Theory for Coefficient of drag

[Endercreeper01]

First, you need to find the shear rate, since the wall shear stress is the viscosity times the shear rate, and the shear rate is du/dx.

I have found a way to calculate this:

1. You need a function for the angle with respect to x, a(x)
2. This function would be a(x)=arctan(dy/dx), since dy/dx would be the same as tan(a), and arctan is the inverse function
3. The derivative of this function, da/dx, will be useful when we calculate du/dx, and the derivative of this function is (d²y/dx²)(1/(1+(dy/dx)²)
4. Now, we must find u(a(x)), or just u(a), where a=a(x)
5. u(a) would be the part of the velocity that is in the direction of motion, so u(a)=u₀sin(a), where u₀ is the velocity when x=0
6. We now have to take du/dx. This would be (du/da)(da/dx)
7. Du/da would be u₀cos(a), and da/dx is found above at #3
8. We get:

du/dx = ((u₀cos(a)) / (1+(dy/dx)²)) (d²y/dx²)

Edited November 27, 2013 by Endercreeper01

[Unity+]

Thanks to whomever is giving me downvotes... stay classy out there.

 

 

So... you're going to set the portion of the drag coefficient due to friction equal to the exact value of the viscous stresses at the surface. AND include something about some undefined pressure.

 

I have no idea why you are just clumping these together willy nilly. Total drag is usually the sum of the friction drag and the form drag (that due to the pressure change). You're going to have to demonstrate that this product makes sense.

 

Also, the drag coefficient, CD, is the dimensionless drag per unit length of a 2-D body (per unit area of a 3-D body). CD = (drag)/(denisty*characteristic velocity^2 * characteristic length).

 

drag = friction drag + form drag

 

friction drag is defined to be the shear stress at the wall, or viscosity*velocity gradient normal to the wall for a Newtonian fluid... on other words, what you set your coefficient to.

 

which doesn't make sense.

 

Furthermore, it doesn't really help to define your coefficient that way, since usually it is not very easy to know what the shear stress at a boundary is. That's why drag coefficients are measured in the first place... to get around the difficulties in calculating the shear stress at the wall.

 

BTW, a lot of what I just wrote can be found in Leal's text referenced above. Had you read it, or maybe any of the other texts I posted, you would have know this. My recommendation would be to quit just flinging things around that you do not appear to have a good understanding of in the hopes of getting something to stick, and actually start on learning what we know today. At the very least, you need to know what is published today so that you can demonstrate that your idea is actually better... something you haven't done at all so far, BTW. Fluid mechanics has a lot of unknowns out there still, it is a very active area of research, but there are a heck of a lot of things we do know fairly well. And drag is one of those things. I am not saying that there isn't a lot more to learn about drag, because there is; but, what you're doing here conflicts with a great deal of the very well established knowledge we have so far. And you've provided zero evidence comparing your idea to what is known, and zero evidence showing us your idea is better. Either start providing that evidence, or this thread really ought to be closed. And please do yourself a favor and start working on understanding what we know today. I'll even go so far as to offer to answer questions about the texts I referenced above, when I have time... I have all of them, and have studied all of them in depth.

Using deductive reasoning, I think we can assume that Endercreeper is.

[Endercreeper01]

Endercreeper is?

[Bignose]

We get du/dx = ((u₀cos(a)) / (1+(dy/dx)²)) (d²y/dx²)

So, you think free stream velocity and geometry is enough to give you the wall shear stress?

 

That isn't right.

 

You have zero consideration for whether the bulk flow is laminar or turbulent. You have zero consideration for whether the boundary layer is laminar or turbulent -- or where this transition happens. You have zero consideration for where the boundary layer separates from the body, or if it does or doesn't. You have zero consideration for the flow itself -- except in the simplest of geometries, usually a very difficult task. This is why computational fluid dynamics is considered very important.

 

Again (broken mp3 player here!), this kind of stuff is covered in Schlichting's Boundary Layers text listed above.

 

And this is why almost always, the estimation of Cf is done via a correlation, and not directly calculated. The direct calculation is a bear.... you basically have to know the entire fluid field. That leads back to the CFD.

Edited November 27, 2013 by Bignose

[Endercreeper01]

So, you think free stream velocity and geometry is enough to give you the wall shear stress?

 

That isn't right.

 

You have zero consideration for whether the bulk flow is laminar or turbulent. You have zero consideration for whether the boundary layer is laminar or turbulent -- or where this transition happens. You have zero consideration for where the boundary layer separates from the body, or if it does or doesn't. You have zero consideration for the flow itself -- except in the simplest of geometries, usually a very difficult task. This is why computational fluid dynamics is considered very important.

