How does the size of a hole affect the rate of water drainage from a barrel?

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The discussion revolves around calculating the rate of water drainage from a 33-gallon barrel with varying hole sizes. To determine the flow rate, the depth of water above the hole is crucial, as it affects the velocity of water draining out, calculated using the formula v = sqrt(2gh). The cross-sectional area of the hole also plays a significant role in the volume flow rate. The conversation highlights the need to match the inflow rate with the outflow rate to keep the barrel full. A reference link was shared for further understanding of fluid dynamics related to pressure and drainage.
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I've got a math problem, that's way over my head, here it is.
Figurativly I have a 33 gallon barrel and it has a one inch hole in the bottom, if the barrel is full of water, how many gallons would escape a minute, the top is open, I need to know to figure out how much water being pumped in would be needed to keep the barrel full?
If it turns out to be a simple answer, what would the result be if it was a two inch hole?
 
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You have to put in the same amount as is leaving.
Unless you know the depth (under water) of the hole,
you can't even start (the speed is 0 if depth is 0).
 
Not sure what your saying , but the barrel is full of water when the spigot at the bottom is opened, and the extra water is being fed into it at the top, at this same rate, from another source.
 
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Although I can't figure it out, I've seen in these forums that it can be calculted how long it would take a cubic mile by a cubic mile container of water to finally drain out of a one inch hole, is this a different mindset?
 
It the barrel remain full you need height of the barrel to find the velocity of water draind given by v = sq.rt.(2gh)
then the volume flow rate will be A*v should be equal to the water to be pumped in where A is the cross-section area of the hole.
 
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Just found this http://www.ap.stmarys.ca/demos/content/fluids/pressure_head/pressure_head.html
Thanks
 
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Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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