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Height to time equation: water cylinder with pinhole

  1. Feb 20, 2016 #1
    Hi guys,

    I was just doing an experiment with relating the weight flow rate of water out of cylinder with a pin hole at the bottom. The equation that I put together was: $$\dot{W}=\rho g C_d A \sqrt{2gh}$$ ##\dot{W}## is the weight flow rate, ##\rho## is the density of water, ##g## is the acceleration due to gravity, ##C_d## is the coefficient of discharge of the orifice, ##A## is the area of the orifice, ##h## is the height of the water above the hole. The problem that I am facing is that the weight flow rate changes with time along with the height of the water. Thus, I thought that if I can find an equation for the height as it relates to time, I can plug it into the equation above, and my equation would be complete. I also need to see how you got to whatever equation makes sense. I am looking for something like: ##h = kt##. As in k can be any function including any variables that make sense. I believe this requires derivatives and integrals so don't be afraid to include and explain them.

    Thank you very much for your help.
    Jonathan
     
  2. jcsd
  3. Feb 20, 2016 #2
    If ##A_x## is the cross sectional area of the tank and h(t) is the height of liquid in the tank, what is the weight of liquid in the tank at any time?
     
  4. Feb 21, 2016 #3
    The weight of the water in the bottle changes in relation to time. At 11.06 seconds, it is 10.51481546 N, at 46.18 seconds, it is 5.507255163 N. The ##A## in the equation above is the area of the hole. I was hoping to include the ##A_x## in the height-time equation. This is what Wikipedia gave:

    Now, 3a8d6169bfcaad15b3368f7709400d75.png where A and a are the cross sections of container and tube respectively, dh is the height of liquid in container corresponding to dx in the tube which decreases in same time dt.
    db7e67a23640989fd765cc4ea061eaf7.png
    2d15cfc92ff3e844b674bca0e0b7da16.png
    843aaa5948420f8c9e3237089a36a50f.png
    25d2e4208b8f845fe67918061931c1e3.png
    c2dafac728b624e1933984f44ec09dba.png
    a3f81f8dd4e5ef2bad91ebce74acbbb0.png
    I guess this makes sense if you are solving for time. I want to solve for height. First of all, I don't know where the ##\frac{Adh}{a\sqrt(2gh)}## came from. Second of all, I don't understand how they integrated ##\int\frac{Adh}{a\sqrt(2gh)}##, did they move the dh in the numerator out to become ##\int\frac{A}{a\sqrt(2gh)}dh##? If what they have done is right, then can you please rearrange and solve the derivative and integral in terms of height and show steps, and if it is not right then can you please propose and explain your solution Thank you for your help,

    Jonathan
     
  5. Feb 21, 2016 #4
    Well, I guess I'll have to answer my own question.

    $$W=\rho g A_xh$$So,$$\frac{dW}{dt}=\rho g A_x\frac{dh}{dt}=-\rho g C_DA\sqrt{2gh}$$
    This gives me:$$\frac{dh}{dt}=-\frac{A}{A_x}C_D\sqrt{2gh}$$
    Do you know how to solve this equation for h as a function of t, subject to the initial condition h = h0 at t = 0?
     
  6. Feb 21, 2016 #5
    Thank you very much for your help, I really do appreciate it but, if I am talking complete nonsense it is because I am in Grade 11 and teaching myself calculus on my own time. So...please bear with me. I'm not sure how to deal with the initial conditions that you set but, I think what you can do is multiply both sides by dt and take the integral of both sides, exposing the height on the left.
    $$dh=-\frac{A}{A_x}dtC_D\sqrt{2gh}$$
    $$\int dh=\int-\frac{A}{A_x}dtC_D\sqrt{2gh}$$ This would give...
    $$h= ...something\ t\ something...$$
    I also wanted to get rid of the h under the root sign, and just have h = some function and variables and t.
    I know that this isn't solved or probably anywhere close to what you wanted but this, this is as far as I have gone, after which I don't know how to proceed. If I am at least on the right track, please say so and if I am not, please correct me. Thank you again for your patience.

