Solving Boat Puncture Problem: Gauge P & Water Velocity

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The gauge pressure at the hole in the boat is calculated to be 1960 Pa, based on the depth of 20 cm. However, the pressure at the hole itself is effectively zero due to atmospheric exposure, leading to a flow of water towards it. Bernoulli's principle indicates that the water velocity entering the hole is influenced by the pressure difference, but calculations initially yielded incorrect results. The discussion clarifies that the pressure is reduced near the hole, causing water to flow in and potentially spray upwards. The upward pressure causing water to enter the hole is related to the pressure differential created by the flowing water.
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The problem statement:

You and a friend are in a boat in a river, the combined mass of this system is 200kg.
The boat hits a rock and a hole of 1-cm diameter forms in the bottom of the boat, which is 20cm below the water line.

1) What is the gauge pressure of the water at the depth of the hole?
2) What is the velocity of the water entering the hole?
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The attempt at a solution:

I got 1) fine, by using P=ρgh, and got 1960Pa.

For 2) I'm trying to use Bernoulli's principle (P + 1/2*ρv2 + ρgy = constant), but I don't know if this is valid seeing as we're talking about vertical (and not horizontal) velocity.

So I made the LHS of Bernoulli apply to the water surface, where P= 0 (gauge pressure is relative to atmospheric pressure), v= 0 and y= 0, so the entire LHS= 0.
Then for the RHS at the hole, P= 1960Pa, y= -0.2m, and v is unknown,
Solving this got me v=0, however. Even making y positive did nothing, it gave imaginary values.

I've even tried using P=F/A to work out the force pushing upwards at the hole, but that gets me nowhere also.

Help would be much appreciated, thanks!
 
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At the location of the hole, the pressure is not the pressure that you found in part (1). The hole is exposed to the atmosphere.
 
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Ah that makes a bit more sense.
Maybe I'm struggling conceptually then.

So is the pressure at -20cm equal to 1960Pa everywhere except for directly underneath the area of the hole where it's equal to 0?
 
The pressure would be 1960 Pa at -20 cm for points where the water is not flowing (say, far from the hole in the boat). In the region underwater near the hole, the water is flowing along streamlines toward the hole. The closer to the hole, the faster the water flows and the pressure is reduced more and more from what it would be if the water were at rest. At the hole, the gauge pressure is reduced back to 0. So, it depends on how close to the hole you get as to what the pressure would be. Hope this makes sense.
 
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Awesome, completely makes sense. So the reduced pressure of the hole causes the water to flow towards it and ultimately 'spray' upwards. (I assume it would reach a max height of 20cm to once again obey Bernoulli's equation)

Really appreciate the help.
 
Yes, according to the result using Bernoulli's equation, if the water is moving vertically upward through the hole, then the water would reach a maximum height equal to the level of the water in the lake. In reality, it would not reach that height due to viscosity, turbulence, etc. Bernoulli's equation neglects these effects.
 
I have a less technical question about the hole in the boat. What is the term for the upward pressure that causes the water to enter the hole? Thanks!
 
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