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Chapter 3: Problem 25

A particular thermal system involves three objects of fixed shape with conduction resistances of \(R_{1}=1 \mathrm{~K} / \mathrm{W}\), \(R_{2}=2 \mathrm{~K} / \mathrm{W}\) and \(R_{3}=4 \mathrm{~K} / \mathrm{W}\), respectively. An objective is to minimize the total thermal resistance \(R_{\text {tot }}\) associated with a combination of \(R_{1}, R_{2}\), and \(R_{3}\). The chief engineer is willing to invest limited funds to specify an alternative material for just one of the three objects; the alternative material will have a thermal conductivity that is twice its nominal value. Which object (1, 2, or 3 ) should be fabricated of the higher thermal conductivity material to most significantly decrease \(R_{\text {tot }}\) ? Hint: Consider two cases, one for which the three thermal resistances are arranged in series, and the second for which the three resistances are arranged in parallel.

Short Answer

Expert verified
In series, the total thermal resistance is \(R_{\text{tot}} = R_{1} + R_{2} + R_{3} = 7\,\text{K/W}\). The new total thermal resistances after changing the material of each object are: 1. \(R_{\text{tot}}' = R_{1}' + R_{2} + R_{3} = \frac{1}{2} R_{1} + R_{2} + R_{3} = 6\,\text{K/W}\), which is a decrease of \(1\,\text{K/W}\). 2. \(R_{\text{tot}}' = R_{1} + R_{2}' + R_{3} = R_{1} + \frac{1}{2} R_{2} + R_{3} = 5\,\text{K/W}\), which is a decrease of \(2\,\text{K/W}\). 3. \(R_{\text{tot}}' = R_{1} + R_{2} + R_{3}' = R_{1} + R_{2} + \frac{1}{2} R_{3} = 5\,\text{K/W}\), which is a decrease of \(2\,\text{K/W}\). In parallel, the total thermal resistance is obtained using \(\frac{1}{R_{\text{tot}}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} = \frac{7}{4}\,\text{K/W}\), so \(R_{\text{tot}} \approx 0.5714\,\text{K/W}\). The new total thermal resistances after changing the material of each object are: 1. \(R_{\text{tot}}' = \left(\frac{1}{R_{1}'} + \frac{1}{R_{2}} + \frac{1}{R_{3}}\right)^{-1} = \left(\frac{1}{\frac{1}{2} R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}\right)^{-1} \approx 0.4\,\text{K/W}\), which is a decrease of \(\approx 0.1714\,\text{K/W}\). 2. \(R_{\text{tot}}' = \left(\frac{1}{R_{1}} + \frac{1}{R_{2}'} + \frac{1}{R_{3}}\right)^{-1} = \left(\frac{1}{R_{1}} + \frac{1}{\frac{1}{2} R_{2}} + \frac{1}{R_{3}}\right)^{-1} \approx 0.3333\,\text{K/W}\), which is a decrease of \(\approx 0.2381\,\text{K/W}\). 3. \(R_{\text{tot}}' = \left(\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}'}\right)^{-1} = \left(\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{\frac{1}{2} R_{3}}\right)^{-1} \approx 0.3636\,\text{K/W}\), which is a decrease of \(\approx 0.2078\,\text{K/W}\). Comparing the most significant decreases in total thermal resistance for both cases (series and parallel), changing the material of object 2 (with a decrease of \(2\,\text{K/W}\) in series and \(\approx 0.2381\,\text{K/W}\) in parallel) results in the most significant decrease in total thermal resistance, \(R_{\text{tot}}\). Therefore, object 2 should be fabricated with the higher thermal conductivity material to most significantly decrease the total thermal resistance.

Step by step solution

01

Case 1: Thermal resistances in series

When the thermal resistances are in series, the total thermal resistance is given by the sum of the individual thermal resistances: \(R_{\text {tot }} = R_{1} + R_{2} + R_{3}\). Now, let's analyze the effect of changing the material of each object: 1. If we change the material of object 1, we get \(R_{1}' = \frac{1}{2} R_{1}\). The new total thermal resistance will be \(R_{\text{tot}}' = R_{1}' + R_{2} + R_{3}\). 2. If we change the material of object 2, we get \(R_{2}' = \frac{1}{2} R_{2}\). The new total thermal resistance will be \(R_{\text{tot}}' = R_{1} + R_{2}' + R_{3}\). 3. If we change the material of object 3, we get \(R_{3}' = \frac{1}{2} R_{3}\). The new total thermal resistance will be \(R_{\text{tot}}' = R_{1} + R_{2} + R_{3}'\). We will calculate \(R_{\text{tot}}\) and \(R_{\text{tot}}'\) for each case, and determine the decrease in total thermal resistance.
02

