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Chapter 5: Problem 45

Copper-coated, epoxy-filled fiberglass circuit boards are treated by heating a stack of them under high pressare as sown in the sketch. The perpose of the pressing-heuting operation is to cure the eposy that bonds the fiberglass sheets, imparting siffness to the boards. The stack. riferted to as a book, is comprised of 10 boards and 11 pressing plates, which prevent epoxy from flowing between the boards and impar a-smooth finish to the cured boards. In order to perform simplified thermal analyses, it is reasonable to approximule the book as having an effective themal conductivity \((k)\) and an effective thermal capacitance \(\left(\rho c_{p}\right)\). Calculate the effective propertics if each of the boards and plates has a tickness of \(2.36 \mathrm{~mm}\) and the following thermophysical properties: board (b) \(\rho_{b}=1000 \mathrm{~kg} / \mathrm{m}^{\prime}, c_{p,}=1500\) \(\mathrm{W} / \mathrm{Kg} \cdot \mathrm{K}, k_{p}=0.30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}_{;}\)plate \((p) \rho_{p}=8\left(000 \mathrm{~kg} / \mathrm{m}^{3}\right.\). \(c_{v}=480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k_{p}=12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Short Answer

Expert verified
Effective thermal conductivity is \(7.696 \text{ W/m路K}\) and effective thermal capacitance is \(2720289 \text{ J/m}^3\cdot\text{K}\).

Step by step solution

01

Calculate Total Thickness

The stack is made up of 10 boards and 11 plates. Each element has a thickness of 2.36 mm. Hence, the total thickness is:\[\text{Total Thickness} = (10 + 11) \times 2.36 \text{ mm} = 21 \times 2.36 \text{ mm} = 49.56 \text{ mm} = 0.04956 \text{ m}\]
02

Calculate Total Volume

Assuming a unit area \(A = 1 \text{ m}^2\), the volume of the stack can be calculated by multiplying the total thickness by unit area:\[\text{Total Volume} = 0.04956 \text{ m}^3\]
03

Calculate Mass for Boards and Plates

Calculate the mass for boards and plates individually using their density and volume:1. Volume of boards: \(V_b = 10 \times 2.36 \text{ mm} = 23.6 \text{ mm} = 0.0236 \text{ m}\) \(m_b = \rho_b \times V_b = 1000 \times 0.0236 = 23.6 \text{ kg}\)2. Volume of plates: \(V_p = 11 \times 2.36 \text{ mm} = 25.96 \text{ mm} = 0.02596 \text{ m}\) \(m_p = \rho_p \times V_p = 8000 \times 0.02596 = 207.68\text{ kg}\)
04

Compute Effective Thermal Conductivity

Using the rule of mixtures for conductivity:\[k_{eff} = \frac{\text{Number of boards} \times k_b \times t_b + \text{Number of plates} \times k_p \times t_p}{\text{Total Thickness}}\]\[k_{eff} = \frac{10 \times 0.30 \times 0.00236 + 11 \times 12 \times 0.00236}{0.04956}\]\[k_{eff} = \frac{0.0708 + 0.31104}{0.04956} = 7.696\text{ W/m}\cdot\text{K}\]
05

Calculate Effective Thermal Capacitance

The effective heat capacity \(\rho c_p\) can be calculated as a weighted average:\[(\rho c_p)_{eff} = \frac{m_b \cdot c_{p,b} + m_p \cdot c_{p,p}}{\text{Total Volume}}\]\[(\rho c_p)_{eff} = \frac{23.6 \times 1500 + 207.68 \times 480}{0.04956}\]\[(\rho c_p)_{eff} = \frac{35400 + 99686.4}{0.04956} = 2720289\text{ J/m}^3\cdot\text{K}\]

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Key Concepts

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

Thermal Capacitance
Thermal capacitance is an important concept in thermal systems. It refers to the ability of a material to store thermal energy. This is defined by the product of the material's density \( \rho \) and its specific heat capacity \( c_p \), giving rise to the term thermal capacitance often represented by \( \rho c_p \). The thermal capacitance tells us how much energy a unit volume of the material can store for a change in temperature of 1 degree Kelvin.
  • Imagine filling a bathtub with water. The water represents stored energy and the size of the bathtub indicates the material's capacity to store this energy.
  • Materials with high thermal capacitance can store a lot of heat energy, acting like a thermal reservoir.
In our circuit board example, the effective thermal capacitance helps us understand the heat storing capacity of the multilayer system during the heating process. Calculating effective thermal capacitance involves determining the weighted average of the capacitances of all materials in the system. This helps ensure optimal curing and structural integrity during manufacturing.
Thermal Analysis
Thermal analysis is used to understand how temperature affects a material or system. This involves studying both the flow and storage of heat within the system. For the circuit board example, thermal analysis determines how heat propagates throughout the assembly to ensure even curing of epoxy between fiberglass sheets.
  • It is crucial in ensuring product quality and prevents defects by maintaining the right temperature controls.
  • Ensures energy-efficient operation by optimizing heating processes.
By using principles of thermal conductivity and capacitance, thermal analysis bridges the gap between material properties and process outcomes. The insights gained help in predicting the behavior under different thermal conditions, allowing for fine-tuning of manufacturing parameters to avoid overheating or under-curing.
Material Properties
Material properties like thermal conductivity and density are crucial in determining how a material will perform in a thermal application.
  • **Thermal Conductivity**: This property signifies a material's ability to conduct heat. High thermal conductivity means heat transfers rapidly through the material.
  • **Density**: It is the mass per unit volume of a material, impacting its thermal capacitance.
In a multilayer material system like the stacked circuit boards, these properties dictate how heat flows across different layers. For instance, the slightly conductive fiberglass and significantly conductive metal plates combine to give an average value of thermal conductivity for the stack. Understanding these material properties allows engineers to engineer composites and assemblies that meet specific temperature control requirements, ensuring reliability and efficiency throughout the product's life span.

