Question

A furnace is shaped like a long equilateral triangular duct where the width of each side is 2 m. Heat is supplied from the base surface, whose emissivity is ε1 = 0.8, at a rate of 800 W/m2 while the side surfaces, whose emissivities are 0.5, are maintained at 500 K. Neglecting the end effects, determine the temperature of the base surface. Can you treat this geometry as a two-surface enclosure?

Answer 1

The geometry is treated as a two surface enclosure because the two surfaces have the same properties.

Let's take the base surface to be surface 1, while the side surfaces are surface 2.

Let's take the heat transfer expression:

`Q_1_2 = (\sigma[(T_1)^4 - (T_2)^4])/((1 - E_1)/(A_1 E_1) + (1)/(A_1 F_1_2) + (1-E_2)/(A_2 E_2))`

Where,

`\sigma = Boltmanz constant = 5.68*10^-^8`

`T_1` = base temperature

`T_2` = surface 2 temperature = 500K

`E_1` = emissivity of surface 1 = 0.8

`E_2` = emissivity of surface 2 = 0.5

`A` = Area

`F_1_2` = shape factor

Substituting figures in the equation, we have:

`800= (5.67*10^-^8[(T_1)^4 - (500)^4])/((1 - 0.8)/(2*0.8) + (1)/(2*1) + (1-0.5)/(4*0.5))`

`[(T_1)^4 - (500)^4] = (700)/(5.67*10^-^8)`

`(T_1)^4 = 1.234*10^1^0 + 6.25*10^1^0`

`T_1 = (7.484*10^1^0)^0^.^2^5`

`T_1 = 523.038 K`

The base temperature is 523.038 k