Question

A 5 mm dia copper cable is insulated with a material of conductivity of 0.16 W/mK and is exposed to air at 30°C with a convection coefficient of 20 W/m^2K. If the surface temperature of the wire can be 120°C, determine the insulation thickness for maximum heat flow and the heat dissipated per m length.

Answer 1

t=5.5 mm

Heat dissipation per unit length = 90.477 W/m

Step-by-step explanation:

Given that

Diameter d = 5 mm ⇒r = 2.5 mm

Conductivity of insulated material K = 0.16 W/mK

Heat transfer coefficient = 20
`(W)/(m^2K)`

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

`r_(c)=(K_(insulation))/(h_(surrounding))`

Now by putting the values

`r_(c)=(0.16)/(20)`

`r_(c)=8 mm`

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