Drag Force Ratio for Flat Plate
 

I did a Civil Engineering course some years ago and this is a question from my textbook that I haven't been able to solve.

A flat plate 1m wide and 4m long moves with a velocity of 4m/s parallel to its longer sides through still air of kinematic viscosity 1.5 * 10^-5 m^2/s and density 1.21 kg/m^3. On one side of the plate an initially laminar boundary layer is formed. On the other side the leading edge is roughened and the boundary layer may be considered to be entirely turbulent. Assuming a critical Reynolds number of 5 * 10^5 determine the ratio of drag forces on the two sides.

For laminar flow CD = 1.46/ (Rex)^1/2 and for turbulent flow CD = 0.074/ (Rex)^1/5

How do you work out Rex ?

and where does critical Reynolds number come into it ?

I really would like to get an understanding on this,
If someone knows about this, it will be much appreciated that you could please reply to.

Also On one side of the plate it says that an initially laminar boundary layer is formed".
Does this mean that this side of the plate has laminar and turbulent flow ?


This Question has now been solved.


Here is the Answer


First Part


First of all workout if the initially laminar side changes (transitions) to turbulent at some point along the length of the plate.


To do this we will use the critical Reynolds Number given of 5 * 10^5.


we can work out x critical ( length of plate from leading edge where laminar changes to a turbulent zone )


Note: Leading edge is the start of the plate where we measure dimensions from


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


where Re critical = Critical Reynolds Number = 5 * 10^5

Rho = Density of air = 1.21 kg /m^3

x critical = length of plate from leading edge where laminar changes to a turbulent zone

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


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


Because the plate is 4 m long, this means that 1.875 m along the plate the flow changes from laminar to turbulent.

The last 2.125 m of the plate is turbulent.


Second Part



We will calculate the Drag Force on the initially laminar side.


Because there is a a turbulent portion on this side of the plate, this is the approach we will take for drag force.


We will treat it as if the plate is fully turbulent, then subtract the turbulent portion for x critical and then add laminar portion for x critical.


To start with we will first calculate Drag Force for the whole plate based on assuming turbulent boundary throughout the whole length of the plate.


AB = Length of plate to the transition point

AC = Length of the whole plate

(FD) for AC = Drag Force for AC

CD = Drag Coefficient


Firstly we need to calculate Drag Coefficient from the formula given.


CD = 0.074 / (Rex)^1/5


For this Rex is the Reynolds Number based on the whole length of the plate.


Rex = (Rho * v * L) / mu


where L = 4 m, as it is the whole length of the plate

and v = velocity of plate


so Rex = (1.21 * 4 * 4) / 0.00001815

Rex = 1,066,666.67


Now we can calculate CD


CD = 0.074 / (1,066,666.67)^1/5

CD = 0.074 / 16.05483

so CD for AC = 0.0046092 (assuming whole length of plate is turbulent)


Lets say (FD) for AC = Drag Force for whole length of plate AC

and A for AC = Area for whole plate covering length AC

Area of plate in AC = 4 * 1 = 4 m^2


(FD) for AC = CD * 0.5 * Rho * v^2 * A for AC

(FD) for AC = 0.0046092 * 0.5 * 1.21 * 4^2 * 4

(FD) for AC = 0.178468 Newtons ( This is Drag Force, assuming whole turbulent length of plate )



Now we will subtract the Drag Force for the turbulent part before the transition zone.

This covers the length AB of the plate.

and we will calculate:

(FD) for AB = Drag Force for AB

Firstly we need to calculate Drag coefficient CD.

CD is different because we are using critical Reynolds Number, which is based on critical length before transition.

