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Just wondering if someone can please can give me some idea, as to how I can solve this question that is not related to current study. I haven't been able to work out where to start with it just yet.

Air flows parallel to the surface of a smooth flat plate 10m long. The boundary layer has zero thickness at the leading edge. The Reynolds number at the trailing edge of the plate is 10^7. Calculate the total drag force due to skin friction on one side of the plate per unit width? Assume that for a laminar boundary layer, up to Rex = 5_10^5, the skin friction coefficient is Cf = 1.328 (Rex)^-1/2 and for turbulent boundary layer Cf = 0.074 (Rex)^-1/5. Take the density of air as 1.2 kg/ m^3 and its dynamic viscosity as 1.8_10^-5 kg/ ms.

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The problem you cite in the question is a pretty standard homework problem in a graduate fluid mechanics course (it might be also an undergraduate course but less likely). Therefore, I would be reluctant to provide the solution.

What is important to remember when dealing with this problem is that it is crucial to attend the lecture at which the subject was covered. Otherwise, it will be difficult to understand how to solve the problem since the books are not very clear on the subject.

Regarding your question - where to start at - I would recommend Robert H. Nunn "Intermediate fluid mechanics" text book. In the 1989 edition (probably the first edition of the book) I would refer you to the paragraph 13.5 "Flat plate boundary layer flows" on p.255.

Brief introduction to the problem.

It was found that when a fluid flows over a flat plate, almost all flow velocity reduction happens in a narrow layer adjacent to the plate. This layer is called boundary layer. That is why in order to find drag force which fluid exerts on the plate, one should consider only narrow layer of the flow adjacent to the plate, i.e. boundary layer.

People have considered it. And revealed that drag force can very conveniently be calculated using "skin friction coefficient" $c_f$ in general and "friction drag coefficient" $C_f$ (or $C_D$ sometimes) in particular.

There are two formulas for friction drag coefficient: one if the ENTIRE boundary layer (over the entire length of the plate) is laminar; the other one is if the ENTIRE boundary layer is turbulent.

It was revealed that in most cases of flows over a plate, boundary layer can start as laminar and then at some length of the plate become turbulent. And then it is turbulent to the end of the plate.

How to deal with it, if we have only formulas for entire laminar and entire turbulent boundary layers? There is one way (not very difficult) of dealing with it, about which one may read in the Nunn's book, for instance.

It is clear from the statement of your problem, that at some point the boundary layer of your flow converts from laminar to turbulent. It is clear because it is stated that at the trailing edge (i.e. at the end of the plate), $Re = 10^7$ which is higher then critical $Re$ at which conversion occurs: $Re_{crit} = 5·10^5$. Hence, you are dealing with a combined boundary layer.

Having found friction drag coefficient for a combined boundary layer (partially laminar - partially turbulent), one can find drag force as: $$F_D = C_f·(0.5·\rho · U_{free stream}^2·A_{plate})$$ Here:

$\rho$ - density of the fluid

$U_{free stream}$ - flow velocity long before it hits the plate

$A_{plate}$ - plate's area

NOTE:

1) $C_f$ - is, actually, called either drag coefficient or friction drag coefficient (sometimes is denoted as $C_D$) - not skin friction coefficient as in the statement of your problem; friction drag coefficient is the coefficient written for the entire length of the plate, whereas skin friction coefficient ($c_f$) is written for a point on the plate; i.e. $C_f$ is an integral of $c_f$

2) You are given critical $Re$ in order to determine at which point the boundary layer converts from laminar to turbulent. You will need it when you'll be calculating combined friction drag coefficient. In order to do it you will need to use the formula: $$x_{crit} = \frac{Re_{crit} · \mu}{\rho · U_{free stream}}$$ Here:

$\mu$ - dynamic coefficient of viscosity of the fluid (the thing which is called "dynamic viscosity" in your question)

$Re_{crit} = 5·10^5$ - $Re$ at which boundary layer becomes turbulent

$x_{crit}$ - length at which boundary layer becomes turbulent

3) You can find $U_{free stream}$, since you know $Re$ at the end of the plate and the length of the plate.

I wish you good luck in your endeavor to write an expression for $C_D$ for a combined boundary layer. While you'll be doing it, I will be glad to answer you theoretical questions if any occur.

