An electrical wire 1.0 mm in diameter is wrapped with teflon 0.2 mm in thickness

Question

An electrical wire 1.0 mm in diameter is wrapped with teflon 0.2 mm in thickness. The thermal conductivity of the teflon is 0.35 W/m ºC. An electrical current flows through the wire such that the steady heat generation in the wire is 1.3 x10^7 W/m3. Now the wire wrapped with the teflon is exposed to a convection environment at 30ºC with the convective heat transfer coefficient of 20 W/m2 ºC. Assume that the wire and the teflon are in perfect contact.

(1) What would be the surface temperature of the wire?

(2) What would be the outside surface temperature of the teflon?

Explanation / Answer

1. let the inner surface temperature is Ti and outer surface temperature is T0.

taking1 m length of the wire.

inner surface area of the wire Aw = 2rl = 2(0.5x10-3)x1 = 3.14 x 10-3 m2

volume of the wire Vw= r2l = (0.5x10-3)2x1 = 7.85 x 10-7 m3

net energy generation by current E= 1.3x107 *7.85 x 10-7 =10.205 W

writinng heat transfer equation from the inner surface to the outer surface by conduction

E = k Aw (To - Ti) / t

=> 10.205 = 0.35*3.14x10-3(To - Ti) /(0.2x10-3)

=> Ti - To = 1.86 -------------------------1

writing equation for the convection

E = hA(To - 30)

A = outer surface area = 2 (0.7 x 10-3) *l =4.4 x 10-3 m2

10.205=20*4.4 x 10-3(To - 30)

=> T0 = 116 C

from 1

Ti = 1.86 + 116 = 117.86

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