Conductivity Calculations

Previous blogs have talked about heat conductivity in very general terms to produce a foundation for this somewhat more technical view for those of you who like formulas and numbers.

Conductive heat transfer can be expressed with “Fourier’s Law”

• q = k A dT / s
• where
• q = heat transfer (W, J/s, Btu/hr)
• A = heat transfer area (m², ft²)
• k = thermal conductivity of material (W/m K or W/m ^oC, Btu/(hr ^oF ft²/ft))
• dT = temperature gradient – difference – in the material (K or ^oC, ^oF)
• s = material thickness (m, ft)

Using this equation, one can easily calculate the amount of heat that will be conducted through a given material in a given period of time knowing its thermal conductivity, its cross section, length of the conductive path (thickness in the case of something more flat) and temperature differential between the heat source and the heat sink.  Let’s look at each in a little more detail – –

First, you need to make sure that the (k) you are using is in the proper units, English or metric.  The heat transfer area (A) is the cross section of the conductive path in either square meters or square feet (inches).  This could be the cross section of a conductive bar or the surface area of an electric heater, for example. The temperature gradient (dT) is the difference in degrees centigrade (or degrees kelvin) or degrees Fahrenheit between the temperature of the heat source and the temperature of the heat destination.  This equation works in the case that the heat source and the heat sink are able to maintain their temperatures despite the removal and addition of heat as heat is conducted from one to the other.

Note – Calculating what will happen if the temperature of the source is lowered by the removal of heat and the temperature of the destination is increased by the addition of conducted heat is a little more complex requiring calculus (which I do as infrequently as possible).

The material thickness (s) is the length of the conductive path in meters or feet between the heat source and the heat destination.

Although this equation can give specific numbers, it can also be useful in getting an intuitive feel for the effect that can be expected when various parameters are changed.  For example, the greater the area of the conductive path, the greater the heat transfer.  Doubling the contact area of a heater will double the ability for it to conduct heat to the heated surface.  In another example, reducing the thickness of a metal by 1/2 will double the amount of heat that can be conducted through it.

The equation also tells us that if you need to increase the temperature differential between the source of heat and the heat sink as in the case of an insulator, you need to either reduce the conductivity of the material between them, increase the length of the conductive path or decrease the cross section of the conductive path.  To decrease the temperature differential as in the case of transferring heat to a heating bath, reverse procedures will produce the desired effect.

It is convenient that everything changes in direct proportion as there are no squared factors.  Doubling or halving any given parameter will either double or halve the result.

 –  FJF  –