Need desperate help on fluid dynamics

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Discussion Overview

The discussion revolves around an experiment involving fluid dynamics, specifically the time it takes for water to drain from a bottle through a slot. Participants explore the relationship between the time taken for the water level to drop and the area of the slot, as well as the mathematical modeling of the flow rate and water height over time.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Mike presents data showing an inverse relationship between the time for the water level to reach the slot and the area of the slot, expressed as T=0.0169/A.
  • Some participants suggest that the problem may involve differential equations and reference Bernoulli's law as a potential framework for understanding the flow dynamics.
  • One participant notes that the data fits a parabolic equation for water height over time, prompting questions about the relationship between the height and time.
  • Another participant introduces Darcy's law, discussing the volume flux and hydraulic conductivity, while cautioning about its applicability in turbulent flow conditions.
  • Mike expresses confusion about deriving a formula for instantaneous flow rate and seeks clarification on how to manipulate existing equations.
  • Participants discuss the influence of various factors, such as slot area, fluid density, and viscosity, on fluid velocity and flow rate.
  • There is a mention of a potential contradiction regarding the relationship between fluid velocity and density, which leads to further clarification requests from Mike.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to derive the necessary formulas or the applicability of different laws. Multiple competing views and methods are presented, and the discussion remains unresolved.

Contextual Notes

Participants note limitations in their understanding of viscosity, pressure gradients, and the specific dynamics of fluid flow, which may affect their ability to fully address the problem.

Who May Find This Useful

This discussion may be of interest to students or individuals studying fluid dynamics, particularly those engaged in experimental work or seeking to understand the mathematical modeling of fluid flow in practical scenarios.

  • #31
Dear god yes! upload it to some webhosting site.
 
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  • #33
Alright, I ahve a differential equation set up that my linear algebra professor at school had set up and i have finished solving it from what I know and it gives a quadratic equation, which is not pretty and gives more info but still lacks some stuff as there are still two unknown constants and I have yet to verify if it corresponds to the data, I wil post it as soon as I fugure out how to tyoe it properly as it is not pretty.
 
  • #34
Well, for one thing, your graph is volume vs time that equation is NOT volume vs time.
 
  • #35
h(t) = -((Surf Area of Bottle)(accel due to gravity)(t^2))/((2)(Surf Area of Slot)) - (accel dur to gravity)(konstant)(t) - ((Surf Area of Bottle)(accel due to gravity)(konstant^2))/((2)(Surf Area of Slot)) + (Constant)/(accel due to gravity)

where Konstant is a constant produced by general integration
and
where Constant is the C in Bernoulli's equation found on http://plus.maths.org/issue2/bottle/index.html

and my linear algebra proffessor set the intial differential equation using that Bernoulli's equation and some crazy stuff...

thing is I do not know if it fits the data, which i wil try soon, and I do not knwo how to find Constant or Konstant; however, it is easy to see that the last two terms have no time variable so they make up initial height h(o) so Constant is no longer a problem, however, Komstant is as it appears in the second term
 
Last edited:
  • #36
Looking at your data, your equation matches very well to what vanesche gave.

Look over your work with more care.
 
  • #37
Wait, which equation, the one provided on the site or the one i have just posted?
 
  • #38
The one on the site. I don't know what that mess you posted is, sorry.
 
  • #39
So you manipulated the equation ont he site into something useful where velocity is not included or did you use the equation as is?
 
  • #40
How can that be, I plug 25seconds into that equationa nd get -2.3meters?
 
  • #41
what do you mean velocity, where do you see velocity anywhere?
 
  • #42
"The symbol u stands for the fluid velocity"

on site
 
  • #43
Yes, and where do you see it in the final equation?

Think about conservation of energy.
 
  • #44
Nowhere, ok so your are using the last equation, I thought you rederived another equation from Bernouilli's equation. Yes, I am using that last equation and it does not fit, perhaps my values for lambda are worng. I solved for them and got around 4.0E-3. Correct?
 
  • #45
I got your graph to match, you better redo those calculations boy, or I am going to smack you!
 
  • #46
OKOKOK i will gimme a sec
 
  • #47
OK, i get lambda is either equal to 9.251E-3 or 3.526E-3 with inconsistent units, correct?
 
  • #48
Nope. Check your work some more. You are close though.
 
  • #49
Oh god, look I know you are trying to do this for my own good but I looked over it and found nothing wrong and I have fired up maple to triple check and here.

{x = 0.003525914211}, {x = 0.009251617087}
 
  • #50
Can you write me your equation your using to solve for lambda?

It might be that its rounding off, but I want to be sure its not what I think it might be.
 
  • #51
0.0251 = 3062.500000 x - 39.13118960 x + 0.125

0.0251 = 3062.500000 x - 39.13118960 x + 0.125

"->"
{x = 0.003525914211}, {x = 0.009251617087}
 
  • #52
No, this is not correct.

Where did you get these numbers?

Intercept is ok, but the rest are wrong.

Where did this come from?

0.0251?
3062.500000 ??
39.13118960 ??
 
  • #53
I used the equation and i plugged 25s into time and the corresponding height 0.0251m from my results and i simplified.
 
  • #54
Ah, this is fine.

BUTTT, you should use data points as far as possible, i.e. use 35.

You will find that you get:

0.0035822
0.00551068
 
  • #55
ok, so wat a couple of decmals :-p so, wat are the units of lumbda?

and now wat? there must be a way to find the values before even doing the experiment?
 
  • #56
You tell me.

Not a couple of decimals, a factor of 2.
 
  • #57
It was a joke, i understand.
 
  • #58
No units, makes me feel better i think
 
  • #59
So now i know lambda has no units and u must do the experient to find it, that's not taht good since its not a general formula yet
 
  • #60
Wait, lambda is not consistent, it has not units is in first term and units in second
 

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