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Laminar Boundary Layer Thickness before Transition1 Attachment(s)
I did a Civil Engineering course some years ago and from my textbook I had a question on this.
Attachment 78253 I have calculated boundary layer thickness based on the boundary layer velocity profile given, The Boundary Layer Thickness = sqrt 180 * [(μ0) / (ρ * Uo)]^1/2 * x critical^1/2 where μ0 = viscosity of water = 1.308 * 10^-3 kg/ ms ρ = density of water at 10 degrees C = 999.7 kg/ m^3 from google search and Uo = 5 m/ s first we have to find x critical (length of plate) at the critical point when it transitions from laminar to turbulent. We have been given the Transition Reynolds number = 4 * 10^5 Now Transition Reynolds Number = (ρ * Uo * x critical) / μ0 Transition Reynolds Number * μ0 = (ρ * Uo * x critical) x critical = (Transition Reynolds Number * μ0) / (ρ * Uo) x critical = (4 * 10^5 * 1.308 * 10^-3) / (999.7 * 5) x critical = 0.10467 m Boundary Layer Thickness = sqrt 180 * [(1.308 * 10^-3) / (999.7 * 5)]^1/2 * 0.10467^1/2 Boundary Layer Thickness = 2.22 mm The textbook answer is 0.906 mm Why is my answer out ? May someone please explain to me what things I need to understand, so that I can work towards the answer and get a better understanding on this ? |

Laminar Boundary Layer Thickness before TransitionQuote:
I haven't had much time to look at this sort of stuff recently because I have been busy with other things. My thinking was that the velocity profile was provided, so that you can come up with the formula for boundary layer thickness. That was how I calculated it , but I didn't get the correct answer. It looks like you have calculated the same critical location as me. From what you have said, you think that Boundary Layer is close to (5.4 x) / (sqrt Rex), I'll go through the calcs to check that out. I just wonder how you arrive at that formula ? the 5.4 is quite different from the sqrt 180 I used in my calculation. 99% criterion is that in relation to the free stream velocity ? |

Dear Rob:
You can perhaps use the velocity profile to derive the boundary layer thickness, not unlike the method used in the train question (link here). But in that case you still need some kind of shear stress. The expression for the skin friction coefficient the laminar boundary layer: is, in fact, the result of the Blasius solution, which in effect makes the velocity profile superfluous, because the Blasius solution also comes with a velocity profile. The 5 you see is actually , a normalized vertical distance and, as you have pointed out, it is related to the freestream velocity. Since the Blasius vertical velocity profile is asymptotic, a commonly accepted criterion for finding is to cut it off when , the happens when . This should have been covered in most text books on fluid dynamics. So, it could be that the textbook author did not like this seemingly arbitrary cut-off, and wanted you to work out the boundary layer yourself. If you end up using the velocity profile to estimate the boundary layer thickness, the coefficient you get should in fact be closer to 5 (i.e., ~ 5.4 based on my calculations) than . Gerry. |

Laminar Boundary Layer Thickness before TransitionQuote:
Using the formula which is similar to yours but in a different format. Putting these values μ0 = viscosity of water = 1.308 * 10^-3 kg/ ms ρ = density of water at 10 degrees C = 999.7 kg/ m^3 from google search and Uo = 5 m/ s into equation Boundary Layer Thickness = 0.906 mm = coefficient * [(1.308 * 10^-3) / (999.7 * 5)]^1/2 * 0.10467^1/2 and I get 5.474 So now its just a matter of working out how to get to that coefficient. You mentioned blassius, I'm not familiar with this Just wondering if you don't mind are you able to please show how you would calculate to get to the coefficient ? This is the part of the question that I am not so sure about, once I get a good understanding on this I think it can get solved. |

Laminar Boundary Layer Thickness before TransitionQuote:
Just wondering if you got the chance to look into my previous reply It went like this I have worked out what coefficient will give the answer of 0.906 mm Using the formula which is similar to yours but in a different format. Putting these values μ0 = viscosity of water = 1.308 * 10^-3 kg/ ms ρ = density of water at 10 degrees C = 999.7 kg/ m^3 from google search and Uo = 5 m/ s into equation Boundary Layer Thickness = 0.906 mm = coefficient * [(1.308 * 10^-3) / (999.7 * 5)]^1/2 * 0.10467^1/2 and I get 5.474 So now its just a matter of working out how to get to that coefficient. You mentioned blassius, I'm not familiar with this Just wondering if you don't mind are you able to please show how you would calculate to get to the coefficient ? This is the part of the question that I am not so sure about, once I get a good understanding on this I think it can get solved. |

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