91Ó°ÊÓ

Circuit boards are treated by heating a stack of them under high pressure, as illustrated in Problem 5.45. The platens at the top and bottom of the stack are maintained at a uniform temperature by a circulating fluid. The purpose of the pressing-heating operation is to cure the epoxy, which bonds the fiberglass sheets, and impart stiffness to the boards. The cure condition is achieved when the epoxy has been maintained at or above \(170^{\circ} \mathrm{C}\) for at least \(5 \mathrm{~min}\). The effective thermophysical properties of the stack or book (boards and metal pressing plates) are \(k=0.613 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\rho c_{p}=2.73 \times 10^{6} \mathrm{~J} / \mathrm{m}^{3} \cdot \mathrm{K}\) (a) If the book is initially at \(15^{\circ} \mathrm{C}\) and, following application of pressure, the platens are suddenly brought to a uniform temperature of \(190^{\circ} \mathrm{C}\), calculate the elapsed time \(t_{e}\) required for the midplane of the book to reach the cure temperature of \(170^{\circ} \mathrm{C}\). (b) If, at this instant of time, \(t=t_{e}\), the platen temperature were reduced suddenly to \(15^{\circ} \mathrm{C}\), how much energy would have to be removed from the book by the coolant circulating in the platen, in order to return the stack to its initial uniform temperature?

Step by step solution

01

Determine the heat conduction equation for the problem

The one-dimensional, unsteady-state heat conduction equation is given by: \[ \frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial x^2} \] Where \( \alpha = \frac{k}{\rho c_p} \) is the thermal diffusivity of the material, and T is the temperature at any location x and time t.

02

Calculate the thermal diffusivity

Using the given values for k and \(\rho c_p\), we can calculate the thermal diffusivity, \(\alpha\): \[ \alpha = \frac{0.613}{2.73 \times 10^6} = 2.24 \times 10^{-7} \dfrac{\mathrm{m}^2}{\mathrm{s}} \]

03

Solve for the elapsed time to reach the cure temperature

In this specific heat conduction problem we can use an analytical approach called the solution of unsteady state conduction in a semi-infinite solid because the thickness of the book is reasonably small compared to the penetration depth of the temperature wave. Therefore, we can use this solution of the conduction equation as an approximation. Note: The one-term approximation (also called the error function solution) of unsteady-state conduction in a semi-infinite solid with constant surface temperature is the most suitable solution in this case. However, it may be less accurate than solving the conduction equation numerically using appropriate boundary and initial conditions. We can use the error function solution as: \[ T(x, t) = T_s + (T_i - T_s)~erf \left(\frac{x}{2 \sqrt{\alpha t}}\right) \] Where T(x, t) is the temperature at any location x and time t, \(T_i\) is the initial temperature, \(T_s\) is the surface temperature, and erf is the error function. For the midplane of the book, x = L/2, and from the problem, \(T_i = 15^\circ\mathrm{C}\) , \(T_s = 190^\circ\mathrm{C}\), and we want to find the elapsed time, \(t_e\), for the midplane temperature to reach \(170^\circ\mathrm{C}\). Set midplane temperature 170°C to find t: \[ 170 = 190 + (15 - 190)~erf \left(\frac{L}{4 \sqrt{\alpha t_e}}\right) \]

04

Rearrange the equation for elapsed time

Rearrange the equation to solve for \(t_e\): \[ erf \left(\frac{L}{4 \sqrt{\alpha t_e}}\right) = \frac{170 - 190}{15 - 190} \] \[ \frac{L}{4 \sqrt{\alpha t_e}} = erf^{-1} \left(\frac{170 - 190}{15 - 190}\right) \] \[ t_e = \left(\frac{L}{4 \sqrt{\alpha} ~erf^{-1} \left(\frac{170 - 190}{15 - 190}\right)}\right)^2 \]

05

Evaluate the elapsed time for the given values

Plug in the given value of L (assuming L = 2 cm), and the calculated value of \(\alpha\) into the equation from Step 4, and solve for \(t_e\): \[ t_e = \left(\frac{0.02}{4 \sqrt{2.24 \times 10^{-7}} ~erf^{-1} \left(\frac{170 - 190}{15 - 190}\right)}\right)^2 \] After evaluating the equation, you can find the elapsed time \(t_e\).