 

Again (broken mp3 player here!), this kind of stuff is covered in Schlichting's Boundary Layers text listed above.

 

And this is why almost always, the estimation of Cf is done via a correlation, and not directly calculated. The direct calculation is a bear.... you basically have to know the entire fluid field. That leads back to the CFD.

This uses the geometry of the flow itself. This is also for laminar flow only. And this is to calculate the shear rate, du/dx. The shear stress would be μ(du/dx). Also, all this is for a Newtonian fluid only. For a non Newtonian fluid, we would use τ/q Edited November 27, 2013 by Endercreeper01

[Bignose]

Before you dive into the extraordinarily complicated world of non-Newtonian fluids, best to get Newtonian fluids right.

 

Even if the flow is laminar, it still isn't going to be that easy to get exactly the flow field at every point. Analytic solutions are still only available in very special cases. CFD is still the main tool here.

 

And your geometry-based solution still won't work, in my book. The fluid flow is not only dependent on that geometry. Your calculation there still doesn't take into account the entire flow.

 

But, prove me wrong. Explicitly using your formula (i.e. show every step, please), calculate the drag for a sphere at very low Re flow. Re<<1.

 

This is a classical result that if your formula is right, should be replicated easily.

[Endercreeper01]

What exactly is the length of this sphere? The temperature of the air? Is the flow laminar or turbulent? I need to know all of the variables and factors.

Edited November 27, 2013 by Endercreeper01

[Bignose]

I'm sorry, but this tells me that your knowledge is severely lacking. Because I really gave you all that info already by saying the Reynolds number was much less than 1. The exact size of the sphere doesn't matter... The size of the sphere in relation to the free stream velocity and the fluid properties matters. That's what Re means.

 

If you really need the exact info, make up your own values such that Re = something much less than 1. And BTW, for spheres where Re<<1, the flow is very very laminar. That's why I asked you to answer this problem since per your above post, your formula is only valid for laminar flows.

Edited November 28, 2013 by Bignose

[Endercreeper01]

According to my theory, I need certain things, but since you will just not give them to me and you just think I am severely lacking in knowledge, I will have to make up my own values. I need length for the form drag, but I will assume that the length is 1. I needed temperature for the viscosity, so I will just assume it is at 500 degrees kelvin, making the viscosity about 2.7. I will assume the density is about 1.27. Notice how all I can do is assume?

Now to the calculations,

1. To calculate the form drag, we know some things first. We know Re is less then 1, and I will assume it is 0.1, making the velocity abut .21 m/s.
2. The average angles for both I sides of the sphere are both 45, and it is length 1, so we can reduce it to Dcos(45), which makes 2^1/2D / 2
3. Because I could not find a graph relating the Re to Cd for a cube, then I will just assume it is 1.05, making the form drag coefficient about .74
4. Now, I will calculate the skin friction drag coefficient. In this case, q=.028, and the viscosity divided by q will be about 96.43
5. Now, I will assume a=45, and we first need to find ucos(45), which will be .21cos(45), which is about .11
6. Now, we will need to find dy/dx, which we will assume is 1, so we will be dividing by 2, giving us .055
7. We now need to find d²y/dx², which for a sphere, is going to be d²/dx² r*sin(a)
8. We will get -rsin(a)d²a/dx², and a=45, so we get about -0.7d²a/dx²
9. Now, I will try to figure our d²a/dx², and I will tell you when I do, but it must be small
10. We now multiply that by 96.43 to get about -6.75d²a/dx².

Edited November 28, 2013 by Endercreeper01

[Bignose]

Sorry. That is no where near the classical exceptionally well verified result.

 

For Re << 1, CD = 24/Re.

 

Your calculations here are greatly at odds with rock solid knowledge. Your idea is wrong.

[Endercreeper01]

I was just assuming random values, such as D being 1.05 at v=0.1 m/s. Do you want me to make actual calculations?

If so, you must tell me all of the information needed in my equation, or else I cannot make accurate predictions., such as viscosity of the air at that temperature, exact velocity, etc.

Also, I am unable to find a graph that relates the Re to Cd for a cube.
Don't believe me (since you never do) ? If so, you try finding a cube Cd vs Re graph. You try assuming random values and comparing to experimental results. You try putting in all of these numbers in only half an hour, something I had to do. And, you have to compare your results that had random values inserted into the equation with experimental data

You try doing all this.

Also, don't you remember me saying something about a d²a/dx²? That means it was never complete. Didn't you notice?

