Question

A laboratory furnace wall is constructed of 0.2 m thick fireclay brick having a thermal conductivity of 1.82 W/m-K. The wall is covered on the outer surface with insulation of thermal conductivity of 0.095 W/m-K. The furnace inner brick surface is at 950 K and the outer surface of the insulation material is at 300 K. The maximum allowable heat transfer rate through the wall of the furnace is 830 W/m^2. Determine how thick in cm the insulation material must be.

Answer 1

Answer:

The appropriate solution will be "6.4 cm".

Explanation:

The given values are:

Length,

l = 0.2 m

Thermal conductivity,

K₁ = 1.82 W/m-K

K₂ = 0.095 W/m-K

Temperature,

T = 950 K

T = 300 K

Heat transfer rate,

Q = 830 W/m²

Now,

⇒  $Q = \frac{\Delta T}{\frac{L_1}{K_1 A} +\frac{L_2}{K_2 A} }=\frac{A \Delta T}{\frac{L_1}{K_1 } +\frac{L_2}{K_2 } }$

⇒ $\frac{Q}{A} =\frac{\Delta T}{\frac{L_1}{K_1} +\frac{L_2}{K_2} }$

On substituting the above given values in the equation, we get

⇒ $830=\frac{(980-300)}{\frac{0.2}{1.82} +\frac{x}{0.095} }$

On applying cross-multiplication, we get

⇒ $\frac{0.2}{1.82} +\frac{x}{0.095} =\frac{950-300}{830}$

⇒ $\frac{0.2}{1.82} +\frac{x}{0.095} =\frac{650}{830}$

⇒                $x =0.639 \ m$

⇒                $x=6.345 \ i.e., 6.4 \ m$