## Boundary Layer Thickness at end of Train Carriage

## Boundary Layer Thickness at end of Train Carriage

(OP)

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.

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.

## RE: Boundary Layer Thickness at end of Train Carriage

I tried to look at the equations, but it's all a soft, lo-resolution blur.

## RE: Boundary Layer Thickness at end of Train Carriage

I think you may have thought the equation is a soft resolution blur because there is no symbols in it.

I don't think the forum allows me to put those symbols like the boundary layer thickness one in this question, but I checked my textbook and believe that I have the correct formula there.

Yes I think that the boundary layer does not increase linearly.

I calculated at the end of the 1st carriage, Boundary Layer Thickness = 0.189 m + 20 % for the gap = 0.2268 m

How does that work with each of the different carriages after this first carriage though ?

So for example if we are doing the second carriage, how do we apply boundary layer thickness equation to this ?

I'm just trying to get a feel of how this works.

May you please explain to me how this works, if you know how this works.

## RE: Boundary Layer Thickness at end of Train Carriage

## RE: Boundary Layer Thickness at end of Train Carriage

Gap-seals within flight controls to structure... and likely the flexible shrouds from car-to-car in high speed trains... minimize thru-flow that disrupts aerodynamics. My dad flew his homebuilt aircraft without FC gap-seals... recorded the data over several flights… then added simple duct tape across the same joints and repeated the tests. At cruise power he calculated a 3-to-4-Kt speed increase with a minor positive change in low/stall-speed controllability.

Such devices like gap-seals become impractical/low-value-added for very low-speed aircraft and ground/wheeled vehicles... but obviously matching surfaces-to-surfaces [cab-to-trailer-to-trailer] close/aligned/similarly-sized... and/or tailored devices... vortex generators and/or flow-reattachment 'fins' on the back trailer... at the front/back-ends of the multi-wheel/van long-haul trucks are obviously worth the fuel savings.

NOTE.

I suspect that [with/without these devices] there is also greater concern for cross-wind effects on multi-trailer semis... Hmmm I wonder if this is a situation where aerodynamicists can help the trucking industry with practical solutions???

BTW.

I live in the mid-west and am constantly on the freeway... Gasoline prices have dropped dramatically due to greatly reduced auto travel... but trucking has NOT seen the same decline in fuel prices. Amazing how gasoline can be $1.10/gal... while at the same station... diesel will be 1.89/gal.

Regards, Wil Taylor

o Trust - But Verify!

o We believe to be true what we prefer to be true. [Unknown]

o For those who believe, no proof is required; for those who cannot believe, no proof is possible. [variation,Stuart Chase]

o Unfortunately, in science what You 'believe' is irrelevant. ["Orion", Homebuiltairplanes.com forum]

## RE: Boundary Layer Thickness at end of Train Carriage

another day in paradise, or is paradise one day closer ?

## RE: Boundary Layer Thickness at end of Train Carriage

I have the question easy to read now.

Yes I think that the boundary layer does not increase linearly.

I calculated at the end of the 1st carriage, Boundary Layer Thickness = 0.189 m + 20 % for the gap = 0.2268 m

How does that work with each of the different carriages after this first carriage though ?

So for example if we are doing the second carriage, how do we apply boundary layer thickness equation to this ?

I'm just trying to get a feel of how this works.

May you please explain to me how this works, if you know how this works.

## RE: Boundary Layer Thickness at end of Train Carriage

TTFN (ta ta for now)

I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg

FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm

## RE: Boundary Layer Thickness at end of Train Carriage

https://forum.altair.com/topic/39250-drag-force-ra...

https://www.cfd-online.com/Forums/main/226461-drag...

Based on these, I must conclude that you are a student.

TTFN (ta ta for now)

I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg

FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm

## RE: Boundary Layer Thickness at end of Train Carriage

I am not a student because I have said up the top of my post that 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.

Can you please keep to the topic of this question.

Do you know much about boundary layer and how it works with this question ?

## RE: Boundary Layer Thickness at end of Train Carriage

## RE: Boundary Layer Thickness at end of Train Carriage

notice that BL is not directly proportional to length, hence not "6*"

how does the BL respond to the 20% increase due to the gap ? At the start of the 2nd carriage it is thicker than it "wants" to be, does it …

1) continue increasing as though nothing's happened ? (but then it is inconsistent with the BL thickness equation?)

2) stay this thickness until "x" along the carriage ? ("x" is the length for the 20% increase)

3) fall to try to match the BL equation ??

how to use the wall shear stress ? (or is this a red herring ?)

note, the question posed is clearly from a text, and this site is intended for professionals and not students.

professionals trying to relearn something can be accepted.

another day in paradise, or is paradise one day closer ?

## RE: Boundary Layer Thickness at end of Train Carriage

It's good to see that some people have a great sense of humour in this forum "another day in paradise, or is paradise one day closer" I say some day we will have a paradise, but things aren't going to be normal for a while yet.

The question is from my Solving Problems in Fluid Mechanics Volume 1 by J F Douglas textbook that I have always kept from when I studied Civil Engineering years ago.

My boundary layer equation comes from the boundary layer profile and wall shear stress given.

I haven't got enough ideas just yet from what you have told me and I still haven't got my head around how this boundary layer increases with the six carriages and locomotive just yet.