From before Re critical = Critical Reynolds = 5 * 10^5

Now we can calculate CD

CD = 0.074 / (Re critical)^1/5

CD = 0.074 / (500,000)^1/5

CD = 0.074 / 13.79729661

so CD for AB = 0.00536337 (assuming turbulent flow in critical length portion before transition)


Lets say (FD) for AB = Drag Force for assumed turbulent length before transition.

and A for AB = Area for whole plate covering length AB

Area of plate in AB = 1.875 * 1 = 1.875 m^2


(FD) for AB = CD * 0.5 * Rho * v^2 * A for AB

(FD) for AB = 0.00536337 * 0.5 * 1.21 * 4^2 * 1.875

(FD) for AB = 0.09734516 Newtons

( This is Drag Force, assuming turbulance in critical length portion before transition )



The actual Drag Force in Turbulent Zone = [ (FD) for AC - (FD) for AB ]


The Drag force in Turbulent zone = 0.178468 - 0.09734516 = 0.081123241 Newtons for B to C


Now we can calculate the actual force in the laminar zone.

(FD) for Laminar or (FD) for AB for Laminar Zone


For this Laminar flow CD = 1.46 / (Re critical)^1/2


CD = 1.46 / (500,000)^1/2

CD = 1.46 / 707.1067812

so CD for AB = 0.002064752 (for actual laminar flow before transition)


Lets say (FD) for AB = Drag Force for actual laminar zone before transition.

and A for AB = Area for whole plate covering length AB

Area of plate in AB = 1.875 * 1 = 1.875 m^2


(FD) for Laminar = CD * 0.5 * Rho * v^2 * A for AC

(FD) for Laminar = 0.002064752 * 0.5 * 1.21 * 4^2 * 1.875

(FD) for Laminar = 0.03747525 Newtons


Total Drag force on plate (for Initially Laminar side) = [ (FD) for AB of Laminar + (FD) for BC of Turbulent ]


Total Drag force on plate (for Initially Laminar side) = 0.03747525 + 0.081123241

Total Drag force on plate (for Initially Laminar side) = 0.118598486 Newtons


Third Part



Now we will calculate the drag force on the other side, where the leading edge is roughened and the boundary layer may be considered to be entirely turbulent.


Firstly we need to calculate Drag Coefficient for turbulent flow from the formula given.


CD = 0.074 / (Rex)^1/5


For this Rex is the Reynolds Number based on the whole length of the plate.


and from before Rex = 1,066,666.67

CD = 0.074 / (1,066,666.67)^1/5

CD = 0.0046092 ( For Fully Turbulent Side)


Now we can calculate the drag force for this fully turbulent side

(FD) for Turbulent of (FD) for A to C Turbulent


From before Area of plate = A for AC = 4 * 1 = 4 m^2


so (FD) for Turbulent = CD * 0.5 * Rho * v^2 * A for AC

(FD) for Turbulent = 0.0046092 * 0.5 * 1.21 * 4^2 * 4

(FD) for Turbulent = 0.178468 Newtons (Drag force for fully turbulent side of plate)


Final Part


Determining the ratio of drag forces on the two sides, is done by the ratio of the highest force to the lowest force for this question.


Fully Turbulent Side has the highest drag force of 0.178468 Newtons


Initially Laminar Side has the lowest drag force of 0.118598486 Newtons


The ratio of the Drag Forces = 0.178468 / 0.118598486


Ratio of Drag Forces on the two sides = 1.5 to 1

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

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.

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.

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.

You seem like you know a lot about this.
Your help will be much appreciated thanks.
I appreciate some of the help you have given so far.

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.

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.

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.

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 ?

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.

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.

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.

Hi Duri and Vesp

I'm only joking about the bullshit, lightin up.

I understand that you are trying to help
You have got to see it from my perspective though.
I did a Civil Engineering some years ago and I still have this textbook.
I have kept an interest in doing the rest of the question that I couldn't do during study. I wouldn't consider them homework type questions because they are not used to assist me with study, you have to see that .

Sometimes I just feel that you seem a little resistant to help because you think it is a homework type question (it's not just remember that).

I came to this forum hoping that I would have some very good help to get these questions sorted easily.

