# Conduction Calculation Example – 1

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”.