Calculating coefficient of drag based on coefficient of drag for plate

[Endercreeper01]

 

 

 

Apply your theory and see how well it matches these measured drag coefficients:

 

How is that picture correct if the drag coefficient depends on Reynolds number?

[doG]

They are average measured values. As evidence by your own previous post there can be quite a bit of variability caused by turbulence. Aside from that how well does your theory match these values?

[Endercreeper01]

They are average measured values. As evidence by your own previous post there can be quite a bit of variability caused by turbulence. Aside from that how well does your theory match these values?

Average? it doesn't say the Reynolds number each shape had when being tested.

[J.C.MacSwell]

Not at all. At the lowest cd in that graph reynolds number is greater than it was for higher values of cd indicating greater turbulence at the minimum value of cd than at the greatest value. Your assertion that the lowest cd was associated with significantly less turbulence is unsupported unless you can show that increasing reynolds numbers indicate reduced turbulence.

No.

 

You are confusing boundary layer turbulence with the much greater turbulence at separation.

 

 

 

The increased turbulence of the energized boundary layer leads to delayed separation, substantially less turbulence overall and a lower Cd

 

Edited September 6, 2013 by J.C.MacSwell

[doG]

No.

 

You are confusing boundary layer turbulence with the much greater turbulence at separation.

 

 

 

The increased turbulence of the energized boundary layer leads to delayed separation, substantially less turbulence overall and a lower Cd

 

No I'm not. You clearly said, "significantly less turbulence'. Not significantly less boundary layer turbulence or significantly less wake turbulence or anything else. Your statement clearly infers turbulence, all of it. I clearly understand the different parts of turbulence and I can clearly read what you wrote and it's incorrect. At the minimum value of Cd in that graph the overall turbulence is greater. The graph does not delineate the different layer of turbulence anyhow so you cannot now say that it indicates anything less than overall turbulence which is greater, not significantly less as you asserted. You said what you said, don't try to claim now that you said something else. You made a statement about an illustration that includes all turbulence there's no reason to believe your remark didn't include all turbulence.

[J.C.MacSwell]

Your statement clearly infers turbulence, all of it...

 

You made a statement about an illustration that includes all turbulence there's no reason to believe your remark didn't include all turbulence.

Good. We agree about what I said.

 

At the minimum value of Cd in that graph the overall turbulence is greater.

This is wrong. Why do you think this is true?

 

Do you believe that at the lower values of Cd the separation stars earlier and has more turbulence? (like the upper smooth golfball)

 

Why (if that is your belief) would you think that would lead to a lower Cd, and the opposite case of delayed separation, (lower dimpled golfball) lead to a higher Cd?

Edited September 6, 2013 by J.C.MacSwell

[J.C.MacSwell]

. Synonymously Cd decreases and increases as turbulence increases because turbulence increases as Re increases.

 

This is generally true (the bold part I bolded) for the boundary layer, and for streamlined objects where shear forces (friction) dominate the drag.

 

It is generally not true for bluff bodies, such as in the example, where pressure drag (form drag) predominates, and for reasons I have explained a number of times,,,the transition to a turbulent boundary layer leads to delayed separation, decreased drag and less turbulence overall.

Edited September 6, 2013 by J.C.MacSwell

[Endercreeper01]

According to the graph, it does both. And also, what do you think of my theory J.C.Maxwell?

Edited September 7, 2013 by Endercreeper01

[doG]

This is generally true (the bold part I bolded) for the boundary layer, and for streamlined objects where shear forces (friction) dominate the drag.

 

It is generally not true for bluff bodies, such as in the example, where pressure drag (form drag) predominates, and for reasons I have explained a number of times,,,the transition to a turbulent boundary layer leads to delayed separation, decreased drag and less turbulence overall.

See the part you're missing is the fact that the scale along the bottom edge of Re is not just the boundary layer flow attached, or unattached, to the body, it is the Re for the total flow condition. I posted a link to the thesis that graph is from and the author clearly states it is the result of a wind tunnel test of oil platform legs. While the boundary layer turbulence attached to the body is reduced the overall turbulence is increasing. I clearly understand that you meant to say "significantly less boundary layer turbulence" but that's not what you said. You simply said "significantly less turbulence" and that does not mean that it only includes boundary layer turbulence.