You imagine, if you were passionate about getting something sorted and it seemed like you weren't getting anywhere with it, you have to do something to get it sorted.

At the moment I am relying on this forum to provide excellent help, which can be hard to come by because some people who don't know the full story may think it is homework (you can't judge a book by it's cover).

Moving forward, if you know how to solve this question, I would appreciate your help to get through this, as I have made a lot of time and effort to get this sorted.

Otherwise, if you don't know how to solve you will have to leave it for someone else to help me out, like an expert in aerodynamics.

Back to the question, I think the ratio may be based on the initially laminar side to turbulent side.

This laminar side is proving a bit of a hassle to work out and holding me back with this question.

Drag Force Ratio for Flat Plate
 

Hi Duri and Vespa

I'm joking about the bullshit.

With all of my questions, I have always stated at the beginning "I did a Civil Engineering course some years ago".

What this means is that these questions are not to help me with current study and I am not trying to pass an "intro to a fluids class".
If you look at a google definition of "what is a homework problem", it is different from mine.

During my Civil Engineering course we didn't cover Flow Past Solid Boundary.
I'm interested in how you solve these questions and.

I came to this forum with the intention of getting excellent help, I'm a good person and can't see why I don't deserve good help.

So far I have felt that some people have wanted to resist giving me help, possibly because of this mindset that it is a homework type.

I think that things would work a lot better if this type of mindset stops because things work a lot better when we all help each other out.
I thought a forum is there to help assist people with questions that they are having difficulty.
There will be times when someone doesn't know where to start with a question and needs some guidance.

I'm a busy man with essential work and also developing a website on the side.

I hope you understand where I am coming from with this now.
I'd appreciate some good help to solve this as quick as possible.

Back to the question:

The question has the answer as a drag force ratio of 1.5 to 1.

I assume that this means the initially laminar side is 1.5 times the drag force on the turbulent side.

I calculated the fully turbulent side, Drag Force as 0.178468 Newtons

If this is correct then this would mean that
Initially Laminar side, Drag Force = 0.178468 * 1.5 = 0.2677 Newtons

Using Drag Force = D = CD * 0.5 * Rho * v^2 * Area of plate

we can find CD

CD = D / (0.5 * Rho * v^2 * Area of plate)
CD= 0.2677 / (0.5 * 1.21 * 4^2 * 4 * 1) = 0.0069137

Now lets see if this works with formula I mentioned
CD = = 1.46/ (Rex)^1/2 - 1440 / Rex

CD = 1.46 / [ (1.21 * 4 * 4) / (0.00001815) ]^1/2 - 1440 / [ (1.21 * 4 * 4) / (0.00001815) ]


CD = 1.46 / [(1,066,667)^1/2] - 1440 / (1,066,667)

CD = 0.0014136 - 0.00135
CD = 0.0000636 This is different

So as you can see from this I am nearly there with working this out, but definitely need help to solve this, otherwise I'll never get the answer.

I would really appreciate some help from anyone that knows how this works.

Thank You

If you have no purpose in solving this other than time pass. Then you should not disturb others. Even assignments or interview preparation is better. To gain knowledge watch online lectures, read books, connect back to university friends or discuss with office colleagues. Wherever you can interact face to face and get your doubts clarified. Forums like this will always have some communication gap.




Many people don't get good help, they only get hints. Check how many questions unanswered or inconclusive in this forum.



No one can stop others in this forum. Anyone can give their opinion and challenge others opinion this is not resistance this is called open mind set. Remember all are busy in their job getting time for this is very difficult and don't expect people to get their hand dirty before reply.




Please write the "BS" comment for yourself. This is never valid on flat plate with attached flow.

If you think a laminar flow has more drag than a turbulent one, you really need to do some reading and brush up on a lot of stuff. There's also plots of them on the same page 466.

