Question

A \(1200-W\) iron is left on the iron board with its base exposed to ambient air at \(26^{\circ} \mathrm{C}\). The base plate of the iron has a thickness of \(L=0.5 \mathrm{~cm}\), base area of \(A=150 \mathrm{~cm}^{2}\), and thermal conductivity of \(k=18 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. The outer surface of the base plate whose emissivity is \(\varepsilon=0.7\), loses heat by convection to ambient air with an average heat transfer coefficient of \(h=\) \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) as well as by radiation to the surrounding surfaces at an average temperature of \(T_{\text {surr }}=295 \mathrm{~K}\). Disregarding any heat loss through the upper part of the iron, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the plate, \((b)\) obtain a relation for the temperature of the outer surface of the plate by solving the differential equation, and (c) evaluate the outer surface temperature.

Short answer

In summary, to determine the outer surface temperature of the iron's base plate, follow these steps: 1. Write the differential equation for steady one-dimensional heat conduction through the plate: \(\frac{d^2 T}{d x^2} =0\) 2. Establish boundary conditions for the inner surface (i.e., heat flux generated by the heaters) and the outer surface (i.e., heat loss due to convection and radiation). 3. Solve the differential equation to find the linear temperature relation within the plate: \(T(x) = C_1x + C_2\) 4. Obtain the relation for the outer surface temperature by substituting the boundary conditions found in Steps 1 and 2: \(T_{out} = T(L) = C_1L + C_2\) 5. Evaluate the outer surface temperature, \(T_{out}\), by plugging in the given parameters. By analyzing the heat transfer processes and applying differential equations for heat conduction, the outer surface temperature of the iron's base plate can be determined.

Step-by-step solution

Write Differential Equation for Steady One-Dimensional Heat Conduction

Since the problem is one-dimensional, we will only consider the heat transfer in the direction normal to the base plate. The differential equation for steady one-dimensional heat conduction in a solid is given by: \[\frac{d^{2} T}{d x^{2}}=0\] In the given problem, heat is conducted from the inner surface of the base plate to the outer surface.

Boundary Conditions

There are two boundary conditions for this problem. The first boundary condition is at the inner surface of the base plate and the second is at the outer surface. At the inner surface (x=0), heat flux is generated by the resistance heaters inside, and it is equal to the power of the iron divided by the base area. So we can write: \[ q_{in} = \frac{1200\,\text{W}}{150\,\text{cm}^2} = - k \left(\frac{dT}{dx}\right)_{x=0}\] At the outer surface (x=L), heat is lost by convection to the ambient air and radiation to the surrounding surfaces. We can write two equations relating the heat flux from the plate to the ambient air: \[q_{conv} = h (T_{out} - T_{\infty})\] \[q_{rad} = \varepsilon \sigma (T_{out}^4 - T_{surr}^4)\] Combined heat loss (\(q_{out}\)) at the outer surface can be written as the sum of convective and radiative heat losses: \[q_{out} = q_{conv} + q_{rad}\] Now, we have another boundary condition (at x=L), which states that the heat flux conducted through the plate is equal to the combined heat loss at the outer surface. This can be written as: \[q_{out} = k\left(-\frac{dT}{dx}\right)_{x=L}\]

Solve the Differential Equation

The differential equation, \(\frac{d^2 T}{d x^2} =0\), implies a linear temperature profile within the plate. Integrating twice with respect to x, we obtain the following linear temperature relation (with arbitrary constants C1 and C2): \[T(x) = C_1x + C_2\] To solve this equation, substitute the boundary conditions from Steps 1 and 2 into the temperature relation and solve for C1 and C2.

Obtain the Relation for the Outer Surface Temperature

After finding the values of C1 and C2, we can substitute them into the temperature relation to obtain the function for the outer surface temperature (at x=L): \[T_{out} = T(L) = C_1L + C_2\]

Evaluate the Outer Surface Temperature

With the temperature relation for the outer surface established, we can plug in the given parameters and evaluate the outer surface temperature, \(T_{out}\).

Key concepts

Thermal Conductivity

Thermal conductivity is a key concept when discussing heat conduction. It is a material-specific trait that tells us how well the material conducts heat. Mathematically, it is denoted by the symbol \(k\) and is measured in watts per meter-kelvin (\(W/m\cdot K\)). The higher the thermal conductivity, the more efficient the material is at transferring heat. In the context of the iron base plate in the exercise, the thermal conductivity is given as \(k=18 W/m\cdot K\), indicating it transfers heat effectively.

One interesting point to be noted here is that when we talk about thermal conductivity in a steady one-dimensional heat transfer situation, it assumes a constant value within the material, implying uniform heat transfer characteristics across the material. Therefore, in the solved problem, the rate of heat flow through the iron's base plate is consistent, leading to the establishment of a steady temperature profile that can be determined by solving appropriate mathematical equations under given boundary conditions.

Steady One-Dimensional Heat Transfer

In the realm of heat transfer, the term 'steady' implies that temperature does not change with time, while 'one-dimensional' specifies that heat flows in only one direction. Steady one-dimensional heat transfer is a simplified scenario that is often used to model heat transfer in systems where temperature gradients exist in only one dimension and remain constant over time.

In our discussed exercise, the heat transfer through the thickness of the iron's base plate represents a classic example of steady one-dimensional heat transfer. This assumption simplifies the problem, allowing us to use the basic heat conduction equation \(\frac{d^{2} T}{d x^{2}}=0\) to describe the system. The solution to this differential equation points to a linear temperature gradient within the base plate, making the analysis far more manageable and the computation of temperature profiles straightforward once the boundary conditions are addressed.

Boundary Conditions

Boundary conditions are essential for solving differential equations in heat transfer problems as they define how the system interacts with its surroundings. They're set based on the physical constraints and forces acting at the boundaries of the system. There are generally two types of boundary conditions in heat transfer: temperature-based (Dirichlet) and flux-based (Neumann).

In the example of the iron's base plate, we have a flux-based boundary condition at the inner surface where heat is being generated by the resistance heater. This is expressed by a uniform heat flux that determines the temperature gradient at that surface. Conversely, at the base plate's outer surface, the heat loss due to convection and radiation to the surroundings sets the boundary condition, which matches the heat flux leaving the plate. Adequately defining and applying these boundary conditions is critical for accurately assessing the temperature distribution within the base plate. Understanding and implementing these conditions correctly allows us to predict how the plate's temperature will react to the internal generation of heat and external cooling influences.