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The drag force equation is described as:

$$ F_D = \frac{1}{2}\rho v^2 C_D A $$

Where the cross-sectional area of the front of the sphere is = πr^2,

so we can deduce that the radius and drag coefficient graph will look similar to this (the radius is on the x-axis and the drag coefficient is on the y-axis) :

enter image description here

But the Reynolds number equation shows that the radius and Reynolds number are directly proportional to each other.

$$ Re = \frac{\rho vL}{\mu} $$

Therefore the relationship between Reynolds number and drag coefficient should be similar to the relationship between radius and drag coefficient. But the graph below is different than the one above. enter image description here

What is the actual relationship between the radius of a sphere and its drag coefficient? should it be deduced from the drag force equation or from the relationship between Reynolds number and drag coefficient?

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  • $\begingroup$ This may be a stupid question.... I'm still in eleventh grade and I'm a bit confused. $\endgroup$
    – Faris W
    Jul 24, 2021 at 20:21
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    $\begingroup$ How did you get your first graph? Did you assume a constant drag force, solve for drag coefficient, and plot drag coefficient vs. radius from that data? Note - as the radius of a sphere goes down, it's cross-sectional area goes down, the drag force on that sphere goes down as a result if all other variables are held constant, and the drag coefficient remains constant based on an unchanging shape. My guess - you have hidden assumptions in your analysis that are invalid. $\endgroup$ Jul 24, 2021 at 20:28
  • $\begingroup$ Yes, I have used the drag force equation to solve for the drag coefficient while keeping drag force and all other variables constant. I am not sure if it's a valid method to find a relationship between radius and drag coefficient. What is the appropriate method to find the relationship between radius and drag coefficient? $\endgroup$
    – Faris W
    Jul 24, 2021 at 20:36
  • $\begingroup$ What quantity do you label as "drag coefficient"? $\endgroup$
    – nasu
    Jul 24, 2021 at 20:59
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    $\begingroup$ @FarisW, the radius of a sphere affects the Reynold's number because the Reynold's number has a characteristic length in it. On another note, exactly what is your hypothesis? $\endgroup$ Jul 27, 2021 at 19:07

3 Answers 3

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FIRST EQUATION

You cannot obtain a graph of $C_D$ vs $r$ by holding the drag force constant, because the drag force increases with increasing r.

$C_D$ is generally assumed to be a constant, at least over a certain range of values (hence the name drag coefficient). This shows that over a range where C_D is relatively constant, drag force is proportional to cross-sectional area. That’s how the first equation would normally be used: you find or look-up a reasonable C_D for your situation and plug it into the equation with the other values to estimate the drag force (also called the drag).

There are some obvious things in that first equation: the amount of drag goes up as velocity goes up, as density goes up, as area goes up, and as drag coefficient goes up. The viscosity is not in the equation and would affect the drag coefficient directly.

SECOND EQUATION

The second equation is a definition equation. Thats the definition of reynolds number. It’s one of several dimensionless parameters used to characterize a flow situation - the most famous one. You can look at a reynolds number and see if it is a laminar or turbulent situation for example without knowing anything else. The $L$ in the equation is the radius of the sphere in your case.

Normally the drag coefficient is relatively constant across changing reynolds number, but not here. Regardless though, the drag force does not appear in that equation. Drag force uses the first equation. The fact that the coefficient is a function of reynolds number means you could make changes that keep reynolds number constant without changing that coefficient (meaning for example you could double both $\mu$ and L and not change the coefficient, because that change would not change reynolds number). But usually you look-up $C_D$ on there and use it in the first.

GETTING A VALUE TO START

To be extra clear, use Reynolds number and the chart you posted. The x-axis of that chart is exponential, so $C_D$ changes very slowly with reynolds number. You could probably assume $C_D$ is constant, depending upon what reynolds numbers you operate at. Just for example, if you conclude that the entire situation will be operating between $Re = 800$ and $Re = 8,000$... Well, $C_D$ at $800$ is maybe $0.6$ and at $8,000$ is about $0.5$. I would just assume constant of $0.55$, especially at first. Maybe later you could let it change with Re if necessary. Partly because fluid dynamics is not exact anyway.

