Calculating coefficient of drag based on coefficient of drag for plate

[doG]

Also note that your reference from the World of Physics - Drag Coefficient...includes a table where the drag coefficient of spheres (golf and tennis balls) increases at lower Reynolds numbers. I don't see where it helps your claim to the opposite.

Yes, and the same table shows that cd = 0.02 for a wing or 0.2 for a wing in turbulent flow where re > 30,000. Even the table at the wiki page shows a cd of 0.001 for a plate in laminar flow and 0.005 for a plate in turbulent flow.

 

In some cases for some shapes cd is higher at lower reynolds numbers and the reverse for others. It is different too for bodies moving through a fluid versus fluid moving through a body. In the case of a short sharp edged orifice versus a tube you can see values for cd go from 0.62 to 0.80. In the article I cited on cleaning fluid conductors it points out:

 

As N_R increases, flow conditions go from laminar, through the critical zone, to turbulent. It has been proven empirically that once N_R exceeds 3,000, resistance to fluid flow is a combination of the effects of turbulence and of viscous drag at the conductor wall.

 

For some conductors it points out that N_R needs to be as high as 25,000 to cause enough turbulence in order to cause enough drag to effectively clean the walls of the conductor.

 

My only point to begin with is the you cannot just calculate drag based on shape and angle of incidence. Velocity is an essential part of the equation. So is overall shape. A sphere will have a higher cd than a hemisphere ( with the round side facing flow) even though both present the same profile to oncoming flow. Drag is a complicated number than cannot be reduced to 2 variables.

[J.C.MacSwell]

Yes, and the same table shows that cd = 0.02 for a wing or 0.2 for a wing in turbulent flow where re > 30,000. Even the table at the wiki page shows a cd of 0.001 for a plate in laminar flow and 0.005 for a plate in turbulent flow.

 

 

As I said earlier:

 

I think that is the key. The friction is greater. But often with the drag of objects the form drag is significantly more important and makes up a higher portion of the drag. If it does not trigger less form drag (a straight pipe would not have any) you would just have the greater friction drag.

In both cases, the wing and a flat plate parallel to the flow, form drag is much less significant than skin friction. A change to a turbulent regime simply means more drag.

[Endercreeper01]

[doG]

In both cases, the wing and a flat plate parallel to the flow, form drag is much less significant than skin friction. A change to a turbulent regime simply means more drag.

Sometimes. At others drag does go down with turbulence as shown by Endercreeper's post below...

 

And your point is? You do realize that this scenario is not true for all profiles?

[J.C.MacSwell]

Sometimes. At others drag does go down with turbulence as shown by Endercreeper's post below...

I guarantee you that in that example their is significantly less turbulence at the lower coefficients of drag in this example.

 

The earlier transition to a turbulent boundary layer, which happens at higher Reynolds Numbers, triggers less turbulence overall and reduces form drag in examples like this one, which is probably a bluff body similar to a sphere.

[doG]

I guarantee you that in that example their is significantly less turbulence at the lower coefficients of drag in this example.

Let me make sure I understand you. You are asserting that they have less turbulence at the lowest cd in that example, at an Re around 4 · 10⁵?

 

FWIW, that exact image is from section 2.4 of a white paper at http://www.diva-portal.org/smash/get/diva2:566139/FULLTEXT01.pdf The dip in drag coefficient is explained by the author:

 

As we can see from figure 2.4, there is a drop in drag at a certain Reynolds number (approximately 4 · 10⁵ Re 6 · 10⁵). This is the critical region, where the boundary layer around the body transitions from laminar to turbulent flow, explained in section 2.3.5. The turbulent mixing that takes place in the boundary layer gives the fluid a higher momentum toward the surface of the body, thus moving the separation point farther back. In the critical range, small variations in the Reynolds number cause considerable changes in the drag coefficient.

 

 

Your claim of significantly less turbulence directly contradicts this paper on Wind Tunnel Model Testing of Offshore Platforms. He clearly states this dip in drag coefficient is caused by the onset of turbulence, not the reduction of it.

[J.C.MacSwell]

 

Let me make sure I understand you. You are asserting that they have less turbulence at the lowest cd in that example, at an Re around 4 · 10⁵?

 

FWIW, that exact image is from section 2.4 of a white paper at http://www.diva-portal.org/smash/get/diva2:566139/FULLTEXT01.pdf The dip in drag coefficient is explained by the author:

 

 

Your claim of significantly less turbulence directly contradicts this paper on Wind Tunnel Model Testing of Offshore Platforms. He clearly states this dip in drag coefficient is caused by the onset of turbulence, not the reduction of it.

 

Read carefully what the Author is describing where you quoted him...more turbulence in the boundary layer (micro turbulence) moves the separation point (the start of much more significant larger turbulence) back further on the body...overall significantly less turbulence and drag

[doG]

Read carefully what the Author is describing where you quoted him...more turbulence in the boundary layer (micro turbulence) moves the separation point (the start of much more significant larger turbulence) back further on the body...overall significantly less turbulence and drag

Read it again. The author says absolutely nothing of the sort, you don't get to randomly just insert words to make it mean what you want.

 

Additionally, Design Aerospace says simply:

 

For low values of Re, flow is laminar. For high values of Re, flow is turbulent.

 

Edited September 2, 2013 by doG

[J.C.MacSwell]

Read it again. The author says absolutely nothing of the sort, you don't get to randomly just insert words to make it mean what you want.

