## Friday, December 9, 2016

### Why the Door Knob?

During fire safety trainings, it is common advice to evaluate the conditions of the next room over by feeling the doorknob with the back of your hand. To explore this advice, I wanted to compare how heat is transferred by conduction through wood (the rest of the door) vs. brass (the doorknob).

If we assume that there is a fire in the next room which has heated the air in the room to 500 degrees Fahrenheit (260 degrees Celsius - average fires can be 1000 to 1500 degrees Celsius themselves), then we can model how much heat would be transferred through an area of the door around 0.01 square meters big. The thermal conductivity coefficient of wood is 0.13 W/m*K)We will also assume the room you're in has only been heated slightly by the fire to 30 degrees Celsius, and that the door is 0.035 meters thick (based on 1 3/8 inch door width):

Q/t = kA(T1 - T2)/l = (0.13 W/m*K)*(0.01 m^2)*(260 - 30 degrees Celsius)/(0.035 m) = 8.54 J/s

For the doorknob, even if we assume a slightly smaller cross-sectional area (ex. the back of your hand is larger than the doorknob, so new area is 0.008 m^2) and a longer l (0.20 meters for the entire length of the doorknob on both sides), the heat that passes through the doorknob to your hand is still much greater. The thermal conductivity coefficient is 109 W/m*K:

kA(T1 - T2)/l  = (109 W/m*K)*(0.008 m^2)*(260 - 30 degrees Celsius)/(0.20 m) = 1002.8 J/s

Despite a smaller cross-sectional area and a longer distance between regions of different temperatures, the doorknob would still transfer much more heat from the other room to your hand than the wooden part of the door, and so it is much easier to gain a sense of how hot the other room is using the doorknob.

Source: http://www.farm.net/~mason/materials/thermal_conductivity.html for thermal conductivity coefficients.