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  • $\begingroup$ Hi Ivan Thanks for reply. I did Civil Engineering course some years ago and from my 1986 Solving Problems in Fluid Mechanics Volume 1 by J F Douglas textbook, I had this question. We didn't cover this question during course, it's more aeronautical, but being interested in this, I wouldn't mind finding an answer. I am doing a website on Cruise Ships at the moment and this is an activity on the side, but maybe down the track I wouldn't mind producing a website on these Engineering calculations and relating it to Engineering. My next comment I will talk about question, no space here $\endgroup$ – Rob Wilkinson Aug 31 '19 at 1:43
  • $\begingroup$ I was confused by trailing and leading edge, but you've clarified that. finding $U_{free stream}$ $Re$ = 10^7 = rho · U_{free stream}}$$ * d / $\mu$ I get 10^7 * 1.8*10^-5 = 1.2 * 10 * v = 12 U_{free stream}}$ so $U_{free stream}$ = 15 m/s. I have x critical as 0.5 m, but where does this come in to finding drag force. We have freesteam velocity, Do we use Cf = 1.328 (Rex)^-1/2 to find drag force for the laminar part (but what Rex do I use)? Then we use Cf = 0.074 (Rex)^-1/5 to find drag force for turbulent part (what is Rex for this part)? Combined Drag Force is Total of these. $\endgroup$ – Rob Wilkinson Aug 31 '19 at 2:35
  • $\begingroup$ I'll write about the matter in more details later. For now I can briefly clarify what I meant in my answer. People have derived expressions for $C_f$ for laminar and turbulent boundary layers under condition that it is laminar or turbulent over the ENTIRE length of the plate. I must have a plate over the ENTIRE physical length of which I have laminar or turbulent boundary layer. It means that if I have partial boundary layer then existing formulas are not valid: your can't add drag forces. It is fundamentally incorrect. I bet your book discusses this question if it provides such a problem. $\endgroup$ – Ivan Nepomnyashchikh Sep 1 '19 at 4:14
  • $\begingroup$ Hi Ivan Nepomnyashchikh A few weeks ago you said that you were going to show me in more detail how this question works. It will be much appreciated if you can do this for me with step by step processes of how to get to the answer of 3.75 N. I just want to make it clear, I have studied Civil Engineering with this textbook in the past, so this is not directed towards current study. We didn't cover boundary layer in the course and I am just a bit confused by this question. Help will be much appreciated. $\endgroup$ – Rob Wilkinson Sep 27 '19 at 5:10
  • $\begingroup$ Hello @RobWilkinson. I apologize for the delay but I'm still extremely busy. I didn't forget about your question - it is on my to do list. As soon as I have a bit more time, I'll write about the matter step by step. Hopefully, after the 18th of October. $\endgroup$ – Ivan Nepomnyashchikh Sep 30 '19 at 14:44
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Solution

