Is my lab missing something in their calculation of drag on a cylinder?

In summary, the researcher is trying to find a function that predicts the pressure on a cylinder. The equation they are using is wrong and the research they have done does not help.

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 $$\left(p-p_{0}\right)_{max} \times cos(\theta)$$

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

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.

Last edited:
"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 don't 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.

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.

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,

$$p_{0}+\frac{1}{2}\rho U^{2} = p + \frac{1}{2}\rho v^{2}_{\theta}$$

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

Rearranging for p-p0

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

Subtituting into the equation for Cp:

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

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

$$C_{p} = 1 - 4\sin^{2}\theta$$

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 $$v = 2U\sin\theta$$, 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, that's 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.

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)

What is the drag on a cylinder?

The drag on a cylinder is a force that resists the motion of the cylinder through a fluid, such as air or water. It is caused by the drag force, which is the result of the interaction between the fluid and the surface of the cylinder.

How is the drag on a cylinder calculated?

The drag on a cylinder can be calculated using the drag equation, which takes into account the fluid density, the velocity of the cylinder, the cross-sectional area of the cylinder, and the drag coefficient. The drag coefficient is a dimensionless quantity that depends on the shape of the cylinder and the properties of the fluid.

What factors affect the drag on a cylinder?

The drag on a cylinder is affected by several factors, including the velocity of the cylinder, the cross-sectional area of the cylinder, the density and viscosity of the fluid, and the shape and roughness of the cylinder's surface. Additionally, the angle of attack, or the angle at which the cylinder is moving through the fluid, can also affect the drag.

How does the drag on a cylinder affect its motion?

The drag on a cylinder can significantly affect its motion, as it is a force that acts in the opposite direction to the cylinder's motion. This means that the drag force can slow down the cylinder and also change its direction of motion. A higher drag force will result in a greater slowing down of the cylinder's motion.

How is the drag on a cylinder different from the drag on a sphere?

The drag on a cylinder is different from the drag on a sphere because of their different shapes. A cylinder has a larger cross-sectional area than a sphere, which means it will experience higher drag forces for the same velocity. Additionally, the drag coefficient for a cylinder is also different from that of a sphere, as it depends on the shape and surface properties of each object.

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