Question

A 1.0-mm diameter wire is maintained at a temperature of 400?C and exposed to a convection environment at 40?C with h = 120 W/m2-?C. Calculate the thermal conductivitywhich will just cause an insulation thickness of 0.2 mm to produce a “critical radius.” How much of this insulation must be added to reduce the heat transfer by 75percent from that which would be experienced by the bare wire?

Answer 1

r_c= (k/h)

r_c= 0.5+0.2 = 0.7 mm and h = 120 W/m²K

solving, k = 0.084 W/mK

heat lost per unit length by the wire,

q = 120*2π*(0.5x10-3)*(400-40) = 135.72 W/m

desired heat loss = 0.25q = 33.93 W/m

now, conduction resistance of insulation per unit length = (1/2πk)*ln(r₂/r₁)

r₁= 0.5 mm, k = 0.084 W/mK

q = 360/(conduction resistance)

and, q = 33.93 W/m

so, conduction resistance = 360/33.93 = 10.61

so,

ln(r₂/0.5) = 10.61*2πk = 5.6

r₂= 135.2 mm

so you need to add 134.71 mm of insulation