Heat Transfer Area
A 200 mm thick brick plane wall has a measured temperature of 250°C on one side and 100°C on the other. If the thermal conductivity is 80 W/m∙°C and the heat transfer rate is 90 kW, what is the surface area (m^2) perpendicular to the direction of heat transfer?
Expand Hint
Fourier’s Law of Conduction:
$$$\dot{Q}=kA\frac{dT}{dx}$$$
where
$$\dot{Q}$$
is the rate of heat transfer,
$$k$$
is the thermal conductivity,
$$A$$
is the surface area perpendicular to the direction of heat transfer,
$$dT$$
is the change in temperature, and
$$dx$$
is the distance.
Hint 2
Solve for
$$A$$
. No need to convert units to Kelvin.
Fourier’s Law of Conduction:
$$$\dot{Q}=kA\frac{dT}{dx}$$$
where
$$\dot{Q}$$
is the rate of heat transfer,
$$k$$
is the thermal conductivity,
$$A$$
is the surface area perpendicular to the direction of heat transfer,
$$dT$$
is the change in temperature, and
$$dx$$
is the distance.
$$$90,000W=80\frac{W}{m\cdot ^{\circ}C}\times A \times \frac{(250^{\circ}C-100^{\circ}C)}{0.2m}$$$
Solving for area:
$$$1,125m= A \times \frac{150}{0.2m}$$$
$$$A=\frac{1,125m}{150}\cdot0.2m=1.5\:m^2$$$
$$$1.5\:m^2$$$
Time Analysis
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