Your question seems to be indeed challenging, thanks. I think that the material property is included in the wall temperature and thus in the film temperature used to evaluate the fluid properties, in particular the fluid thermal conductivity appearing in the expression of the Nusselt number.

The fluid properties are calculated with film temperature which is the average of surface temperature and surrounding bulk temperature and h depends on the type of fluid and fluid velocity.

For same wall temperatures, shapes, and same fluid operating conditions, there is no effect for the wall material on h. The wall material affects only on the diffusion through the wall.

The typical boundary conditions that describe heat transfer between a solid (S) and a fluid (L) (ignoring radiation effects) are (i) continuity of temperature T_S=T_L, (ii) continuity of the normal heat flux q_S=q_L, which can also be expressed as k_S(\nabla T)_S . n=k_L(\nabla T)_L . n, where n is the unit normal vector to the surface, k_S, k_L the thermal conductivities of the solid and fluid, and \nablaT the temperature gradient evaluated at the interface (\nabT)_S in the solid side and (\nabT)_L at the interface in the fluid phase. Thus the temperature gradient is not continuous across the interface but the heat flux is.

When the fluid is in motion, one has to solve a heat conduction problem in the solid coupled to the thermal forced convection (with possibly natural convection) problem on the fluid side. This is often a challenging problem. The introduction of a heat transfer coefficient is a method to decouple the heat conduction problem from thermal convection problem. Thus one assumes that q_L=h(T_L-T_R), where T_R is a suitable reference temperature in the fluid. In dimensionless form the heat transfer coefficient is expressed as a Nusselt number and is typically expressed in terms of the Rayleigh number and Prandtl number for natural convection or Reynolds number and Prandtl number for forced convection. For simple flows these correlations can be found analytically but often suitable correlations are developed from experimental data.

So in summary h does not depend on the physical properties of the solid.

@ Prof. Brian

(Cc: all others)

As mentioned, _L=h(T_L-T_R), shouldn't here be T_S-T_L (where T_S is solid temperature in contact with the fluid and T_L is temperature of liquid coming in contact) Even if I assume some ref. temperature for fluid then still temperature of wall from where heat is transferred has to come into picture.

and since it does, then owing to different surface temperature distribution won't h be different for both materials.

If I assume that h is constant and doesn't depend on solid properties, then does my assertion that h along with thermal conductivity will affect temperature distribution in the solid as well as its surface hold true??

Deewalkar, As I noted in my post, the heat transfer coefficient h is a stand-in parameter for the the unknown heat flux, and usually the reference temperature T_R is selected so that h can be measured experimentally. Normally it is the bulk flow temperature of the fluid. The physics of heat flux requires that the temperature be continuous across the solid-liquid interface. If this were not the case, then there would be an infinite temperature gradient at the interface, which would give rise to an infinite heat flux which is aphysical.

As I mentioned in my earlier post, correlations for the heat transfer coefficient expressed as a Nusselt number, requires knowledge of physical parameters such as fluid density, fluid viscosity, fluid thermal conductivity, etc, all of which are functions of temperature. The specifics of how the correlation is computed dictates what temperature should be used to evaluate the temperature dependent parameters. It could be the wall temperature, but often it is the bulk temperature of the fluid.

There is a nice problem that you can work out analytically that illustrates the points I made above. Consider a solid sphere with a simple heat source suspended in a stagnant fluid. For this problem you can solve for the steady temperature distribution in the solid and the fluid ( it is a pure conduction problem) with the appropriate interface BCs : continuity of temperature and continuity of heat flux. After you have found the temperature distribution, you can then determine the equivalent heat transfer coefficient for this problem that avoids solving for the temperature distribution in the fluid.

Deewakar, I realized I did not address your last question:

"If I assume that h is constant and doesn't depend on solid properties, then does my assertion that h along with thermal conductivity will affect temperature distribution in the solid as well as its surface hold true??"

Yes you are correct. The heat transfer coefficient, and the thermal conductivity of the solid will affect the temperature distribution in the solid and the interface temperature T_S which is equal to T_L. Thus to determine the steady state temperature of sphere of radius R_o in a fluid infinite in extent would be:

\nabla^2T=0 for 0

Normally, the heat transfer coefficient, h, does not depend upon the properties of the wall. There are three exceptions, two of which can be encountered in normal practice.

