An aluminum electrical cable is 20 mm in diameter is covered by a plastic insulation (k = 1 w/m-k) of critical thickness. this wire is placed in a room with an air flow heat transfer coefficient of 50 w/m^2-k. compared to the bare aluminum wire, the heat loss from this insulated wire will be a) less b) more c) same d) zero

the heat loss from this insulated wire is less

Explanation:

Given data in question

diameter of cable (d)  =  20 mm

( K ) = 1 W/m-k

heat transfer coefficient (h) = 50 W/m²-K

To find out

the heat loss from this insulated wire

solution

we will find out thickness of wire

heat loss is depend on wire thickness also

we have given dia 20 mm

so radius will be d/2 = 20/ 2 = 10 mm

Now we find the critical thickness i.e.

critical thickness = K / heat transfer coefficient

critical thickness = 1 / 50 = 0.02 m i.e. 20 mm

now we can see that critical thickness is greater than radius 10 mm

so our rate of heat loss will be decreasing

so we can say our correct option is (a) less

i think i backed my dad

a true.