I managed to solve my drag force ratio question through a youtube Gate Academy Plus Fluid Mechanics Boundary Layer video.

I just need to find a good video that explains how boundary layer increases with the length, this is the key to getting through this question I think.

Thanks

Robert

## RE: Boundary Layer Thickness at end of Train Carriage

this is wrong as …

1) train + 6 carriages = 7 (not 6)

2) BL thickness is not linear with length (from your equation, BL depends on "x^0.8")

3) can you show your work, derivation of BL thickness ?

4) how does the BL respond after a 20% jump ?

another day in paradise, or is paradise one day closer ?

## RE: Boundary Layer Thickness at end of Train Carriage

The length of the train and it's carriages = 18 * 7 = 126 m , but I know this is not based on total length.

The gap in between each carriage is causing me a bit of difficulty with understanding how it works with this question.

They say it increases by 1/5 of the boundary layer thickness at the end of the carriage upstream of the gap.

Looking for an understanding on how the boundary layer thickness increases throughout the whole train going from one carriage to another ?

## RE: Boundary Layer Thickness at end of Train Carriage

## RE: Boundary Layer Thickness at end of Train Carriage

The step growth may or may not be a real factor, but just to create the basis for setting up the boundary condition at the leading edge of each following car. So whatever the rate of growth there is for a boundary layer of that thickness begins there and the layer grows per that until the end of the car and the next step.

It's been too long and I've been too lazy to see how the shear force is translated into a boundary layer thickness rate of change, but I expect there is at least one stage of integration or differentiation to solve this problem.

## RE: Boundary Layer Thickness at end of Train Carriage

Are you able to please show me the process of how this boundary layer works for this question.

I just still haven't got an understanding of how it should be calculated for the end of the sixth carriage.

Robert

## RE: Boundary Layer Thickness at end of Train Carriage

at the end of the train the height of the BL is "easily" calc'd. In my experience the BL height asymptotes to some height, based on free stream velocity, shear stresses, blah, blah (i'm not an aero guy).

The key question, and one I don't think we can easily answer, is how does the BL respond to the 20% step due to gap between the carriages ? Should the "x" for the BL height be the distance from the nose of the train ? Is the gap a known distance (1m) ? But the 20% bump causes the BL to be out of step with what "x" would want it to be. So does the BL …

1) keep growing as a function of x ? (not likely IMO)

2) stablise at the 1.20 height until the BL equation exceeds this ?

3) fall to what the BL equation wants ? Clearly not immediately ('cause then the 20% would disappear from the problem)

But wait, if over the length of the next carriage the BL exceeds the 20% bump then you don't care about it ('cause your equation for BL height will exceed this bump, if BL height is a function of "x")

Why do you know the BL is not based on a length of 127m ? (this is not BL height at 18m *7)

another day in paradise, or is paradise one day closer ?

## RE: Boundary Layer Thickness at end of Train Carriage

It seems that you are not completely sure how this boundary layer is worked out like me.

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.

You mentioned x being the distance along the carriage, makes me think that the 1/5 part is just added on at the end and is not part of x.

The confusing part is when it says "increase suddenly by 1/5 of the boundary layer thickness at the end of the carriage upstream of the gap".

I think you are and I am confused by this part because the gap would have a bit of a distance, but I think you have to treat it as if it is negligible, so it is not much distance.

I worked out boundary layer thickness at end of first carriage = 0.189 m and with 20 % added on for the gap that gives you 0.2268 m.

What I am confused about is how does it work from there, going to the second carriage and the ones after that.

I need to find a good website that discusses how boundary layer thickness changes over distance.

## RE: Boundary Layer Thickness at end of Train Carriage

I can scan from my textbook an example of how I would get to that formula for boundary layer thickness based on the boundary layer profile and wall shear stress given in question.

## RE: Boundary Layer Thickness at end of Train Carriage

"I worked out boundary layer thickness at end of first carriage = 0.189 m and with 20 % added on for the gap that gives you 0.2268 m." … understood

"What I am confused about is how does it work from there, going to the second carriage and the ones after that." … yep, that's the crux of the matter ! How does the BL respond if it is "too big" (due to the 20% bump) ?

If the train were one long carriage, you could easily calc the BL height (for x = 127m) and it's not 7*0.189.

Perhaps you calc BL for 14m and if greater than 0.2268, then you use that; else BL is 0.2268. Then the 2nd gap jumps the BL to 0.272m. etc

another day in paradise, or is paradise one day closer ?

## RE: Boundary Layer Thickness at end of Train Carriage

Yes this is a tricky question, I think it's just a case of finding a good website on the internet that explains how boundary layer changes over distance and when there is a gap (or a bump).

I see how you got 0.272 m that's 0.2268 * 1.2. The 0.2268 m was the first carriage plus the gap.

That 0.272 m must be for the second gap. Although you'd have the length of the carriage before that gap to take into consideration as well.

How come you said BL for 14 m, the carriages are 18 m long.

Just got to find a good resource that gives a good understanding on how this is calculated.

## RE: Boundary Layer Thickness at end of Train Carriage

you're right, how does the BL respond to a step change from an outside influence (like a gap or a shock) ?

You got an equation to calc height as a function of x (you've done x = 7, how about x = 14 ? … it's not linear)

another day in paradise, or is paradise one day closer ?