Conduction of heat: A concrete wall of a cold storage room measures 3.0 m high, 5.0 m wide, and 20 cm thick. The room temperature is maintained at -10°C and the outside temperature is 20°C The inside wall is to be covered by a layer of wood in order to reduce the rate of heat loss through the wall BY 90 percent. The thermal conductivities of concrete and wood are 0.80 W/m · K and 0.040 W/m · K, respectively. Under steady state conditions, the thickness of the layer of wood required is closest to

Question

Quizplus · Accepted Answer

To reduce the heat loss by 90%, the new system (concrete + wood) must conduct only 10% of the original heat. The rate of heat transfer is given by $q = \frac{kA(T_2 - T_1)}{d}$, where $q$ is the heat transfer rate, $k$ is the thermal conductivity, $A$ is the area, $T_2 - T_1$ is the temperature difference, and $d$ is the thickness. For the concrete wall, $q_{concrete} = \frac{0.80 \times A \times (20 - (-10))}{0.2}$. For the combined system to transfer only 10% of this, if $d_{wood}$ is the thickness of the wood, then $q_{new} = \frac{0.10 \times 0.80 \times A \times 30}{0.2} = \frac{0.040 \times A \times 30}{d_{wood}}$. Solving for $d_{wood}$ gives approximately 90 mm.

Conduction of Heat: a Concrete Wall of a Cold Storage