/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 142 Finned passages are frequently f... [FREE SOLUTION] | 91Ó°ÊÓ

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Finned passages are frequently formed between parallel plates to enhance convection heat transfer in compact heat exchanger cores. An important application is in electronic equipment cooling, where one or more air-cooled stacks are placed between heat-dissipating electrical components. Consider a single stack of rectangular fins of length \(L\) and thickness \(t\), with convection conditions corresponding to \(h\) and \(T_{\infty}\). (a) Obtain expressions for the fin heat transfer rates, \(q_{f, o}\) and \(q_{f, L}\), in terms of the base temperatures, \(T_{o}\) and \(T_{L}\). (b) In a specific application, a stack that is \(200 \mathrm{~mm}\) wide and \(100 \mathrm{~mm}\) deep contains 50 fins, each of length \(L=12 \mathrm{~mm}\). The entire stack is made from aluminum, which is everywhere \(1.0 \mathrm{~mm}\) thick. If temperature limitations associated with electrical components joined to opposite plates dictate maximum allowable plate temperatures of \(T_{o}=400 \mathrm{~K}\) and \(T_{L}=350 \mathrm{~K}\), what are the corresponding maximum power dissipations if \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=300 \mathrm{~K} ?\)

Short Answer

Expert verified
(a) Expressions for the fin heat transfer rates in terms of base temperatures are: \(q_{f, o} = (T_o - T_{\infty}) / R_{conv}\) \(q_{f, L} = (T_L - T_{\infty}) / R_{conv}\) (b) In the specific application, the maximum power dissipations are: \(P_{max, o} = 3355.44 W\) \(P_{max, L} = 1677.72 W\)

Step by step solution

01

Obtain expressions for heat transfer rates, \(q_{f, o}\) and \(q_{f, L}\)

To obtain the expressions for heat transfer rates, consider a fin and analyze the heat transfer via conduction through the fin, and convection from the fin to the fluid. We can start by analyzing the heat transfer between \(T_{o}\), \(T_{L}\) and \(T_{\infty}\) using the thermal resistance network: conduction resistance along the fin length, and convection resistance on the surface of the fin. (a) First, the thermal resistance network can be written as: \(R_{cond} = \frac{L}{(hA_c)}\) for conduction resistance \(R_{conv} = \frac{1}{(hA_s)}\) for convection resistance where, \(A_c\) is the cross-sectional area, and \(A_s\) is the surface area of the fin. The total heat transfer rate, \(q_f = q_{f, o} + q_{f, L}\) can be expressed as: \(q_f = (T_o - T_{\infty}) / (R_{cond} + R_{conv})\) We can also write individual heat transfer rates for each temperature: \(q_{f, o} = (T_o - T_{\infty}) / R_{conv}\) \(q_{f, L} = (T_L - T_{\infty}) / R_{conv}\)
02

Calculate the maximum power dissipations

For part (b), we are given the specific application parameters. We can use these values and the expressions obtained in Step 1 to calculate the maximum power dissipations for \(T_{o}\) and \(T_{L}\). Given values: Width (w) = 200 mm Depth (d) = 100 mm Number of fins (n) = 50 Fin length (L) = 12 mm Fin thickness (t) = 1 mm Temperatures: \(T_{o} = 400\) K, \(T_{L} = 350\) K, \(T_{\infty} = 300\) K Convective heat transfer coefficient (h) = 150 W/(m²·K) Calculate the areas: \(A_c = wt = (200 mm)(1 mm) = 200 mm^2\) \(A_s = 2(wh) + 2(wd) = 2(200 mm)(12 mm) + 2(200 mm)(100 mm) = 44,800 mm^2\) Convert areas to m²: \(A_c = 0.02 m^2\) \(A_s = 0.00448 m^2\) Calculate thermal resistances: \(R_{cond} = \frac{L}{(hA_c)} = \frac{0.012 m}{(150 \frac{W}{m^2 K})(0.02 m^2)} = 0.004 K/W\) \(R_{conv} = \frac{1}{(hA_s)} = \frac{1}{(150 \frac{W}{m^2 K})(0.00448 m^2)} = 1.48819 K/W\) Calculate heat transfer rates \(q_{f, o}\) and \(q_{f, L}\): \(q_{f, o} = (T_o - T_{\infty}) / R_{conv} = (400 K - 300 K) / 1.48819 K/W = 67.1088 W\) \(q_{f, L} = (T_L - T_{\infty}) / R_{conv} = (350 K - 300 K) / 1.48819 K/W = 33.5544 W\) Finally, the maximum power dissipations are: \(P_{max, o} = nq_{f, o} = 50 \times 67.1088 W = 3355.44 W\) \(P_{max, L} = nq_{f, L} = 50 \times 33.5544 W = 1677.72 W\)

