[Rob Wilk]

Originally Posted by vesp

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.

[Rob Wilk]

Originally Posted by vesp

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 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

[duri]

Originally Posted by Rob Wilk

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.

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.

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.

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

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.

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.

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

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

[LuckyTran]

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.

[Rob Wilk]

Originally Posted by LuckyTran

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.

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.

[LuckyTran]

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.

[Rob Wilk]

Originally Posted by LuckyTran

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.

[Rob Wilk]

Originally Posted by LuckyTran

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.

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.

[Rob Wilk]

Originally Posted by LuckyTran

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.

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.

[LuckyTran]

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.

Originally Posted by Rob Wilk

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

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.

[duri]

Originally Posted by Rob Wilk

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.

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

[Rob Wilk]

Originally Posted by LuckyTran

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.

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.

[vesp]

not sure who is trolling whom here anymore...

[LuckyTran]

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.

[Rob Wilk]

Originally Posted by LuckyTran

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.

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.

[vesp]

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!

[Rob Wilk]

Originally Posted by LuckyTran

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.

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.

[Rob Wilk]

Originally Posted by vesp

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.

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.

[Rob Wilk]

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

[Rob Wilk]

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.

with the b****s , you are saying here that I should not expect others to prove my formula that has a sarcastic tone to it.

The CD formula can be similar to the book.


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.

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