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    $\begingroup$ Thanks a lot for your answer. I am currently working on a lab report that determines a relationship between the radius of a sphere and its drag coefficient. In my hypothesis, should I predict the relationship using the Reynolds number relationship with drag coefficient? Based on what do you think should I make the hypothesis for my experiment? I really appreciate your help. $\endgroup$
    – Faris W
    Jul 27, 2021 at 17:24
  • $\begingroup$ Yes I would use reynolds number and the chart you posted. The x-axis of that chart is exponential, so C_D changes very slowly with reynolds number. You could probably assume C_D is constant, depending upon what reynolds numbers you operate at. Just for example, if you conclude that the entire situation will be operating between Re = 800 and Re = 8,000... Well, C_D at 800 is maybe 0.6 and at 8,000 is about 0.5. I would just assume constant of 0.55, especially at first. Maybe later you could let it change with Re if necessary. Partly because fluid dynamics is not exact anyway. $\endgroup$
    – Al Brown
    Jul 27, 2021 at 23:40
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    $\begingroup$ Thanks a lot for helping $\endgroup$
    – Faris W
    Jul 28, 2021 at 20:07
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The drag coefficient depends on an object's shape, the material on its surface, and any particular quirks about how the fluid interacts with the object and itself. It's made to do the best it can at being constant even if you change $A, v^2$ or $\rho$, but in reality it's not quite constant especially if you change one of those variables a lot.

The graph of reynolds number vs. $C_D$ is basically showing the imperfection of trying to represent those complicated relationships between the sphere and drag force as a function of velocity and/or length of the sphere (since $L$ in reynolds number refers to longitudal length and not cross sectional area)

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In theory the drag coefficient should remain roughly the same as you change the radius of the sphere. The whole point of non-dimensionalisation is to allow scaling of problems, so if you calculate the drag coefficient of a model sphere, then a large sphere's drag force can be calculated using the drag equation.

However, other non-dimensional parameters also need to be kept constant. So the drag coefficient would only remain the same if you also kept the Reynolds Number and the Mach Number roughly constant. If the drag coefficient was not a function of the Reynolds Number or the Mach Number, you would get a constant Cd regardless of size.

On that graph you have shown of Cd against Re, you may be wondering why there is a sudden dip in drag. This is called the 'drag crisis'. What has happened is that the flow regime has turned turbulent. Generally a high Reynolds Number means the flow is unstable and will quickly turn turbulent when perturbed. One of the main sources of drag is due to flow separation. To sum it up very quickly, this happens when the flow has insufficient energy to be able to make the bend around the sphere on the downstream side. When the flow becomes turbulent, the flow near the wall (the boundary layer) becomes more energetic, so it can make the turn now. This stops the flow from separating, which reduces the pressure drag created. You may like to search "adverse pressure gradients" to learn more about this.

A typical Reynolds number for flow over a sphere might be:

$$Re = \frac{\rho u D}{\mu} = \frac{(1.28)(20)(0.05)}{1.81 \times 10^{-5}} = 66000$$

That assumes flow at $20 ms^{-1}$, the density of air is about $1.2 kgm^{-3}$, the diameter is 5 cm, and the dynamic viscosity of air is $1.81 \times 10^{-5} Pa \cdot s$. Now, holding the freestream velocity $u$ constant, you might vary the diameter of your sphere to maybe $30 cm$ at a maximum. This would give an Re of about $400000$. From your graph, this is in the range of Re from $10^4$ to $10^5$.

For a smooth sphere, you might see a dramatic decrease in drag as you raise the sphere diameter, because the flow regime becomes turbulent. For a rough sphere the Cd should stay fairly constant.

If you want things to stay relatively uneventful, you could reduce the freestream air velocity to keep the Re in the range $10^3$ to $10^4$ where the Cd is fairly constant. The more interesting variation in Cd with radius will happen at very low Reynold's Numbers and high Re. At a low Re, you have Stokes' Flow, where there is no flow separation. Flow separation causes pressure drag, so the only source of drag is viscous (skin friction drag on the surface of the sphere). As you increase the radius of your sphere, the Re will increase, and the relative magnitude of the viscous forces will decline, so the Cd decreases. This is because the Re is a ratio of the inertial forces in the fluid to the viscous forces - at a low Re, the viscous forces are dominant and the fluid's inertia can be neglected.

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