Drag Coefficient and Reynolds Number Related to free fall

  • Thread starter lucas
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Homework Statement


Hi, I am doing a theoretical investigation which will be compared to an experimental I'll do later. I am trying to calculate how much time will it take an object to fall a heigh H. the object is a 5x5x5 cm cube. I have the Reynolds number as this Re=3546*Velocity; and I need both the drag coefficient and the terminal velocity.

Air density ([itex]\rho[/itex]): 1.25 kg/m3
Air Dynamic Viscosity ([itex]\mu[/itex]): 1.76E-05
Area: 25 cm3
Cube side length: 0.05 m or 5 cm
Height (H): 30m

Homework Equations


Re=[itex]\frac{\rho*V*L}{\mu}[/itex]

Cd= Fdrag/0.5*ρ*Asurface*V2*Cdrag


The Attempt at a Solution


Re=3546*V
I used classic mechanics to estimate the avarage of V for a height H of 30m, and got a value of 12.15 m/s so Re=3546*12.15.
I'm stuck from here on.
I need the Drag Coefficient, and the terminal V, and at what time does the object reach the velocity.
 

Answers and Replies

  • #2
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You need an equation for the drag coefficient as a function of the Reynolds number. Then you solve by trial and error to determine the velocity for which the drag force is equal to the weight of the cube.

Chet
 
  • #3
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Thank you Chet, but it is that equation that I can't seem to find, as there are a variety of equations; but no specification for which parameters.
For example: Cd=[itex]\frac{0.664}{\sqrt{Re}}[/itex] or Cd=[itex]\frac{1.33}{\sqrt{Re}}[/itex]; and even Cd=0.0742 / Re1/5.
 
  • #4
21,646
4,918
Thank you Chet, but it is that equation that I can't seem to find, as there are a variety of equations; but no specification for which parameters.
For example: Cd=[itex]\frac{0.664}{\sqrt{Re}}[/itex] or Cd=[itex]\frac{1.33}{\sqrt{Re}}[/itex]; and even Cd=0.0742 / Re1/5.
The equations you have are for drag over a flat plate or for pressure drop in a tube. Look up drag over a sphere. This will get you closer to what you want. I don't think you will be able to find an equation for the drag coefficient for a cube, especially since it will vary with angle of attack.

Chet
 

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