The inner surface of a 400 mm thick brick wall of a furnace is kept at 707°C and it is found that the outer surface temperature is 460 Kelvin Assuming that the conductivity is 0.75 W/m-K, the heat loss per square metre of wall area will be:

Question

Prepp Team · Accepted Answer

Understanding heat transfer through a material like a brick wall is crucial in many engineering applications, especially for furnaces. This problem involves calculating the heat loss per unit area through a furnace wall, which is a classic application of Fourier's Law of Heat Conduction.

Brick Wall Heat Loss Problem Overview

The question asks us to determine the rate of heat loss per square meter of wall area for a thick brick wall of a furnace. We are provided with the wall's thickness, the temperatures of its inner and outer surfaces, and the thermal conductivity of the brick material. The key is to correctly apply the formula for one-dimensional steady-state heat conduction to find the heat loss.

Given Parameters for Heat Loss Calculation

Let's list down all the important information provided in the question that is relevant to calculating the heat loss:

Thickness of the brick wall (L): $400 \text{ mm}$
Inner surface temperature (T₁): $707^\circ\text{C}$
Outer surface temperature (T₂): $460 \text{ Kelvin}$
Thermal conductivity of the wall material (k): $0.75 \text{ W/m-K}$

Unit Conversions for Consistent Calculation

Before proceeding with the calculation, it is essential to ensure all units are consistent. The standard units for these types of heat transfer calculations are meters for length and Kelvin for temperature. The thermal conductivity is already in W/m-K, which is suitable.

Converting Wall Thickness: The thickness of the brick wall is given in millimeters. We convert it to meters for consistency: $$L = 400 \text{ mm} = 400 \times 10^{-3} \text{ m} = 0.400 \text{ m}$$
Converting Inner Surface Temperature: The inner surface temperature is given in degrees Celsius. We need to convert it to Kelvin by adding 273: $$T_1 = 707^\circ\text{C} + 273 \text{ K} = 980 \text{ K}$$ The outer surface temperature $T_2$ is already provided in Kelvin, so no conversion is needed: $T_2 = 460 \text{ K}$.

Applying Fourier's Law of Heat Conduction

For one-dimensional, steady-state heat conduction through a plane wall, Fourier's Law of Heat Conduction is used to calculate the heat transfer rate. It states that the heat transfer rate (Q) is directly proportional to the area (A) perpendicular to heat flow and the temperature difference ($\Delta T$) across the wall, and inversely proportional to the thickness (L). The general formula for heat transfer rate is:

$$Q = k \cdot A \cdot \frac{\Delta T}{L}$$

Where:

$Q$ is the total heat transfer rate in Watts (W).
$k$ is the thermal conductivity of the material in W/m-K.
$A$ is the cross-sectional area perpendicular to heat flow in square meters (m²).
$\Delta T$ is the temperature difference across the wall in Kelvin (K).
$L$ is the thickness of the wall in meters (m).

The problem specifically asks for the heat loss per square metre of wall area, which means we need to find the value of $\frac{Q}{A}$.

Rearranging the Fourier's Law formula to find heat loss per unit area, we get:

$$\frac{Q}{A} = k \cdot \frac{\Delta T}{L}$$

Heat Loss Calculation Steps

Now, let's substitute the converted values of thermal conductivity ($k$), temperature difference ($\Delta T$), and wall thickness ($L$) into the rearranged formula to calculate the heat loss per square metre.

First, calculate the temperature difference $\Delta T$ across the brick wall:

$$\Delta T = T_1 - T_2 = 980 \text{ K} - 460 \text{ K} = 520 \text{ K}$$

Next, substitute the values into the formula for heat loss per unit area:

$$\frac{Q}{A} = 0.75 \text{ W/m-K} \cdot \frac{520 \text{ K}}{0.400 \text{ m}}$$

Performing the multiplication and division:

$$\frac{Q}{A} = 0.75 \cdot \frac{520}{0.4} \text{ W/m}^2$$

$$\frac{Q}{A} = 0.75 \cdot 1300 \text{ W/m}^2$$

$$\frac{Q}{A} = 975 \text{ W/m}^2$$

Final Result for Heat Loss

The calculated heat loss per square metre of the furnace brick wall is $975 \text{ W}$. This indicates that for every square meter of the wall surface, 975 Watts of thermal energy are lost due to conduction from the hot inner side to the colder outer side.