The formula for drag over a transitional plate should be something that looks like 0.074/Re^(1/5)- ####/Re. The first part of the formula should be identical to the drag on a wholly turbulent plate and not the laminar one. The magical numbers in the numerator (the 1440) depends on which correlations you started with for laminar and turbulent cases. Since White uses specific formulas for the laminar and turbulent plate, they get 1440. You should get a value that is not 1440, but it will be similar. If you want to reproduce exactly the answer in your textbook, you need to hunt down the formula in your textbook.

Or you could use the formula from White (the correct one) and not what you have written and proceed and you'll get a similar answer. But you do need to copy and use the correct formula. There is also a worked example on the same page 466 going into 467 of a very similar problem with different wording. Everyone in this forum will be hesitant to answer your question because there's a worked example done for you in White.

Drag Force Ratio for Flat Plate
 

Hi LuckyTran

i have some good news.
I have found this youtube video https://www.youtube.com/watch?v=OKsvTwWdu-I 47 and a half minutes in, that I think is great help for me to understand this.
It talks about drag force on a plate for mixed laminar and turbulent flow.

You have to calculate drag force for turbulent flow for the whole plate, subtract drag force for turbulent flow before transition and then add drag force for laminar flow and that is where the laminar CD = 1.46/ (Rex)^1/2 comes into it.

I will work my way through the calculations.

Hopefully I can get the answer of 1.5 to 1, I'll put my calculations up when I have gone through it.

You should get 0.074/Re^(1/5)- 1650/Re

1650 is rounded, it's not exactly 1650.

I'm surprised your book doesn't present it this way.

The book is called Solving Problems in Fluid Mechanics Volume 1 by J F Douglas.

How did you get that formula 0.074/Re^(1/5)- 1650/Re, are you able to show me how you got that ?

I'm just working on it the way shown in the video and see how I go.

Drag Force Ratio for Flat Plate
 

Hi LuckyTran

It seems to me that this formula you have for the initially laminar side CD = 0.074/Re^(1/5)- 1650/Re

is based on initially treating the plate as fully turbulent, taken away the turbulent part before the transition change and then adding the laminar part.

However you have done it with Drag Coefficient instead of drag forces like I have.

I have put up my answer up to this question.

Are you able to please show or explain how you got that CD = 0.074/Re^(1/5)- 1650/Re for the initially laminar side.

I'd just like to know because I want to broaden my knowledge on this type of subject and I feel that understanding it from a different way is beneficial for learning.

Hope to hear from you LuckyTran.
Thanks for the help so far, much appreciated.

Drag Force Ratio for Flat Plate
 

Hi LuckyTran

Just wondering if you have seen my last message to you, because I haven't heard back from you.

Are you able to please show or explain how you got that CD = 0.074/Re^(1/5)- 1650/Re for the initially laminar side.

I'd just like to know because I want to broaden my knowledge on this type of subject and I feel that understanding it from a different way is beneficial for learning.

Thanks for the help so far, much appreciated.

I did the drag coefficient yes. The drag force is the drag coefficient times the dynamic head which is just a number. It's a fixed conversion factor (for this problem) to go from drag coefficient to drag force. Consequently, the ratio of drag coefficients is equivalent to the ratio of drag forces and arguing that I "did it for drag coefficient" is pedantic.

There is no right way, but the approach in the youtube video you linked of subtracting the turbulent part and artificially adding back in a laminar part is the most kindergarten way to solve this problem. This method, consistently overpredicts the turbulent boundary layer thickness because it sets the artificial starting point of the turbulent boundary layer to be the start of the plate (at x=0) instead of at an intermediate location.

The A/Re follows the boundary layer theory of Schlichting. Actually the A is determined from the drag of the laminar plate. The difference is that this approach generally matches what we expect empirically the drag on a plate undergoing laminar-to-turbulent transition to look like.

This isn't the end. There are yet even more complicated methods where the exponent of Re is not unity. Schlichting's is just really easy to apply and one of the most commonly used.


Thanks, I'll keep it in mind to not recommend this text to others. Any text that covers boundary layers over flat plates and doesn't introduce Schlichting's formula or some version of it is probably dogshit.

