Answer to Question #183119 in Mechanical Engineering for luke

Question

A 2mm diameter electrical wire is insulated by a 2mm thick rubberized sheath (k=0.13W/m-K). The convection heat transfer coefficient at the outer surface of the sheath is 10W/m2-K and the temperature of the ambient air is 20C. If the temperature of the insulation may not exceed 50C, what is the maximum allowable power that may be dissipated per meter length of the conductor (watts) 

Answer 1

Diameter of wire is 2 mm= $2\times 10^{-3}$ m, Thermal heat conduction = k=0.13W/m-K,

thickness of sheath= t= $2\times 10^{-3}$ m, Ambient temperature= $T_a$ = $20^o C$ , Temperature of insulation= 50$^o C$

Now net thermal resistance,

$R=\frac{ln \frac{r+t}{r}}{2 \times \pi k} + \frac{1}{2 \pi rh}$

$R=\frac{ln \frac{1+2}{1}}{2 \times \pi k} + \frac{1}{2 \pi 1\times 10}$

R= 6.650 m²K/W

Maximum allowable power per unit length

Q'=$\frac{T_{in}- T_{\infty}}{6.650}=\frac{50-20}{6.65}= 4.511 \frac{W}{m}$