Can anyone explain me why the thermal boundary layer develops faster for viscous fluids? I would just say it would develop more slowly because due to high viscosities there are low reynoldsnumbers and thus less turbulence or mixing. This causes a slow homogenization of temperature (assume a cold fluid flowing over a hot plate). Since low mixing, the fluid will remain cold for a large part, and thus the thermal boundary layer is thin?
Imagine a cylinder that starts to rotate in a fluid (similarly to a rotational viscosimeter with concentric cylinders geometry) reaching quickly a fixed velocity. If the fluid is very viscous, only a thin layer near the cylinder will reach appreciable velocity, so stationary state will be reached very quickly. If the fluid has low viscosity, the movement may involve all the available volume, and this may take quite some time especially if the volume is large. You can experiment stirring the contents of a bucket full of say cement+water, and a bucket full of water.
This is not correct. If the flow is laminar, the velocity boundary layer will grow in proportion to the square root of the product of kinematic viscosity times time. Eventually the boundary layer will span the entire gap between the cylinders, and the velocity profile will then be fully developed. The amount of time for this to happen is on the order square of the gap width divided by the kinematic viscosity. The fully developed velocity profile will be independent of the fluid viscosity, and will be the same for all fluids, as long as the flow is laminar. So, the higher the fluid viscosity, the faster the boundary layer grows, and the sooner the fully developed profile is achieved.
thanks, but this has to do with the hydrodynamic boundary layer, but why does the thermal boundary layer also develops faster for a viscous fluid?
Can you please site a specific example? This would be very helpful. Are you saying that the thermal boundary layer grows more quickly for a viscous fluid than for an inviscid fluid flowing through the same channel at the same bulk flow rate? Are you thinking of flow past an obstacle, or flow through a channel? Chet
I don't really have a specific example, the question was asked me in this way 'does the thermal boundary layer develops faster for a viscous fluid than for an inviscid fluid'? I could make an example: imagine you have a hot flat plate at 40°C and there flows a liquid over it, initially at a temperature of 20°C.. where does the thermal boundary develops faster: for a viscous or inviscid fluid?
OK. This is very helpful. I have a specific problem in mind that's even simpler than the one you describe, and that has an analytic solution for both an inviscid fluid and a viscous fluid. We can then compare the equations for the heat transfer coefficients or the thermal boundary layer thicknesses as a function of distance along the surface. How does that sound? Chet
OK. The problem involves steady heat transfer for flow in a tube. You have either an inviscid fluid or a viscous fluid flowing in a tube. The viscous flow is laminar. The tubes are identical and the volumetric flow rates are identical. In the case of the viscous laminar flow, the flow is fully developed, so that the velocity profile is parabolic, with zero velocity at the wall, and maximum velocity (equal to 2x the mean velocity) at the center. In the case of the inviscid fluid, the velocity profile is flat, and equal to the mean velocity at all radial locations. At a certain axial location along both tubes (x = x_{0}), the wall temperature changes abruptly from a value of T_{0} at x<x_{0} to a value of T_{1} at x > x_{0}. The effect of the temperature change on the viscosity of the fluid is neglected (for the viscous laminar flow case). The objective is to solve for the heat transfer rate and the thermal boundary layer development in the tubes at locations x > x_{0}. Attention will be confined to the thermal entrance region, where the boundary layer thickness is small compared to the radius of the tubes. Near the tube wall, the velocity profile in the viscous case is v = γy, where y is the distance from the wall, and γ is the radial velocity gradient at the wall: [tex]γ=\frac{8\overline{v}}{D}=\frac{32Q}{\pi D^3}[/tex] where Q is the volumetric flow rate and D is the tube diameter. In the case of inviscid flow, the velocity near the wall is: [tex]v=\frac{4Q}{πD^2}[/tex] So the velocity profiles near the wall, where the boundary layer development takes place, are very different in the two cases. This is the reason that the rate of thermal boundary layer growth with distance along the wall is so different in the two cases. Are you satisfied with this problem description and its ability to address the question you are asking?
I am confused. Surely an inviscid flow has infinite Reynolds number and is inherently turbulent, assuming there is some sort of perturbation to excite the instability. If there is no perturbation (i.e. theoretical scenario and not real) then there is no mechanism for transverse heat conduction and hence there will be no thermal boundary layer development. Are you comparing a fully developed turbulent boundary layer with free slip BC to a viscous boundary layer with no slip BC?