    Jonathan
     
  7. Feb 21, 2016 #6
    Suppose I rewrote the equation as $$h^{-1/2}dh=-\frac{A}{A_x}C_D\sqrt{2g}dt$$
    Could you integrate each side of the equation then, at least as an indefinite integral (i.e., with an arbitrary constant)?
     
  8. Feb 21, 2016 #7
    Would this be:
    $$\int \frac{dh}{\sqrt{h}} =\int -\frac{A}{A_x}C_D\sqrt{2g}dt$$ By factoring constants:
    $$\int \frac{dh}{\sqrt{h}} = -\frac{A}{A_x}C_D\sqrt{2g}\int dt$$ Solving:
    $$ 2\sqrt{h} = -\frac{A}{A_x}C_Dt\sqrt{2g}$$ Solving for h:
    $$ h = \frac{2gt^2A^2C_D^2}{4A_x^2}$$
    Did I do this correctly?
     
  9. Feb 21, 2016 #8
    Not quite. You left out the constant of integration:
    $$ 2\sqrt{h} = -\frac{A}{A_x}C_Dt\sqrt{2g}+C$$
    To determine the constant of integration, you need to substitute the initial condition into this equation. What do you get for C? When you substitute this back into the equation for arbitrary h, what do you get then?
     
  10. Feb 21, 2016 #9
    When ##h = h_0##, ##t = 0##, and ##h_0 = 0.14## m, I get ##C = 2\sqrt{0.14}## or ##\frac{\sqrt{14}}{5}## or ##0.7483314774##
     
  11. Feb 21, 2016 #10
    This is correct, but I has hoping that you would substitute ##C=2\sqrt{h_0}## algebraically into the equation for h, and then solve for h. Alternately, you could leave the equation in terms of ##\sqrt {h}##. This would show that ##\sqrt {h}## is a linear function of t, with an initial value of ##\sqrt {h_0}##. What would you get if you plotted your data on a graph of ##\sqrt {h}## versus t? What would be the slope and what would be the intercept? What would you get if you plotted ##\sqrt{W}## vs t?
     
  12. Feb 21, 2016 #11
    With time on the x axis and the ##\sqrt{h}## on the y-axis, I get the slope to be ##-0.002625## and the intercept is ##0.3670##. The correlation is -0.9998.

    upload_2016-2-21_12-54-32.png
     
  13. Feb 21, 2016 #12
    Excellent. Now the slope should be equal to some combination of the parameters in your model. Algebraically, from your final equation, what is the slope equal to? If you know the pinhole area, you should be able to use the slope of your graph to determine the discharge coefficient.
     
  14. Feb 21, 2016 #13
    um... I got ##9014.33573##
    ##A = 0.0000001781393481## ##m^2## (area of the pin hole)
    ##A_x=0.0094717086## ##m^2## (cross sectional area of the cylinder)
    ##g = 9.81## ##ms^-2##
    ##C=\frac{\sqrt{14}}{5}##
    ##m = -0.002625## (slope of the graph)
    $$C_D=\frac{m-C}{-\frac{A}{A_x}\sqrt{2g}}$$
    I don't think that is the right answer because when I looked up the coefficient of discharge for a hole like mine, I got 0.98... is there anything that I did wrong?
     
  15. Feb 21, 2016 #14
    The C shouldn't be in there.
     
  16. Feb 21, 2016 #15
    Without the C I get 31.50999077, it's still way off.
     
  17. Feb 21, 2016 #16
    How accurate is the pin hole area?
     
  18. Feb 21, 2016 #17
    I used a ##\frac{3}{16}## inch diameter drillbit.
     
  19. Feb 21, 2016 #18
    For a 3/16" hole, 1 get 100 x your area.
     
  20. Feb 21, 2016 #19
    True, you are right, the coefficient that I get is 0.315, thank you very much.
     
  21. Feb 21, 2016 #20
    Not so fast. That's still pretty far off. Is there a factor of ##\pi## missing somewhere?
     
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