1) Upstream velocity is not given, but it goes into the drag force expression. Hence, free stream velocity is necessary to find: $$Re_L=\frac{\rho · U_{free stream} · L}{\mu}=10^7$$ (Re at the end of the plate where x=L is given: $10^7$). From here free stream velocity can be found as: $$U_{free stream} = \frac{Re_L · \mu}{\rho · L}=\frac{10^7 · 1.8·10^{-5}\frac{kg}{m·s}}{1.2\frac{kg}{m^3} · 10m}=15\frac{m}{s}$$ 2) Just to get appreciation of the boundary layer structure (and to formally prove that we do in fact deal with a mixed laminar + turbulent flow), let's find $x_{crit}$ - position along the plate where initially laminar boundary layer converts to a turbulent boundary layer. $Re_{crit}$ is given, upstream velocity is found in the item 1, hence $x_{crit}$ can be found from the expression for $Re_{crit}$: $$x_{crit} = \frac{Re_{crit} · \mu}{\rho · U_{free stream}}=\frac{5·10^5 · 1.8·10^{-5}\frac{kg}{m·s}}{1.2\frac{kg}{m^3} · 15\frac{m}{s}}=0.5m$$ One can see on the schematic that conversion occurs pretty close to the leading edge. Hence, the effect of initial laminar boundary layer can be neglected in general. The entire boundary layer can be assumed turbulent and expression for turbulent boundary layer (which is given in the statement of the problem) can be safely used. Still, we shall proceed without this assumption.
3) Drag coefficient for the mixed laminar-turbulent boundary layer needs to be found now. If $Re_{crit}$ is assumed to be equal $5·10^5$ (which is given in the statement of the problem) and neglecting effects of the surface roughness, drag coefficient can be found as follows: $$C_{f_{total}}=C_{f_{turb_{0->L}}}-\frac{1750}{Re_L}$$ Here: $C_{f_{total}}$ - the drag coefficient of interest, $C_{f_{turb_{0->L}}}$ - drag coefficient as if the entire boundary layer were turbulent, $\frac{1750}{Re_L}$ - amendment taking into account half a meter of the laminar boundary layer at the trailing edge. $C_{f_{turb_{0->L}}}$ can be found according to the expression given in the statement of the problem: $$C_{f_{turb_{0->L}}}=\frac{0.074}{Re_L^{0.2}}$$ Also, $Re_L$ - Re number at the end of the plate - is given in the statement of the problem. Overall, drag coefficient is equal to: $$C_{f_{total}}=\frac{0.074}{Re_L^{0.2}}-\frac{1750}{Re_L}=\frac{0.074}{(10^7)^{0.2}}-\frac{1750}{10^7}=0.00278$$ 4) Now, drag force acting on one side of the plate can be found as follows: $$F_D = C_{f_{total}}·(0.5·\rho · U_{free stream}^2·A_{plate})$$ Here, $A_{plate} = L·b$ is the plate's area, where $L$ is the plate's length and $b$ is the plate's width. Therefore, drag force per unit width can be found as follows: $$\frac{F_D}{b} = C_{f_{total}}·(0.5·\rho · U_{free stream}^2·L)=0.00278·0.5·1.2\frac{kg}{m^3}·(15\frac{m}{s})^2·10m=3.753\frac{N}{m}$$

Answer: $3.75\frac{N}{m}$

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  • $\begingroup$ Hi Ivan Nepomnyashchikh Thanks very much for showing me how this works. I follow most of what you have showed me. I'm just wondering how do you get 1750/ ReL ? With Cf Total = 0.074/ ReL^0.2 - 1750/ ReL from what I see, you are subtracting because you are taking into account laminar boundary layer as well as turbulent. Just wondering why that is subtracted ? I don't have the textbook on me at the moment, but I seem to remember that the freestream velocity is 99% of the upstream velocity. But as you said we don't need to take this into account for this question. $\endgroup$ – Rob Wilkinson Jan 7 '20 at 3:59
  • $\begingroup$ Hello @RobWilkinson. The coefficient 1750 - there is a relatively long derivation behind this coefficient. I am sorry but I don't have time to show it here. As I wrote, if your critical Reynolds number is 500000 - you can use this coefficient. And people do it. And yes - the last term on the right hand side is a correction for the initial $0.5m$ laminar boundary layer. Also those 99% - this thing does relate to boundary layer theory but is not relative for the purpose of the present discussion. $\endgroup$ – Ivan Nepomnyashchikh Jan 9 '20 at 2:44
  • $\begingroup$ Hi Ivan Just when you get the time, may you please show me how you got 1750 / ReL. I understand all the other calculations. $\endgroup$ – Rob Wilkinson Jan 10 '20 at 11:24
  • $\begingroup$ Hi Rob, I'm sorry I can't - I am extremely busy and it doesn't seem to change in a few upcoming years. The topic is not that easy and I can't just sit and write the explanation of $1750/Re_L$ - I will need some time to prepare for it and a lot of time to explain it in writing. Maybe this video will help you youtube.com/watch?v=SpkYBNdSOhQ - mixed BL start from the 12th minute. But bear in mind that mixed boundary layers in general and more detailed boundary layer theory which is necessary for understanding $1750/Re_L$ is a matter of graduate level fluid mechanics course. $\endgroup$ – Ivan Nepomnyashchikh Jan 11 '20 at 20:10
  • $\begingroup$ I did a quick Internet search and this video is one of the few resources I could find. I would also suggest Nunn's textbook "Intermediate fluid mechanics" and White's textbook "Fluid mechanics". These are popular books and you can find cheap used ones - at least, I found a used copy of my Nunn's book on Ebay for 50$. $\endgroup$ – Ivan Nepomnyashchikh Jan 11 '20 at 20:10
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Answer

First Part

We need to find the free stream velocity using the equation for the Reynolds Number at the trailing edge of the plate :

ReL = (Rho * U freestream * L) / mu = 10^7

where ReL = Reynolds Number at the trailing edge of the plate = 10^7

Rho = Density of air = 1.2 kg /m^3

L = length of plate from leading to trailing edge.