OK. For the thermal boundary layer problems that I outlined previously, the boundary layer equations (for the so-called "thermal entrance region" of the heated section) are as follows: [tex]\overline{v}\frac{\partial T}{\partial x}=α\frac{\partial^2T}{\partial y^2}[/tex] for the inviscid flow case and [tex]\left(\frac{8\overline{v}}{D}\right)y\frac{\partial T}{\partial x}=α\frac{\partial^2T}{\partial y^2}[/tex] for the viscous flow case, where α is the thermal diffusivity. Boundary and initial conditions on these problems are as follows: [itex]T=T_0[/itex] at y = 0, x < 0 [itex]T=T_1[/itex] at y = 0, x > 0 [itex]T=T_0[/itex] at y → ∞, all x [itex]T=T_0[/itex] at x < 0, all y The solution to these equations for both cases can be found in Transport Phenomena by Bird, Stewart, and Lightfoot. From these solutions, the local heat flux at the wall for each of these cases is given by: [tex]q(x)=\frac{k(T_1-T_0)}{δ(x)}[/tex] where δ can be recognized as the effective boundary layer thickness. This is given by: [itex]\frac{δ(x)}{D}=0.806\left(\frac{αx}{\overline{v}D^2}\right)^{1/3}[/itex] for the viscous flow case and by [itex]\frac{δ(x)}{D}=1.77\left(\frac{αx}{\overline{v}D^2}\right)^{1/2}[/itex] for the inviscid case. These equations are valid for [itex]\left(\frac{αx}{\overline{v}D^2}\right)[/itex] less than 0.01. Why don't you plot them up on a log-log plot and see what you get. Even though the slope is higher for the inviscid case than for the viscous case, the viscous case starts out higher at low values of x. The only real difference between the viscous and the inviscid cases with respect to the cause of the difference in boundary layer growth rate is the nature of the velocity profile near the boundary, where in the inviscid case, the fluid velocity is constant, while, in the viscous case, the velocity is zero at the wall, and increases linearly with distance from the wall (far from the center of the tube).
Inviscid flow is a highly idealized situation, and is not representative of the flow behavior of any real fluids except in certain limiting situations. The basic equations of motion for inviscid flow are the Euler equations, which are the same as the Navier-Stokes equations with the viscosity set equal to zero. Solutions to the Euler equations for inviscid flow do not include turbulence, even though the Reynolds number is infinite. So, if we have no qualms about talking about inviscid flow even though the solution to the equations neglects the possibility of turbulence, why should we have any qualms about doing the same for heat transfer to these hypothetical materials. Inviscid flow solutions to the Euler equations merely provide a basis for reference and comparison for the response of real fluids. If consideration of turbulence had been included, the velocity variations would have been significantly different, although, in the regions away from solid boundaries, the inviscid solutions might not provide too bad an approximation. Another feature of solutions to the Euler equations for inviscid flow is that they implicitly do not satisfy the no-slip boundary condition at solid boundaries. These are the regions where, as a result of the non-slip conditions, significant vorticity generation occurs in real fluids. And the latter contributes significantly to turbulence generation. To answer your second point, the presence of turbulence is not necessary for transverse heat transfer to occur. Heat conduction into the fluid can and does take place without turbulence being involved, for example, in the case of laminar viscous flow. This leads to the development of a thermal boundary layer, as you can see from the examples I presented in my previous post. Thermal boundary layer development also occurs in cases of idealized inviscid fluids where the velocity at the boundary is not zero.
The Euler equations can absolutely include turbulence, as the formation of turbulence is through the non linear advection terms, not the diffusive viscous terms. The diffusion terms merely serve to act as a dissipative mechanism for the end of the turbulent energy cascade. In absence of the viscous terms, it simply means there is no smallest length scale (komolgorov) and so the energy cascade goes on indefinitely. In a numerical simulation, The Euler equations can be used as an approximation to NS as numerical dissipation can fulfill this role. I still fail to see how an inviscid , laminar flow can transmit heat (without advection) perpendicular to a boundary layer unless it is radiative!
I assume you can see how heat can be conducted into a solid cylinder from the outside to the inside in a transient situation without any flow whatsoever (either perpendicular to the wall or parallel to the axis). Now imagine that the cylinder is translating along its axis and you are applying heat to the region x > 0 in a steady state situation. Now, situate yourself in a frame of reference traveling with the axial velocity of the cylinder. Now, in this frame of reference, you are again observing the original transient heat conduction into the cylinder (at the cross section that you are following). The flat velocity profile for the solid cylinder is the same as the flat velocity profile for an inviscid fluid flowing axially. Chet
If you model inviscid flow you use Euler equations. The energy equation does not contain heat conduction or any viscous stress terms, only the pressure work and energy advection. Using Euler equations for compressible inviscid flow there is no thermal conductivity unless you include that term in the energy equation.
Who says that the energy equation does not include conduction? The mechanical energy balance equation, of course, does not include heat conduction, but the overall energy balance equation certainly does. You need to learn the distinction between these two. See Transport Phenomena by Bird, Stewart, and Lightfoot, Chapter 11. They also cover the thermal energy balance equation, which is the difference between the two. Chet
I am fully aware of the difference between thermal and mechanical energy, and that the Euler equations effectively only consider mechanical energy balance, not diffusivity