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Chapter 3: Problem 174

Determine the parallel plate separation distance \(L\), above which the thermal resistance associated with molecule-surface collisions \(R_{t, m-s}\) is less than \(1 \%\) of the resistance associated with molecule-molecule collisions, \(R_{t, m-m}\) for (i) air between steel plates with \(\alpha_{t}=0.92\) and (ii) helium between clean aluminum plates with \(\alpha_{t}=0.02\). The gases are at atmospheric pressure, and the temperature is \(T=300 \mathrm{~K}\).

Short Answer

Expert verified
The minimum plate separation distance for air between steel plates with \(\alpha_{t}=0.92\) is approximately \(L \geq 23.6 \,mm\), and for helium between clean aluminum plates with \(\alpha_{t}=0.02\) is approximately \(L \geq 2.88 \,mm\).

Step by step solution

01

Case (i): Air between steel plates with \(\alpha_{t}=0.92\)

First, we need to find the thermal resistance associated with molecule-surface collisions, \(R_{t, m-s}\). For this, we can use the following equation: \(R_{t, m-s} = \frac{1}{\alpha_{t}} \frac{L}{k}\) Where \(L\) is the parallel plate separation distance and \(k\) is the thermal conductivity of the gas. For air, the thermal conductivity at room temperature is approximately \(k = 0.0257\frac{W}{mK}\). Now, we need to find the thermal resistance associated with molecule-molecule collisions, \(R_{t, m-m}\). This can be obtained using the equation: \(R_{t, m-m} = \frac{L}{k}\) Now we can write the inequality we want to solve for \(L\): \(R_{t, m-s} = \frac{1}{0.92} \frac{L}{0.0257} \leq 0.01 \frac{L}{0.0257}\) Solving the inequality for \(L\), we get: \(L \geq 0.0257 \times 0.92\) Calculating the value of \(L\): \(L \geq 0.0236\, meters\)
02

Case (ii): Helium between clean aluminum plates with \(\alpha_{t}=0.02\)

Similar to the previous case, we need to find the thermal resistance associated with molecule-surface collisions, \(R_{t, m-s}\), and molecule-molecule collisions, \(R_{t, m-m}\). For helium, the thermal conductivity at room temperature is approximately \(k = 0.144 \frac{W}{mK}\). Now we can write the inequality we want to solve for \(L\): \(R_{t, m-s} = \frac{1}{0.02} \frac{L}{0.144} \leq 0.01 \frac{L}{0.144}\) Solving the inequality for \(L\), we get: \(L \geq 0.144 \times 0.02\) Calculating the value of \(L\): \(L \geq 0.00288\, meters\) Thus, the minimum plate separation distance for case (i) is approximately 23.6 mm, and for case (ii) is approximately 2.88 mm.

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Key Concepts

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

Molecule-Surface Collisions
When molecules collide with surfaces, such as the walls of a container or, in this exercise, the surface of steel or aluminum plates, they transfer energy. This energy transfer happens through direct interaction between the molecule and the surface. The efficiency of this transfer depends on the thermal accommodation coefficient \(\alpha_{t}\), which indicates how readily the molecules transfer their kinetic energy to the surface.
For air and steel plates, the value of \(\alpha_{t} = 0.92\), suggesting a high efficiency in energy transfer between molecules and the surface. In contrast, for helium between aluminum plates, \(\alpha_{t} = 0.02\), reflecting much lower efficiency. This disparity shows that not all gases behave the same when interacting with surfaces, which is a crucial nuance to understand in thermal dynamics.

Understanding how molecule-surface collisions contribute to thermal resistance helps us grasp why different materials and gases have varying thermal conductive properties. The less energy molecules lose to the surface, the lower the thermal resistance, making some materials ideal for certain applications, like insulation or heat dissipation.
Molecule-Molecule Collisions
Molecule-molecule collisions are crucial for understanding how gases transfer heat. When gas molecules collide, they transfer energy among themselves. This energy exchange is what primarily contributes to thermal conductivity in gases. Unlike molecule-surface collisions, molecule-molecule collisions do not depend on a surface but only on the properties of the gas itself.

The thermal conductivity \(k\) of a gas is a measure of its ability to transfer heat through molecule-molecule collisions. It varies based on the type of gas and its temperature. For example, in this exercise, the thermal conductivity of air is \(k = 0.0257 \frac{W}{mK}\), while for helium, it is \(k = 0.144 \frac{W}{mK}\). These values indicate that helium conducts heat much more effectively than air due to more frequent or energetic collisions among its molecules.

When analyzing thermal systems, comparing the thermal resistances from both molecule-surface and molecule-molecule collisions can tell us which mechanism dominates the heat transfer. This understanding helps in designing systems that can either enhance or minimize heat transfer by choosing the appropriate materials and gases.
Parallel Plate Separation
The distance between two parallel plates, referred to as the plate separation \(L\), plays a significant role in the thermal resistance of both molecule-surface and molecule-molecule collisions. In this context, thermal resistance is the opposition to the flow of heat, depending on the properties of the medium between the plates and the distance over which heat must travel.
  • For molecule-surface collisions, increasing \(L\) reduces the frequency of these collisions, thus lowering the thermal resistance associated with this pathway.

