Conduction: One Dimensional - 2

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^oC and 90^oC. The thermal conductivity of the wall material is 1.28 W/(m·^oC). Find out the rate of heat transfer through the wall?

Solution 2.1

Assumptions

• Steady-state one-dimensional conduction.
• Thermal conductivity is constant for the temperature range of interest.
• Heat loss through the edge (side) surface is insignificant.
• The wall faces are in perfect thermal contact with the temperature reservoirs.

Given

• Thickness of wall, L = 2 mm = 0.002 m
• Cross-sectional area, A = 0.2 m²
• Temperatures, T_cold = 30^oC and T_hot = 90^oC
• Thermal conductivity, k = 1.28 W/(m·^oC)

Fig. 2.4: Illustration 2.1

Method

The one-dimensional steady conduction heat transfer rate through a uniform plane wall is given by Fourier's law in the integrated form:

q̇ = k A (ΔT) / L

Calculation (step-wise)

Compute the temperature difference.

ΔT = T_hot - T_cold = 90 - 30 = 60 °C

Substitute known values into the conduction expression.

q̇ = k × A × ΔT / L

q̇ = 1.28 × 0.2 × 60 / 0.002

Calculate the numerator first: 1.28 × 0.2 × 60 = 15.36

Divide by thickness: 15.36 / 0.002 = 7680

Therefore, the rate of heat transfer through the wall is q̇ = 7.68 × 10³ W (or 7.68 kW).

Illustration 2.2

Solution 2.2

Assumptions

• Steady-state one-dimensional conduction.
• Thermal conductivity is constant for the temperature range of interest.
• Heat loss through the edge (side) surface is insignificant.
• Layers are in perfect thermal contact (no contact resistance unless specified).

On putting all the known values,

Fig. 2.5: Illustration 2.2

Thus,

Composite Wall and Equivalent Thermal Resistance

The preceding discussion showed the resistances of individual layers. To understand the concept of an equivalent resistance for a composite wall, consider the geometry shown in the figure below.

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 figure. The equivalent electrical circuit of the above composite is shown in the next figure.

Fig.2.6. (a) Composite wall, and (b) equivalent electrical circuit

Equivalent thermal resistance

For conduction through a plane layer, the thermal resistance is defined as the ratio of the layer thickness to the product of its thermal conductivity and cross-sectional area:

R_i = L_i / (k_i A_i)

For layers in series (one-dimensional heat flow perpendicular to the layers), the total or equivalent resistance is the sum of the resistances of individual layers:

R_eq = Σ R_i = Σ (L_i / k_i A_i)

where,

At steady state the rate of heat transfer through the composite can be represented by an expression analogous to Ohm's law:

q̇ = ΔT / R_eq

where R (or R_eq) is the equivalent thermal resistance between the two boundary temperatures.

Remarks and Applications

• The series-resistance model applies when heat flows normal to flat layers and the same heat flux passes through each layer.
• If layers have different cross-sectional areas, use the appropriate A_i for each layer in R_i = L_i/(k_iA_i).
• For composite systems involving convection at the outer surfaces, include convective resistances as R = 1/(hA) in series with conductive resistances.
• The thermal resistance analogy to electrical resistance helps assemble complex multi-layer systems and compute global heat transfer rates efficiently.