A current-carrying 10 gage (.1019 in. diameter) copper wire is insulated with a
ID: 1848386 • Letter: A
Question
A current-carrying 10 gage (.1019 in. diameter) copper wire is insulated with a .05 in thick layer of rubber. The maximum temperature of the rubber is 200F when the temperature of the surrounding are is 120F. The external heat transfer coefficient can be taken as 4 Btu/hr-ft^2-F. For the copper wire, Kw=220 Btu/hr-ft-F.Assume the electrical resistance of the wire is 1.288 ohms/1000 linear feet. (a)Determine the temperature of the outside surface of the insulation. (b)Assuming uniform energy generation inside the wire, determine the temperature at the center of the wire. (c)Determine the electrical current in the wire.
Explanation / Answer
Further to your 1st question, both you and "some people" are saying the same thing. The heat transfer through insulation (and just through insulation) is governed by conduction. Q = 2*pi*L*k*DeltaT/ln(d2/d1) Where: Q = heat flow through insulation [W] k = insulation thermal conductivity [W/(m°C)] DeltaT = temperature difference between outer and inner surface of the insulation [°C] d2 = outer insulation diameter [m]. d2 is equal to the pipe diameter + twice the insulation thickness d1 = inner insulation diameter [m]. d1 is equal to the outer pipe diameter This is an increasing function (just differentiate with respect to d1) and so heat increases as pipe diameter increases.