Question about the Derivation of the Stream-function for a Doublet

  • Thread starter Thread starter Master1022
  • Start date Start date
  • Tags Tags
    Derivation
Click For Summary
SUMMARY

The discussion centers on the derivation of the stream function \(\psi\) for a doublet in fluid dynamics. Participants explore the necessity of dividing by the distance between two sources, denoted as \(s\), before taking the limit, emphasizing that this step is crucial for accurately representing the behavior of the system as the sources approach each other. Additionally, the conversation addresses the implications of swapping the positions of the source and sink, leading to different signs in the resulting equations, which reflects the physical reality of fluid velocities being in opposite directions. The participants confirm that the mathematical treatment aligns with the expected physical outcomes.

PREREQUISITES
  • Understanding of fluid dynamics concepts, specifically doublets
  • Familiarity with stream functions in fluid mechanics
  • Knowledge of limits and calculus, particularly in the context of approaching singularities
  • Basic algebraic manipulation skills for handling trigonometric functions
NEXT STEPS
  • Study the derivation of the stream function for doublets in fluid dynamics
  • Learn about the physical interpretation of fluid velocities derived from stream functions
  • Explore the mathematical implications of limits in calculus, particularly in fluid mechanics
  • Investigate the effects of source and sink placement on flow patterns in fluid dynamics
USEFUL FOR

Students and professionals in fluid dynamics, physicists, and engineers interested in understanding the mathematical foundations of fluid flow behavior, particularly in relation to doublets and stream functions.

Master1022
Messages
590
Reaction score
116
TL;DR
This is a question about the derivation of the stream-function of a doublet using potential flow theory
Hi,

I just wanted to ask a question about the derivation of the stream function \psi for a doublet. In the pictures below is a derivation (in this one the source is on the left and the sink is on the right). I understand everything in the left photo, however my questions are:

1) Why do we divide by the distance between two sources s before taking the limit? I cannot really understand the reason provided. I understand that the sources will cancel out, but am unclear as to why that means that we need to counter that by including a \frac{1}{s} term.

2) Why does the derivation come out differently if we swap the placements of the source and sink? Intuitively, I feel that I have made an algebraic error, but I fail to see it. If we place the source on the right and the sink on the left, then we will get:
\psi = \frac{m}{2 \pi} \theta_{source} - \frac{m}{2 \pi} \theta_{sink} = \frac{m}{2 \pi} \left( atan(\frac{y}{x-s}) - atan(\frac{y}{x+s}) \right). Now we will get the same expression as before (except we have +m instead of -m), except we will end up with a different sign at the end. Surely, the end result should be the same irrespective of the original geometry.

I would appreciate any help. Thanks in advance.

Screen Shot 2020-01-12 at 12.03.18 PM.png
Screen Shot 2020-01-12 at 12.03.24 PM.png
 
Engineering news on Phys.org
If you swap the locations, the fluid velocities (calculated from the derivatives of the stream function) are, as expected physically, in the opposite direction. So what's the problem?
 
Chestermiller said:
If you swap the locations, the fluid velocities (calculated from the derivatives of the stream function) are, as expected physically, in the opposite direction. So what's the problem?
Ah yes, that is true. Thank you.

Do you know the reason behind the division by \frac{1}{s}?
 
Master1022 said:
Ah yes, that is true. Thank you.

Do you know the reason behind the division by \frac{1}{s}?
You are determining the limit as the two charges are very close together.
 
Chestermiller said:
You are determining the limit as the two charges are very close together.
I understand the intuition, but why does that mean that we need a \frac{1}{s} term
 
I have encountered a vertically oriented hydraulic cylinder that is designed to actuate and slice heavy cabling into sections with a blade. The cylinder is quite small (around 1.5 inches in diameter) and has an equally small stroke. The cylinder is single acting (i.e. it is pressurized from the bottom, and vented to atmosphere with a spring return, roughly 200lbs of force on the spring). The system operates at roughly 2500 psi. Interestingly, the cylinder has a pin that passes through its...

Similar threads

Replies
0
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
732
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 34 ·
2
Replies
34
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
5
Views
4K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K