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HELP! Investigating the rate of discharge of water from a hole in a bucket

  1. Aug 18, 2012 #1
    Hey guys, um i'm currently in grd 11, and for my physics assignment, ive chosen to investigate the flow rate of water from a leaky bucket. It is relatively simply however, I've been trying to find a suitable formula/s to equate the flow rate at different heights, ie, the volume of water over time, (the rate) and the water height. plz note i dont want to cheat or anything, but a formula or a reference or something would be of great help, thanks in advance... :D
    (I might add, the hole is at the bottom of the bucket, on the base, and also, i am aware that the size of the hole will make a difference, lol)(also, its the height of water in the bucket, not the height of the bucket itself, and the bucket is cylindrical)
     
    Last edited: Aug 18, 2012
  2. jcsd
  3. Aug 18, 2012 #2
    I am trying to investigate how the rate of the water leaving the bucket will change as the water level decreases, as less downward force is acting upon the hole,
     
  4. Aug 19, 2012 #3
    plz help lol
     
  5. Aug 19, 2012 #4

    K^2

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    Final result should look something like rate=c*height^n. You'll have to figure out what values for c and n work by taking measurements and trying to fit the data to the equation.

    Once you have data, if you need help fitting it, feel free to ask.
     
  6. Aug 19, 2012 #5
    Thanx so much, and btw, whats c, and n, i mean what do the represent
     
  7. Aug 19, 2012 #6

    K^2

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    They basically represent all of the variables you don't look at explicitly. There are a whole bunch of things going on here, so the formula will be mostly empirical.

    Reason why it's probably going to be a power law dependence like this is because all of the limiting cases are simple power laws. For example, lets say the viscosity isn't a factor at all. In that case, rate=v*A, work done on fluid per unit time is v*A*P. Pressure P=rho*g*h. And energy lost to water flow is (1/2)rho*rate*v^2. So the equation to balance work being done v*A*rho*g*h=(1/2)rho*A*v^3. Or v=sqrt(2*g*h). That gives you rate ~ height^(1/2). On the other hand, if you force fluid through a long, narrow tube, you will get rate ~ height. (I'm not going to go through derivation, as it involves fluid mechanics.) In a realistic case, depending on which factors win out, the actual power will vary. And the actual dependence will be more complex, of course, but I doubt you'll have precision in the experiment to warrant a more complicated fit.
     
  8. Aug 19, 2012 #7
    THANKYOU VERY MUCH!!! lol, big help
     
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