Centerline Temperature of a Resistance Heater
A 2-kW resistance heater wire whose thermal conductivity is k = 15 W / m \cdot{ }^{\circ} C , has a diameter of D = 4 mm and a length of L = 0.5 m, and is used to boil water (Fig). If the outer surface temperature of the resistance wire is T_{s} = 105^{\circ} C , determine the temperature at the center of the wire.
SOLUTION The surface temperature of a resistance heater submerged in water is to be determined.
Assumptions 1 Heat transfer is steady since there is no change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no change in the axial direction. 3 Thermal conductivity is constant. 4 Heat generation in the heater is uniform.
Properties The thermal conductivity is given to be k = 15 W / m \cdot{ }^{\circ} C
Analysis The 2-kW resistance heater converts electric energy into heat at a rate of 2 kW. The heat generation per unit volume of the wire is
\dot{g} = \frac{\dot{Q}_{\text {gen }}}{V_{\text {wire }}} = \frac{\dot{Q}_{\text {gen }}}{\pi r_{o}^{2} L} = \frac{2000 W }{\pi(0.002 m )^{2}(0.5 m )} = 0.318 \times 10^{9} W / m ^{3}
Then the center temperature of the wire is determined from Eq. \Delta T_{\max , \text { cylinder }} = T_{o} – T_{s} = \frac{g r_{o}^{2}}{4 k} to be
T_{o} = T_{s} + \frac{\dot{g} r_{o}^{2}}{4 k} = 105^{\circ} C + \frac{\left(0.318 \times 10^{9} W / m ^{3}\right)(0.002 m )^{2}}{4 \times\left(15 W / m \cdot{ }^{\circ} C \right)} = 126^{\circ} C
Discussion Note that the temperature difference between the center and the surface of the wire is 21^{\circ} C .