Bernoulli Principle and fluid velocity

  • Thread starter gloo
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  • #26
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Sorry, @gloo, I've been pretty busy lately. Anyway...



Yes. Mathematically, since the force is the rate of change of momentum, you are looking at ##\dot{m}v## to get force. Since ##\dot{m}## is constant through the tube, it means the force that could be generated by the stream increases proportionally to ##v##. That said, this is just the potential to exert a force with the exit stream compared to the conditions at the inlet. If you increased your pulling force, that wouldn't necessarily follow the same pattern since force and velocity are not necessarily linear (and a very likely to not be).



I couldn't really tell you off the top of my head. That's an enormously complex problem, to be honest. You could measure it experimentally to get some empirical answer, but solving it analytically would be difficult without CFD software. You may be able to get a decent answer by adapting the concept of head loss for pipe constrictions. The reason here is that Bernoulli's equation is not really applicable in situations like this and only gives you an approximate answer in situations like this where viscosity is likely to be quite important. In order to use a Bernoulli-like analysis, you need to incorporate the concept of head loss to correct the Bernoulli equation. You could therefore likely come up with a relative efficiency be comparing the head loss associated with a smooth taper compared to that of an abrupt area change.



Keep in mind that kinetic energy in a moving fluid is typically represented by dynamics pressure, ##q = \frac{1}{2}\rho v^2##. The stream with the higher dynamic pressure (kinetic energy) will certainly travel farther, but keep in mind that a design with a closed end with a hole in it can still produce the same outlet velocity as the tapered design, albeit with a greater pull force required. So for the same velocity, both designs will travel the same distance and be capable of the same impact force.

So, your goal in determining efficiency is to determine either the the difference in force required to achieve the same distance or else apply the same force and compare the distance. The head loss method approach is effectively doing the first of these options. Doing either option experimentally in a pool would be nearly impossible, however, as you would need a way to make sure you are pulling with the same force both times and/or find a way to measure the force accurately, neither of which you could do manually.

Again, thank you for taking the time and explaining things to me with my layman physics knowledge. Ultimately, my aim is searching for the ability to drive a stream of water upward (head) as hard, or high as possible with a given input force (like the Geyser guys cannon toy). My end goal is for the stream of water to be able to have a large enough upward impact force, or to be able to push through a body of water of a certain depth. But my confusion started with the explanation of venturi effect, and applying Bernoulli (still not strong on it) principle.

You said force and velocity increase is not linear, but is the force increase greater, or less in proportion to the velocity?
 
  • #27
boneh3ad
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Generally, the pressure (head) loss due to the shape goes up with the square of the velocity, so the force required is going to rise faster than the velocity being output. If I remember my undergraduate course well enough, the head loss is something like
[tex]h_L = K\dfrac{V^2}{2g}[/tex]
where ##h_L## is the head loss (has units of length), ##K## is a discharge coefficient that depends on the shape, ##V## is the velocity and ##g## is the acceleration due to gravity. Generally speaking, and abrupt area change like a hole in the end of a capped tube is going to have a higher value of ##K## than a smooth taper will. You could further reduce the value of ##K## if you designed the taper with certain curves such that the boundary layer did not separate at any point.
 
  • #28
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Generally, the pressure (head) loss due to the shape goes up with the square of the velocity, so the force required is going to rise faster than the velocity being output. If I remember my undergraduate course well enough, the head loss is something like
[tex]h_L = K\dfrac{V^2}{2g}[/tex]
where ##h_L## is the head loss (has units of length), ##K## is a discharge coefficient that depends on the shape, ##V## is the velocity and ##g## is the acceleration due to gravity. Generally speaking, and abrupt area change like a hole in the end of a capped tube is going to have a higher value of ##K## than a smooth taper will. You could further reduce the value of ##K## if you designed the taper with certain curves such that the boundary layer did not separate at any point.
Ok...so interpreting.... you are saying (will probably get it wrong)

with the less efficient (bigger pressure loss) shape , the force required to achieve the target velocity or head will be higher by some power of 2?
 

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