Case 2: Thermal resistances in parallel

When the thermal resistances are in parallel, the total thermal resistance is given by the reciprocal sum of the individual thermal resistances: \(\frac{1}{R_{\text {tot }}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}\). Now, let's analyze the effect of changing the material of each object: 1. If we change the material of object 1, we get \(R_{1}' = \frac{1}{2} R_{1}\). The new total thermal resistance will be \( \frac{1}{R_{\text{tot}}'} = \frac{1}{R_{1}'} + \frac{1}{R_{2}} + \frac{1}{R_{3}} \). 2. If we change the material of object 2, we get \(R_{2}' = \frac{1}{2} R_{2}\). The new total thermal resistance will be \( \frac{1}{R_{\text{tot}}'} = \frac{1}{R_{1}} + \frac{1}{R_{2}'} + \frac{1}{R_{3}} \). 3. If we change the material of object 3, we get \(R_{3}' = \frac{1}{2} R_{3}\). The new total thermal resistance will be \( \frac{1}{R_{\text{tot}}'} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}'} \). We will calculate \(R_{\text{tot}}\) and \(R_{\text{tot}}'\) for each case, and determine the decrease in total thermal resistance.
03

Calculations and Conclusion

Calculate the total thermal resistance for each case before and after changing the material, and compare the decreases in thermal resistance to determine the most significant change: 1. In series, \(R_{\text{tot}} = R_{1} + R_{2} + R_{3} = 1+2+4=7\,\text{K/W}\). Calculate the new total thermal resistances when changing the material of each object, and find the most significant decrease. 2. In parallel, calculate \(R_{\text{tot}}\) using \(\frac{1}{R_{\text {tot }}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}\), then compute the new total thermal resistances when changing the material of each object and find the most significant decrease. Compare the most significant decreases in total thermal resistance for both cases (series and parallel) to determine which object should be fabricated with the higher thermal conductivity material to most significantly decrease \(R_{\text {tot }}\).

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Key Concepts

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

Conduction Resistance
Conduction resistance is a crucial concept in thermal systems. It describes how resistive an object is to the flow of heat. Less resistance allows for more efficient heat transfer, while higher resistance prevents heat from moving freely. This resistance is analogous to electrical resistance in a circuit. The unit used is Kelvin per Watt (K/W), representing the thermal resistance in terms of temperature difference per unit of power transferred. The equation for thermal conduction through a material is given by Fourier's Law:\[ q = -kA \frac{dT}{dx} \]Where:- \( q \) is the heat transfer rate,- \( k \) is the thermal conductivity,- \( A \) is the cross-sectional area perpendicular to heat flow,- \( \frac{dT}{dx} \) is the temperature gradient.Understanding conduction resistance is key to optimizing thermal systems, like our given problem where different materials with different resistances are analyzed.
Thermal Conductivity
Thermal conductivity is a measure of a material's ability to conduct heat. It is often denoted by the symbol \( k \) and typically measured in Watts per meter Kelvin (W/mK). High thermal conductivity materials, such as metals, allow heat to pass through quickly. Low thermal conductivity materials, like insulation, resist heat flow.In the context of our original exercise, improving the thermal conductivity of one of the objects means selecting a material that allows heat to travel more efficiently through it. By doubling the thermal conductivity, we effectively halve the conduction resistance for that object. The improved material's reduced resistance is calculated using:\[ R' i = \frac{R i}{2} \]Where:- \( R' i \) is the new resistance of the modified object,- \( R i \) is the initial resistance.The goal here is to select the object that, when altered, reduces the total thermal resistance the most, facilitating better overall heat transfer in the system.
Series and Parallel Circuits
Series and parallel circuits are ways of organizing components that impact how the total resistance is calculated. Similarly, thermal systems can have resistances arranged in these configurations.When resistances are in **series**, they simply add up:- Total resistance, \( R_{\text{tot}} = R_{1} + R_{2} + R_{3} \).- Each resistance directly impacts the total, and changing one affects the sum.Conversely, for resistances in **parallel**, the total resistance is given by:\[ \frac{1}{R_{\text{tot}}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} \]- In parallel, reducing one resistance significantly affects the total because the inverses sum.In our problem, understanding these arrangements is crucial. When thermal resistances are in series, the best choice would be the component contributing the highest resistance. For parallel setups, focusing on any of the components significantly lowers total resistance, often demanding more strategic analysis.

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Most popular questions from this chapter

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?

Consider two long, slender rods of the same diameter but different materials. One end of each rod is attached to a base surface maintained at \(100^{\circ} \mathrm{C}\), while the surfaces of the rods are exposed to ambient air at \(20^{\circ} \mathrm{C}\). By traversing the length of each rod with a thermocouple, it was observed that the temperatures of the rods were equal at the positions \(x_{\mathrm{A}}=0.15 \mathrm{~m}\) and \(x_{\mathrm{B}}=0.075 \mathrm{~m}\), where \(x\) is measured from the base surface. If the thermal conductivity of rod \(\mathrm{A}\) is known to be \(k_{\mathrm{A}}=70 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), determine the value of \(k_{\mathrm{B}}\) for rod B.

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.

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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