Navier Stokes Thm Homework: Equations & Solutions

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Discussion Overview

The discussion revolves around a homework problem related to the Navier-Stokes theorem, specifically focusing on the equations and solutions involved. Participants are examining the presence of a variable "h" in the equations and its implications on the solution process.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant questions the introduction of the variable "h" in the equations, noting that it was not present in earlier steps and suggesting that it should not be added as a constant.
  • Another participant points out an oversight in an earlier equation, proposing a corrected expression for a constant involving "h" and other variables.
  • A third participant connects the variable "h" to the potential energy equation, indicating that it factors in the sine of an angle.
  • One participant mentions following the mathematical process until noticing a transcription error regarding the variable "h".

Areas of Agreement / Disagreement

Participants express differing views on the introduction and role of the variable "h" in the equations, indicating that the discussion remains unresolved with multiple competing interpretations of the equations.

Contextual Notes

There are indications of potential oversights in transcription and variable inclusion, but these issues remain unaddressed in terms of resolution.

yecko
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Homework Statement


螢幕快照 2018-04-25 下午8.44.18.png
[/B]
螢幕快照 2018-04-25 下午8.44.41.png


Homework Equations


Navier strokes theorem

The Attempt at a Solution


May I ask why would there suddenly a "h" in the highlighted part?
"h" wasnt existed in the previous steps, which C2=0 shouldn't add height of the liquid as a constant in the formula...
thanks
 

Attachments

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  • 螢幕快照 2018-04-25 下午8.44.18.png
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The oversight occurs earlier, where it should say:

So that the constant in Eqn (2) is
##C_1\ =\ \dfrac{\rho g}\mu h\sin α##
 
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Likes   Reactions: yecko and jedishrfu
That's following the ##U## being potential energy and so extending the ##U = mgh## to factor in the ##sin (\alpha)## right?
 
I followed the maths until I saw where the author had forgotten to transcribe an h[/color].
 

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