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Old   April 29, 2020, 03:18
Smile Boundary Layer Thickness at end of Train Carriage
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I did a Civil Engineering course some years ago and from my textbook I have an attached scan of the question on boundary layer for a train, that I have been trying to solve, but haven't.



Based on boundary layer profile and wall shear stress given, I have come up with an expression for boundary layer thickness. I can go into the calculations later for that, if needed.

I don't unfortunately have a way of putting the symbols up, but I'll explain what I have so far and why I can't get to the answer of 1.089 m.

Boundary Layer Thickness for one carriage = (0.233 * 5/4)^4/5 * (viscosity / Air Density * Train Speed)^1/5 x^4/5 where x = carriage length = 18 m

Boundary Layer Thickness for one carriage = (0.233 * 5/4)^4/5 * (0.000019 / (1.18 * 180/3.6))^1/5 * 18^4/5

Boundary Layer Thickness for one carriage = 0.189 m

There are 6 carriages and a locomotive and there is a gap in between them.

From the question it says that the gap between the carriages was found to cause the boundary layer thickness to increase suddenly by 1/5 of the boundary layer thickness at the end of the carriage upstream of the gap.

Now how do I go from here to get 1.089 m ? If I multiply 0.189 * 6 and take into account the gaps, it is more than 1.089 m

Do I need to use some sort of integration between some limits to work this out ?

I am stumped as to how to get the answer from here, need some help.

Last edited by Rob Wilk; May 4, 2020 at 12:36.
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Old   April 29, 2020, 04:37
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I guess this and your other posts are typical homework problems. We had a discussion here a while ago if / how to answer them. I would suggest you askthe moderators to create a subforum for homework / assignment questions and post them there. This willavoud conflicts and increase your chances of getting answers.

This is not directed at you personally, but I would suggest establishing some rapport in the forum before posting a bunch of questions - answer some yourself first, contribute to the discussions etc. „quid pro quo“. If I see someone who signed up and starts posting HW questions on the same day right away, my motivation for answering is honestly not great.
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Old   April 29, 2020, 06:34
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Get the boundary layer thickness for an 18m long plate. I 'm not going to check your work, but presumably it's 0.189 m. Then you add a factor of 1/5 to this, so the thickness at the beginning of the 2nd train car is 0.189*1.2=0.227 m.

So now you need to figure out how a much thicker this boundary layer becomes after an additional length of 18m. There's LOTS of ways to interpret this. One way to think about it is, back-calculate how long a plate is needed to generate a BL thickness of 0.227m and then add 18m to that length.

For example (and this is definitely not the answer), but if you know that an 18m long plate has a thickness of 0.189m, what was the BL thickness at 9m? at 12m? Try calculating the BL thickness at different locations of this plate and get a feel for how the BL develops.


Yet another way is to take the derivative of the BL thickness correlation, that tells you the rate at which the BL layer grows. And then you set up an initial value problem and march your boundary layer by integrating the thing you just differentiated. It's all the same, it's just what works best for you works best for you.
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Old   May 4, 2020, 04:45
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Quote:
Originally Posted by vesp View Post
I guess this and your other posts are typical homework problems. We had a discussion here a while ago if / how to answer them. I would suggest you askthe moderators to create a subforum for homework / assignment questions and post them there. This willavoud conflicts and increase your chances of getting answers.

This is not directed at you personally, but I would suggest establishing some rapport in the forum before posting a bunch of questions - answer some yourself first, contribute to the discussions etc. „quid pro quo“. If I see someone who signed up and starts posting HW questions on the same day right away, my motivation for answering is honestly not great.
If I saw someone posting my type of questions and I knew how to solve them, I would only be happy to help them.
I would be happy to help others also, if I knew how to solve them.

I would not be judging someone in a negative way like you have and showing them disrespect.
Honestly I just cannot understand why my questions should cause conflict, it's just bloody ridiculous.
You should be able to come to a forum to get help and there should not be a problem at all.
I have discussed about answering questions in my other replies I have sent you.
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Old   May 4, 2020, 05:39
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less entitlement, more humility would go a long way.

