Transport Phenomena - Heat Transfer Problem : Maximum current in an electric wire

Problem. An electric wire with radius r0 of 0.50 mm is made of copper [electrical conductivity = 5.1 x

107 ohm-1 m-1 and thermal conductivity = 380 W/(m K)]. It is insulated (see figure) to an outer radius r1 of 1.50 mm with plastic [thermal conductivity = 0.350 W/(m K)].

Figure. Heating of an insulated electric wire. The ambient air is at 38.0oC and the heat transfer coefficient from the outer insulated surface to the surrounding air is 8.500 W/(m2 K). Determine the maximum current in amperes that can flow at steady-state in the wire without any portion of the insulation getting heated above its maximum allowable temperature of 93.0oC.

Solution. Step. Thermal resistance representation for insulation and airIn general, the heat flow is given by Q = T/Rth, where T is the temperature driving force (thermal potential difference). The thermal resistance for a cylindrical annulus is Rth = ln (r1/r0)/(2 kL) and the thermal resistance for a fluid film at a solid-fluid interface is Rth = 1/(hA). Here, k is the thermal conductivity, h is the heat transfer coefficient and A is the surface area for convection. The thermal resistances for the insulation and air film are in series as shown in the figure below.

Figure. Thermal resistance representation of insulation and air film. Based on the above thermal resistance representation, the heat flow is

(1)

where k is the thermal conductivity of the plastic insulation.

Step. Heat flow due to current in wireThe flow of an electric current results in some electrical energy getting converted to thermal energy irreversibly. The heat generation by electrical dissipation per unit volume is given by

S = I 2/ke where I is the current density (in amp/m2) and ke is the electrical conductivity (in ohm-1 m-1). The total heat generated within the wire is simply the product of S and the volume of the wire. At steady-state, all this heat generated within the wire by electrical dissipation must leave through the wire surface and therefore the heat flow is given by

(2)

Step. Expression for currentOn eliminating Q from the above two equations, the current density is

(3)

On multiplying the current density by the cross-sectional area of the wire, the current is obtained from

(4)

For the maximum current, the temperature T0 must be maximized.

Step. Substitution of numerical valuesOn setting the temperature T0 to 93.0oC (i.e., the maximum allowable temperature for the insulation), the maximum current that can flow through the wire may be calculated as 13.027 amp. The numerical values substituted in the equation are given below. The values below may be changed and the problem solution recalculated with the new values provided in consistent units.

Variable name Symbol Value Unit

electrical conductivity ke ohm-1 m-1

maximum temperature T0oC

ambient temperature T2oC

outer radius r1 m

wire radius r0 m

plastic thermal conductivity k W/(m K)

heat transfer coefficient h W/(m2 K)

Calculated Variable Symbol Value Unit

Maximum current 13.027 amp

5100000

93.0000

38.0000

0.001500

0.000500

0.35000

8.50000