Note: Leading edge is the start of the plate where we measure length from Trailing edge is the end of the plate where length is measured to

where mu = dynamic Viscosity = 1.8 * 10^-5

ReL at the end of the plate where x = L is given as 10^7

From here free stream velocity can be found as:

U freestream = (ReL * mu) / (Rho * L)

U freestream = (10^7 * 1.8 * 10^-5 kg/ms) / (1.2 kg/m^3 * 10 m)

U freestream = 15 m/s

Second 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 )

Re critical = (Rho * U freestream * 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 = 1.8 * 10^-5 or 0.000018 kg/ ms

Re critical * mu = (Rho * U freestream * x critical)

x critical = (Re critical * mu) / (Rho * U freestream)

x critical = (5 * 10^5 * 0.000018) / (1.2 * 15)

x critical = 0.5 m

Because the plate is 10 m long, this means that 0.5 m along the plate the flow changes from laminar to turbulent.

The next 9.5 m of the plate is turbulent.

Final Part

We will calculate the total drag force due to skin friction on one side of the plate per unit width.

As a small part of it is laminar and most of it is turbulent, 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

Cf = Skin Friction Coefficient

Firstly we need to calculate Skin Friction Coefficient from the formula given.

Cf = 0.074 / (Rex)^1/5

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

From before Rex = 10^7 for the whole plate.

Now we can calculate Cf

Cf = 0.074 / (10,0000,000)^1/5

Cf = 0.074 / 25.118864

so Cf for AC = 0.00294599 (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 per metre span

Area of plate in AC = 10 * 1 = 10 m^2 per metre span

(FD) for AC = Cf * 0.5 * Rho * (U freestream)^2 * A for AC

(FD) for AC = 0.00294599 * 0.5 * 1.2 * 15^2 * 10

(FD) for AC = 3.9770906 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 Skin Friction Coefficient Cf.

Cf 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 Cf

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

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

CD = 0.074 / 13.79729661

so Cf 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 = 0.5 * 1 = 0.5 m^2 per metre span

(FD) for AB = Cf * 0.5 * Rho * (U freestream)^2 * A for AB

(FD) for AB = 0.00536337 * 0.5 * 1.2 * 15^2 * 0.5

(FD) for AB = 0.362027 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 = 3.9770906 - 0.362027 = 3.6150636 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 = 1.328 / (Re critical)^1/2

CD = 1.328 / (500,000)^1/2

CD = 1.328 / 707.1067812

so CD for AB = 0.00187807561 (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 = 0.5 * 1 = 0.5 m^2 per metre span

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

(FD) for Laminar = 0.00187807561 * 0.5 * 1.2 * 15^2 * 0.5

(FD) for Laminar = 0.12677010373 Newtons

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

Total Drag force on plate = 0.12677010373 + 3.6150636

so Total Drag Force due to Skin Friction on one side of the plate per unit width = 3.7418 Newtons per metre span

Book answer is 3.75 Newtons because it rounded up to the nearest 0.05 Newtons

There is another way of doing this, it involves calculating the combined Skin Friction Coefficient Cf and then calculating Drag force from that formula FD = 0.5 * Rho * (U freestream)^2 * A for the whole plate

However I am not sure how to calculate the combined Skin friction Cf doing it this way.

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  • $\begingroup$ Back again, thought you told us you were off to pastures greener? $\endgroup$ – Solar Mike May 8 '20 at 10:09
  • $\begingroup$ @SolarMike Thanks for your friendly welcome again. At least I have managed to find an answer this time and you can't complain about that. However the purpose of this post how do I this the other way I mentioned using Combined Skin Friction and drag force formula after that. $\endgroup$ – Rob Wilkinson May 8 '20 at 10:38
  • $\begingroup$ @SolarMike Hi how are you doing, if you know of anyone that can help me understand how this is calculated with the combined skin friction coefficient way, then I will be really grateful for that. I know we haven't always seen eye to eye with some of the posts that I have put up, can we put this behind us and move on. I can still write messages, but I can't put any more posts up due to the very strict standards on this forum. $\endgroup$ – Rob Wilkinson May 12 '20 at 2:14

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