Site pages

Current course

Participants

General

Module 1. Basic Concepts, Conductive Heat Transfer...

Module 2. Convection

Module 3. Radiation

Module 4. Heat Exchangers

Module 5. Mass Transfer

## Lesson 5. Electrical analogy and Numerical Problems related to conduction

**Electrical Analogy For Conduction Problems**

Consider heat flowing through a slab of thickness ‘L’ and area “A’ and T_{1} and T_{2} are the temperatures on the two faces of the slab as shown in Figure 1. Heat transfer from high temperature side to low temperature side is expressed as

Where (T_{1} –T_{2}) is the thermal potential

is the thermal resistance

Now, consider an electric circuit having resistance ‘R’ and electric potential E_{1} and E_{2} at the ends as shown in Figure 2. Current ‘I’ passing through the circuit can be expressed as

Equations (1) and (2) are found to be symmetrical on comparison. ‘Q’ amount of heat flows through the slab having thermal resistance when a thermal potential (T_{1} –T_{2}) exits. Similarly, ‘I’ amount of current passes through the circuit having resistance ‘R’ when an electric potential (E_{1} – E_{2}) exists. Therefore, flow of heat through the slab can be represented by an electric circuit as shown in Figure 3.

If a hot gas at temperature T_{g} is in contact with one side of the slab and air at temperature T_{a} at the other side, then heat transfer from the hot gas to air through this slab of thickness can be represented by an electric circuit as shown in Figure 4

Heat transfer from hot gas to air can be expressed as

For a slab made of three material having thermal conductivities K_{A}, K_{B} and K_{C} respectively and is exposed to a hot gas on one side and atmospheric air on the other side as shown in Figure 5, an equivalent electric circuit has been shown in Figure 6. Heat transfer from the hot gas to atmospheric air is expressed as

Similarly for the composite slab shown in Figure 7, an equivalent electric circuit has been shown in Figure 8.

**Example 3.14 The walls of heating unit in cold region comprise three layers**

**10cm outer brick work (k =0.75 W/m-deg) **

**1.5cm inner wooden paneling (k =0.75 W/m-deg)**

**8cm intermediate layer of insulating material**

**The insulation layer is stated to offer resistance twice the thermal resistance of brick work. If the inside and outside temperatures of the composite wall are 30°C and -10°C respectively, determine the rate of heat loss per unit area of the wall and the thermal conductivity of the insulating material.**

**Solution: **Thermal resistance for a plane wall of thickness δ, area A and thermal conductivity k is prescribed by the relation R_{t} =δ/kA

** **

**Example 3.15 A furnace wall is a made up of steel plate 10 mm thick (k=62.8 kj/m-hr-deg) lined on inside with silica briks 150 mm thick (k =7.32 kj/m-hr-deg) and on the outside with magnesia bricks 200 mm thick (k = 18.84 kj/m-hr-deg). The inside and outside surfaces of the wall are at temperature 650° C respectively. Make calculations for the heat loss from unit area of the wall.**

**It is required that the heat loss be reduced to 10Mj/hour by means of air gap between steel and magnesia bricks. Estimate the necessary width of air gap if thermal conductivity for is 0.126 kj/m-hr-deg.**

**Solution: **Thermal resistance for a plane wall of thickness δ, area A and thermal conductivity k is prescribed by the relation R_{t} =δ/kA.

**Example 3.16 A furnace wall comprises three layers: 13.5 cm thick inside layer of fire brick, 7.5 cm thick middle layer of insulating brick and 11.5 cm thick outside layer of red brick. The furnace operates at 870°C and it is anticipated that the outside of this composite wall can be maintained at 40°C by the circulation of air. Assuming close bonding of layers at their interfaces, find the rate of heat loss from the furnace and the wall interface temperature. The wall measures 5m****2m and the data on thermal conductivities is:**

**Fire brick k _{1} = 2.4 W/m-deg**

**Insulating brick k _{2 }= 0.14 W/m-deg**

**Red brick k _{3} = 0.85 W/m-deg**

**Solution: **The wall area (5m 2m) = 10 m^{2} is constant for all layers of the composite wall. The thermal resistance R_{t} of a slab (thickness δ, conductivity k and area A ) is given by