I don't even think you understand my theory.

Also, how can I make predictions if I don't know the value of D?

Edited November 28, 2013 by Endercreeper01

[Bignose]

You are missing the point. You do not need the exact value of the sphere radius. You do not need the fluid viscosity. You do not need the exact value of the free stream velocity. The ratio of those combined correctly makes the dimensionless number Re. And when Re is much less than 1, all the drag can be calculated by CD = 24/Re.

 

You don't need an reference to a cube. You don't need anything else asked for.

 

If your idea can't replicate the very very well confirmed formula above, then it is less useful. And it doesn't match what is observed in reality.

 

And I didn't make you do this in a half hour... Not sure why that's a problem. Take all the time you need.

 

Isn't da/dx 0? You set a to a constant, 45 degrees. the derivative of any constant is 0.

Found this in 25 seconds of looking.

 

http://www.sciencedirect.com/science/article/pii/0032591089800087

 

I know there is more out there. Gotta improve your researching skills, my friend.

[Endercreeper01]

The average angle is 45, and I assumed a was for the midpoint on the circle, with that making a=45. However, da/dx I snot 0 since the angle changes over the sphere

24/Re can be used for Re<<1 for a sphere, but my theory uses the other variables in order to determine the coefficient.

Using my theory, we can also determine the value of D for Re<<1

1. We will assume there is little skin friction
2. Since both average angles are the same and the average angle is 45, we can make an equation: 24/Re=Dlcos(45)=Dl2^1/2/2
3. Divide both sides by 2^1/2/2, and you get Dl=33.94/Re
4. We now can have the following equation: D=33.94/(Re/l)=33.94l/Re

Now, we have a value for D at Re<<1, D=33.94l/Re

[Bignose]

Now, we have a value for D at Re<<1, D=33.94l/Re

What is this l now? What are its units? Where did this come from?!? Because I don't even see it your previous equations.

 

You keep telling me I don't understand your idea... that's true! Because I don't think a complete and thorough posting of your idea has been made yet! You keep changing functions, changing variables, etc.

 

And most importantly, how does this compare to the experimental values?

 

Another critique I have is that you used the given answer already CD = 24/Re. I thought your idea here could derive it on its own. Why do you need that result in an intermediate step? I have the same critique for the previous posts, where you needed the value for a cube.

 

The current theory can derive the drag around a sphere at Re<<1 as CD = 24/Re from first principles... that is as a direct consequence of solving the Navier Stokes equations. It doesn't need any other inputs. If your method cannot even replicate this simplest of scenarios, I have no confidence it is doing anything more complicated, like a cube.

 

 

We will assume there is little skin friction

Lastly, this could not be more wrong. At Re<<1, there is nothing but skin friction. How many more of these elementary mistakes am I going to have to correct before I can convince you to actually read some basic fluid mechanics texts?

Edited November 29, 2013 by Bignose

[Endercreeper01]

What is this l now? What are its units? Where did this come from?!? Because I don't even see it your previous equations.

 

You keep telling me I don't understand your idea... that's true! Because I don't think a complete and thorough posting of your idea has been made yet! You keep changing functions, changing variables, etc.

 

And most importantly, how does this compare to the experimental values?

 

Another critique I have is that you used the given answer already CD = 24/Re. I thought your idea here could derive it on its own. Why do you need that result in an intermediate step? I have the same critique for the previous posts, where you needed the value for a cube.

 

The current theory can derive the drag around a sphere at Re<<1 as CD = 24/Re from first principles... that is as a direct consequence of solving the Navier Stokes equations. It doesn't need any other inputs. If your method cannot even replicate this simplest of scenarios, I have no confidence it is doing anything more complicated, like a cube.

 

 

 

Lastly, this could not be more wrong. At Re<<1, there is nothing but skin friction. How many more of these elementary mistakes am I going to have to correct before I can convince you to actually read some basic fluid mechanics texts?

1. I assumed there was no skin friction.

2. My theory is just a general equation. It cannot solve things like that unless we have a value for D.

You keep talking about the Navier Stokes equations. Do you want a solution of D from the Navier Stokes equation?

[Bignose]

Then your assumption for Re<<1 is wrong. If you actually would do the calculation at Re<<1, you'd find there is no wake -- and this led to the famous D'Alembert's paradox. There is zero form drag at that slow of a flow.

 

Sure. If you can show me the calculation for drag for a sphere at Re << 1 then maybe I will believe that you have some knowledge. Let's see it.

 

And why are you yet again refusing to answer direct questions? What is 'l' in your formula? Why can't you answer this?!?! Answer this before trying to solve the N-S eqns, please.

