# Drag Force Ratio for Flat Plate

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 April 28, 2020, 11:29 #2 Senior Member   Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,730 Rep Power: 66 One side of the plate is initially turbulent and will stay turbulent. The other side is initially laminar but if the plate is long enough, can eventually become turbulent and that's why you need the critical Reynolds number. You need to check whether or not it becomes turbulent. Btw, if it does become turbulent in the middle of the plate then you would need a different correlation for the drag (hint hint, wink wink). You can just calculate ReL where L is 4m and you'll know right away. The x in Rex is distance from leading edge of the plate. So the Reynolds number is 0 at the leading edge of the plate and increases with distance from the leading edge.

April 28, 2020, 22:47
#3
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Rob Wilkinson
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Quote:
 Originally Posted by LuckyTran One side of the plate is initially turbulent and will stay turbulent. The other side is initially laminar but if the plate is long enough, can eventually become turbulent and that's why you need the critical Reynolds number. You need to check whether or not it becomes turbulent. Btw, if it does become turbulent in the middle of the plate then you would need a different correlation for the drag (hint hint, wink wink). You can just calculate ReL where L is 4m and you'll know right away. The x in Rex is distance from leading edge of the plate. So the Reynolds number is 0 at the leading edge of the plate and increases with distance from the leading edge.

Hi LuckyTran

I have just worked out ReL like this.

ReL = (Rho * v * x) / mu

where mu = dynamic Viscosity = kinematic viscosity * density
mu = 1.5 * 10^-5 * 1.21 = 0.00001815 kg / ms

Rho = air density = 1.21 kg / m^3
v = plate velocity = 4 m /s
x = plate length = 4 m

ReL = (1.21 * 4 * 4) / 0.00001815 = 1,066,667

and this is over critical Reynold Number of 5 * 10^5

Initially Laminar Side

So does this mean that there is initially a laminar boundary layer and then it changes to turbulent ?

Now is ReL the same as Rex

With the critical Reynolds number = 5 * 10^5
we can work out x critical (critical x dimension from leading edge where laminar to turbulent change occurs)

Re critical = (Rho * v * x critical) / mu

Re critical * mu = (Rho * v * x critical)

x critical = (Re critical * mu) / (Rho * v)

x critical = (5 * 10^5 * 0.00001815) / (1.21 * 4)

x critical = 1.875 m

This is the part that I find difficult to understand how to deal with a mixed boundary layer of laminar and turbulent.
This for finding Drag Coefficient for Drag Force

Are you able to show me how to work out the drag force on that initially laminar side from here, I'm just confused with it ?

Turbulent Side

Because this side is all turbulent do we only need to use CD = 0.074/ (Rex)^1/5 to workout drag force ?

Is Rex the same as ReL in this situation ?

If Rex = ReL

then CD = 0.074 / (1,066,667)^1/5 = 0.0046092

Drag Force = D = CD * 0.5 * Rho * v^2 * Area of plate = 0.0046092 * 0.5 * 1.21 * 4^2 * 4 * 1

so Drag Force = 0.178468 Newtons for fully turbulent side

Last edited by Rob Wilk; April 28, 2020 at 23:17. Reason: Drag Force including 0.5 in multiplication

 April 29, 2020, 04:35 #4 Senior Member   Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,730 Rep Power: 66 The turbulent side is easy. I guess the hint hint wink wink wasn't enough. The mixed laminar & turbulent side is done the same way, you just need to find the right formula for it. There is a formula for handling mixed cases.

April 29, 2020, 10:27
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Rob Wilkinson
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Quote:
 Originally Posted by LuckyTran The turbulent side is easy. I guess the hint hint wink wink wasn't enough. The mixed laminar & turbulent side is done the same way, you just need to find the right formula for it. There is a formula for handling mixed cases.
Hi LuckyTran

So is my way of calculating the drag force on the turbulent side correct ?

As for the initially laminar turbulent side formula,
I have found from Fluid Mechanics Eighth Edition by Frank M White https://www.academia.edu/31936164/Fl...Eighth_Edition, that maybe for this question,

drag coefficient could be CD = = 1.46/ (Rex)^1/2 - 1440 / Rex

is this correct ?
I am not sure how 1440 is worked out in the formula.
This is the part that I really need to get an understanding on.

Your hint hint wink wink are gradually giving me clues how to solve this, just need to figure this out with you as we discuss this.

 April 29, 2020, 19:07 #6 Member     Rob Wilkinson Join Date: Apr 2020 Location: Wellington, New Zealand Posts: 69 Rep Power: 6 Hi LuckyTran Just wondering if you got the chance to see my previous reply about the formula for the initially laminar and turbulent side. It's very important that I get an understanding on this.

