Question

PROBLEM 2.1. Determine the steady-state temperature of the surface of the wire of 0.25 in. diameter and 2, 1 ft long, carrying a current of 50 A. The convective heat transfer coefficient is 4.0 BTUhr-ft-oF, temperature of the air surrounding the wire is 32°F, the electrical resistivity of the material of the wire is 1.5 μΩ-ft and its thermal conductivity is 11 BTU/hr-ft-up,

Answer 1

Diameter of the wire is d = 0.25 in cross-sectional area of the wire is A = pi(d/2)^2 = pi d^2/4 A = 3.14 times (0.25)^2/4 = 0.0490625 in^2 because 1 in^2 = 0.006944 ft^2 a = 3.407 times 10^-4 ft^2 length of the wire is L = 2.1 ft resistance of the wire is R = rho L/A where rho be the resistivity of the wire given that rho = 1.54 mu ohm - ft rho = 1.5 times 10^-6 ohm ft R = 1.5 times 10^-6 times 2.1/3.407 times 10^-4 R = 0.9245 times 10^-2 ohm current in the wire is I = 50 Amp. Heat dissipated through the wire is form ohm�s law P = I^2R = (20)^2 times 0.9245 times 10^-2 P = 23.1125 watt. Or Jouel/second P = 23.1125 times 0.000947817/(13600) BTU/hr P = 78.863 BTU/hr Surrounding temperature is T_0 = 32 degree F let T be the steady temperature of the surface of the wire heat transfer from the wire is P =