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Chapter 3: Problem 115

A bonding operation utilizes a laser to provide a constant heat flux, \(q_{o}^{\prime \prime}\), across the top surface of a thin adhesivebacked, plastic film to be affixed to a metal strip as shown in the sketch. The metal strip has a thickness \(d=1.25 \mathrm{~mm}\), and its width is large relative to that of the film. The thermophysical properties of the strip are \(\rho=7850 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=435 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The thermal resistance of the plastic film of width \(w_{1}=40 \mathrm{~mm}\) is negligible. The upper and lower surfaces of the strip (including the plastic film) experience convection with air at \(25^{\circ} \mathrm{C}\) and a convection coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The strip and film are very long in the direction normal to the page. Assume the edges of the metal strip are at the air temperature \(\left(T_{\infty}\right)\). (a) Derive an expression for the temperature distribution in the portion of the steel strip with the plastic film \(\left(-w_{1} / 2 \leq x \leq+w_{1} / 2\right)\). (b) If the heat flux provided by the laser is 10,000 \(\mathrm{W} / \mathrm{m}^{2}\), determine the temperature of the plastic film at the center \((x=0)\) and its edges \(\left(x=\pm w_{1} / 2\right)\). (c) Plot the temperature distribution for the entire strip and point out its special features.

Short Answer

Expert verified
The temperature distribution in the steel strip with plastic film is given by \(T(x) = -\frac{q^{\prime \prime}}{2k}x^2 + C_1x + C_2\), where we can find the constants \(C_1\) and \(C_2\) by applying boundary conditions. Using the given heat flux, \(q^{\prime \prime} = 10000\,\mathrm{W} / \mathrm{m}^{2}\), we can calculate the temperature at the center and edges of the plastic film. By plotting the temperature distribution for the entire strip, the special features may include symmetry, peaks, or other points of interest.

Step by step solution

01

Derive the expression for temperature distribution

To find the temperature distribution of the metal strip, we'll have to write the steady-state heat conduction equation: \(-k\frac{d^{2} T}{d x^{2}} + q^{\prime \prime} = 0\) Since the thermal resistance of the plastic film is negligible, and the portion of the steel strip that we're analyzing is not affected by the boundary conditions, we can simplify the equation and write: \(\frac{d^{2} T}{d x^{2}} = -\frac{q^{\prime \prime}}{k}\) Now, we'll integrate the equation twice to find the temperature distribution: \(\frac{dT}{dx} = -\frac{q^{\prime \prime}}{k}x + C_1\) \(T(x) = -\frac{q^{\prime \prime}}{2k}x^2 + C_1x + C_2\)
02

Apply the boundary conditions

The edges of the metal strip are at the air temperature (\(T_\infty\)) and can be used as boundary conditions. The strip dimensions are large, so we assume that the temperature at the edges is roughly the same. When\(x = -w_1/2\), \(T(-w_1/2) = T_\infty\): \(T_\infty = -\frac{q^{\prime \prime}}{2k}\left(-\frac{w_1}{2}\right)^2 + C_1 \frac{-w_1}{2} + C_2\) When\(x = w_1/2\), \(T(w_1/2) = T_\infty\): \(T_\infty = -\frac{q^{\prime \prime}}{2k}\left(\frac{w_1}{2}\right)^2 + C_1 \frac{w_1}{2} + C_2\) By solving these two equations, we can find the values for the constants \(C_1\) and \(C_2\).
03

Calculate the temperature at the center and edges of the plastic film

Given the heat flux provided by the laser, \(q^{\prime \prime} = 10000\,\mathrm{W} / \mathrm{m}^{2}\), we can now find the temperature of the plastic film at the center and its edges. At the center of the film, \(x = 0\). Plugging this value into the temperature distribution equation, we get: \(T(0) = -\frac{q^{\prime \prime}}{2k}(0)^2 + C_1 (0) + C_2\) \(T(0) = C_2\) Similarly, at the edges of the film, \(x = \pm w_1/2\), we can find the temperature by plugging in the corresponding values into the temperature distribution equation.
04

Plot the temperature distribution for the entire strip

To plot the temperature distribution for the entire strip, we need to find the temperature as a function of the position \(x\), using the temperature distribution equation derived earlier, along with the values of \(q^{\prime \prime}\), \(k\), \(C_1\), and \(C_2\). The plot will show the temperature distribution along the length of the strip, from \(x = -w_1/2\) to \(x = w_1/2\). The special features of the plot may include the symmetry, peaks, or additional points of interest along the distribution. The temperature distribution plot could be created using any graphing software or tools, like Excel, MATLAB, or Python libraries like matplotlib.