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Most popular questions from this chapter

Consider a thin electrical heater artached to a plate and hacked by insulation. Initially, the heater and plate are at the temperafure of the ambient air, \(T\) - Suddenly, the power to the heater is activated, yielding a constant heat fux \(q_{e}^{*}\left(\mathrm{~W} / \mathrm{m}^{2}\right)\) at the inner surface of the plate. (a) Sketch and label, on \(T-x\) cocrdinates, the temperature distributions: initial, steady-state, and at two intermediate times. (b) Sketch the heat flux at the outer surface \(q_{z}^{\prime \prime}(L, t)\) as a function of time.

A very thick plate with thermal diffusivity \(5.6 \times\) \(10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) and thermal conductivity \(20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is initially at a uniform temperature of \(325^{\circ} \mathrm{C}\). Suddenly, the surface is exposed to a coolant at \(15^{\circ} \mathrm{C}\) for which the convection heat transfer coefficient is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Using the finite- difference method with a space increment of \(1 \mathrm{x}=15 \mathrm{~mm}\) and a time increment of \(18 \mathrm{~s}\), determine temperatures at the surface and at a depth of \(45 \mathrm{~mm}\) after 3 min have elapsed.

Two large blocks of different materials, such as copper and concrete, have been sitting in a room \(\left(23^{\circ} \mathrm{C}\right)\) for a very lone time. Which of the two blocks, if either, will feel colder to the touch? Assume the blocks to be seminfinite solids and your hand to be at a temperaturt of \(37^{\circ} \mathrm{C}\).

Heat transfer is not an intuitive process muses the Curious Cock. Does doubling the thickness of a hamburger approximately double the cooking time? What effect does the initial temperature have on cooking time? To answer these questions, develop a model to do virtual cooking of meat of thickness \(2 L\) in a doublesided grill. The meat is initially at \(20 \% \mathrm{C}\) when it is placed in the grill and both sides experience convection heat transfer characterized by an ambient temperature of \(100^{\circ} \mathrm{C}\) and a convection coefficient of 5000 W/m \({ }^{2} \cdot \mathrm{K}\). Assume the meat to have the properties of liquid water at \(300 \mathrm{~K}\) and to be properly cooked when the center temperature is \(605 \mathrm{C}\). (a) For hamburgers of thickness \(2 L=10,20\), and 30 \(\mathrm{mm}\), calculate the time for the center to reach the required cooking temperature of \(60 \mathrm{C}\). Determine a relationship between the cooking time and the thickness. For your solution, use the finite-element method of FEHT, the ready-to-solve model in the ModelSTransient Conduction/Plane Wall section of \(I H T\), or a numerical procedure of your choice. For one of the thicknesses, use an appropriate anaIytical solution to validate your numerical results. (b) Without performing a detailed numerical solution, but drawing on the recults of part (a), what can you say about the effect on the cooking time of changing the initial temperature of the meat from \(20{ }^{\circ} \mathrm{C}\) to \(5^{\circ} \mathrm{C}\) ? You may use your numerical model from part (a) to confirm your assessment.

A cold air chamber is proposed for quenching steel ball bearings of dinmeter \(D=0.2 \mathrm{~m}\) and initial temperature \(T_{i}=400^{\circ} \mathrm{C}\). Air in the chamber is maintained at \(-15^{\circ} \mathrm{C}\) by a refrigeration system, and the steel balls pass through the chamber on a conveyor belt. Optimum bearing production requires that \(70 \%\) of the initial thermal energy content of the ball above \(-15^{\circ} \mathrm{C}\) be removed. Radiation effects muy be neglected, and the convection heut transfer coefficient within the chamber is \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Estimute the residence time of the balls within the chamber, und recommend a drive velocity of the conveyor. The following properties may be used for the steel: \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=2 \times 10^{-3} \mathrm{~m}^{2} / \mathrm{s}\), and \(c=450 \mathrm{~J} / \mathrm{kg}-\mathrm{K}\).
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