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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

A rod of diameter \(D=25 \mathrm{~mm}\) and thermal conductivity \(k=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) protrudes normally from a furnace wall that is at \(T_{w}=200^{\circ} \mathrm{C}\) and is covered by insulation of thickness \(L_{\text {ins }}=200 \mathrm{~mm}\). The rod is welded to the furnace wall and is used as a hanger for supporting instrumentation cables. To avoid damaging the cables, the temperature of the rod at its exposed surface, \(T_{o}\), must be maintained below a specified operating limit of \(T_{\max }=100^{\circ} \mathrm{C}\). The ambient air temperature is \(T_{\infty}=\) \(25^{\circ} \mathrm{C}\), and the convection coefficient is \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Derive an expression for the exposed surface temperature \(T_{o}\) as a function of the prescribed thermal and geometrical parameters. The rod has an exposed length \(L_{o}\), and its tip is well insulated. (b) Will a rod with \(L_{o}=200 \mathrm{~mm}\) meet the specified operating limit? If not, what design parameters would you change? Consider another material, increasing the thickness of the insulation, and increasing the rod length. Also, consider how you might attach the base of the rod to the furnace wall as a means to reduce \(T_{o}\).

Consider the conditions of Problem \(3.149\) but now allow for a tube wall thickness of \(5 \mathrm{~mm}\) (inner and outer diameters of 50 and \(60 \mathrm{~mm}\) ), a fin-to-tube thermal contact resistance of \(10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), and the fact that the water temperature, \(T_{w}=350 \mathrm{~K}\), is known, not the tube surface temperature. The water-side convection coefficient is \(h_{w}=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the rate of heat transfer per unit tube length \((\mathrm{W} / \mathrm{m})\) to the water. What would be the separate effect of each of the following design changes on the heat rate: (i) elimination of the contact resistance; (ii) increasing the number of fins from four to eight; and (iii) changing the tube wall and fin material from copper to AISI 304 stainless steel \((k=20\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\) ?

Consider the flat plate of Problem \(3.112\), but with the heat sinks at different temperatures, \(T(0)=T_{o}\) and \(T(L)=T_{L}\), and with the bottom surface no longer insulated. Convection heat transfer is now allowed to occur between this surface and a fluid at \(T_{\infty}\), with a convection coefficient \(h\). (a) Derive the differential equation that determines the steady-state temperature distribution \(T(x)\) in the plate. (b) Solve the foregoing equation for the temperature distribution, and obtain an expression for the rate of heat transfer from the plate to the heat sinks. (c) For \(q_{o}^{\prime \prime}=20,000 \mathrm{~W} / \mathrm{m}^{2}, T_{o}=100^{\circ} \mathrm{C}, T_{L}=35^{\circ} \mathrm{C}\), \(T_{\infty}=25^{\circ} \mathrm{C}, k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, h=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(L=100 \mathrm{~mm}, t=5 \mathrm{~mm}\), and a plate width of \(W=\) \(30 \mathrm{~mm}\), plot the temperature distribution and determine the sink heat rates, \(q_{x}(0)\) and \(q_{x}(L)\). On the same graph, plot three additional temperature distributions corresponding to changes in the following parameters, with the remaining parameters unchanged: (i) \(q_{o}^{\prime \prime}=30,000 \mathrm{~W} / \mathrm{m}^{2}\), (ii) \(h=200\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and (iii) the value of \(q_{o}^{\prime \prime}\) for which \(q_{x}(0)=0\) when \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A spherical Pyrex glass shell has inside and outside diameters of \(D_{1}=0.1 \mathrm{~m}\) and \(D_{2}=0.2 \mathrm{~m}\), respectively. The inner surface is at \(T_{s, 1}=100^{\circ} \mathrm{C}\) while the outer surface is at \(T_{s, 2}=45^{\circ} \mathrm{C}\). (a) Determine the temperature at the midpoint of the shell thickness, \(T\left(r_{m}=0.075 \mathrm{~m}\right)\). (b) For the same surface temperatures and dimensions as in part (a), show how the midpoint temperature would change if the shell material were aluminum.

Consider an extended surface of rectangular cross section with heat flow in the longitudinal direction. In this problem we seek to determine conditions for which the transverse ( \(y\)-direction) temperature difference within the extended surface is negligible compared to the temperature difference between the surface and the environment, such that the one-dimensional analysis of Section 3.6.1 is valid. (a) Assume that the transverse temperature distribution is parabolic and of the form $$ \frac{T(y)-T_{o}(x)}{T_{s}(x)-T_{o}(x)}=\left(\frac{y}{t}\right)^{2} $$ where \(T_{s}(x)\) is the surface temperature and \(T_{o}(x)\) is the centerline temperature at any \(x\)-location. Using Fourier's law, write an expression for the conduction heat flux at the surface, \(q_{y}^{\prime \prime}(t)\), in terms of \(T_{s}\) and \(T_{a^{+}}\) (b) Write an expression for the convection heat flux at the surface for the \(x\)-location. Equating the two expressions for the heat flux by conduction and convection, identify the parameter that determines the ratio \(\left(T_{o}-T_{s}\right) /\left(T_{s}-T_{\infty}\right)\). (c) From the foregoing analysis, develop a criterion for establishing the validity of the onedimensional assumption used to model an extended surface.
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