 April 29, 2020, 19:40 #7 Senior Member   Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,730 Rep Power: 66 I saw it. But if you don't understand it then you don't understand it. What is a yes/no from me going to change about that?

April 29, 2020, 20:30
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Rob Wilkinson
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Quote:
 Originally Posted by LuckyTran I saw it. But if you don't understand it then you don't understand it. What is a yes/no from me going to change about that?
Hi LuckyTran

I'm just confirming that I am on the correct track with this.

Look I have not been able to find the answer for this for a while now and would appreciate some help to get me on the right track to solve this.

If I see that there is an area that is important for me to understand to get this solved then I will want to clarify this with someone who knows a lot about this.
Why wouldn't I.

I can understand this type of question, once I know what the best approach is.
I have this in my Civil Engineering textbook, but we didn't cover Flow Past a Solid Boundary much. As I am interested in this, it is a dream of mine to get this solved.

Your help will be much appreciated thanks.
I appreciate some of the help you have given so far.

Last edited by Rob Wilk; April 29, 2020 at 23:18.

April 30, 2020, 07:37
#9
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Rob Wilkinson
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Quote:
 Originally Posted by Rob Wilk Hi LuckyTran Just wondering if you got the chance to see my previous reply about the formula for the initially laminar and turbulent side. It's very important that I get an understanding on this.
Just wondering if you got the chance to look at my previous reply.

I feel that I need some help from you on how we do this formula for the combined laminar and turbulent side.

We are already given CD = 1.46/ (Rex)^1/2 for laminar flow, but don't we have to subtract something from this involving Rex to give this new formula.

LuckTran, I think once I have got that sorted, then I can use CD in the drag force formula to calculate that and then the ratio, provided I have the turbulent side correct, which I was trying to confirm with you before.

I hope it makes sense to you, where I am with this and I'd kindly appreciate some more help to get through this.

April 30, 2020, 08:14
#10
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duri
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Quote:
 Originally Posted by Rob Wilk Hi LuckyTran So is my way of calculating the drag force on the turbulent side correct ? As for the initially laminar turbulent side formula, I have found from Fluid Mechanics Eighth Edition by Frank M White https://www.academia.edu/31936164/Fl...Eighth_Edition, that maybe for this question, drag coefficient could be CD = = 1.46/ (Rex)^1/2 - 1440 / Rex is this correct ?

Check the formula again turbulence cd should be more than laminar cd. So the formula have to use Rex^(1/5) or (1/7) instead of (1/2). More over all the turbulence relations are empirical and hence it is better to use as per recommended Reynolds number range.

April 30, 2020, 09:35
#11
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Rob Wilkinson
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Quote:
 Originally Posted by duri Check the formula again turbulence cd should be more than laminar cd. So the formula have to use Rex^(1/5) or (1/7) instead of (1/2). More over all the turbulence relations are empirical and hence it is better to use as per recommended Reynolds number range.
Hi Duri

I'm not sure what you mean by this.

Laminar CD = 1.46/ (Rex)^1/2

but as mentioned before the initially laminar side turns to turbulent 1.875 m from leading edge.

The formula for CD related to this is what I am have difficulty with.

it seems that you have to take away something from CD = 1.46 / (Rex)^1/2

As for turbulent side, I'm confirming that I have my calculation for Drag Force correct here.
I used CD = 0.074/ (Rex)^1/5 because it is entirely turbulent.

May 1, 2020, 10:57
#12
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duri
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Quote:
 Originally Posted by Rob Wilk The formula for CD related to this is what I am have difficulty with. it seems that you have to take away something from CD = 1.46 / (Rex)^1/2

On the side which has laminar to turbulent why do you think that something has to be taken away from laminar CD. You could also add something to laminar CD or take away something from turbulent CD. Multiple options are there. Since all these relations are based on experiments you can't assume some equations for transitional flows. Just take what is available in text books or journals with given limitations and apply as required.

May 1, 2020, 11:28
#13
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Rob Wilkinson
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Quote:
 Originally Posted by duri On the side which has laminar to turbulent why do you think that something has to be taken away from laminar CD. You could also add something to laminar CD or take away something from turbulent CD. Multiple options are there. Since all these relations are based on experiments you can't assume some equations for transitional flows. Just take what is available in text books or journals with given limitations and apply as required.
Remember how I said

"As for the initially laminar turbulent side formula,
I have found from Fluid Mechanics Eighth Edition by Frank M White https://www.academia.edu/31936164/Fl...Eighth_Edition on page 466 the formula CD = 0.031 / ReL^1/7 - 1440 / ReL. for Re Trans = 5 * 10^5

So for this question,

could drag coefficient CD = = 1.46/ (Rex)^1/2 - 1440 / Rex. ?