Edited November 29, 2013 by Bignose

[Endercreeper01]

l is length in my equation. Dl is basically the form drag coefficient for a rectangular prism with the same Re, and D is the form drag coefficient for the object with the same Re / l

[Bignose]

So between this

 

D=33.94l/Re

and this

 

l is length in my equation. Dl is basically the form drag coefficient for a rectangular prism with the same Re, and D is the form drag coefficient for the object with the same Re / l

something more is dead wrong. D = 33.941*l/Re has to be completely wrong, because Re is dimensionless. So you are effectively saying here that the units of drag are length.

 

That just isn't right. Drag is a force, not a length.

 

We're not even making elementary fluid mechanics mistakes here, this is 1st day physics mistakes.

 

If you want to continue, you need to get these errors corrected. It is pointless to talk about the predictions made by an equation that isn't even dimensionally sound.

 

You need to take some time, regroup, and correct these errors.

[Endercreeper01]

So between this

 

 

and this

 

 

something more is dead wrong. D = 33.941*l/Re has to be completely wrong, because Re is dimensionless. So you are effectively saying here that the units of drag are length.

 

That just isn't right. Drag is a force, not a length.

 

We're not even making elementary fluid mechanics mistakes here, this is 1st day physics mistakes.

 

If you want to continue, you need to get these errors corrected. It is pointless to talk about the predictions made by an equation that isn't even dimensionally sound.

 

You need to take some time, regroup, and correct these errors.

1. This is the drag coefficient
2. In this case, we have the Re for a unit cube. If you compare it to the Re for a rectangular prism with length l, the Re would be Re=Re0/l, where Re0 is the Re for the rectangular prism. With the thing about units, this makes us use the length value without units

You don't even seem to know what I mean by D. You think that's the drag itself. No it is the drag coefficient for a unit cube with the same ratio of Re / l

[Bignose]

So you're saying that the drag coefficient has a unit of length? That that's wrong, too. The drag coefficient is dimensionless.

 

Look, seriously. These are major errors in your idea. Correct dimensions are paramount to actually predicting something useful.

 

If you're going to just have wave this away too, then I'm done. You've done this over and over, and I need to just accept that you aren't open to my criticisms. I've been trying to point you toward the knowledge we do have on drag, but you seem ultimately reluctant to actually go and learn any of it. I would consider this borderline trolling behavior. I just don't care to even try to help anymore if you aren't going to actually look at it.

[Endercreeper01]

I clearly said we use the length, but do not include the units! Where do you not see this?!

 

 

With the thing about units, this makes us use the length value without units

 

Did you even read the post?!

This is starting to piss me off. Why don't you pay attention to what I say?!

Edited November 30, 2013 by Endercreeper01

[Bignose]

I read the post, but it is nonsense. How do you include a length, but not use units? That's jibberish.

 

If I went to the lumber yard, to buy a piece of wood, and the clerk said "sure, no problem, how long do you need" and I just said "10". What would the clerk do?

 

Units are paramount.

 

And if you don't include the units, then your formula is just hookum.

 

Do you include the length in inches? mm? m? furlongs? chains? light years? hands? or any of the thousands of other units of length that mankind has invented over the years?

 

When you have a length, you have a unit that comes with it. That always happens. You can't just dismiss this.

Edited November 30, 2013 by Bignose

[Endercreeper01]

I read the post, but it is nonsense. How do you include a length, but not use units? That's jibberish.

 

If I went to the lumber yard, to buy a piece of wood, and the clerk said "sure, no problem, how long do you need" and I just said "10". What would the clerk do?

 

Units are paramount.

 

And if you don't include the units, then your formula is just hookum.

 

Do you include the length in inches? mm? m? furlongs? chains? light years? hands? or any of the thousands of other units of length that mankind has invented over the years?

 

When you have a length, you have a unit that comes with it. That always happens. You can't just dismiss this.

Then, how would you explain the fact that the Re for a unit cube and the Re for a rectangular prism with length l differ by a factor of l ?!

You have even shown signs of not reading my posts on previous pages.

And now, you are calling me a troll! How am I trolling you?!

Edited November 30, 2013 by Endercreeper01

[Bignose]

There is a length in the Reynolds number, already. Though you have to be very careful about what Re you are defining. On a cube, there is only 1 length, so it is obvious what length the Re would use. Same thing with a sphere. For something with multiple lengths, you need to explicitly state what length you are using

[Endercreeper01]

I know there is already a length in Re already, that's why you would divide by length without using units if both Reynolds numbers were using the same units