a one brick thick wall of 10cm thickness has temperatures 40deg C and 5deg C at the surfaces. if the thermal conductivity of the brick is 0.6W per m per K, what is the rate of heat flow per unit area W per m square through the bricks if steady state conditions apply?
explain why under these conditions, the temperature of the outer surface of the wall must be greater than the air temperature. At what rate must the outer surface be losing heat?
q = α•ΔT/Δx =0.6•35/0.1 = 210 W/m²
To calculate the rate of heat flow through the brick wall, we can use Fourier's Law of Heat Conduction, which states that the rate of heat transfer (Q) is equal to the thermal conductivity (k) multiplied by the temperature difference (ΔT) divided by the thickness (L) of the material:
Q = (k * ΔT) / L
In this case, we are given the following values:
- Thermal conductivity of the brick (k) = 0.6 W/mK
- Temperature at the surface of the wall facing the hot side (T1) = 40°C
- Temperature at the surface of the wall facing the cold side (T2) = 5°C
- Thickness of the wall (L) = 10 cm = 0.1 m
Using the given values in the formula, we can calculate the rate of heat flow per unit area (Q/A) through the brick:
Q/A = (k * ΔT) / L
Q/A = (0.6 W/mK * (40°C - 5°C)) / 0.1 m
Q/A = (0.6 W/mK * 35°C) / 0.1 m
Q/A = 6 W/m²
Therefore, the rate of heat flow per unit area through the brick wall is 6 W/m².
Now, let's explain why, under these conditions, the temperature of the outer surface of the wall must be greater than the air temperature.
The rate of heat flow from a warmer object to a cooler object follows the second law of thermodynamics, which states that heat always moves from areas of higher temperature to areas of lower temperature until thermal equilibrium is reached.
In this case, since the hot surface of the wall (T1 = 40°C) is in contact with the hotter environment, heat will naturally flow from the hotter surface to the cooler environment. Thus, the temperature of the outer surface of the wall must be greater than the air temperature to facilitate this heat transfer.
To calculate the rate at which the outer surface is losing heat, we need to know the temperature of the surrounding air. Once we have that information, we can multiply it by the surface area to calculate the convection heat transfer rate using the heat transfer equation:
Q_conv = h * A * (T_surface - T_air)
Where:
- Q_conv is the heat transfer rate by convection
- h is the convective heat transfer coefficient (depends on the medium and conditions)
- A is the surface area of the wall
- T_surface is the temperature of the outer surface of the wall
- T_air is the temperature of the surrounding air
Without the value for T_air or the convective heat transfer coefficient (h), we cannot calculate the exact rate at which the outer surface is losing heat in this particular scenario.