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Ideal fluids, bernouilli's law

  • Thread starter lovelyrwwr
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  • #1
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Screenshot_2013-10-16-23-16-47.png


I do not understand why B is not true. There is much greater fluid molecules above point A, and given P = density x g x depth, pressure should be greater at point A.....making B true. But then again if u use poiseuilles' law for nonideal fluids you see that in order to keep flow rate Q constant, if radius is larger, then pressure must be smaller at point A....making B untrue. Could you pls explain how to recincile these cinflictinf cinclusions? Thank u so much !
 
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Answers and Replies

  • #2
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If you wanna get mathematical, pressure can be defined as force per unit area, so

[tex]P = \frac{F}{A}[/tex]

In the case of your picture, the cross sectional area of A is larger than B, so that would mean the pressure is less at A than at B
 
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  • #3
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Thank u for your reply :) but how come we can assume that force is equal at points A @nd B?
 
  • #4
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If you wanna get mathematical, pressure can be defined as force per unit area, so

[tex]P = \frac{F}{A}[/tex]

In the case of your picture, the cross sectional area of A is larger than B, so that would mean the pressure is less at A than at B
The explanation is misleading. One should not analyze the situation from this perspective because doing so will cause you to hold F constant, which doesn't make sense.
 
  • #5
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View attachment 63024

There is much greater fluid molecules above point A, and given P = density x g x depth, pressure should be greater at point A.....making B true.
You are using an equation from fluid statics to analyze a dynamic situation.

In dynamic fluid, pressure is never simply density x g x depth. You can study bernoulli's equation and realize that there are static pressure, dynamic pressure and pressure caused by gravity.
 
  • #6
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Okay I guess it's a little more complicated than that, let's see:

You have your continuity equation for fluids

[tex]\rho_{1}A_{1}v_{1}=\rho_{2}A_{2}v_{2}[/tex]

The problem says that this is an ideal fluid, so the fluid isn't compressible, so

[tex]\rho_{1}=\rho_{2}[/tex]

and

[tex]A_{1}v_{1}=A_{2}v_{2}[/tex]

And since A = F/P

[tex]\frac{F_{1}}{P_{1}}v_{1}=\frac{F_{2}}{P_{2}}v_{2}[/tex]

From a simple thought experiment we can conclude that the velocities are different, think about putting your thumb over a running water hose. And we have already seen that the pressures are the same. This leaves us with

[tex]F_{1}v_{1}=F_{2}v_{2}[/tex]

Since the velocities are different, we can conclude that in fact, the forces must be different if both sides of the equation are to be equal.
 
  • #7
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[tex]\frac{F_{1}}{P_{1}}v_{1}=\frac{F_{2}}{P_{2}}v_{2}[/tex]

From a simple thought experiment we can conclude that the velocities are different, think about putting your thumb over a running water hose. And we have already seen that the pressures are the same. This leaves us with

[tex]F_{1}v_{1}=F_{2}v_{2}[/tex]
##P_{1} = P_{2}## ? How is it possible?
 
  • #8
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Oh wow I just realized I wrote that, allow me to slowly walk away, as I clearly don't know fluid mechanics :sly: I'll go ahead and blame the time of night
 
  • #9
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Thank you both the disvussion was very helpful!!!!
 

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