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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
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
Now by putting the values
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