A 1000 W iron utilizes a resistance wire which is 20 inches long and has a diameter of 0.08 inches. Determine the rate of heat generation in the wire per unit volume, in Btu/hrft3, and the heat flux on the outer surface of the wire, in Btu/hrft2, as a result of this generation.

The rate of heat generation in the wire per unit volume is 5.79*10^7 Btu/hrft^3

Heat flux is 9.67*10^7 Btu/hrft^2

Explanation:

Rate of heat generation = 1000 W = 1000/0.29307 = 3412.15 Btu/hr

Area (A) = πD^2/4

Diameter (D) = 0.08 inches = 0.08 in * 3.2808 ft/39.37 in = 0.0067 ft

A = 3.142*0.0067^2/4 = 3.53*10^-5 ft^2

Volume (V) = A * Length

L = 20 inches = 20 in * 3.2808 ft/39.37 in = 1.67 ft

V = 3.53*10^-5 * 1.67 = 5.8951*10^-5 ft^3

Rate of heat generation in the wire per unit volume = 3412.15 Btu/hr : 5.8951*10^-5 ft^3 = 5.79*10^7 Btu/hrft^3

Heat flux = 3412.15 Btu/hr : 3.53*10^-5 ft^2 = 9.67*10^7 Btu/hrft^2