A 2-kW resistance heater wire whose thermal conductivity is k = 15 W/m·K has a diameter of D = 4 mm and a length of L = 0.5 m, and is used to boil water (Fig. 2–58). If the outer surface temperature of the resistance wire is T_{s} = 105°C, determine the temperature at the center of the wire.
Question : A 2-kW resistance heater wire whose thermal conductivity is ...
SOLUTION The center 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 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}.
Analysis The 2-k W resistance heater converts electric energy into heat at a rate of 2 \mathrm{~kW}. The heat generation per unit volume of the wire is
\dot{e}_{\text {gen }}=\frac{\dot{E}_{\text {gen }}}{V_{\text {kire }}}=\frac{\dot{E}_{\text {gen }}}{\pi r^{2} \bar{\sigma}}=\frac{2000 \mathrm{~W}}{\pi(0.002 \mathrm{~m})^{2}(0.5 \mathrm{~m})}=0.318 \times 10^{9} \mathrm{~W} / \mathrm{m}^{3}Then the center temperature of the wire is determined from Eq. 2-71
\Delta T_{\max , \text { cylinder }}=T_{0}-T_{s}=\frac{\dot{e}_{\operatorname{gen}} r_{o}^{2}}{4 k} to be
T_{0}=T_{s}+\frac{\dot{e}_{\mathrm{gen}} r_{0}^{2}}{4 k}=105^{\circ} \mathrm{C}+\frac{\left(0.318 \times 10^{9} \mathrm{~W} / \mathrm{m}^{3}\right)(0.002 \mathrm{~m})^{2}}{4 \times\left(15 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C}\right)}=126^{\circ} \mathrm{C}Discussion Note that the temperature difference between the center and the surface of the wire is 21^{\circ} \mathrm{C}. Also, the thermal conductivity units \mathrm{W} / \mathrm{m} \cdot{ }^{\circ} \mathrm{C} and W / m \cdot K are equivalent.