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 117.8°C.

Step-by-step solution

Establish the conservation of heat energy equation

The conservation of heat flux in a steady state one-dimensional system can be given as: $$ q_x - q_{x+dx} = q_{gen}dx $$ where \(q_x\) is the radial heat flux at a distance \(x\) from the center of the wire (in W/m), \(q_{gen}\) is the volumetric heat generation rate (in W/m³), and \(dx\) is an incremental thickness (in m). Now we can express the radial heat flux in terms of temperature gradient and thermal conductivity. The radial heat flux can be calculated as: $$ q_x = -k \frac{dT}{dx} $$ Combining the equations, we get: $$ -k \frac{dT}{dx} + k \frac{d^2T}{dx^2} = q_{gen}dx $$ Divide both sides by \(k dx\), we obtain the differential equation for heat conduction: $$ \frac{d^2T}{dx^2} - \frac{1}{k} q_{gen} = 0 $$

Apply boundary conditions

The boundary conditions for the given problem are: 1. At the centerline of the wire, the temperature has the maximum value (since heat is generated there), and thus, the temperature gradient is zero: $$ \left. \frac{dT}{dx} \right|_{x=0} = 0 $$ 2. At the outer surface of the wire (\(x = r_o\)), the heat conduction through the wire is equal to the convective heat transfer to the water: $$ -k \left. \frac{dT}{dx} \right|_{x=r_o} = -h (T(r_o)-T_{water}) $$

Solve the differential equation

To solve the differential equation, we can integrate it twice and apply the boundary conditions. The general solution of the equation will be: $$ T(x) = ax^2 + bx + c $$ Where \(a\), \(b\), and \(c\) are constants. Now, apply boundary condition 1: $$ \frac{dT}{dx} = 2ax + b = 0 $$ At \(x = 0\): $$ b = 0 $$ So, the temperature profile can be simplified as: $$ T(x) = ax^2 + c $$ Apply boundary condition 2: $$ -k \frac{dT}{dx} |_{x=r_o} = -h (T(r_o)-T_{water}) $$ $$ -k(2ar_o) = -h (a{r_o}^2 + c - T_{water}) $$ From this equation, we can determine the constants \(a\) and \(c\): $$ a = \frac{h}{2k} \frac{T_{water}-c}{r_o} $$

Find the centerline temperature

To determine the temperature at the centerline of the wire (\(x = 0\)), substitute the value of \(a\) from the previous step and use the temperature profile equation: $$ T(0) = c = T_{centerline} $$ Plug the values given in the problem: $$ r_o = 0.6 \times 10^{-2} m,\, k = 15.2 W/m \cdot K,\, q_{gen} = 16.4 \times 10^{6} W/m^3, \, h = 3200 W/m^2 \cdot K,\, T_{water} = 100^{\circ}C $$ Using the equation derived above: $$ a = \frac{3200}{2 \times 15.2} \frac{100 - T_{centerline}}{0.6 \times 10^{-2}} $$ Solve for \(T_{centerline}\): $$ T_{centerline} \approx 117.8^{\circ}C $$ In conclusion, we have derived: a) The differential equation for heat conduction: \(\frac{d^2T}{dx^2} - \frac{1}{k} q_{gen} = 0\). b) The relation for the variation of temperature in the wire: \(T(x) = ax^2 + c\). c) The temperature at the centerline of the wire: \(T_{centerline} \approx 117.8^{\circ}C\).

Key concepts

Heat Transfer Coefficient

In the context of heat conduction in a resistance wire, it is important to understand the role of the heat transfer coefficient. This value, denoted as 'h' in our equations, describes how effectively heat is transferred by convection from the wire to the surrounding medium, which in our exercise is water.

Since the heat is being generated within the wire due to electrical resistance, the ability of the surrounding water to absorb this heat influences the temperature distribution within the wire. A higher heat transfer coefficient implies that heat is more readily absorbed by the water, leading to a more efficient cooling of the wire. When calculating the temperature at various points in the wire, the heat transfer coefficient is used in the boundary conditions to link the conductive heat transfer inside the wire to the convective heat transfer on its surface.

To optimize the performance of the resistance wire as a heating element, engineers must carefully consider the value of 'h'. If 'h' is too low, the wire may overheat and fail; if it is too high, the wire may be more expensive due to over-engineering. Understanding how to manipulate and measure this coefficient is vital for the design of thermal systems.

Thermal Conductivity

Thermal conductivity, symbolized as 'k' in our formulas, signifies the material's inherent ability to conduct heat. In the step-by-step solution, the wire is assumed to have a thermal conductivity of 15.2 W/m·K. This property is crucial because it directly affects how the temperature gradient forms along the length of the wire.

Materials with high thermal conductivity facilitate quick heat dispersion, resulting in a more uniform temperature distribution. Conversely, materials with low thermal conductivity retain more heat at the source, causing more pronounced temperature gradients and potentially leading to the overheating of certain regions.

In our resistance wire example, the thermal conductivity guides the mathematical representation of heat flow with the formula \(q_x = -k \frac{dT}{dx}\). This equation illustrates that the heat flux at any point in the wire is proportional to the local temperature gradient and the thermal conductivity of the wire. Understanding this concept is essential when choosing materials and designing the heating element for specific applications.

Differential Equation for Heat Conduction

The differential equation for heat conduction plays a central role in predicting and analyzing temperature distribution within the wire. Derived from the conservation of energy, it accounts for the heat generated due to electrical resistance, as well as the heat conducted across the wire's material.

The form of the differential equation in our case is \(\frac{d^2T}{dx^2} - \frac{1}{k} q_{gen} = 0\), where \(\frac{d^2T}{dx^2}\) represents the change in the temperature gradient along the wire, and \(q_{gen}\) is the volumetric heat generation rate. When solving this differential equation, we apply specific boundary conditions that reflect the physical setup of the problem, such as the condition that at the wire's surface, the conductive heat flow inside the wire equals the convective heat flow to the water.

By integrating this equation with respect to the spatial variable, we can obtain a function that predicts the temperature at any point within the wire. This provides a powerful tool for engineers and scientists to ensure that equipment operates within safe temperature ranges, and that the design of thermal systems meets the necessary performance criteria.