  • Meanwhile, the resistance due to molecule-molecule collisions increases with \(L\) because molecules have to travel further, which means more energy is required to maintain the same heat transfer rate.
In the exercise scenario, the necessary goal was to find the minimum separation where molecule-surface collision resistance becomes less than 1% of molecule-molecule collision resistance. The calculations provided an essential insight that as \(L\) increases, the effects of direct molecule-surface interaction diminish, favoring molecule-molecule interactions.
This demonstrates a key design consideration in thermal management: knowing the critical separation distance can significantly improve heat transfer efficiency in engineered systems.

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Most popular questions from this chapter

A composite cylindrical wall is composed of two materials of thermal conductivity \(k_{\mathrm{A}}\) and \(k_{\mathrm{B}}\), which are separated by a very thin, electric resistance heater for which interfacial contact resistances are negligible. Liquid pumped through the tube is at a temperature \(T_{\infty, i}\) and provides a convection coefficient \(h_{i}\) at the inner surface of the composite. The outer surface is exposed to ambient air, which is at \(T_{\infty, o}\) and provides a convection coefficient of \(h_{o^{*}}\) Under steady-state conditions, a uniform heat flux of \(q_{h}^{n}\) is dissipated by the heater. (a) Sketch the equivalent thermal circuit of the system and express all resistances in terms of relevant variables. (b) Obtain an expression that may be used to determine the heater temperature, \(T_{h+}\). (c) Obtain an expression for the ratio of heat flows to the outer and inner fluids, \(q_{o}^{\prime} / q_{i}^{\prime}\). How might the variables of the problem be adjusted to minimize this ratio?

As a means of enhancing heat transfer from highperformance logic chips, it is common to attach a heat \(\sin k\) to the chip surface in order to increase the surface area available for convection heat transfer. Because of the ease with which it may be manufactured (by taking orthogonal sawcuts in a block of material), an attractive option is to use a heat sink consisting of an array of square fins of width \(w\) on a side. The spacing between adjoining fins would be determined by the width of the sawblade, with the sum of this spacing and the fin width designated as the fin pitch \(S\). The method by which the heat sink is joined to the chip would determine the interfacial contact resistance, \(R_{t, c^{*}}^{n}\) Consider a square chip of width \(W_{c}=16 \mathrm{~mm}\) and conditions for which cooling is provided by a dielectric liquid with \(T_{\infty}=25^{\circ} \mathrm{C}\) and \(h=1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat \(\operatorname{sink}\) is fabricated from copper \((k=400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and its characteristic dimensions are \(w=0.25 \mathrm{~mm}\), \(S=0.50 \mathrm{~mm}, L_{f}=6 \mathrm{~mm}\), and \(L_{b}=3 \mathrm{~mm}\). The prescribed values of \(w\) and \(S\) represent minima imposed by manufacturing constraints and the need to maintain adequate flow in the passages between fins. (a) If a metallurgical joint provides a contact resistance of \(R_{t, c}^{\prime \prime}=5 \times 10^{-6} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) and the maximum allowable chip temperature is \(85^{\circ} \mathrm{C}\), what is the maximum allowable chip power dissipation \(q_{c} ?\) Assume all of the heat to be transferred through the heat sink. (b) It may be possible to increase the heat dissipation by increasing \(w\), subject to the constraint that \((S-w) \geq 0.25 \mathrm{~mm}\), and/or increasing \(L_{f}\) (subject to manufacturing constraints that \(L_{f} \leq 10 \mathrm{~mm}\) ). Assess the effect of such changes.

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?

A spherical vessel used as a reactor for producing pharmaceuticals has a 10 -mm-thick stainless steel wall \((k=17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and an inner diameter of \(1 \mathrm{~m}\). The exterior surface of the vessel is exposed to ambient air \(\left(T_{\infty}=25^{\circ} \mathrm{C}\right)\) for which a convection coefficient of \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) may be assumed. (a) During steady-state operation, an inner surface temperature of \(50^{\circ} \mathrm{C}\) is maintained by energy generated within the reactor. What is the heat loss from the vessel? (b) If a 20 -mm-thick layer of fiberglass insulation \((k=0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is applied to the exterior of the vessel and the rate of thermal energy generation is unchanged, what is the inner surface temperature of the vessel?

A high-temperature, gas-cooled nuclear reactor consists of a composite cylindrical wall for which a thorium fuel element \((k \approx 57 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is encased in graphite \((k \approx 3\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\) and gaseous helium flows through an annular coolant channel. Consider conditions for which the helium temperature is \(T_{\infty}=600 \mathrm{~K}\) and the convection coefficient at the outer surface of the graphite is \(h=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If thermal energy is uniformly generated in the fuel element at a rate \(\dot{q}=10^{8} \mathrm{~W} / \mathrm{m}^{3}\), what are the temperatures \(T_{1}\) and \(T_{2}\) at the inner and outer surfaces, respectively, of the fuel element? (b) Compute and plot the temperature distribution in the composite wall for selected values of \(\dot{q}\). What is the maximum allowable value of \(\dot{q}\) ?
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