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

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

Step-by-step 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 = (2\cdot k\cdot L)/(\ln \left((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 = (\dot Q)/(2\cdot L\cdot (T_(i)-T_(o)))\cdot \ln\left((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[(40.8\,W)/(2\cdot (0.6\,m)\cdot (50\,^(\circ)C-20\,^(\circ)C))\right]\cdot \ln \left((0.04\,m)/(0.01\,m) \right)`

`k \approx 1.571\,(W)/(m\cdot ^(\circ)C)`

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