 

FWIW, hear it is from the Engineering Toolbox:

 

The Reynolds Number can be used to determine if flow is laminar, transient or turbulent. The flow is

• laminar when Re < 2300
• transient when 2300 < Re < 4000
• turbulent when Re > 4000

 

Notice there is no mention of any bodies in or around the flow. Re is independent of any bodies in the flow and turbulence increases with Re, period. That a body submerged in the flow may or may not experience reduced drag as a result of reduced boundary layer turbulence has no effect on the value of Re the graph is plotted as. The scale of Re represents the total Re of the total flow and the corresponding turbulence associated with that flow at that point on the graph, not simply the boundary layer turbulence of a submerged body. Next time you want someone to hear "significantly less boundary layer turbulence" then that's what you need to say, not I said, "significantly less turbulence" but intended you to hear "significantly less boundary layer turbulence".

[John Cuthber]

How is that picture correct if the drag coefficient depends on Reynolds number?

OK, that looks like a fair point.

Pick a Reynolds number and answer the question.

Apply your theory and see how well it matches these measured drag coefficients:

[J.C.MacSwell]

Next time you want someone to hear "significantly less boundary layer turbulence" then that's what you need to say, not I said, "significantly less turbulence" but intended you to hear "significantly less boundary layer turbulence".

No.

 

Can you not keep it straight? I described it correctly. There is significantly less turbulence at the lower Cd's.

 

Why would I say "significantly less boundary layer turbulence". THAT would not be correct. THAT is the case at higher Cd's on the left of the graph where the skin friction is significantly lower....but due to the earlier separation and increased turbulence....higher Cd's

 

 

 

Let's stay with what I said. There is significantly less turbulence at the lower Cd's.

 

 

 

Now can you explain why you believe otherwise?

 

What makes you think the turbulence is greater at the lower Cd's in the graph?

[Endercreeper01]

OK, that looks like a fair point.

Pick a Reynolds number and answer the question.

Apply your theory and see how well it matches these measured drag coefficients:

I can't find any graphs for a plate

[John Cuthber]

Chose an arbitrary starting value and see if the relative answers come out right.

[Endercreeper01]

Chose an arbitrary starting value and see if the relative answers come out right.

Fine, Ill pick Re=10³. Now show me what it would be for a plate with that Reynolds number.

[J.C.MacSwell]

According to the graph, it does both. And also, what do you think of my theory J.C.Maxwell?

Are you suggesting the forces on a plate in a fluid stream are somewhat equivalent to those from a jet of fluid impacting on the plate?

Edited September 7, 2013 by J.C.MacSwell

[Endercreeper01]

Are you suggesting the forces on a plate in a fluid stream are somewhat equivalent to those from a jet of fluid impacting on the plate?

No, because in the second case, then you would add that extra force to get the total air resistance when you want air resistance. Im talking about the coefficient of drag. Edited September 7, 2013 by Endercreeper01

[doG]

I described it correctly. There is significantly less turbulence at the lower Cd's.

No you didn't. At the lowest Cd in that graph, about 0.1-0.2, reynolds number are around 2 · 10⁵ but at the highest value of Cd, >1.5 at the beginning of the graph, reynolds number is only around 2 · 10¹. Where Cd drops to it's lowest value on the graph there is significantly more turbulence. The submerged body certainly has reduced boundary layer turbulence but that's not what is graphed. What is graphed is the amount of drag on a body submerged in a flow that is increasingly more turbulent, as depicted by the increase in reynolds numbers. If you want to claim otherwise then you need to cite a credible source that shows less turbulence at values of Re > 20,000 versus Re = 20. My own citation shows flow is turbulent for Re > 4000.

I have made a way to calculate a coefficient of drag based on what it would be if it was just a plate. I will give the coefficient of drag of a plate a name. I will just refer to it as just D. So, lets start simple. Lets say that we had a 2D plate and it was tilted at an angle θ from the x axis (axis orthogonal to direction of velocity. Then the drag coefficient would be Dcosθ. Now lets say it was not 1, but 2 plates and they are connected somehow. Now you have 2 angles, θ1 and θ2. If θ1=θ2, then it is still Dcosθ, but if θ1#θ2, then you need to find the average angle, or (θ1+θ2)/2 and so it is Dcos(1/2(θ1+θ2)) In fact, you can do this with a shape with any number of sides, you just need to find the average angle θavg,and just find the cosine of that angle and then multiply by D. So we can just write it as Cd=Dcos((Σθ)/n), where n is the number of angles. So then is this correct?

And also, what would the value of D be?

 

 

I can't find any graphs for a plate

 

You shouldn't need one. Your stated theory from your first post says the drag coefficient is Dcosθ where θ is the angle the plate is tilted relative to the x axis. This is the same as saying that Cd varies from D ·1 to D ·0 as θ varies from 0° to 90°. This implies that the coefficient of drag would approach 0 as the angle of the plate approaches 90° relative to the x axis.

