Question

A long homogeneous resistance wire of radius \(r_{o}=0.6 \mathrm{~cm}\) and thermal conductivity \(k=15.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is being used to boil water at atmospheric pressure by the passage of electric current. Heat is generated in the wire uniformly as a result of resistance heating at a rate of \(16.4 \mathrm{~W} / \mathrm{cm}^{3}\). The heat generated is transferred to water at \(100^{\circ} \mathrm{C}\) by convection with an average heat transfer coefficient of $h=3200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. Assuming steady one-dimensional heat transfer, \)(a)$ express the differential equation and the boundary conditions for heat conduction through the wire, \((b)\) obtain a relation for the variation of temperature in the wire by solving the differential equation, and \((c)\) determine the temperature at the centerline of the wire.

Short answer

Answer: The temperature at the centerline of the wire is approximately 100.309 °C.

Step-by-step solution

Differential equation for heat conduction

The rate of heat generation per unit volume in the wire is given as \(16.4 \mathrm{~W} / \mathrm{cm}^{3}\), so we will denote it by \(q_{v}\). For steady one-dimensional heat conduction, we can use the following equation: $$ - k \frac{d^2T}{dr^2} = q_{v} $$ Where \(T\) is the temperature, \(r\) is the radial distance from the centerline, \(k\) is the thermal conductivity of the wire, and \(q_{v}\) is the volume heat generation rate.

Applying boundary conditions

We need to apply boundary conditions for this exercise. This involves two conditions: Condition 1 - At the centerline of the wire \((r=0)\), the temperature gradient is zero. Following Newton's law of cooling for convection, we have: $$ \frac{dT}{dr} \Bigg|_{r=0} = 0 $$ Condition 2 - At the outer surface of the wire \((r = r_{o})\), the rate of heat transfer from the wire to water is given by: $$ -k \frac{dT}{dr} \Bigg|_{r=r_{o}} = h [T(r_{o}) - T_{w}] $$ where \(h\) is the heat transfer coefficient, and \(T_w\) is the water temperature.

Solving the differential equation

To solve the differential equation and obtain the temperature profile in the wire, we can first solve the homogeneous heat conduction equation: $$ - k \frac{d^2T}{dr^2} = q_{v} $$ Integrating once, we have: $$ - k \frac{d T}{d r} = q_{v} r + C1 $$ We can apply the first boundary condition: $$ \frac{dT}{dr} \Bigg|_{r=0} = 0 $$ This gives: $$ C1 = 0 $$ Integrating once again, we have: $$ - k T = \frac{q_{v}}{2}r^2 + C2 $$ We can now apply the second boundary condition to find \(C2\): $$ -k \frac{dT}{dr} \Bigg|_{r=r_{o}} = h [T(r_{o}) - T_{w}] $$ Substituting the expression for \(\frac{dT}{dr}\), we find: $$ h \left[T(r_o) - T_w\right] = q_vr_o $$ Now, we can substitute the expression for \(T(r)\) back into this equation: $$ h\left[-\frac{k}{q_v}\left(\frac{q_v}{2}r_o^2 +C2\right) - T_w\right] = q_vr_o $$ Rearranging and solving for \(C2\), we get: $$ C2 = -\left(\frac{k}{2}r_o + hT_w\right) $$ Therefore, the temperature profile in the wire is given by: $$ T(r) = -\frac{q_v}{2k}r^2 + \frac{q_{v}r_{o}}{2h} + T_w $$

Finding the temperature at the centerline

Finally, we can find the temperature at the centerline of the wire by setting \(r = 0\): $$ T(0) = -\frac{q_v}{2k}(0)^2 + \frac{q_{v}r_{o}}{2h} + T_w $$ Plugging in the given values, we obtain: $$ T(0) = -\frac{16.4}{2(15.2)}(0)^2 + \frac{(16.4)(0.6)}{2(3200)} + 100 $$ Calculating the result, we find the temperature at the centerline of the wire: $$ T(0) = 100.309 ^{\circ}C $$