Question

A furnace consists of a 100 cm thick brick wall chamber in which gaseous material is combusted. The air surrounding the furnace chamber is at ambient temperature (25°C) and circulates due to natural convection with a value of the convection coefficient at 15 W/(m2.K). Assuming a thermal conductivity of 1 W/(m.K) for brick wall, which can be assumed a blackbody, calculate the of the furnace wall if the outer wall inner temperature of the furnace wall if the oute temperature reaches 80°C at steady-state.

Answer 1

Step-by-step explanation:

The heat transferred through conduction = heat transfer through radiation + heat transfer through convection

It is given that thermal conductivity of wall is 1 W/(m.K)

Boltzmann constant (
`\sigma_(e)`) =
`5.68 * 10^(-8) W/T^(4) m^(2)`

Formula to calculate the inner temperature of the furnace is as follows.

kA(T - 80) =
`\sigma_(e) * A * (80 + 273)^(4) - (25 + 273)^(4) + hA(80 - 25)`

It is known that for black body emissitivity = 1

Now putting the given values into the above formula as follows.

kA(T - 80) =
`\sigma_(e) * A * (80 + 273)^(4) - (25 + 273)^(4) + hA(80 - 25)`

`1 * A(T - 80)` = A (
`\sigma_(e) * (80 + 273)^(4) - (25 + 273)^(4) + h(80 - 25)`)

T - 80 =
`5.68 * 10^(-8) * [(353)^(4) - (298)^(4)] + 1 * 55`

T =
`1338.25^(o)C`

Thus, we can conclude that the inner temperature of the furnace wall is
`1338.25^(o)C`.