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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

Consider the configuration of Example \(3.8\), where uniform volumetric heating within a stainless steel tube is induced by an electric current and heat is transferred by convection to air flowing through the tube. The tube wall has inner and outer radii of \(r_{1}=25 \mathrm{~mm}\) and \(r_{2}=\) \(35 \mathrm{~mm}\), a thermal conductivity of \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), an electrical resistivity of \(\rho_{e}=0.7 \times 10^{-6} \Omega \cdot \mathrm{m}\), and a maximum allowable operating temperature of \(1400 \mathrm{~K}\). (a) Assuming the outer tube surface to be perfectly insulated and the airflow to be characterized by a temperature and convection coefficient of \(T_{\infty, 1}=\) \(400 \mathrm{~K}\) and \(h_{1}=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the maximum allowable electric current \(I\). (b) Compute and plot the radial temperature distribution in the tube wall for the electric current of part (a) and three values of \(h_{1}\left(100,500\right.\), and \(\left.1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\). For each value of \(h_{1}\), determine the rate of heat transfer to the air per unit length of tube. (c) In practice, even the best of insulating materials would be unable to maintain adiabatic conditions at the outer tube surface. Consider use of a refractory insulating material of thermal conductivity \(k=1.0\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) and neglect radiation exchange at its outer surface. For \(h_{1}=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the maximum allowable current determined in part (a), compute and plot the temperature distribution in the composite wall for two values of the insulation thickness \((\delta=25\) and \(50 \mathrm{~mm})\). The outer surface of the insulation is exposed to room air for which \(T_{\infty, 2}=300 \mathrm{~K}\) and \(h_{2}=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). For each insulation thickness, determine the rate of heat transfer per unit tube length to the inner airflow and the ambient air.

A nanolaminated material is fabricated with an atomic layer deposition process, resulting in a series of stacked, alternating layers of tungsten and aluminum oxide, each layer being \(\delta=0.5 \mathrm{~nm}\) thick. Each tungsten-aluminum oxide interface is associated with a thermal resistance of \(R_{t, i}^{\prime \prime}=3.85 \times 10^{-9} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The theoretical values of the thermal conductivities of the thin aluminum oxide and tungsten layers are \(k_{\mathrm{A}}=1.65 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{\mathrm{T}}=6.10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively. The properties are evaluated at \(T=300 \mathrm{~K}\). (a) Determine the effective thermal conductivity of the nanolaminated material. Compare the value of the effective thermal conductivity to the bulk thermal conductivities of aluminum oxide and tungsten, given in Tables A.1 and A.2. (b) Determine the effective thermal conductivity of the nanolaminated material assuming that the thermal conductivities of the tungsten and aluminum oxide layers are equal to their bulk values.

A probe of overall length \(L=200 \mathrm{~mm}\) and diameter \(D=\) \(12.5 \mathrm{~mm}\) is inserted through a duct wall such that a portion of its length, referred to as the immersion length \(L_{i}\), is in contact with the water stream whose temperature, \(T_{\infty, i}\) is to be determined. The convection coefficients over the immersion and ambient-exposed lengths are \(h_{i}=1100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{o}=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. The probe has a thermal conductivity of \(177 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and is in poor thermal contact with the duct wall. (a) Derive an expression for evaluating the measurement error, \(\Delta T_{\mathrm{err}}=T_{\text {tip }}-T_{\infty, i}\), which is the difference between the tip temperature, \(T_{\text {lip }}\), and the water temperature, \(T_{\infty, i \cdot}\) Hint: Define a coordinate system with the origin at the duct wall and treat the probe as two fins extending inward and outward from the duct, but having the same base temperature. Use Case A results from Table 3.4. (b) With the water and ambient air temperatures at 80 and \(20^{\circ} \mathrm{C}\), respectively, calculate the measurement error, \(\Delta T_{\mathrm{er}}\), as a function of immersion length for the conditions \(L_{i} / L=0.225,0.425\), and \(0.625\). (c) Compute and plot the effects of probe thermal conductivity and water velocity \(\left(h_{i}\right)\) on the measurement error.

Circular copper rods of diameter \(D=1 \mathrm{~mm}\) and length \(L=25 \mathrm{~mm}\) are used to enhance heat transfer from a surface that is maintained at \(T_{s, 1}=100^{\circ} \mathrm{C}\). One end of the rod is attached to this surface (at \(x=0\) ), while the other end \((x=25 \mathrm{~mm})\) is joined to a second surface, which is maintained at \(T_{s, 2}=0^{\circ} \mathrm{C}\). Air flowing between the surfaces (and over the rods) is also at a temperature of \(T_{\infty}=0^{\circ} \mathrm{C}\), and a convection coefficient of \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained. (a) What is the rate of heat transfer by convection from a single copper rod to the air? (b) What is the total rate of heat transfer from a \(1 \mathrm{~m} \times 1 \mathrm{~m}\) section of the surface at \(100^{\circ} \mathrm{C}\), if a bundle of the rods is installed on 4 -mm centers?

A radioactive material of thermal conductivity \(k\) is cast as a solid sphere of radius \(r_{o}\) and placed in a liquid bath for which the temperature \(T_{\infty}\) and convection coefficient \(h\) are known. Heat is uniformly generated within the solid at a volumetric rate of \(\dot{q}\). Obtain the steadystate radial temperature distribution in the solid, expressing your result in terms of \(r_{o}, \dot{q}, k, h\), and \(T_{\infty}\).
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