(1) At high temperatures radiant heat transfer is important and thus the emissivity of the surface can be important. It is only important for heat transfer to the fluid if the fluid is semitransparent to infrared radiation. If the fluid is opaque, there will be no radiant transfer because the surface is at the same temperature as the wall. If the fluid is transparent, there will be radiant transfer through the fluid to other surfaces but not directly to the gas. In the intermediate case of a semitransparent fluid, commonly called the "dusty gas problem" both the optical properties of the fluid and the emissivity of the solid must be considered

(2) For boiling heat transfer, there is a transition from nucleate boiling to film boiling. The heat flux for both the initiation of nucleate boiling and the transition from nucleate boiling to film boiling will depend upon the surface finish, but generally not upon the material of the surface.

(3) The heat transfer coefficient will also depend upon the surface properties at pressures so low that the continuum assumptions are invalid. Under these conditions, a "sticking coefficient" comes into play where there is a time factor for the thermal equilibration of the adsorbed molecule with the surface as opposed to the billiard ball collision of the molecule with the surface.

Where the material comes into play, in normal practice, is in calculation of the overall heat transfer coefficient, U. The overall heat transfer coefficient for transfer of heat from one fluid, through a wall, to a second fluid involves a heat transfer coefficient from the first fluid to the wall, a conductive heat transfer coefficient incorporating geometry and wall material properties and a second heat transfer coefficient from the wall to the second fluid.

At the solid-fluid interface, the convective heat transfer coefficient depends on the thermal conductivity of the fluid, temperature gradient at the interface and temperature difference between the wall and the fluid. Thermal conductivity , temperature gradient at the interface and temperature difference all depending on the type of fluid. Therefore, convective heat transfer coefficient is surely independent of the type of solid material.

Anil,

What you say is generally true, but not for low pressure gases when the characteristic distances are comparable or less than the mean free path in the gas phase. Although then you might claim that there is no convection occurring.

The heat transfer coefficient usually does not depend on the physical properties of the solid, but it depends by the thermo-physical proprieties of the fluid. Only in case of two-phase convection the surface characteristics play an important role in the heat transfer. For example in the pool boiling correlations the main target of the researchers is to include into it parameters such as surface roughness or at least the solid-fluid coefficient.

Your question is confusing. You state that the boundary and ambient conditions are identical, that implies identical Tw for both metals. Th heat transfer coefficient h (free or forced) is a function of the film temperature ..normally (Tw+Ta)/2 where (Ta is ambient fluid bulk temperature) as well as the properties of the fluid. The h will be identical for same conditions i.e Tw on any surface and Ta and a given fluid. Patil is correct.

La nature de la paroi n'intervient pas dans le coefficient d'échange, seul la rugosité peut avoir de l'importance, par contre elle intervient dans le calcul global des flux.

The original question treated the problem as a convective heat transfer problem and set the boundary of the control volume at the interface of fluid and solid. The Tw is treated as a known boundary condition. If the property of the solid material is to be considered, then the problem bbecome convection-conduction problem and Tw is nolonger a known parameber. The boundary of the control volume needs to be set at a place where the boundary condition is known (known T or known gratidant of T or known heat flux). The convective heat transfer coefficient h and temperature Tw can be solved for iteratively.

The heat transfer coefficient is only function of fluid velocity, thermal properties of fluid, Tw, Ta and geometrical shape of the body. However, thermal conductivity of solid surface  affects the teperature gradient near the surface.

For steady state heat transfer conditions, the heat transfer by conduction from Tw to the interior of the solid or from the interior to the surface,  will of course depend on the solid’s/wall’s thermal conductivity. That will establish the value of Tw.  In turn Tw will affect the film temperature and hence the convection coefficient.  Therefore no matter what material the solid is made off under your assumption of Tw being identical for the two materials the wall is made up, the convective coefficient will be identical for both situations. The properties of the solid do not affect the convection coefficient. The convective coefficient will be different for natural compared to forced convection conditions.