Why is Δv Replaced by du/dx in Series Expansion?

Click For Summary

Discussion Overview

The discussion revolves around the use of Taylor series expansion in fluid dynamics, specifically addressing the transition from Δv to du/dx in a given equation. Participants are exploring the mathematical reasoning behind this substitution and its implications in the context of fluid velocity components.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant seeks clarification on the series expansion and the specific equation where Δv is replaced by du/dx.
  • Another participant explains that the Taylor expansion is used to derive the expression, noting that higher-order terms are considered negligible as Δx approaches zero.
  • Some participants mention the swapping of terms and the approximation of Δu and Δv in relation to their partial derivatives with respect to x and y.

Areas of Agreement / Disagreement

There is no clear consensus on the reasoning behind the substitution of Δv with du/dx, as participants are still seeking clarification and further explanation of the mathematical steps involved.

Contextual Notes

The discussion does not resolve the assumptions made in the Taylor expansion or the conditions under which the higher-order terms are deemed negligible.

mech-eng
Messages
826
Reaction score
13
Hi, I stamped at a series expansion. It is probably Taylor. Would you explain it? It's in the vid.

https://confluence.cornell.edu/disp...mics+-+Differential+Form+of+Mass+Conservation

derivative.png


I understand equation 1 in the picture but I do not understand 2.
I cannot understand how Δv=du/dx.

Thank you.
 
Physics news on Phys.org
U is the velocity in the x direction. And they used a Taylor expansion at 2:38 into the video to get the expression you see. They drop the other terms of the series as negligible when ##\Delta x## tends to zero along with the other deltas t and y.
 
jedishrfu said:
U is the velocity in the x direction. And they used a Taylor expansion at 2:38 into the video to get the expression you see. They drop the other terms of the series as negligible when ##\Delta x## tends to zero along with the other deltas t and y.

In 2 Δv disappears instead du/dx and Δy appears.

Thank you.
 
mech-eng said:
In 2 Δv disappears instead du/dx and Δy appears.

Thank you.

They swapped the order of the first two terms in addition to using \Delta u \approx \frac{\partial u}{\partial x} \Delta x and \Delta v \approx \frac{\partial v}{\partial y} \Delta y
 
  • Like
Likes   Reactions: mech-eng and jedishrfu

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
2
Views
1K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 9 ·
Replies
9
Views
4K