chestermiller said:
Even for incompressible flows, the velocity profile affects the thermal boundary layer but, if the temperature difference between the free stream and the wall is small, the temperature profile does not feed back into the velocity profile. So the temperature profile is coupled to the velocity profile, but the velocity profile is not coupled to the temperature profile.
Fair enough. I should be more careful with my language.
chestermiller said:
I don't think that this is the case. As long as the temperature profile does not feed back into the velocity profile, the dimensionless temperature profile within the thermal boundary layer doesn't depend on the temperature difference between the free stream and the wall.
You may be correct here. For a zero-pressure-gradient flat plate, that holds true. The ratio of the two thicknesses for the Blasius boundary layer is approximately \delta_u/\delta_T \approx Pr^{0.4}. I am not 100% sure this holds true for all incompressible flows, but I am at least leaning your direction now.
chestermiller said:
This is the case for liquid metals. But, if the velocity boundary layer is thicker than the thermal boundary layer, then the Prandtl number is high. This is the case with ordinary liquids.
The OP was specifically asking about the thermal boundary layer being larger than the velocity boundary layer, thus the reason I limited discussion to small Pr.
HuskyNamedNala said:
1) I think it is important to state that losses due to heat contribute to the form of the boundary layer. Otherwise, where does the kinetic energy go? At high speeds (even subsonic Mach numbers based on my experience with a wind tunnel at GE) surface heating has a noticeable impact on engineering systems.
The actual heat generated by viscous dissipation is typically quite small unless the flow velocity is high and/or the boundary layer is turbulent. The dissipation term in the energy equation is ##O(U^2)##, while the conduction terms are ##O(c_p\Delta T)##. The latter is usually much higher and dissipation is negligible. This idea is encapsulated in the dimensionless Brinkman number. See
Viscous Flow by White for a good reference source on that sort of thing.
HuskyNamedNala said:
2) I know laminar flow does not strictly come into the definition of the boundary layer. The fact of the matter is that very near the wall the flow must be laminar and that is why the temperature profile takes the basic concave up shape; the rate of heat transfer is proportional to the velocity and flow regime. Without complicating the answer to the question with a turbulent boundary layer or more esoteric cases, I feel the answer I gave suffices.
I disagree. You can define the thermal boundary layer without any sort of discussion of laminar or turbulent flow. It's simply a matter of the temperature profile. Now, there is certainly a discussion that can be had about how much convection and conduction contribute to the shape of the curve as a function of the wall-normal coordinate and how laminar versus turbulent flow affect that answer, but none of that discussion is necessary to simply define the thermal boundary layer. You also have to be careful about how you describe "the smallest scales," as to many people, that kind of wording carries a certain turbulence-related baggage (i.e. Kolmogorov scales).
HuskyNamedNala said:
3) I agree, but in reality they are coupled. Incompressible flow is merely an assumption used to make calculations easier. I think sometimes taking these assumptions for reality gives a false intuition for newcomers.
I think
@Chestermiller clarified my answer quite thoroughly in that the two boundary layers are only decoupled in one direction, but you are correct that it is just an approximation. The problem is, though, that the approximation for low speeds so so nearly exact that you get very small error using it and it is very useful for gaining insight into an otherwise very complex physical situation.
HuskyNamedNala said:
Is this true even for high temperature gradients? For example T_inf = 298K but T_wall = 400K?
This dissipation function is not a function of temperature and depends only on viscosity and the velocity gradients, so yes, his statement is true so long as the viscosity can reasonable be assumed to be constant. For air, viscosity scales roughly with ##\sqrt{T}## (e.g. Sutherland's law), so while the viscosity will change, it is not usually a very noticeable effect and is often ignored (though it does still exist as a factor). For other liquids or gases, the effect may be greater.