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Drag on a Cylinder - Confused

  • Thread starter saad87
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  • #1
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Homework Statement


I'm writing a lab report on drag on a non-rotating cylinder. The drag coefficient is calculated using the Pressure co-efficient. The problem I'm facing is that my lab states that the pressure on the cylinder can be predicted by [tex]\left(p-p_{0}\right)_{max} \times cos(\theta) [/tex]



The Attempt at a Solution


The research I've done on the subject leads me to conclude that the pressure on a cylinder can be predicted by [tex]\left(p-p_{0}\right)_{max} \times \left(1-4sin^{2}(\theta)[/tex] instead of the above mentioned equation.

The following is a plot I did in pylab with the experimental plot in blue and the therotical plot in green (using the 2nd equation). The y-axis represents the coeffiecent of pressure while the x-axis is the position from the stagnation point on the cylinder, in degrees.

drag-cylinder.png


My question is, am I missing something? Or is my lab wrong on this? I've been literally reading every material I can get my hands on for the past 3 or 4 hours and I'm still terribly confused.

EDIT: The following is the text from my lab notebook which confuses me.

Drag acts in the plane of motion so the pressure acting in this plane on an element at θ degrees to the cylinder is (p-p0)cosθ. The area it acts upon is small and equal to Rδθ for a unit length of the cylinder. The force on the element in the plane of the drag is the pressure multiplied by the area of the element = (p - p0)Rcosθδθ.
 
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Answers and Replies

  • #2
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"Drag acts in the plane of motion so the pressure acting in this plane on an element at θ degrees to the cylinder is (p-p0)cosθ."

The wording might be a bit confusing.

It should read something like: The contribution of pressure to Drag is (p-p0) cosθ at an angle θ. Since force is calculated by the following equation

F = - INTEGRAL (p * n )dS , wheren is the normal vector.

it is easy to see that the contribution to drag (Fx: force in the x-direction) is in fact (p-p0) cosθ. (simply by using Cartesian coordinates)


**sorry, i dont know how to add all the fancy math symbols.

following through your calculations you should get
[text]\left(p-p_{0}\right)_{max} \times \left(1-4sin^{2}(\theta)[/tex]

"The force on the element in the plane of the drag is the pressure multiplied by the area of the element = (p - p0)Rcosθδθ."

basically the equation i wrote above for force, but this is for a differential force (dFx).

hopefully this was of some use.
 
  • #3
4
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I'm sorry,
this is wrong:

\left(p-p_{0}\right)_{max} \times \left(1-4sin^{2}(\theta)

you should solve for p-p0 through Bernoulli's, but the rest i wrote should be okay.
 
  • #4
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To get 1-4sin^2(x), I did use Bernoulli's. The following page gives the derivation, you'll have to scroll down a bit.

http://www.ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&topic=fl&chap_sec=07.4&page=theory

Since,

[tex]p_{0}+\frac{1}{2}\rho U^{2} = p + \frac{1}{2}\rho v^{2}_{\theta}[/tex]

where U is the velocity of the air flow and v is the velocity of the airflow at [tex]\theta[/tex]. v is given by v = [tex]2U\sin\theta[/tex]

Rearranging for p-p0

[tex]p-p_{0} = \frac{1}{2}\rho\left(U^{2} - v^{2}_{\theta}\right)}[/tex]

Subtituting into the equation for Cp:

[tex]C_{p} = \frac{\frac{1}{2}\rho\left(U^{2} - v^{2}_{\theta}\right)}{\frac{1}{2}\rho U^{2}}[/tex]

If I substitute [tex] v = 2U\sin\theta[/tex] in the above equation and cancel various terms, I end up with the following:

[tex]C_{p} = 1 - 4\sin^{2}\theta[/tex]

Does this make sense? I have not studied this material into this much depth, I knew Bernoulli but I don't know how we end up with [tex] v = 2U\sin\theta[/tex], so I just took it on faith from the above mentioned website and did the derivation.

I just want to show a theoretical pressure distribution i.e pressure distribution around a cylinder in a ideal flow, thats why I'm hunting for a function (and ended up with the above mentioned one) for Cp in terms of the position of the cylinder.
 
  • #5
4
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your equations are correct. It seems to me that if you read the section "Flow Past a Fixed Circular Cylinder" in the link you provided, you will know how v = 2 U sin(theta)
 

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