Maybe take a hint from how the ppl here react to your tone and questions (if they react at all, that is.)

Ppl are volunteering their time here, just treat them accordingly.

In fact, this would make a nice tutorial on how not to interact with ppl in a forum

a) register and start posting without so much as a greeting / introducing oneself

b) filling the main forum up with 8 or so new threads at once

c) getting into fights with / berating others who are trying to help

d) being pushy by repeatedly posting „did you read what I said“ to keep the thread up in the list

e) using foul language


If you look to other questions / posts here, you will find that the community is quite active and helpful. If you find that that is not what you experience, it might be due to a number of things, among them 1) nobody has a good answer, 2) one does not feel inclined to answer.

I‘m sure you are a nice and courteous person IRL, just also be one online.

Cheers and back to topic!
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Old   May 4, 2020, 12:47
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Quote:
Originally Posted by vesp View Post
less entitlement, more humility would go a long way.

Maybe take a hint from how the ppl here react to your tone and questions (if they react at all, that is.)

Ppl are volunteering their time here, just treat them accordingly.

In fact, this would make a nice tutorial on how not to interact with ppl in a forum

a) register and start posting without so much as a greeting / introducing oneself

b) filling the main forum up with 8 or so new threads at once

c) getting into fights with / berating others who are trying to help

d) being pushy by repeatedly posting „did you read what I said“ to keep the thread up in the list

e) using foul language


If you look to other questions / posts here, you will find that the community is quite active and helpful. If you find that that is not what you experience, it might be due to a number of things, among them 1) nobody has a good answer, 2) one does not feel inclined to answer.

I‘m sure you are a nice and courteous person IRL, just also be one online.

Cheers and back to topic!
This question is a lot easier to read now since I have put in a new scan.

Do you know how this boundary layer works with the carriages ?
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Old   May 12, 2020, 03:35
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Quote:
Originally Posted by vesp View Post
I guess this and your other posts are typical homework problems. We had a discussion here a while ago if / how to answer them. I would suggest you askthe moderators to create a subforum for homework / assignment questions and post them there. This willavoud conflicts and increase your chances of getting answers.

This is not directed at you personally, but I would suggest establishing some rapport in the forum before posting a bunch of questions - answer some yourself first, contribute to the discussions etc. „quid pro quo“. If I see someone who signed up and starts posting HW questions on the same day right away, my motivation for answering is honestly not great.
Seeing I have gone through a tough experience so far for help on this forum, I think in moving forward the best way to sort this out, is to put my Questions (Fresh from scratch) into a subforum like you mentioned.
However I think it would be disrespectful if it was called homework problems, a better name I think would be "Past Textbook Questions".

I am not sure where to locate the moderators on this website, if it can still be done.

I am giving a second chance at this to get these questions solved through this forum and I hope that this forum will not let me down.
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Old   July 25, 2020, 22:03
Default Boundary Layer Thickness at end of Train Carriage
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I am still not getting a good understanding on how I can solve this question on Boundary Layer Thickness at end of Train Carriage.
The part that I am having difficulty getting an understanding on is how you work out the boundary layer thickness at the end of the sixth carriage.

I have worked out boundary layer thickness for one carriage and adding on the gap of 20 percent more, but I am not sure how this works for the other carriages.

I would really be appreciative of getting some help from someone who knows how this works.
Once I have got a good understanding on how this works then I can go ahead and calculate it.
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Old   July 27, 2020, 10:39
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Rob,
I think that if you re-read Lucky Tran's post you will see that someone did answer your question. I will restate the answer in slightly different terms below.

Step one: Determine the type of geometry the roof of a railroad carriage most resembles.
(Lucky Tran gave you a big hint)
Step two: Find the equations that would tell you what the boundary layer thickness at a known distance (you seem to have done that)

Now do the following
A. Calculate the thickness at the end of carriage 1, add the delta
B. Calculate the thickness at the end of carriage 2 based on the starting thickness
Repeat Step B for each carriage.

