Deriving Bernoulli's Equation Using Euler's Law of Motion

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Discussion Overview

The discussion revolves around the derivation of Bernoulli's equation using Euler's law of motion, specifically focusing on the pressure differences acting on a particle in motion as illustrated in a referenced figure. Participants explore the implications of pressure changes and net forces in this context.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions why the pressure at the top of the particle is assumed to be higher than at the bottom, suggesting that the net force should align with the particle's motion.
  • Another participant seeks clarification on the assumption that the change in pressure (dP) is greater than zero, indicating uncertainty about the conditions under which this assumption holds.
  • A participant points out that the expression (P+dP) does not necessarily imply that dP is positive, noting that dP could also be negative, which introduces ambiguity regarding pressure changes.

Areas of Agreement / Disagreement

Participants express differing views on the assumptions regarding pressure differences and the implications for net forces, indicating that multiple competing interpretations exist without a clear consensus.

Contextual Notes

The discussion highlights the dependence on the interpretation of pressure changes and the conditions under which these interpretations apply, without resolving the underlying assumptions or mathematical implications.

yjl
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Hi all,

I have attached an image of a page out of the book I am using for context. The blue arrow in Figure 12-3 describes the motion of the particle. I figured the net force would need to be in the same direction, but apparently the net force opposes the motion. So, in Figure 12-3 the pressure at the top of the particle is higher than the pressure at the bottom (P + dP versus P). I am wondering why this is the case.
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Hello yjl, :welcome: !

Where does it say ##dp > 0 ## ?
 
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BvU said:
Hello yjl, :welcome: !

Where does it say ##dp > 0 ## ?

Thank you :-)

Nowhere explicitly. But surely there must be a good reason for assuming the top part of the particle to have a greater pressure.

Is it not true that the net force in Figure 12-3 would make the particle accelerate towards the x-axis?
 
What net force ?
 
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BvU said:
What net force ?

The net force due to the pressure difference when dP > 0.
 
The expression ##(P+dP)## does not mean that dP is positive. It only means that pressure changes from one end of the free body to the other. But dP can also be negative.
 
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Chestermiller said:
The expression ##(P+dP)## does not mean that dP is positive. It only means that pressure changes from one end of the free body to the other. But dP can also be negative.

Ah.
Well, thank you :-).
I see now that that was what BvU was hinting at.
 

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