Answer:
The critical radius of the plastic insulation is 0.72 inches.
Explanation:
Given that,
Diameter = 0.091 in
Thickness = 0.02 in
Initial temperature = 90°F
Final temperature = 50°F
Heat transfer coefficient = 2.5 Btu/h.ft²°F
Material conductivity = 0.075 Btu/h.ft °F
We need to calculate the critical radius of the plastic insulation
Using formula of critical radius
[tex]r_{cr}=\dfrac{2K}{h}[/tex]
Where, k = Material conductivity
h = Heat transfer coefficient
Put the value into the formula
[tex]r_{cr}=\dfrac{2\times0.075}{2.5}[/tex]
[tex]r_{cr}=0.06\ ft[/tex]
[tex]r_{cr}=0.72\ inches[/tex]
Hence, The critical radius of the plastic insulation is 0.72 inches.
