
[Sponsors] 
April 28, 2020, 10:02 
Drag Force Ratio for Flat Plate

#1 
Member
Rob Wilkinson
Join Date: Apr 2020
Location: Wellington, New Zealand
Posts: 69
Rep Power: 5 
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 Last edited by Rob Wilk; May 8, 2020 at 04:04. Reason: Question Answered 

April 28, 2020, 11:29 

#2 
Senior Member
Lucky
Join Date: Apr 2011
Location: Orlando, FL USA
Posts: 5,665
Rep Power: 65 
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  
Member
Rob Wilkinson
Join Date: Apr 2020
Location: Wellington, New Zealand
Posts: 69
Rep Power: 5 
Quote:
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: 5,665
Rep Power: 65 
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 

#5  
Member
Rob Wilkinson
Join Date: Apr 2020
Location: Wellington, New Zealand
Posts: 69
Rep Power: 5 
Quote:
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: 5 
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: 5,665
Rep Power: 65 
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 

#8  
Member
Rob Wilkinson
Join Date: Apr 2020
Location: Wellington, New Zealand
Posts: 69
Rep Power: 5 
Quote:
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. Last edited by Rob Wilk; April 29, 2020 at 23:18. 

April 30, 2020, 07:37 

#9  
Member
Rob Wilkinson
Join Date: Apr 2020
Location: Wellington, New Zealand
Posts: 69
Rep Power: 5 
Quote:
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  
Senior Member
duri
Join Date: May 2010
Posts: 245
Rep Power: 16 
Quote:
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  
Member
Rob Wilkinson
Join Date: Apr 2020
Location: Wellington, New Zealand
Posts: 69
Rep Power: 5 
Quote:
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  
Senior Member
duri
Join Date: May 2010
Posts: 245
Rep Power: 16 
Quote:
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  
Member
Rob Wilkinson
Join Date: Apr 2020
Location: Wellington, New Zealand
Posts: 69
Rep Power: 5 
Quote:
"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: 16 
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 

#15  
Member
Rob Wilkinson
Join Date: Apr 2020
Location: Wellington, New Zealand
Posts: 69
Rep Power: 5 
Quote:
"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  
Senior Member
duri
Join Date: May 2010
Posts: 245
Rep Power: 16 
Quote:
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  
Member
Rob Wilkinson
Join Date: Apr 2020
Location: Wellington, New Zealand
Posts: 69
Rep Power: 5 
Quote:
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  
Senior Member
duri
Join Date: May 2010
Posts: 245
Rep Power: 16 
Quote:
So you need to use this equation then. This should be your question. Quote:
Quote:
Quote:


May 3, 2020, 12:08 

#19  
Member
Rob Wilkinson
Join Date: Apr 2020
Location: Wellington, New Zealand
Posts: 69
Rep Power: 5 
Quote:
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: 7 
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. 

Tags 
aerodynamics, boundary layer 
Thread Tools  Search this Thread 
Display Modes  


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
The CoP Does not exist: Validating Aerodynamic forces through a "line of action"  ds4719  Main CFD Forum  14  February 18, 2022 18:05 
Drag Coefficient on flat plate normal to flow  Bin  STARCCM+  19  June 26, 2017 11:03 
pressure eq. "converges" after few time steps  maddalena  OpenFOAM Running, Solving & CFD  69  July 21, 2011 07:42 
Force vectors for drag during sweeping motion  aamer  FLUENT  0  April 18, 2011 08:17 
Drag force on scaled models.  arunjingade  Main CFD Forum  6  July 1, 2010 08:54 