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Key Concepts

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

steady-state heat conduction
Imagine a situation where heat flows continuously through a material, but the temperature at every point remains constant over time. That's steady-state heat conduction. It occurs when all incoming heat into a system is balanced with the outgoing heat, leading to no changes in temperature.

In the context of our exercise, the metal strip's temperature does not fluctuate with time, as the system has reached thermal equilibrium. This creates a steady condition where the heat flux introduced by the laser is evenly distributed across the surface of the plastic-backed film.

In mathematical terms, steady-state heat conduction can be represented through an equation. For one-dimensional cases, like the one described, it is simplified to:
  • \[-k \frac{d^{2} T}{d x^{2}} + q^{\prime \prime} = 0\]
This equation tells us that the rate at which heat enters a section is equal to the rate at which it leaves, ensuring constancy in temperature distribution.
thermal resistance
Thermal resistance is like the hurdle to heat flow through a material, much like electrical resistance is to electric current. It's a measure of a material's ability to resist heat flow across it.

In this exercise, the emphasis is on the thermal properties of the metal strip and the negligible resistance of the plastic film. Since the film's thermal resistance is negligible, the major resistance to heat flow occurs within the metal strip itself.

Thermal resistance becomes crucial when calculating how temperature changes across different sections of the strip, knowing:
  • The thicker the material, the greater its thermal resistance.
  • Conversely, a higher thermal conductivity (\(k\) value) results in lower thermal resistance.
For efficient heat conduction, it's ideal to minimize thermal resistance, allowing heat to distribute evenly, as observed in the steady-state condition.
boundary conditions
Boundary conditions are crucial in understanding how heat behaves at the limits of a system, providing insight into how systems interact with their environment.

In this specific problem, the metal strip's edges are at the air temperature, set to be the boundary condition. Additionally, both the upper and lower surfaces of the strip experience convection with air, maintaining the metal pieces' surface temperatures close to the ambient temperature of 25掳C.

Boundary conditions allow us to apply a real-world context; they help solve equations representing heat conduction by providing initial or fixed values.
  • In this exercise, the assumption that the strip edges are at air temperature forms the Neumann boundary condition.
  • This concept is instrumental in determining constants in the temperature distribution formula created in the steady-state heat conduction equations.
By defining what happens at these boundaries, one determines how internal temperatures adjust accordingly throughout the system.
temperature distribution
Temperature distribution describes how temperatures vary at different positions within a body under study, and it is essential for determining heat flow efficiency.

In this exercise, it's the predicted variation of temperature from the strip's center to its edges, illustrating how the heat flux affects the material. The derived formulae help predict these values:
  • The equation \[T(x) = -\frac{q^{\prime \prime}}{2k}x^2 + C_1x + C_2\] represents how heat distributes through the strip.
When peak temperatures arise at specific points, they provide insights into heat conduction efficiency and material stability.
  • Here, knowing temperatures at the center and edges helps adjust parameters in practical applications, like ensuring the plastic film undergoes optimal laser bonding without damage.
Ultimately, understanding temperature distribution allows engineers to design systems that manage and utilize heat effectively, ensuring safety and performance.