Look I don't want to go around in circles with trying to work this out, if you happen to know this formula and how it's arrived at, can you please let me know.

 May 1, 2020, 12:18 #14 Senior Member   duri Join Date: May 2010 Posts: 245 Rep Power: 17 I have already mentioned multiple times how the formula have arrived. If you still didn't got check the reference in the book.

May 1, 2020, 12:25
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Rob Wilkinson
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Quote:
 Originally Posted by duri I have already mentioned multiple times how the formula have arrived. If you still didn't got check the reference in the book.
I think you are referrring to you mentioning this earlier on

"Check the formula again turbulence cd should be more than laminar cd. So the formula have to use Rex^(1/5) or (1/7) instead of (1/2). More over all the turbulence relations are empirical and hence it is better to use as per recommended Reynolds number range."

You haven't clearly stated what the formula is for the initially laminar side though.

How do you get the formula for CD on that side that starts off initially laminar ?

Last edited by Rob Wilk; May 2, 2020 at 06:40.

May 2, 2020, 14:07
#16
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duri
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Quote:
 Originally Posted by Rob Wilk I think you are referrring to you mentioning this earlier on "Check the formula again turbulence cd should be more than laminar cd. So the formula have to use Rex^(1/5) or (1/7) instead of (1/2). More over all the turbulence relations are empirical and hence it is better to use as per recommended Reynolds number range." You haven't clearly stated what the formula is for the initially laminar side though. How do you get the formula for CD on that side that starts off initially laminar ?

I checked the book. It has the equation you need 7.49a in page 466. This can be used directly for combined laminar and turbulent side of the plate. Check the example 7.4 in the same page it solves similar problem (part c).

May 3, 2020, 03:16
#17
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Rob Wilkinson
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Quote:
 Originally Posted by duri I checked the book. It has the equation you need 7.49a in page 466. This can be used directly for combined laminar and turbulent side of the plate. Check the example 7.4 in the same page it solves similar problem (part c).

Where does the CD = 1.46/ (Rex)^1/2 for laminar flow that is given in the question, come into this then ?

To me it seems that the formula on pg 466 in the textbook is based on a different CD.
I think you can see now why I mentioned the subtraction coming into it.
I think that is just due to the fact that it is not entirely laminar

It just seems to me that for this question maybe it should be
CD = 1.46/ (Rex)^1/2 - 1440 / Rex for initially laminar side.

Surely I have to use CD = 1.46 / (Rex)^1/2 somewhere in this question.

I don't know if you are an expert on this topic, but I'd appreciate your thoughts on what you think about this.

May 3, 2020, 11:07
#18
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duri
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Quote:
 Originally Posted by Rob Wilk Where does the CD = 1.46/ (Rex)^1/2 for laminar flow that is given in the question, come into this then ?

So you need to use this equation then. This should be your question.

Quote:
 To me it seems that the formula on pg 466 in the textbook is based on a different CD.
You gave this reference and you should know what you are referring.

Quote:
 It just seems to me that for this question maybe it should be CD = 1.46/ (Rex)^1/2 - 1440 / Rex for initially laminar side.
So, this is your own formula. Then you should not expect others to prove your formula. Anyway this is not correct from my understanding.

Quote:
 Surely I have to use CD = 1.46 / (Rex)^1/2 somewhere in this question.
Yes you can use this if you had approached the problem differently. Assume instant transition and use turbulent full length CD - turbulent transition length CD + laminar transition length CD.

May 3, 2020, 12:08
#19
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Rob Wilkinson
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Quote:
 Originally Posted by duri So you need to use this equation then. This should be your question. You gave this reference and you should know what you are referring. So, this is your own formula. Then you should not expect others to prove your formula. Anyway this is not correct from my understanding. Yes you can use this if you had approached the problem differently. Assume instant transition and use turbulent full length CD - turbulent transition length CD + laminar transition length CD.

Enough of the bullshit.

Try working it out yourself and see if you can get the answer of 1.5 to 1 for the ratio of the drag forces.

Then you can tell me how it is worked out.

 May 3, 2020, 15:00 #20 Member   Join Date: Aug 2018 Posts: 77 Rep Power: 8 b...s..t? a bit rich for a guy who floods this forum with his homework problems... insulting ppl who are trying to help... classy. Hopefully others will take note and think twice before helping you pass your intro to fluids class... smh.

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