Question

The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin metal sheets. The oven air is at T[infinity],????= 300℃, and the corresponding convection coefficient is ℎ???? = 30 W/ m2 °K. The inner wall absorbs a radiant flux of ????′′????????????= 100 W/ m2 from hotter objects within the oven. The room air at T[infinity],o= 25℃, and overall coefficient for convection and radiation from the outer surface is ℎo = 10 W/ m2 °K. What insulation thickness L is required to maintain the outer wall surface at a safe-to-touch temperature of To = 40 degree C?

Answer 1

86 mm

Step-by-step explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

`\frac {T_ \infty, i-T_(i)}{ R^(` Equation 1

Similarly, the equation for outer node “o” will be

`\frac {T_( i)-T_(o)}{ R^(` Equation 2

The conventive thermal resistance in i-node will be

`R^(` Equation 3

The conventive hermal resistance per unit area is

`R^(` Equation 4

The conductive thermal resistance per unit area is

`R^(` Equation 5

Since
`q_(rad)` is given as 100,
`T_(o)` is 40
`T_ \infty` is 300
`T_(\infty, o)` is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

`\frac {300-T_(i)}{0.033} +\frac {40-T_(i)}{L/0.05} +100=0` Equation 6

`\frac {T_( i)-40}{L/0.05}+ \frac {25-40}{0.100}=0`

`T_(i)-40= \frac {L}{0.05}*150`

`T_(i)-40=3000L`

`T_(i)=3000L+40` Equation 7

From equation 6 we can substitute wherever there’s
`T_(i)` with 3000L+40 as seen in equation 7 hence we obtain

`\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0`

The above can be simplified to be

`\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0`

`\frac {260-3000L}{0.033}=50`

-3000L=1.665-260

`L= \frac {-258.33}{-3000}=0.086*10^(-3)m= 86mm`

Therefore, insulation thickness is 86mm