Question

A common method to measure thermal conductivity of a biomaterial is to insert a long metallic probe axially into the center of a long cylindrical container of the material. The probe is maintained at some uniform temperature Ti, and the outside of the container is maintained at a temperature To. Inside the metallic probe is an electrical heater for which the electrical power is measured. If the diameter of the probe maintained at 50°C is 1 cm, the outer diameter of the container maintained at 20°C is 4 cm, the length of the cylinder container is 60 cm, and the power input is 40.8 W, calculate the thermal conductivity of this material.

Answer 1

Answer:

The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.

Explanation:

Let suppose that thermal conduction is uniform and one-dimensional, the conduction heat transfer ($\dot Q$), measured in watts, in the hollow cylinder is:

$\dot Q = \frac{2\cdot k\cdot L}{\ln \left(\frac{D_{o}}{D_{i}} \right)}\cdot (T_{i}-T_{o})$

Where:

$k$ - Thermal conductivity, measured in watts per meter-Celsius.

$L$ - Length of the cylinder, measured in meters.

$D_{i}$ - Inner diameter, measured in meters.

$D_{o}$ - Outer diameter, measured in meters.

$T_{i}$ - Temperature at inner surface, measured in Celsius.

$T_{o}$ - Temperature at outer surface, measured in Celsius.

Now we clear the thermal conductivity in the equation:

$k = \frac{\dot Q}{2\cdot L\cdot (T_{i}-T_{o})}\cdot \ln\left(\frac{D_{o}}{D_{i}} \right)$

If we know that $\dot Q = 40.8\,W$, $L = 0.6\,m$, $T_{i} = 50\,^{\circ}C$, $T_{o} = 20\,^{\circ}C$, $D_{i} = 0.01\,m$ and $D_{o} = 0.04\,m$, the thermal conductivity of the biomaterial is:

$k = \left[\frac{40.8\,W}{2\cdot (0.6\,m)\cdot (50\,^{\circ}C-20\,^{\circ}C)}\right]\cdot \ln \left(\frac{0.04\,m}{0.01\,m} \right)$

$k \approx 1.571\,\frac{W}{m\cdot ^{\circ}C}$

The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.