# Conduction Calculation Example – 2

In daily life, we will encounter various type of conductive surface. It may a plate with different surface area on both side or application on pipes. Or, we will encounter insulation in series and parallel arrangement. The conduction calculation basis is still the same. However, we need to transform to suit the application.

### Conduction through Solid with Variable Surface Area.

As illustrated above, conduction occurs in solid material with changing surface area. In this case, we may calculate the area as an arithmetic average or logarithmic average. The logarithmic average is calculated as follows:

For a cylindrical body such as pipe, we may transform the equation into:

When the ratio of or is not exceeding 2, we can use arithmetic average with small range difference of about 4% compared to logarithmic average. Note should be taken that arithmetic average is always yielding bigger result than logarithmic average. Arithmetic average for the area can be equated as follows:

### Conduction Calculation in Serial Arrangement

Heat transfer flow for material in series can be calculated by two methods. We can perform calculation by dividing temperature difference for each material with respective resistance or by calculating total temperature difference then divide it by total resistance from each material. Referring to the basic equation below:

Application to each layer of material are as following equations:

We may apply assumption that the heat transferred through each layer is the same:

We can get a final equation which represents second method stated above, as follows:

with:
Ka, Kb, Kc = thermal conductivity for each material
Aa, Ab, Ac = average surface area for each material
La, Lb, Lc = thickness for each material

If the material is in cylindrical form, it is better to use the logarithmic average of surface area for each material.

### Conduction Calculation through Parallel Arrangement

In this arrangement, the easiest way is to assume that all the surface is exposed to heat equally. And, we may assume that total heat transfer is equal as a total of heat transferred from each material.

The equation used is as following:

Based on the above assumption, T1 = T3 and T2 = T4 then:

This equation is similar to the electrical equation for parallel resistance.