A 2-kW resistance heater wire whose thermal conductivity is 15 W/m.°C has a diameter of 4 mm and a length of 0⋅5 m, is used to boil the water. If the outer surface temperature of the resistance wire is 105°C, what is the temperature at the centre of the wire?

Concept:

Steady State One Dimensional Heat conduction equation with Heat Generation for a Cylinder is given by

\(T-T_S = \frac{q_g}{4k}(R^2-r^2) \)

where T = Temperature at radius r from the centerline, T_S = Surface Temperature, q_g = Heat Generation per Unit Volume, k = Thermal Conductivity,R = Radius of Cylinder

for the centerline temperature of the wire, we have to put r = 0

So, \(T_o-T_S = \frac{q_gR^2}{4k}\)

Calculation:

Given:

T_S = 105°C, q_g = 2000 W, R = 2 mm, k = 15 W/m.°C, L = 0.5 m

\(T_o-105 = \frac{(2000)0.002^2}{4\times 15}(\frac{1}{\pi R^2L})\)

\(T_o-105 = \frac{(2000)0.002^2}{4\times 15}(\frac{1}{\pi \times 0.5\times 0.002^2})\)

T_o -105 = 21.23

T_o = 126.23°C

The temperature at the centre of the wire is 126.23°C.