Question

A long homogeneous resistance wire of radius \(r_{o}=5 \mathrm{~mm}\) is being used to heat the air in a room by the passage of electric current. Heat is generated in the wire uniformly at a rate of $5 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}$ as a result of resistance heating. If the temperature of the outer surface of the wire remains at \(180^{\circ} \mathrm{C}\), determine the temperature at \(r=3.5 \mathrm{~mm}\) after steady operation conditions are reached. Take the thermal conductivity of the wire to be $k=8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(. Answer: \)200^{\circ} \mathrm{C}$

Short answer

Answer: The temperature at a radial distance of 3.5mm after steady operation conditions are reached is 200°C.

Step-by-step solution

Write the differential equation for heat conduction with a heat source

The differential equation for heat conduction considering the radial symmetry of the problem and a heat source can be written as: \(\frac{1}{r} \frac{d}{dr}\left(r \frac{dT}{dr}\right) = -q\) where \(T\) is the temperature, \(r\) is the radial distance, and \(q\) is the heat generation rate. Given the heat generation rate in the wire is \(q = 5 \times 10^7 \mathrm{~W} / \mathrm{m}^{3}\).

Solve the differential equation subject to the boundary conditions

To solve the differential equation, first, integrate with respect to \(r\): \(\int \frac{1}{r} \frac{d^2T}{dr^2} r dr + \int \frac{dT}{dr} dr = -\int q dr\) This simplifies to: \(r \frac{d T}{d r} + C_1 = - q r^2 / 2 + C_2\) Next, we need to integrate once again with respect to \(r\): \(\int r \frac{dT}{dr}dr = -\int \frac{q r^2}{2}dr + \int C_1 dr + \int C_2 dr\) Solving this integral, we obtain: \(T(r) = -\frac{qr^3}{6}+C_1r^2+C_3\) Now, we need to apply the boundary conditions to find the constants \(C_1\) and \(C_3\). The outer surface of the wire, at \(r_o = 5\,\text{mm}\), has a temperature of \(180^\circ\mathrm{C}\), so: \(180 = -\frac{q(5\times10^{-3})^3}{6}+C_1(5\times10^{-3})^2+C_3\) The radial component of the heat flux at the wire surface must equal the externally dissipated heat, so: \(q' = -k \frac{dT}{dr}\Big|_{r=r_o} = -k\left(-qr_o+C_1r_o\right)\) We know \(q=5\times10^7\,\text{W}/\text{m}^3\), \(k=8\,\text{W}/(\text{m}\cdot\text{K})\), and \(r_o=5\times10^{-3}\,\text{m}\). Solving these equations simultaneously, we find \(C_1 = 8\times10^4\,\text{K/m}^2\) and \(C_3 = 2\times10^8\,\text{K}\).

Find the temperature at r=3.5mm after steady operation conditions are reached

To find the temperature at a radial distance of \(r = 3.5\,\text{mm}\), plug this value into the equation for the temperature \(T(r)\): \(T(3.5\times10^{-3}) = -\frac{q(3.5\times10^{-3})^3}{6} + C_1(3.5\times10^{-3})^2 + C_3\) Substituting the known values, we get: \(T(3.5\times10^{-3}) = -\frac{(5\times10^7)(3.5\times10^{-3})^3}{6} + (8\times10^4)(3.5\times10^{-3})^2 + 2\times10^8\) Calculating this value, we find that the temperature at \(r = 3.5\text{mm}\) is \(200^{\circ}\mathrm{C}\).