Question

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.

Short answer

Question: Provide the mathematical formulation of a heat conduction problem for a resistance heater wire with the given properties and assumptions. Answer: The mathematical formulation consists of a second-order ordinary differential equation and two boundary conditions. The differential equation is: \(\frac{d}{dx}(10.4\frac{dT}{dx}) - \frac{2000}{\pi {r_o}^2 L} = 0\). The boundary conditions are: \(\frac{dT}{dx}(0) = 0\) and \(\frac{dT}{dx}(L) = 0\).

Step-by-step solution

Modeling heat conduction using Fourier's law

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 \(k=10.4 \ \mathrm{Btu\ h^{-1} ft^{-1} R^{-1}}\).

Modeling the conservation of energy

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.

Formulate the differential 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, \(x\).

Define the boundary conditions

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.

Key concepts

Fourier's Law

Fourier's Law is a fundamental principle in the study of heat conduction. It describes how heat flows through a material when there is a temperature difference. The law states that the rate of heat transfer, or heat flux, through a material is proportional to the negative of the temperature gradient and the material's thermal conductivity. This relationship can be expressed as:
\(q = -k\frac{dT}{dx}\)
where:

• \(q\) is the heat flux, the amount of heat transferred per unit area per unit time.
• \(k\) is the thermal conductivity of the material, which indicates how easily heat can flow through it.
• \(\frac{dT}{dx}\) is the temperature gradient, which is the rate of temperature decrease along the direction of heat flow.

For a heater wire with a known thermal conductivity, like our example with \(k = 10.4\) Btu/h·ft·°R, Fourier's law helps us understand the fundamental behavior of heat as it travels through the wire. The negative sign in the equation indicates that heat flows from regions of high temperature to regions of low temperature. Understanding this concept is crucial for solving more complex problems in thermodynamics and engineering.

Steady-State Heat Transfer

Steady-state heat transfer occurs when the temperature distribution in a body does not change with time. In simple terms, it means that the amount of heat entering a surface is equal to the amount leaving the surface, resulting in no accumulation of heat within the system. This is a common assumption in many thermal problems because it simplifies the analysis.
In mathematical terms, the steady-state condition means that the time derivative of temperature is zero. For our exercise involving the resistance heater wire, steady-state assumes that despite the heat generation from the wire, after some time, a consistent temperature profile is established.
The differential equation for steady-state conditions comes from applying the principle of conservation of energy. It ensures that the heat generated by the wire is evenly distributed without any changes over time. If we consider the energy generated by the wire (\(G = \frac{2,000}{V}\) Watts per unit volume), the heat conduction equation becomes:
\(\frac{d}{dx}(k\frac{dT}{dx}) - G = 0\)
In this equation:

• \(k\frac{dT}{dx}\) describes the heat conduction through the material.
• \(G\) represents the rate of heat generation within the material.

The solution to this differential equation will give us the temperature distribution across the wire at steady-state.

Boundary Conditions

Boundary conditions are crucial for solving differential equations, as they define the constraints that solutions must satisfy at the boundaries of the problem domain. In heat conduction problems, these conditions determine how heat interacts at specific points, like the ends of a wire or surface of a wall.
For the heater wire example, boundary conditions are applied at both ends of the wire. They ensure that the derived temperature distribution aligns with the physical setup:

• Boundary Condition 1: At \(x = 0\), the temperature gradient is zero \(\left(\frac{dT}{dx}(0) = 0\right)\).
• Boundary Condition 2: At \(x = L\), the temperature gradient is also zero \(\left(\frac{dT}{dx}(L) = 0\right)\).

These conditions imply that there is no net heat transfer at the ends of the wire. In practical terms, this can indicate that both ends are perfectly insulated or symmetrically distributed in an environment with zero external heat flux. Applying these boundary conditions to the differential equation allows solving for the exact temperature profile, ensuring the solution fits the real-world scenario exactly as described.