 

Additionally, Design Aerospace says simply:

 

"As we can see from figure 2.4, there is a drop in drag at a certain Reynolds number (approximately 4 · 10⁵ Re 6 · 10⁵). This is the critical region, where the boundary layer around the body transitions from laminar to turbulent flow, explained in section 2.3.5. The turbulent mixing that takes place in the boundary layer gives the fluid a higher momentum toward the surface of the body, thus moving the separation point farther back. In the critical range, small variations in the Reynolds number cause considerable changes in the drag coefficient."

 

Can I just bold his words and hope you can understand them?

[doG]

"As we can see from figure 2.4, there is a drop in drag at a certain Reynolds number (approximately 4 · 10⁵ Re 6 · 10⁵). This is the critical region, where the boundary layer around the body transitions from laminar to turbulent flow, explained in section 2.3.5. The turbulent mixing that takes place in the boundary layer gives the fluid a higher momentum toward the surface of the body, thus moving the separation point farther back. In the critical range, small variations in the Reynolds number cause considerable changes in the drag coefficient."

 

Can I just bold his words and hope you can understand them?

No you can't. Nowhere, even in the bolded section is there any mention that any turbulence has decreased, only the the point of separation has moved, In fact, in your bolded section he clearly states that their is higher momentum toward the surface of the body, not laminar to it, which indicates increased turbulence, not significantly reduced turbulence as you have tried to suggest by putting words in his mouth. Nowhere is your assertion that turbulence is decreased at higher reynolds numbers supported.

[studiot]

 

According to my graph i showed you before, the coefficient of drag decreases with a higher reynolds number

 

 

I looked, but could not find any reference to what your graph referred to as it had no title or explanatory key.

 

Oh and thank you for your response to my post#11.

[J.C.MacSwell]

No you can't. Nowhere, even in the bolded section is there any mention that any turbulence has decreased, only the the point of separation has moved, In fact, in your bolded section he clearly states that their is higher momentum toward the surface of the body, not laminar to it, which indicates increased turbulence, not significantly reduced turbulence as you have tried to suggest by putting words in his mouth. Nowhere is your assertion that turbulence is decreased at higher reynolds numbers supported.

What do you think happens at separation? Do you believe it is not turbulence? Turbulent, but less so than a turbulent boundary layer?

 

For bluff bodies earlier separation without reattachment is significantly more turbulent than delayed separation and lower drag

[Endercreeper01]

I put the graph to show what happens. It is a sphere. It goes down and up, but the derivative is a lot less when it is going down

[doG]

What do you think happens at separation?

More turbulence, not significantly less turbulence as you asserted.

I put the graph to show what happens. It is a sphere. It goes down and up, but the derivative is a lot less when it is going down

Why then does it match the image in the paper I cited. In that paper it is the leg of an oil platform.

[studiot]

 

I put the graph to show what happens. It is a sphere. It goes down and up, but the derivative is a lot less when it is going down

 

 

Was that a reply to my question?

 

How about something like

 

The flow under gravity of golden syrup round polished steel balls showing drag coefficients for a range of Reynolds numbers, at 50degC.

Edited September 3, 2013 by studiot

[Endercreeper01]

 

Was that a reply to my question?

 

How about something like

 

The flow under gravity of golden syrup round polished steel balls showing drag coefficients for a range of Reynolds numbers, at 50degC.

it was an example using spheres.

More turbulence, not significantly less turbulence as you asserted.

 

Why then does it match the image in the paper I cited. In that paper it is the leg of an oil platform.

In that picture, it was an example using spheres

[J.C.MacSwell]

More turbulence, not significantly less turbulence as you asserted.

Perhaps you can point out where you thought I asserted that.

 

It's getting harder and harder to believe that your confusion is genuine.

[doG]

I guarantee you that in that example their is significantly less turbulence at the lower coefficients of drag in this example.

 

The earlier transition to a turbulent boundary layer, which happens at higher Reynolds Numbers, triggers less turbulence overall and reduces form drag in examples like this one, which is probably a bluff body similar to a sphere.

 

 

Perhaps you can point out where you thought I asserted that.

 

It's getting harder and harder to believe that your confusion is genuine.

 

Wow, can't remember your own posts.

 

BTW, the bolding was yours...

[J.C.MacSwell]

 

 

 

Wow, can't remember your own posts.

 

BTW, the bolding was yours...

 

 

 

 

 

Wow, can't remember your own posts.

 

BTW, the bolding was yours...

...and the context was for the the opposite case. So you really are genuinely confused?

Edited September 4, 2013 by J.C.MacSwell

[Endercreeper01]

 

 

 

Wow, can't remember your own posts.

 

BTW, the bolding was yours...

Show us the post then

[doG]

 

 

...and the context was for the the opposite case. So you really are genuinely confused?

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.

[Endercreeper01]

In the graph, it both increases and decreases, but it decreases much greater.

You can continue debating, but what do you think of my theory?

[doG]

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

 

I think your theory is insufficient because it does not consider all of the factors that affect drag. The unpredictability of drag is the reason for wind tunnel testing.

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

 

I think your theory is insufficient because it does not consider all of the factors that affect drag. The unpredictability of drag is the reason for wind tunnel testing.

[Endercreeper01]

I think your theory is insufficient because it does not consider all of the factors that affect drag. The unpredictability of drag is the reason for wind tunnel testing.

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

What do you think I am missing? Re is covered in the part about the coefficient of drag for a 2d plate Edited September 5, 2013 by Endercreeper01

[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?

 

 

What do you think I am missing? Re is covered in the part about the coefficient of drag for a 2d plate

 

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