Drag Coefficient and Reynolds Number Related to free fall

[lucas]

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

Homework Equations

Re=$\frac{\rho*V*L}{\mu}$

Cd= F_drag/0.5*ρ*A_surface*V²*C_drag

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.

 

[Chestermiller]

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

 

[lucas]

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: C_d=$\frac{0.664}{\sqrt{Re}}$ or C_d=$\frac{1.33}{\sqrt{Re}}$; and even C_d=0.0742 / Re^1/5.

 

[Chestermiller]

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: C_d=$\frac{0.664}{\sqrt{Re}}$ or C_d=$\frac{1.33}{\sqrt{Re}}$; and even C_d=0.0742 / Re^1/5.

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

 

[rezeew]

Hi there,

First of all, great job on your theoretical investigation! It's always good to compare your theoretical results with experimental data to validate your calculations.

To solve for the drag coefficient (Cd), we can use the formula you provided: Cd = Fdrag / (0.5 * ρ * Asurface * V^2 * Cdrag). Since we know the air density (ρ) and the area (Asurface), we just need to find the drag force (Fdrag) and the cube's velocity (V) to solve for Cd.

To find the drag force, we can use the formula Fdrag = 0.5 * ρ * V^2 * Cd * Asurface. We already know the air density and the area, so we just need to find the cube's velocity (V) at a height of 30m.

To do this, we can use the free fall equation: H = 1/2 * g * t^2, where H is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time it takes the object to fall. We can solve for t by rearranging the equation to t = √(2H/g).

Plugging in our values (H = 30m and g = 9.8 m/s^2), we get t = √(2*30/9.8) = 2.73 seconds. This is the time it takes for the cube to reach a velocity of 12.15 m/s (which you calculated earlier).

Now, we can plug in our values for Fdrag and V into the drag force formula to solve for Cd: Cd = Fdrag / (0.5 * ρ * Asurface * V^2 * Cdrag). This will give us the drag coefficient for our cube.

To find the terminal velocity, we can use the formula Vterminal = √(2mg/ρAsurfaceCd), where m is the mass of the cube. Since we know the dimensions of the cube (5x5x5 cm), we can calculate its volume (V = 125 cm^3) and use the density of the cube (ρcube = 1.25 kg/m^3) to find its mass (m = ρcube * V = 1.25*10^-3 kg).

Plugging in our values, we get Vterminal =