A5 mm dia copper cable is insulated with a material of conductivity of 0.16 w/mk and is exposed to air at 30â°c with a convection coefficient of 20 w/m^2k. if the surface temperature of the wire can be 120â°c, determine the insulation thickness for maximum heat flow and the heat dissipated per m length.

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