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**Illustration 2.1****The two sides of a wall (2 mm thick, with a cross-sectional area of 0.2 m2) are maintained at 30 ^{o}C and 90^{o}C. The thermal conductivity of the wall material is 1.28 W/(mÂ·^{o}C). 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

Given,

**Fig. 2.4: Illustration 2.1**

**Illustration 2.2**

**Solution 2.2**

Assumptions:

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**

Thus,

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 T_{1} to T_{5} 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,

where,

Therefore, at steady state the rate of heat transfer through the composite can be represented by,

where, R is the equivalent resistance.

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