Jul 6, 2015 #2
    dV/dt = - Q (note the negative sign)

    The volume is decreasing. Also, the starting volume is not zero, it is V0.

    Chet
     
    Last edited: Jul 6, 2015
  4. Jul 7, 2015 #3
    Thank you, Chestmiller! I understand your remark, but I intended ##V## to be the volume of water leaked out of the barrel.
    The flow rate of the volume leaving the barrel is ##-Q## and the flow rate of the the volume filling the exterior is ##Q##. I mathematically see that ##\int_{0}^{V_0/2}\frac{1}{Q}dV=+\infty\ne-\int_{V_0}^{V_0/2}\frac{1}{Q}dV## but I think that ##-\int_{V_0}^{V_0/2}\frac{1}{Q}dV## is the time needed for the barrel to "go" from the full volume ##V_0## to its half ##V_0/2## (this integral is finite and I think it to be the correct answer to the exercise) while ##\int_{0}^{V_0/2}\frac{1}{Q}dV## should be the time needed for the exterior to "go" from a dry state to host half the full volume: ##V_0/2##. I have some trouble to understand the reason why the two integral are not identical... Thank you so much again!!!
     
    Last edited: Jul 7, 2015
  5. Jul 7, 2015 #4

    haruspex

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    But that's not the way you used it when substituting for ##\Delta p##.
     
  6. Jul 7, 2015 #5
    I think I have understood now: the variable ##V=\pi R^2 h## in the integral is the volume of the water in the barrel, it is not the volume gone out of the pipe! :headbang:
    Thank you both!!!
     
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