# Calculating Thermal Convection Coefficient

I am designing a heat sink, the diameter of the hole is 6mm. and the diameter of the fibre rod is 1.2mm.

The material I am using is aluminium, with thermal conductivity of 230. The ambient temperature is 25 degrees celsius. Assume that the boundary condition on the fiber rod wall is at a temperature of 70 degrees celsius.

I am trying to determine the temperature of thermal equilibrium. I am facing trouble calculating the Convection Coefficient ( h ).

I first calculate the heat flux through the entire object through conduction, Using Fourier's Law, $$\vec q = -k ( T_s - T_\infty)$$. $$\vec q = 230 \times ( 70 - 25 )$$ which gives $$10350W/m^2$$.

Next, I calculate the convection from the surface to the ambient air, Using Newton's law of cooling. The heat Transfer equation gives $$\vec q=h(T_s - T_\infty)$$. By re-arranging the equation gives $$h=10350/ (70-25)$$ . I feel that my calculations are inaccurate. Can anyone enlighten and help me with this? I know that there are much more to convection than this, such as nusselt number etc? Any help will be greatly appreciated!

• Heat flux (or thermal flux, or heat flux density, or heat-flow density or heat flow rate intensity) is a flow of energy per unit of area per unit of time. It is expressed in W/m^2 (not in temperature degrees). Please explain the heat source more explicitly. Are you pumping some fluid in there at 70C? Also, what is the other larger annular opening, what is it passing through there and at what temperature is it? Aug 24 at 7:43
• The heat source is a fibre rod, that is heated to 70 degrees Celcius. It is a solid rod, not a liquid! The larger annular opening is a hinge, and can be ignored! Thank you for your reply! @NMech Aug 24 at 7:49
• The temperature on the outside is irrespective of the power (the intensity of the light) pumped in the fibre rod? Because, 10 kW per m^2 is a substantial quantity, and it should come from somewhere. Aug 24 at 7:51
• For a simplified analysis, I would assume that the temperature on the outside is constant at room temperature (25 degrees Celsius). The 10kW per m^2 comes from a power source, which supplies up to 50W, The 70 degrees celsius is an estimated temperature through the fibre rod! @NMech Aug 24 at 7:59
• if you mean the outer surface of the object, I would assume the difference between the inner surface(the one in contact with the fiber) and the outer surface ( the one in contact with the ambient air )to be 0.1 degrees Celcius because it has a thickness of 1cm. @NMech Aug 24 at 8:03

I will try and go through the calculation. This is not a solution however, because what I'll be providing you is ok for flat plates/walls, and the irregular geometry that you have will need some numerical tools to simulate and get an reasonably accurate estimate.

Assuming that you have a constant wall temperature of 70oC at the wall of the fiber wall, then the heat power (energy in the unit of time) $$\dot{Q}$$ ending up in the environment will be:

$$\dot{Q} = \frac{\Delta T_{tot}}{R_{tot}}$$

where:

• $$\Delta T_{tot}= 70-25 [^oC]$$ is the total temperature difference
• $$R_{tot}$$ is the total thermal resistance (see MIT lectures on combined conduction and convection)

The total thermal resistance can be found (for the flat wall case) as:

$$R = \frac{L}{kA} + \frac{1}{h_{\infty}A}$$

where:

• k is the Thermal Conductivity 230 (W/m K)
• $$h_{\infty}$$ is the convective coefficient of air side.
• L is the thickness of the aluminium
• A is the cross-sectional area of the heat exchange.

The reason you will need a numerical method (FEM or similar), is that the last two parameters (L and A) provide a problem, because:

• the thickness L of the aluminium is not constant
• the cross-sectional area A is not constant.

However for a simplified analysis you can calculate an average quantity for both L, and A and use that to get a ball park figure.

## convective heat transfer coefficient of the process

The convective heat transfer coefficient is expressed in $$\frac{W}{m^2.^oC}$$. This heat coefficient is very much depended on the velocity. The easiest way (avoiding Nusselt and Prandtl number which are required for a detailed calculation) the easiest way would be to use the following equation (ref)):

$$h_\infty = 12.12 - 1.16 V + 11.6 \sqrt{V} \tag{eq.1}$$

where:

• $$V$$ is the velocity of air as it approaches the exchange surface.
• $$h_\infty$$ is the convectivity heat transfer coefficient in $$\frac{W}{m^2. C}$$

Make a note, that although in your case is zero, you can use a fan to provide faster air and thus increase the convective coefficient and thus the overall heat transfer. (that is why computers use fans).

## quick calculation of required power per area

Assuming you have a 50 W source for the rod of $$d_f=1.2 mm$$ diameter, with a fiber length of $$L_f=60mm$$, then the exchange area of the rod would be:

$$A= \pi\cdot d_r \cdot L_f=226 mm^2$$

That means that per $$m^2$$ the energy is in the order of $$\frac{P}{A} = 221\frac{kW}{m^2}$$ (which is quite a lot). You might need to increase the dimensions (also I don't know the fraction of the 50W that ends up on the heat sink, however even if its only 10%, you'd still have 22 $$\frac{kW}{m^2}$$). Just to give a measure of reference direct sun on a clear sky is about 1 $$\frac{kW}{m^2}$$.

assuming the fibre is not touching then the $$R_{tot}$$ would be modified as follows (in red is the additional quantity):

$$R =\color{red}{\frac{1}{h_{fr}A_{fr}}}+ \frac{L}{kA} + \frac{1}{h_{\infty}A}$$

where:

• $$h_{fr}$$ is the heat convectivity coefficient between the fibre rod wall and the hole
• $$A_{fr} = \pi d_{fr} L_{fr}$$ the exchange are between the fibre rod wall and the hole
• $$d_{fr}$$ the diameter of the fiber rod
• $$L_{fr}$$ the length of the fiber rod
• Your equation (1) doesn’t seem to use SI units, and its origin is unclear. Please add any pertinent information for its use in case the link goes bad, as often happens. Aug 25 at 14:23
• @chemomechanics Thanks for that I updated. Hopefully now its in order. Aug 25 at 14:27
• The equation is from Siple and Passel's "Measurements of Dry Atmospheric Cooling in Subfreezing Temperatures," Proceedings of the American Philosophical Society 89 1 (1945), which should be reviewed before using the equation to ensure applicability. Aug 25 at 16:06
• @Chemomechanics TBH you taught me something (I didn;t know about the limitation of this equation). In this case though, given that there is so little information, and irregular geometry, I think it would be a exercise with diminishing returns to try and calculate the $h$ with the use of Nusselt and Prandlt numbers. Therefore, I think that the 10 W/m2.K is a reasonable approximation (to my knowledge its between 1 and 25 W/m2.K for air convection). Of course, since the question is primarily for $h$ you could offer another relationship which is more appropriate as an answer. Aug 26 at 12:01