As electric current moves through a wire (Fig. P7.14), heat generated by resistance is conducted through a layer of insulation and then convected to the surrounding air. The steady-state temperature of the wire can be computed as

Determine the thickness of insulation r_i (m) that minimizes the wire’s temperature given the following parameters: q = heat generation rate = 75 W/m, r_ω = wire radius = 6 mm, k = thermal conductivity of insulation = 0.17 W/(m K), h = convective heat transfer coefficient = 12 W/(m² K), and T_air = air temperature = 293 K.

Transcribed Image Text:

q rw ri 1 1 x + 2²/17 [ / ¹0 (²W +²²) + = √²+₁] In 2л k hrw ri T = Tair +