Category: Technology

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.

After getting a brief knowledge of conduction heat transfer, we can go further to understand how we can calculate the conduction heat transfer. We also may modify the equation thus the thickness of medium or insulation can be calculated. For better understanding conduction calculation, we provide several examples of heat transfer calculation. Disclaimer: This calculation is for rough estimation only. Do not attempt to use the calculation for severe applications. Contact professional if critical and severe application is required.

Conduction Calculation through Wall or Thick Plate:

From the above illustration, we have a wall with heat transfer surface area (A), zone with higher temperature (T1) and zone with lower temperature (T2). In this calculation, we assume that thermal conductivity remains constant. Rearrange basic equation obtained before, we can get the following equation:

To find insulation thickness, we may assume that the calculated heat flow, Q, is the same with Q that flowing through the insulation. Thus, we can calculate the thickness of the insulation with the following equation:

As example, we assume that a storage room with managed temperature of 20 °C surrounded by 20 cm thick brick wall. If wall’s thermal conductivity is 2,6 W/m.K and outside temperature 35 °C, calculate heat transferred from outside into the storage room.

Since the room temperature is managed using an air condition system, the duty of the system is as calculated. If less duty is expected, an additional insulation might be added. When 40% reducing is expected, what is the insulation thickness needed?
First, we need to calculate temperature at wall surface if heat transfer rate is reduced.

After that, we can calculate insulation thickness based on that temperature. Assuming that insulation has thermal conductivity of 0.24 W/(m.K), the insulation thickness is:

We can select thicker insulation to meet market availability of insulation material. The standard thickness is usually in the increment of 10 mm or 0.5”.