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

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SUMMARY

The discussion focuses on calculating the rate of water drainage from a barrel with varying hole sizes. A 33-gallon barrel with a one-inch hole was analyzed, emphasizing the importance of the water depth for determining drainage velocity using the formula v = √(2gh). The volume flow rate, represented as A*v, where A is the cross-sectional area of the hole, must equal the inflow rate to maintain a full barrel. The conversation also touches on the implications of using a two-inch hole for 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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