Is this not Bullshit? Check the post for which you gave this comment. I asked you to do the same.

Drag Force Ratio for Flat Plate
 

So you have worked this out yourself, but don't want to explain to me how you got this, not very helpful.
I have asked you about 3 times how you got this and when you have finally replied to me you say "Consequently, the ratio of drag coefficients is equivalent to the ratio of drag forces and arguing that I "did it for drag coefficient" is pedantic."

I find that very insulting.

i've said before that I'd just like to know because I want to broaden my knowledge on this type of subject and I feel that understanding it from a different way is beneficial for learning.

You also criticise my textbook a lot and say that the youtube way is a kindergarten way to solve this problem.
I know that some methods can be more accurate than others, but I am trying to look at it from the other way you worked it out as well.

I've spent time on this.

I would of thought better of your last reply if you had shown me how you arrived at that formula CD = 0.074/Re^(1/5) - 1650/Re, surely you can do that.

That's what I am interested in now.

not sure who is trolling whom here anymore...

What are you on about!? There is a worked example on pg 466 of White but using different formula.

I criticize your text because everbody knows you shouldn't just subtract the turbulent drag and add back a laminar drag. Yet the fact that you can't find the equivalent Prandtl-Schlichting equation in your book suggests that it indeed wants you to approach the problem this crude way. A pedagogical text should not do take this approach for the same reason that we shouldn't teach kids that pi is 22/7, it's not.

How you get the A/Re in the Prandtl-Schlichting equation is non-trivial and requires you to pick up the book by Schlichting. There is a reason most texts just give the formula and don't give the methodology to get the A coefficient in the numerator. I was kind enough to give you the A coefficient so that you can move along and follow the worked example in White. I am sorry you are insulted by my kindness. I won't be kind anymore. If you want to learn how to get the A coefficient, it's in Schlichting. If that's not enough, it comes from taking the laminar and turbulent drag coefficients at the critical Reynolds number. If you had calculated the drag coefficient using the laminar and turbulent formula, their difference is A*500 000. The difference is that here we are taking the difference according to Schlichting w/ Reynolds number dependence and not a bloody constant.

Drag Force Ratio for Flat Plate
 

Alright lets's put the past behind us and be kind to each other now.
Move on to get this resolved.

I have found this Schlichting book.

You mentioned it comes from taking the laminar and turbulent drag coefficients at the critical Reynolds number.
So that would be 1.46/ (Rex)^1/2 and 0.074/ (Rex)^1/5

I just haven't found yet where it talks about a CD formula for combined laminar and turbulent flow in the book.

So I haven't worked out yet how to get that 0.074/Re^(1/5)- 1650/Re.
I'm bloody determined to get it though because I am interested how the question is approached from this way.

it is honestly a shame to post rip-offs of the great book. I will let Springer know, hopefullythey can do sth about it.

Please remove the link!

Drag Force Ratio for Flat Plate
 

May you please show me the way you have calculated this to get the 0.074/Re^(1/5)- 1650/Re, the administrator doesn't like me showing a link.

Drag Force Ratio for Flat Plate
 

Assuming you are the administrator this doesn't look like a very nice manner to talk to your users.
very sarcastic especially saying floods this forum with homework problems and pass your intro to fluids class... smh.
I could write a forum about my experience with this forum.

You don't know the full story of what's happened eg rudeness encountered my way.

I would of thought that an administrator is there to help create a welcoming experience for a user.

You say I didn't introduce myself to this forum.
I have a picture of myself and you don't.

I may be looking to go to another forum that is more welcoming.
I am on another forum as well as this, that I have felt is a lot more welcoming, the people are nice and not rude like here.

Drag Force Ratio for Flat Plate
 

I think this was the start of you being tricky with me.
You didn't clearly mention it at this stage.

Yes this is related to my answer you are correct about that, I hadn't discovered this just at that time though.