The two sides of a wall (2 mm thick, with a cross-sectional area of 0.2 m2) are maintained at 30oC and 90oC. The thermal conductivity of the wall material is 1.28 W/(m·oC). Find out the rate of heat transfer through the wall?
1. Steady-state one-dimensional conduction
2. Thermal conductivity is constant for the temperature range of interest
3. The heat loss through the edge side surface is insignificant
4. The layers are in perfect thermal contact
Fig. 2.4: Illustration 2.1
1. Steady-state one-dimensional conduction.
2. Thermal conductivity is constant for the temperature range of interest.
3. The heat loss through the edge side surface is insignificant.
4. The layers are in perfect thermal contact.
On putting all the known values,
Fig. 2.5: Illustration 2.2
The previous discussion showed the resistances of different layers. Now to understand the concept of equivalent resistance, we will consider the geometry of a composite as shown in fig.2.6a.
The wall is composed of seven different layers indicated by 1 to 7. The interface temperatures of the composite are T1 to T5 as shown in the fig.2.6a. The equivalent electrical circuit of the above composite is shown in the fig 2.6b below,
Fig.2.6. (a) Composite wall, and (b) equivalent electrical circuit
The equivalent resistance of the wall will be,
Therefore, at steady state the rate of heat transfer through the composite can be represented by,
where, R is the equivalent resistance.