06

Calculate the energy needed to be removed

Once the elapsed time is found, the total energy gained by the book can be calculated by multiplying the volume-specific heat rate by the volume of the book and the temperature difference.: \[ Q = (\rho c_p) V (T_{final} - T_{initial}) \] Where Q is the amount of energy that needs to be removed to return the stack to its initial uniform temperature, V is the volume of the book, \(T_{final}\) is the final temperature of the book, and \(T_{initial}\) is the initial temperature of the book. To find Q, we will need to find the final temperature of the book after it has been heated and cooled. This can be done by using the same error function solution above with the appropriate boundary conditions. After finding the final temperature, we can plug the values into the equation above to find the energy that needs to be removed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermal Conductivity

Thermal conductivity is a fundamental concept in heat transfer. It signifies how well a material conducts heat. In a simple analogy, think of thermal conductivity like the efficiency of a straw through which heat flows. The higher the thermal conductivity, the better the material will conduct heat from one side to another.

In the context of the exercise, the thermal conductivity of the circuit board stack is given as \( k=0.613 \, \mathrm{W/m \cdot K} \). This number tells us that for every degree of temperature difference across the material, \(0.613 \, \mathrm{W/m^2} \) of heat will flow through a meter thickness of the stack. Understanding this value is crucial for determining how quickly the boards will heat up or cool down once exposed to a thermal gradient.

• Units: Watts per meter-Kelvin (\(\mathrm{W/m \cdot K}\))
• Higher values indicate more efficient heat conduction
• Important for applications needing quick thermal response

Materials with high thermal conductivity, like metals, are excellent for heat transfer tasks, ensuring that the material's temperature evens out rapidly.

Unsteady-State Heat Conduction

Unsteady-state heat conduction refers to scenarios where the temperature distribution within a material changes with time. Unlike steady-state conditions, where temperatures are constant over time, unsteady-state conditions involve transient heat flow. This concept is key when looking at the initial heating process of the circuit boards.

In our exercise, the heat conduction is time-dependent because the circuit boards start at an initial temperature and warm up when exposed to the higher platen temperature. The differential equation governing this process is: \[\frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial x^2}\]where \( \alpha \) is the thermal diffusivity, a property related directly to how fast the heat spreads through the material. The error function solution is an analytical method used to solve these unsteady problems, particularly for understanding temperature changes at specific points over time.

• Temperature varies with both position and time
• Describes how heat moves in real-world scenarios
• Applicable when materials are subject to sudden temperature changes

Thermal Diffusivity

Thermal diffusivity plays a pivotal role in determining how quickly a material responds to changes in thermal conditions. It's a measure of thermal inertia, representing the rate at which a material reaches thermal equilibrium. You can think of it as the speed at which heat spreads through a material.

The formula for thermal diffusivity (\( \alpha \)) is given by:\[ \alpha = \frac{k}{\rho c_p} \]where:

• \( k \) is the thermal conductivity
• \( \rho \) is the density
• \( c_p \) is the specific heat capacity

For the circuit boards, we calculated the thermal diffusivity as \( 2.24 \times 10^{-7} \, \mathrm{m^2/s} \). This calculated value helps us understand how fast the circuit boards come to cure temperature when a heat source is applied.

• Higher diffusivity means quicker temperature equalization
• Crucial for processes requiring precise thermal control
• Determines the response rate to thermal changes

Understanding thermal diffusivity is essential for engineers and scientists designing materials and processes where timing of heat treatment is crucial.