As to the homework issue. It sure looks like a homework problem to me. And you have not explicitly said that it is not. If you are trying to work out CFD on your own from a text book, you should say so, but forgive some skepticism on the part of this forums membership. We get lots of students looking for the answer to homework problems. Even grad students looking for project help. I for one do not think I improve the CFD community if I help someone, who then passes the work of this forum off as their own.
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Old   July 30, 2020, 23:35
Default Boundary Layer Thickness at end of Train Carriage
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Quote:
Originally Posted by AndyR View Post
Rob,
I think that if you re-read Lucky Tran's post you will see that someone did answer your question. I will restate the answer in slightly different terms below.

Step one: Determine the type of geometry the roof of a railroad carriage most resembles.
(Lucky Tran gave you a big hint)
Step two: Find the equations that would tell you what the boundary layer thickness at a known distance (you seem to have done that)

Now do the following
A. Calculate the thickness at the end of carriage 1, add the delta
B. Calculate the thickness at the end of carriage 2 based on the starting thickness
Repeat Step B for each carriage.

As to the homework issue. It sure looks like a homework problem to me. And you have not explicitly said that it is not. If you are trying to work out CFD on your own from a text book, you should say so, but forgive some skepticism on the part of this forums membership. We get lots of students looking for the answer to homework problems. Even grad students looking for project help. I for one do not think I improve the CFD community if I help someone, who then passes the work of this forum off as their own.
From before, I have worked out that the boundary layer thickness for the first carriage = 0.189 m plus the gap takes it to 0.227 m

The question says that the train consists of a locomotive and six carriages, which can be regarded as smooth slab-sided boxes 18 m long, so that is the type of geometry it resembles.

Now from what I have been given as a clue before, we cannot work it out like this
Boundary layer thickness at end of sixth carriage = 0.227 * 7 = 1.589 m

The answer is 1.089 m

So just to be clear, what I am asking is what method do we use to get to the answer, is it like differentiation or integration and what way is it done.
Once I get a clear understanding on that then I can work towards that answer.
At the moment I'm not getting enough to go by.

It's also understanding how the boundary layer develops.


The other point off topic that I want to mention is up the top of my post I have put this
"I did a Civil Engineering course some years ago and from my textbook I have an attached scan of the question on boundary layer for a train, that I have been trying to solve, but haven't."

So I think you can clearly see from this that my question is not a homework problem.
To me a homework problem is for people who are currently studying, which I am not. I can understand why you might be on your guard against people like that who might try to use a forum to cheat to get an answer which would help them with their study.

I'm not a cheater and I am a trustworthy person just wanting to solve this because I have a side interest in it.


I think I at least deserve some good help here to make me have a good understanding on this.
I'm a busy man and don't want to spend forever on this.
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Old   July 31, 2020, 04:40
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Dear Rob:

I am curious ... which text book is that?

I did not browse through the other responses, nor did I try to work out the question, but I think for this you would need to first work out the BL thickness over the continuous length of the six carriages (18 m x 7, no gaps), and then apply the 1/5 increase at the end the train to get the BL thickness there.

In case you missed it, the 1/5 factor compounds itself with each carriage, so I imagine your calculation would end up being something like this:

[BL at the end of train w/ gaps] = [BL at end of train w/o gaps] x [ 1.2 ] ^ [ # gaps between carriages ].

I don't know if they also count the open end at the end of the train as a gap so give or take a gap for this. It looks like a "busy work" type of problem. If you can do this for two cars, you can do this for six, or a hundred.

Hope that helps, Gerry.

P.S. - Homework or not, if you have questions, then you have questions. There is nothing wrong asking for opinions here. Others might think you should be doing this completely on your own, but they can choose not to answer at all.

