Does a Styrofoam Ball Rise in a 30 ft Water-Filled Pipe?

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A Styrofoam ball placed in a 30 ft water-filled pipe will rise due to buoyancy, as the constant water flow does not create a pressure differential. The ball's ascent will be influenced by drag from the downward-moving water, which will slow its speed. If the downward water flow is faster than the ball's rise speed, the ball will descend; if slower, it will rise. The ball's rise speed will be determined by its buoyancy and the size of the gap between it and the pipe walls. Ultimately, the interaction between buoyancy and drag will dictate the ball's behavior in the flowing water.
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i have a 30 ft pipe - 10" diameter. there is a constant water flow at the top, keeping the water level constant ( the pipe is always full to the top). at the bottom the water just flows out of the basin and away. the pipe is is just 1" underwater in the pool underneath.
the yellow object is a styrofoam ball - diameter 9" so it will easily fit into the pipe.
will the ball rise up due to buoyancy and density differencial, or will the pressure from the water column above supress these forces and keep it at the bottom.
if the ball does rise, will its speed be slowed by the downward flow of water around it?
what will its estimate spped be, compared to the clasiccla setup where an object is released in a static water column, closed at the bottom .

thanks
 

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I could be wrong, but:

I believe the ball would rise. The water level in the pipe is always the same, so it is as though the ball were placed in a water-filled pipe with a bottom. Therefore there is not a pressure differential because water is added and replaced at the same rate. If the ball were placed at the top of the pipe it would not be sucked downward either.

As to the velocity of its rise compared to a static column: I believe based on the above that it would be equivalent.

Great question, BTW.
 
Last edited:
thanks for the reply.
anybody else?
 
I believe the ball would rise and its speed would be slowed by the water moving past it, but this is a very difficult question to answer mathematically. It isn't an issue of static water pressure, it is an issue of drag.
 
in experimenting with something like this a while back (mine was ping pong balls in tubes)

the ball will try to rise at some relatively slow speed, based on the buoyancy of the ball and the orifice formed between the ball and the side walls of the tube. the tighter the size of the ball to the tube the slower it will rise.

now given that rise speed (which will be constant regardless of water column above it) if the water is flowing down the tube faster than that rise speed, the ball will travel down, if the water is flowing slower than that speed, the ball will rise. if the two speeds match the ball will hold its relative position in the tube.

:) hope this helps
 
I agree with Russ, it's a matter of drag. I imagine that if the drag force was equal to the buoyant force it would not rise. On the other hand if the drag force wasn't large enough it would rise. Dunno for sure...

CS
 
I think it's easist first to watch a short vidio clip I find these videos very relaxing to watch .. I got to thinking is this being done in the most efficient way? The sand has to be suspended in the water to move it to the outlet ... The faster the water , the more turbulance and the sand stays suspended, so it seems to me the rule of thumb is the hose be aimed towards the outlet at all times .. Many times the workers hit the sand directly which will greatly reduce the water...
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