A 2-kW resistance heater wire whose thermal conductivity is \(k=10.4 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot \mathrm{R}\) has a radius of \(r_{o}=0.06\) in and a length of \(L=15\) in, and is used for space heating. Assuming constant thermal conductivity and one-dimensional heat transfer, express the mathematical formulation (the differential equation and the boundary conditions) of this heat conduction problem during steady operation. Do not solve.

Fourier's law states that the heat flux, \(q\), in a material is proportional to its thermal conductivity, \(k\), and the temperature gradient: \(q = -k\frac{dT}{dx}\) In this case, the heater wire's thermal conductivity is given as \(k=10.4 \ \mathrm{Btu\ h^{-1} ft^{-1} R^{-1}}\).

The conservation of energy, also known as the continuity equation, requires that the rate of energy generation equals the net rate at which energy is entering and/or leaving the system. Since this is a steady-state process, we can express the continuity equation as: \(\frac{d}{dx}(k\frac{dT}{dx}) - G = 0\) Where \(G\) is the rate of energy generation per unit volume. In this case, a 2-kW resistance heater wire is used, so: \(G = \frac{2,000\ W}{V}\), where \(V\) is the volume of the wire. To find the volume, we will use the given radius \(r_o=0.06\) in and length \(L=15\) in: \(V = \pi {r_o}^2 L\) Now that we have all the necessary quantities, we can plug them into the continuity equation.

Plugging the values of thermal conductivity and energy generation rate into the continuity equation, we get: \(\frac{d}{dx}(10.4\frac{dT}{dx}) - \frac{2000}{\pi {r_o}^2 L} = 0\) This is a second-order ordinary differential equation, since it includes the second derivative of temperature with respect to position, \(x\).

Since the problem is one-dimensional, we need two boundary conditions to solve the differential equation. In this case, we can use the heat conduction equation at the two ends of the wire, where \(x=0\) and \(x=L\). At both ends of the wire, the rate of heat transfer must be zero because there is no heat transferred horizontally at these points: Boundary condition 1: At \(x=0\), \(\frac{dT}{dx}(0) = 0\) Boundary condition 2: At \(x=L\), \(\frac{dT}{dx}(L) = 0\) #Conclusion# The mathematical formulation for this heat conduction problem consists of the following differential equation and boundary conditions: Differential equation: \(\frac{d}{dx}(10.4\frac{dT}{dx}) - \frac{2000}{\pi {r_o}^2 L} = 0\) Boundary condition 1: \(\frac{dT}{dx}(0) = 0\) Boundary condition 2: \(\frac{dT}{dx}(L) = 0\) These equations can be used to solve for the temperature distribution within the wire under the given assumptions.