P.Ss - For the record, I calculated this using a length of 126 m (18 x 7), the answer provided by the textbook (1.089 m) corresponds exactly to three gaps over the same distance (I got would be 1.88 m for 6 gaps). Whether it is pure luck, or my approach is wrong, or the textbook answer is actually incorrect, I don't know.

This is the approach I took:

1) Start with von Karman's integral relation: d[theta]/dx = 1/2 Cf
2) Use [tau_w] to determine what Cf is
3) Use the 1/6 velocity profile to solve for [theta] (momentum thickness)
4) Substitute (2) and (3) to (1) to get an expression for delta(x), the gap-less boundary layer thickness

I got the following expression at the end of (4):

delta(x) = [ (0.8) (0.23333) (0.0238185) x ] ^ (0.8)

or

delta(x) = (0.0131351) x^(0.8)

From your initial post (now I have a bit of time to read) it looks like you have reached this far. I recognize some of the numbers you have, though not all.

... and finally:

5) evaluate delta at the end of the train and compound the 1.2 factor for each gap.

I have spent enough time on this but it is good to get back to my theory, which I must admit is getting rusty after all these years.

Last edited by Gerry Kan; July 31, 2020 at 09:11.
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Old   July 31, 2020, 09:21
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Rob:

I found one tiny mistake from your original post ...

Quote:
... (0.233 * 5/4)^4/5
should be

Quote:
... (0.233 * 4/5)^4/5
Gerry.

P.S. - You know it's Friday afternoon!
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Old   July 31, 2020, 10:18
Default Boundary Layer Thickness at end of Train Carriage
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Quote:
Originally Posted by Gerry Kan View Post
Dear Rob:

I am curious ... which text book is that?

I did not browse through the other responses, nor did I try to work out the question, but I think for this you would need to first work out the BL thickness over the continuous length of the six carriages (18 m x 7, no gaps), and then apply the 1/5 increase at the end the train to get the BL thickness there.

In case you missed it, the 1/5 factor compounds itself with each carriage, so I imagine your calculation would end up being something like this:

[BL at the end of train w/ gaps] = [BL at end of train w/o gaps] x [ 1.2 ] ^ [ # gaps between carriages ].

I don't know if they also count the open end at the end of the train as a gap so give or take a gap for this. It looks like a "busy work" type of problem. If you can do this for two cars, you can do this for six, or a hundred.

Hope that helps, Gerry.

P.S. - Homework or not, if you have questions, then you have questions. There is nothing wrong asking for opinions here. Others might think you should be doing this completely on your own, but they can choose not to answer at all.

P.Ss - For the record, I calculated this using a length of 126 m (18 x 7), the answer provided by the textbook (1.089 m) corresponds exactly to three gaps over the same distance (I got would be 1.88 m for 6 gaps). Whether it is pure luck, or my approach is wrong, or the textbook answer is actually incorrect, I don't know.

This is the approach I took:

1) Start with von Karman's integral relation: d[theta]/dx = 1/2 Cf
2) Use [tau_w] to determine what Cf is
3) Use the 1/6 velocity profile to solve for [theta] (momentum thickness)
4) Substitute (2) and (3) to (1) to get an expression for delta(x), the gap-less boundary layer thickness

I got the following expression at the end of (4):

delta(x) = [ (0.8) (0.23333) (0.0238185) x ] ^ (0.8)

or

delta(x) = (0.0131351) x^(0.8)

From your initial post (now I have a bit of time to read) it looks like you have reached this far. I recognize some of the numbers you have, though not all.

... and finally:

5) evaluate delta at the end of the train and compound the 1.2 factor for each gap.

I have spent enough time on this but it is good to get back to my theory, which I must admit is getting rusty after all these years.
Hi Gerry

The textbook is Solving problems in Fluid Mechanics Volume 1 by J F Douglas and this is question 8 from chapter 14 Flow past solid boundary.

Thanks for the help very appreciated, I am going to bed now, but when I get some spare time, I will follow through what you have explained and workout how I can get to the answer.
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