Question

A 3-mm-diameter electrical wire is insulated by a 2-mm-thick rubberized sheath with k = 0.13 W/m·K, and the wire/sheath interface is characterized by a thermal contact resistance of R ’’ t,c=3 × 10−4 m2 ·K/W. The convection heat transfer coefficient at the outer surface of the sheath is 15 W/m2 ·K, and the ambient air temperature is 20°C. If the temperature of the sheath may not exceed 50°C, what is the maximum allowable electrical power that may be dissipated per unit length of the conductor? What is the critical radius of the insulating sheath?

Answer 1

Answer:

$Q_{max} = 6,3928 W/m$

Explanation:

The thermal resistances in this case act as series resistances, so to get the total resistance you need to add al resitances:

$R_{tot} = R_{contact} + R_{conductive} + R_{convective}$

To get the contact resistence per unit length we have to multiply the contact resistance by the perimeter of the contact area:

$P_{wire-sheath} = \pi * 0,003m = 9,4248 * 10^{-3} m$

$R_{contact} = \frac{3*10^{-4} m2 K/W}{9,4248 * 10^{-3} m} = 0,03183 m K/W$

For the conductive resistance on the sheath:

$R_{conductive} = \frac{thickness}{k*P_{wire-sheath}} = \frac{ 0,002m}{9,4248 * 10^{-3} m * 0,13 W/mK} = 1,63 mK/W$

Convective resistance:

The perimeter of the auter surfice is:

$P_{out-sheath} = \pi * 0,007m = 2,199* 10^{-2} m$

$R_{convective} = \frac{1}{Convection coeficient * P_{out-sheath} } = \frac{ 1}{15 W/m^{2}K * 2,199* 10^{-2} m} = 3,031 mK/W$

Total Resistance:

$R_{tot} = 4,6928 mK/W$

Heat transfer:

$Q = dT/R_{tot}$

$dT = T_{wire} - T_{amb} = 323 K - 293 K = 30 K$

$Q_{max} = \frac{30K}{4,6928 mK/W} = 6,3928 W/m$