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HELP! Investigating the rate of discharge of water from a hole in a bucket |
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| Aug18-12, 07:59 AM | #1 |
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HELP! Investigating the rate of discharge of water from a hole in a bucket
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) |
| Aug18-12, 08:11 AM | #2 |
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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,
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| Aug19-12, 01:08 AM | #3 |
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plz help lol
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| Aug19-12, 02:52 AM | #4 |
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Recognitions:
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HELP! Investigating the rate of discharge of water from a hole in a bucket
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. |
| Aug19-12, 02:54 AM | #5 |
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| Aug19-12, 03:50 AM | #6 |
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Recognitions:
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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. |
| Aug19-12, 03:58 AM | #7 |
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THANKYOU VERY MUCH!!! lol, big help
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| flow rate, water flow rate |
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