Calculate the drag force but I dont know the drag coefficient?

AI Thread Summary
To calculate drag force without a known drag coefficient, one can refer to tables for simple shapes, such as a sphere with a drag coefficient of 0.47. However, this value may not be accurate for low Reynolds number scenarios, such as a sphere moving through oil. The drag coefficient is highly dependent on the Reynolds number, which in this case is calculated to be approximately 9.9. For low Reynolds numbers, alternative formulas like Stokes' law may apply, where the drag coefficient is given by C_D = 24/Re_D. The frontal area used in the drag equation should be the sphere's cross-sectional area, calculated as (pi/4) d^2.
skaboy607
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Hi

In a Drag problem, I'm trying to calculate the drag force but I don't know the drag coefficient? Is there any way to calculate it?

Thanks
 
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Ok, I am trying to do a tutorial sheet on it, how do I calculate it say for sphere?

thanks
 
If you click the link there is a table that lists the Drag Coefficient for a sphere as C_d = 0.47
 
Yea I have used that but it doesn't give me the required answer.
 
Do you mean how do you calculate that it is 0.47 for a sphere, or how do you calculate the drag for a sphere in given conditions?

The drag equation ( propertional Area * velocity^2) is an approximation for high Reynolds number flow (eg air) it isn't necessarily correct for low speed or high viscosity cases.
 
Using the Drag force equation I am trying the force on a sphere as it moves through an oil. The only unknown that I have is the drag coefficient? And if I use 0.47, it doesn't give me the right answer.

Thanks
 
mgb_phys said:
The drag equation ( propertional Area * velocity^2) is an approximation for high Reynolds number flow (eg air) it isn't necessarily correct for low speed or high viscosity cases.

The drag coefficient is quite a function of Reynolds, and potentially other factors. Man, I must be in a good mood today. Let's see what I can find. For REALLY low Reynolds numbers, (Re < 1), we have

C_f = \frac{24}{R^*} \left( 1 + \frac{3}{16}R^* - \frac{7k}{48}R^* \right)\,\,R^*=2R
Not sure why it's written like that, but oh well. R\equiv Reynolds number of course. k = V^* / U_\infty where V* is the radial velocity of blowing through the surface...which I assume you can take to be zero in your case.

There is also a "famous" Oseen's (1910) drag coefficient forumula for a sphere in uniform stream:
C_D = \frac{24}{{Re}_D}\left(1+\frac{3}{16}{Re}_D\right)

Stokes gave an exact solution in the limit as Re->0, such as creeping flow, where:
C_D = \frac{24}{{Re}_D}
However, that's only valid where Reynolds is less than 0.2.

What type of Reynolds are you looking at?
 
Well I calculate my Reynolds number to be 9.9, the velocity of the sphere is 0.08m/s, the density is 850, diameter of the sphere is 14.7(10)^-3, and the viscosity is 0.1 which yields 9.996?

Also from the drag equation, where I use area, will it be the surface area of the sphere, i.e. d^2*pi.

Thanks
 
  • #10
skaboy607 said:
Also from the drag equation, where I use area, will it be the surface area of the sphere, i.e. d^2*pi.

Thanks
The area you use is the frontal area, which is pi r^2 or (pi/4) d^2.
 
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