[Endercreeper01]

You shouldn't need one. Your stated theory from your first post says the drag coefficient is Dcosθ where θ is the angle the plate is tilted relative to the x axis. This is the same as saying that Cd varies from D ·1 to D ·0 as θ varies from 0° to 90°. This implies that the coefficient of drag would approach 0 as the angle of the plate approaches 90° relative to the x axis.

Its not relative to the x axis, its relative to the surface perpendicular to the direction of motion. Just think about it. When that angle is 90 degrees, it is in 2 dimensions, and no part of it is interacting with the air at 90 degrees, so therefore, the coefficient is zero, and the total drag is zero.

Edited September 7, 2013 by Endercreeper01

[John Cuthber]

Fine, Ill pick Re=10³. Now show me what it would be for a plate with that Reynolds number.

It's not my theory so it's not my job to test it.

That chart gives values of 1.05 and 0.8 for a cube depending on whether it's face on or edge on.

Does the proposed model also show a 30% difference?

 

If not, you can give up on it.

Edited September 7, 2013 by John Cuthber

[doG]

I have made a way to calculate a coefficient of drag based on what it would be if it was just a plate. I will give the coefficient of drag of a plate a name. I will just refer to it as just D. So, lets start simple. Lets say that we had a 2D plate and it was tilted at an angle θ from the x axis (axis orthogonal to direction of velocity. Then the drag coefficient would be Dcosθ. Now lets say it was not 1, but 2 plates and they are connected somehow. Now you have 2 angles, θ1 and θ2. If θ1=θ2, then it is still Dcosθ, but if θ1#θ2, then you need to find the average angle, or (θ1+θ2)/2 and so it is Dcos(1/2(θ1+θ2)) In fact, you can do this with a shape with any number of sides, you just need to find the average angle θavg,and just find the cosine of that angle and then multiply by D. So we can just write it as Cd=Dcos((Σθ)/n), where n is the number of angles. So then is this correct?

And also, what would the value of D be?

 

 

Its not relative to the x axis, its relative to the surface perpendicular to the direction of motion. Just think about it. When that angle is 90 degrees, it is in 2 dimensions, and no part of it is interacting with the air at 90 degrees, so therefore, the coefficient is zero, and the total drag is zero.

 

First, which one is it? Relative to the x axis or not? First you say it is then you say it's not.

 

Secondly, a submerged plate in a fluid flow still has boundary layer drag even when it is parallel to flow. See Boundary Layer at Thermopedia. Note: The coefficient of drag is referred to as the coefficient of friction in their analysis.

[doG]

First, by x axis i meant axis perpendicular to direction of motion, and second, im talking about a 2 dimensional plate, not a 3 dimensional one

Are you saying you are talking about an imaginary plate? All plates in the real world are 3 dimensional.

[Endercreeper01]

 

 

 

First, which one is it? Relative to the x axis or not? First you say it is then you say it's not.

 

Secondly, a submerged plate in a fluid flow still has boundary layer drag even when it is parallel to flow. See Boundary Layer at Thermopedia. Note: The coefficient of drag is referred to as the coefficient of friction in their analysis.

First, by x axis i meant axis perpendicular to direction of motion, and second, im talking about a 2 dimensional plate, not a 3 dimensional one

It's not my theory so it's not my job to test it.

That chart gives values of 1.05 and 0.8 for a cube depending on whether it's face on or edge on.

Does the proposed model also show a 30% difference?

 

If not, you can give up on it.

It it about a 30% difference in the cosine of 0 and 45 degrees, so it is correct. And according to my theory, they would have Reynolds numbers that are roughly equal

Edited September 7, 2013 by Endercreeper01

[Endercreeper01]

Are you saying you are talking about an imaginary plate? All plates in the real world are 3 dimensional.

That imaginary plate is the basis for my theory

[doG]

That imaginary plate is the basis for my theory

Well then, perhaps you should seek out an imaginary forum to seek out imaginary answers to your imaginary problem. Then again, since it's imaginary you could just make up the answers that you want.

[Endercreeper01]

Well then, perhaps you should seek out an imaginary forum to seek out imaginary answers to your imaginary problem. Then again, since it's imaginary you could just make up the answers that you want.

I cant just make the answers what I want, I have to use theory. It is 2 dimensional, and the closer that angle is to 90 degrees, then the less drag coefficient there is. It then must be dcosθ and d is the drag coefficient it would have at 0 degrees.