Drag Force at very very high Renolds Number ?

AI Thread Summary
The discussion centers on the expression for drag force at very high Reynolds numbers, specifically those exceeding 10^6. While the classic drag force equation, 0.5kpAv^2, is applicable for Reynolds numbers between 1 and 10^5, its validity at higher values is debated. Participants highlight that the drag coefficient can vary significantly based on environmental factors and geometries, complicating the calculation. A specific example involving a sphere moving at high speed illustrates the need for precision in determining drag force, suggesting that while the basic formula can provide estimates, more complex calculations may be necessary for accuracy. Ultimately, the conversation emphasizes the importance of context and the variability of drag coefficients in high Reynolds number scenarios.
sunnyguha
Messages
5
Reaction score
0
What is the expression for Drag Force at very very High Renolds Number flow (>10^6)
as we know the expression
for drag force between Re1 and Re10^5 = 0.5kpAv^2
but this isn't valid aft 10^5 ..
so ...
 
Physics news on Phys.org
There isn't any general equation for drag at any value of Reynolds number to be honest. Even the equation you quoted depends on a a coefficient that varies wildly for different environments and geometries.

Let me instead pose this question: what exactly is the situation that you are in? What are the details of the system? How high do you consider "very very high," because 10^6 is actually pretty low for quite a few situations.
 
as in let's take an example for a sphere of radius 0.1 m moking with the speed of 100 m/sec
Renolds = pv/viscosity which gives 10^7

so what wud be the drag force calculating expression for this situation
 
It depends on how accurate you want to be. You could use the old formula for a ballpark estimate, or you could go through the long, complicated process of finding pressure drag including the separated regions and viscous drag. It all depends on what you want.
 
sunnyguha said:
for drag force between Re1 and Re10^5 = 0.5kpAv^2
but this isn't valid aft 10^5 ..
so ...

Since when is this not valid for Re > 10^5?

As long as you know your drag coefficient this equation is completely valid.
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top