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Drag on a Cylinder - Confused

  1. Apr 10, 2009 #1
    1. The problem statement, all variables and given/known data
    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]



    3. 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.

     
    Last edited: Apr 10, 2009
  2. jcsd
  3. Apr 12, 2009 #2
    "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.
     
  4. Apr 12, 2009 #3
    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.
     
  5. Apr 12, 2009 #4
    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.
     
  6. Apr 12, 2009 #5
    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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