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Most popular questions from this chapter

A firefighter's protective clothing, referred to as a turnout coat, is typically constructed as an ensemble of three layers separated by air gaps, as shown schematically. The air gaps between the layers are \(1 \mathrm{~mm}\) thick, and heat is transferred by conduction and radiation exchange through the stagnant air. The linearized radiation coefficient for a gap may be approximated as, \(h_{\text {rad }}=\sigma\left(T_{1}+T_{2}\right)\left(T_{1}^{2}+T_{2}^{2}\right) \approx 4 \sigma T_{\text {avg }}^{3}\), where \(T_{\text {avg }}\) represents the average temperature of the surfaces comprising the gap, and the radiation flux across the gap may be expressed as \(q_{\text {rad }}^{\prime \prime}=h_{\text {rad }}\left(T_{1}-T_{2}\right)\). (a) Represent the turnout coat by a thermal circuit, labeling all the thermal resistances. Calculate and tabulate the thermal resistances per unit area \(\left(\mathrm{m}^{2}\right.\). \(\mathrm{K} / \mathrm{W}\) ) for each of the layers, as well as for the conduction and radiation processes in the gaps. Assume that a value of \(T_{\mathrm{avg}}=470 \mathrm{~K}\) may be used to approximate the radiation resistance of both gaps. Comment on the relative magnitudes of the resistances. (b) For a pre-ash-over fire environment in which firefighters often work, the typical radiant heat flux on the fire-side of the turnout coat is \(0.25 \mathrm{~W} / \mathrm{cm}^{2}\). What is the outer surface temperature of the turnout coat if the inner surface temperature is \(66^{\circ} \mathrm{C}\), a condition that would result in burn injury?

Consider a composite wall that includes an 8-mm-thick hardwood siding, 40 -mm by 130 -mm hardwood studs on \(0.65-\mathrm{m}\) centers with glass fiber insulation (paper faced, \(28 \mathrm{~kg} / \mathrm{m}^{3}\) ), and a 12 -mm layer of gypsum (vermiculite) wall board. What is the thermal resistance associated with a wall that is \(2.5 \mathrm{~m}\) high by \(6.5 \mathrm{~m}\) wide (having 10 studs, each \(2.5 \mathrm{~m}\) high)? Assume surfaces normal to the \(x\)-direction are isothermal.

Rows of the thermoelectric modules of Example \(3.13\) are attached to the flat absorber plate of Problem 3.108. The rows of modules are separated by \(L_{\text {sep }}=0.5 \mathrm{~m}\) and the backs of the modules are cooled by water at a temperature of \(T_{w}=40^{\circ} \mathrm{C}\), with \(h=45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the electric power produced by one row of thermoelectric modules connected in series electrically with a load resistance of \(60 \Omega\). Calculate the heat transfer rate to the flowing water. Assume rows of 20 immediately adjacent modules, with the lengths of both the module rows and water tubing to be \(L_{\text {row }}=20 W\) where \(W=54 \mathrm{~mm}\) is the module dimension taken from Example 3.13. Neglect thermal contact resistances and the temperature drop across the tube wall, and assume that the high thermal conductivity tube wall creates a uniform temperature around the tube perimeter. Because of the thermal resistance provided by the thermoelectric modules, it is no longer appropriate to assume that the temperature of the absorber plate directly above a tube is equal to that of the water.

The walls of a refrigerator are typically constructed by sandwiching a layer of insulation between sheet metal panels. Consider a wall made from fiberglass insulation of thermal conductivity \(k_{i}=0.046 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thickness \(L_{i}=50 \mathrm{~mm}\) and steel panels, each of thermal conductivity \(k_{p}=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thickness \(L_{p}=3 \mathrm{~mm}\). If the wall separates refrigerated air at \(T_{\infty, i}=4^{\circ} \mathrm{C}\) from ambient air at \(T_{\infty, o}=25^{\circ} \mathrm{C}\), what is the heat gain per unit surface area? Coefficients associated with natural convection at the inner and outer surfaces may be approximated as \(h_{i}=h_{o}=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Copper tubing is joined to a solar collector plate of thickness \(t\), and the working fluid maintains the temperature of the plate above the tubes at \(T_{o}\). There is a uniform net radiation heat flux \(q_{\text {rad }}^{\prime \prime}\) to the top surface of the plate, while the bottom surface is well insulated. The top surface is also exposed to a fluid at \(T_{\infty}\) that provides for a uniform convection coefficient \(h\). (a) Derive the differential equation that governs the temperature distribution \(T(x)\) in the plate. (b) Obtain a solution to the differential equation for appropriate boundary conditions.
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