Bernoulli's Equation: Explained for Non-Experts

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Bernoulli's Equation describes the relationship between static, dynamic, and positional pressures in a fluid flow, stating that their sum remains constant along a streamline. The discussion highlights confusion regarding the application of the equation and the various forms it can take. Key points include the understanding that static pressure arises from molecular interactions, dynamic pressure from fluid movement, and positional pressure from height differences. The participants clarify that the zero gravitational energy level can be set arbitrarily and explain why dynamic pressure can be negligible in certain contexts. Overall, the thread emphasizes the importance of recognizing how these pressure components interact to solve fluid dynamics problems effectively.
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Ok this stuff is confusing. Can someone give me a dumbified overview of this? I don't get when/how to use the equation. There are so many different forms and variations of pressure, bernoulli's equations it gets really confusing. I don't know what to equate what to what... ugh i don't know how to explain. I looked it up in wolfram and wikipedia and I read through the section numerous times and i still do not get it... :mad:

http://k.ebzlo.com/prob.htm is an example of some problem i don't know how to solve. It shows I got #1 right but i don't know how to get it, i just copied it from the textbook.

help is greatly appreciated.
 
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If you consider a current flow starting from the liquid surface through the straw, you must have that the sum of the static, dynamic and positional pressures must be the same along the flow. At the liquid surface, in the glass, you have
p_0+\rho g h_0
considering the lower end of the straw as zero gravitational energy level. The dynamic pressure can be neglected here because the initial velocity v_{surface}=0.

At the lower straw end you have the total pressure
p_0+\frac{1}{2}\rho v^2.

These two terms must be equal (Bernoulli's law) and v=...
 
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Bernoulli says that there are 3 terms that can add up to give you the total pressure (energy) of the fluid at a given point. That sum should be the same no matter where in your system you are looking. Those 3 items are:

1) P_{o} The fluid's static pressure (pressure created by molecules just bumping into each other)

2) \frac{1}{2}\rho v^2 The fluid's dynamic pressure (pressure created by the fluid moving)

3) \gamma zThe potential or positional pressure (similar to potential energy, it's pressure due to height above a selected level)

You can use this to calculate pressures, velocities, etc... at other points in a a system if you know that data in another part.

Does that help?
 
how do you know where to set height=0?
why is the dynamic pressure in the cup 0? there's fluid going into the straw isn't there?
why doesn't the lower end of the straw have any positional pressure in it? isn't it falling down so there should be force pulling the fluid down the straw? even both sides?

thanks for the explanation thogh.
 
1. you can set the zero gravitational energy level where you want (because in uniform gravitational field you deal with \Delta E_p and not with the absolute value of E_p)
2. Because of S_{glass}\cdot v_1=S_{straw}\cdot v_2 when S_{glass}\gg S_{straw} you have v_1\ll v_2.
 
ohhh i see.. thanks. that made sense.

can you help me with this one? I got the answer to this question already but I don't really get it.
http://k.ebzlo.com/prob2.htm

Can I get an explanation of the summary below part A's answer? the part about how fluid in part A is not pushed up by kenetic energy.. etc. Why doesn't the pressure involve the dynamic pressure of the flow beneath both tubes?
 
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Now you must be able to solve this problem yourself...the sum of the three pressures must be the same in all the points of each streamline of the flow. If you chose the horizontal central streamline, you have the same positional pressure but different dynamic (\frac{1}{2}\rho v_1^2 and \frac{1}{2}\rho v_2^2) and static pressures (p_0+\rho g h_1 and p_0+\rho g h_2).

ALL YOU NEED TO SOLVE THIS PROBLEM IS IN THIS THREAD!
 
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