An oven is insulated with 8 inches of fire brick, 4 inches
of insulating brick, and 4 inches of outside refractory. An air gap of 0.25-inch is left between fire
brick and the insulating brick. Thermal
conductivity data for these materials are provided as 0.68, 0.0265, 0.15, and
0.4 BTUs per hour per foot per degrees Fahrenheit. How much heat is lost from the oven, per square foot of the
surface, when inside surface is at 1600 degrees Fahrenheit while outside
surface is maintained at 125 degrees Fahrenheit.
Solution:
Heat is being transferred through conduction. According to Fourier’s law, heat flux is
directly proportional to the temperature gradient and inversely proportional to
the thermal resistance. This resistance
in the rectangular coordinate system (x, y, and z), is given by thickness of
the insulation (L) divided by thermal conductivity (k) and divided by area (A)
normal to heat flow. In a composite
block, these resistances are added arithmetically to obtain total resistance. It is just like when a current is flowing
through resistances that are arranged in series.
In our case, R1, R2, R3,
and R4 are found to be 0.98, 0.7876, 2.222, and 1.25,
respectively. Thermal resistance has a
unit of hour degree Fahrenheit per BTU.
The total resistance is calculated to be 5.239.
Total driving force is 1600 minus 125 equal 1475 degrees
Fahrenheit. So, the heat flowing
through the system is delta T over R and is found to be 1475 over 5.239 or
281.6 BTUs per hour. This heat is
flowing through the system.
We can find interface temperatures by using appropriate
resistance values. Let us calculate T2,
temperature at the interface of firebrick and air. First we need to find delta T1. This can be calculated by multiplying Q with
R1. Plugging in their
values, we obtain delta T1 as 281.6 times 0.98 or 276 degrees
Fahrenheit. That produces T2
as 1600 minus 276 or 1324 degrees delta Fahrenheit.
We can repeat these calculations and get delta T2
and delta T3, as 221, and 626 degrees Fahrenheit. These delta values give T3 and T4
of 1102 and 477 degrees Fahrenheit. And
if we calculate delta T4, it will be found to be 352 degrees
Fahrenheit, thus confirming T5 as 125 degrees F.