Answer:
t=5.5 mm
Heat dissipation per unit length = 90.477 W/m
Explanation:
Given that
Diameter d = 5 mm ⇒r = 2.5 mm
Conductivity of insulated material K = 0.16 W/mK
Heat transfer coefficient = 20 [tex]\frac{W}{m^2K}[/tex]
When thickness reaches up to critical radius of insulation then heat dissipation will be maximum
We know that critical radius of insulation of wire is given as follow
[tex]r_{c}=\dfrac{K_{insulation}}{h_{surrounding}}[/tex]
Now by putting the values
[tex]r_{c}=\dfrac{0.16}{20}[/tex]
[tex]r_{c}=8 mm [/tex]
So the thickness of insulation
t=8-2.5 mm
t=5.5 mm
As we know that heat transfer due to convection given as follows
Q = hAΔ T
Q=20 x 2 x π x 0.008 x (120-30)
Q = 90.477 W/m
So heat dissipation per unit length = 90.477 W/m
