Fluid Mechanics: Dimensionless Groups Question, explain an outlier

In summary, the conversation revolves around explaining an outlier point in an experiment involving passing air through a wind tunnel with a ball in the center. The parameters V, D, drag, and omega are defined and it is mentioned that the drag coefficient is a function of both the Reynolds number and ωV/D. One person suggests calculating the Reynolds number for each case and another person questions the validity of a theory based on the Reynolds number. The conversation ends with the suggestion to explore a different theory.
  • #1
Kushwoho44
25
1

Homework Statement


Hi guys, I need to explain the outlier point here, which has been shaded in the excel spread sheet when comparing the two dimensionless groups,

The dimenionsless group, drag-coefficient is given by Drag/(density*V^2*D^2)
and dimensionelss group, spin parameter, is given by omega/VD

2lcsqis.png


2n7m53t.png


The Attempt at a Solution


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The question explicitly asks to, "explain the outlier(s)". My friend thinks that it is because the flow for the outlier point, shaded, is extremely turbulent, whereas the others all have a roughly similar reynolds number of ~20,000.

I can't explain it but I don't think this is correct. I think it has to do with the way we vary the product of the velocity and diameter of the ball.
 
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  • #2
What is the experiment, and what are the definitions of the parameters V, D, drag, and omega (including units).

Chet
 
  • #3
The experiment is passing air through a wind-tunnel and keeping a ball of diameter D in the center and then plotting the relevant dimensionless groups against each other.

The parameters have units:
V = ms^-1
D = m
Drag = Newtons
omega = rad/s
 
  • #4
I guess the ball is rotating? Did you calculate the Reynolds number for each case?

Chet
 
  • #5
Kushwoho44 said:
My friend thinks that it is because the flow for the outlier point, shaded, is extremely turbulent, whereas the others all have a roughly similar reynolds number of ~20,000.

I can't explain it but I don't think this is correct. I think it has to do with the way we vary the product of the velocity and diameter of the ball.

What length and speed scales are you using to define the reynolds number?

If you use the diameter of the ball and the velocity of the oncoming fluid, then for the same fluid the reynolds number is proportional to the product of the velocity and diameter (with the constant of proportionality being the fluid density divided by the dynamic viscosity). For the run with V = 10.34 and D = 0.03 this reynolds number is indeed of the order of 20,000, whereas for the run with V = 50 and D = 0.4 the reynolds number is of the order of 1,300,000.
 
  • #6
You should calculate the Reynolds number for each and every case. The drag coefficient is a function of both the Reynolds number and ωV/D.

Chet
 
  • #7
Thanks a lot Chestermiller, this makes sense.

Thanks for the help!
 
  • #8
Chestermiller said:
You should calculate the Reynolds number for each and every case. The drag coefficient is a function of both the Reynolds number and ωV/D.

Chet
what should i do after finding reynolds number? what is the meaning of re number in this case?
 
  • #9
william14835 said:
what should i do after finding reynolds number? what is the meaning of re number in this case?

Well, your friend had a theory about explaining the outlier based on the Reynolds No. of the flow.

Kushwoho44 said:

The question explicitly asks to, "explain the outlier(s)". My friend thinks that it is because the flow for the outlier point, shaded, is extremely turbulent, whereas the others all have a roughly similar reynolds number of ~20,000.

I can't explain it but I don't think this is correct.

If you didn't calculate the Reynolds No. originally, then how do you know Re ~ 20,000?

If you can eliminate your friend's theory as an explanation, then you are free to explore a different theory.
 

Related to Fluid Mechanics: Dimensionless Groups Question, explain an outlier

1. What is the purpose of using dimensionless groups in fluid mechanics?

Dimensionless groups are used in fluid mechanics to simplify and generalize equations and relationships by removing units and focusing on the underlying physical principles. They allow for easier comparisons between different systems and scenarios.

2. How do dimensionless groups help in solving fluid mechanics problems?

Dimensionless groups provide a way to scale and compare different physical quantities in fluid mechanics, making it easier to analyze and solve problems. They also help to identify important parameters and relationships that govern the behavior of fluids.

3. Can you give an example of an important dimensionless group in fluid mechanics?

The Reynolds number (Re) is a commonly used dimensionless group in fluid mechanics, which relates the ratio of inertial forces to viscous forces in a fluid flow. It is used to determine whether a flow is laminar or turbulent, and can help predict the behavior of the flow.

4. What is an outlier in dimensionless groups and why is it important to identify?

An outlier in dimensionless groups refers to a data point that deviates significantly from the expected results or trend. It is important to identify outliers as they can indicate errors in measurements or assumptions, and can affect the accuracy of the analysis and predictions.

5. How can an outlier in dimensionless groups be explained?

An outlier in dimensionless groups can be explained by factors such as experimental errors, incorrect assumptions, or the presence of unexpected physical phenomena. Further investigation and analysis may be needed to fully understand the cause of the outlier and its impact on the results.

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