How high can water rise in a stream when a hand is placed in it?

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Discussion Overview

The discussion revolves around the question of how high water can rise on a hand placed in a stream, considering factors such as the speed of the stream and the angle of the arm. Participants explore theoretical approaches using Bernoulli's principle and energy conservation, while also questioning the conditions of the scenario.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant proposes using Bernoulli's equation to relate the pressure and velocity of the water around the arm, suggesting that the height to which water can rise depends on the speed of the stream and the angle of the arm.
  • Another participant questions the conversion of kinetic energy to potential energy, suggesting that if all kinetic energy were converted, the height could be calculated as h = U²/2g.
  • There is a clarification sought regarding whether the hand is in or out of the stream, which influences the applicability of Bernoulli's principle versus simple force resolution.
  • A participant expresses uncertainty about the correctness of their initial approach and seeks guidance on how to proceed with the problem.

Areas of Agreement / Disagreement

Participants express uncertainty and seek clarification on the conditions of the problem, particularly regarding the position of the hand in the stream. There is no consensus on the correct approach or solution, as different methods are proposed and questioned.

Contextual Notes

The discussion includes assumptions about the flow of water and the effects of the arm's angle, which may not be fully resolved. The dependence on the cross-sectional area of the stream and the area occupied by the arm is also mentioned but not elaborated upon.

SimbaTheLion
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"If you put your arm in a stream flowing at U metres per second, how far can water rise up it?"

Not really sure if what I'm doing is right. Presumably it depends on the angle of your arm in the water. If we let the speed of the water at your arm be V, we get, by Bernoulli, taking the height up your arm to be h, the pressure to be p_atm before and p afterwards:

p_atm/ρ + U²/2 = p/ρ + V²/2.

And we know p = p_atm + ρgh.

Which gives:

U²/2 = gh + V²/2.

I'm trying to use mass conservation now, but then don't we need the cross-sectional area of the stream, and the area that is occupied by your arm?

Am I going about the question in completely the wrong way?

Help please :) !
 
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If all of the kinetic energy in the moving fluid were converted to potential (by increasing elevation) how high could it go?
 
gmax137 said:
If all of the kinetic energy in the moving fluid were converted to potential (by increasing elevation) how high could it go?

Well, suppose the fluid had kinetic energy E = mU²/2. Then we need to find h such that mgh = mU²/2. So h = U²/2g. Is that all I have to do? :S
 
Is the hand in the stream or out of it?

If it's in, there there's sense to apply Bernoulli's principal...otherwise it's simple force resolution.

I think.
 
dE_logics said:
Is the hand in the stream or out of it?

If it's in, there there's sense to apply Bernoulli's principal...otherwise it's simple force resolution.

I think.

Yeah, the hand is in the stream. Hmm, so is what I've done in the first post correct? If so, how do I continue?
 

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