Answer:
86 mm
Explanation:
From the attached thermal circuit diagram, equation for i-nodes will be
[tex]\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0[/tex] Equation 1
Similarly, the equation for outer node “o” will be
[tex]\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0[/tex] Equation 2
The conventive thermal resistance in i-node will be
[tex]R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w[/tex] Equation 3
The conventive hermal resistance per unit area is
[tex]R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w[/tex] Equation 4
The conductive thermal resistance per unit area is
[tex]R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w[/tex] Equation 5
Since [tex]q_{rad}[/tex] is given as 100, [tex]T_{o}[/tex] is 40 [tex]T_ \infty[/tex] is 300 [tex]T_{\infty, o}[/tex] is 25
Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain
[tex]\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0[/tex] Equation 6
[tex]\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0[/tex]
[tex]T_{i}-40= \frac {L}{0.05}*150[/tex]
[tex]T_{i}-40=3000L[/tex]
[tex]T_{i}=3000L+40[/tex] Equation 7
From equation 6 we can substitute wherever there’s [tex]T_{i}[/tex] with 3000L+40 as seen in equation 7 hence we obtain
[tex]\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0[/tex]
The above can be simplified to be
[tex]\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0[/tex]
[tex]\frac {260-3000L}{0.033}=50[/tex]
-3000L=1.665-260
[tex]L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm[/tex]
Therefore, insulation thickness is 86mm