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Key Concepts

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

Fin Heat Transfer Rate
When discussing heat transfer in fins, we focus on the efficiency of these fins in transferring thermal energy from a surface to the environment or vice versa. A fin's heat transfer rate, often denoted as \(q_f\), can be seen as the measurement of how much thermal energy can be transported per unit time.

In the context of finned passages in heat exchangers, heat must be transferred through the fin material by conduction, and then dissipated into the surrounding fluid, typically air, via convection. The heat transfer rate at the base of the fin, \(q_{f, o}\), and at the end of the fin, \(q_{f, L}\), are critical points to consider, as they represent the areas of highest and lowest temperatures, respectively.

To optimize heat transfer, the fin material and dimensions are chosen to be highly conductive and to increase surface area without significantly raising resistance. In conduction, resistance is proportional to material length and inversely proportional to cross-sectional area and conductivity. With fins, the objective is to maximize the surface area for convection—the mechanism through which the majority of the heat is dissipated—while carefully configuring the fin's dimensions to balance between conductive and convective resistances.
Thermal Resistance Network
When analyzing heat exchangers and finned surfaces, the concept of a thermal resistance network becomes an invaluable tool. This network is similar to an electrical resistance network, where each resistor represents a 'hindrance' or 'resistance' to the flow of heat rather than the flow of electrical current.

The thermal resistance network allows us to model and quantify two primary types of resistance encountered in fin heat transfer: conduction resistance along the fin and convection resistance at the surface of the fin. Conduction resistance is determined by the material's thermal conductivity and the dimensions of the fin. On the other hand, convective resistance is dependent on the surface area available for the convection process and the convective heat transfer coefficient, \(h\).

In the given exercise, the conduction resistance is determined by the fin's length and cross-sectional area while ignoring the thermal conductivity (since it cancels out when calculating the heat transfer rate). The convective resistance, however, directly involves the convective heat transfer coefficient and the surface area of the fin. By summing these resistances, we can devise a clear pathway to visualize and calculate the heat transfer through finned surfaces.
Convective Heat Transfer Coefficient
The convective heat transfer coefficient, \(h\), is a measure of the convective heat transfer capability of a fluid past a surface. It is a pivotal factor in understanding and predicting how efficiently heat is transferred between a solid surface and the fluid moving over it.

This coefficient's units are power per unit area per unit temperature difference, typically expressed as \(W/m^2\cdot K\). A larger \(h\) signifies a more effective transfer of thermal energy, resulting from the fluid's properties, velocity, and the nature of the flow—whether it is turbulent or laminar.

In heat exchanger design and optimization, we often seek to maximize \(h\) to enhance the rate of convective heat transfer to or from the fins. Various methods such as increasing the fluid velocity, altering the fin's geometry to induce turbulence, or using fins with greater surface roughness can increase \(h\) and, as a result, the fin heat transfer rate. In our specific exercise, the convective heat transfer coefficient has been provided, and it serves as an integral part in the calculation of thermal resistances and the maximum power dissipation allowed by the fins, given the temperature limitations of the associated electrical components.

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Most popular questions from this chapter

Consider uniform thermal energy generation inside a one-dimensional plane wall of thickness \(L\) with one surface held at \(T_{s, 1}\) and the other surface insulated. (a) Find an expression for the conduction heat flux to the cold surface and the temperature of the hot surface \(T_{s, 2}\), expressing your results in terms of \(k, \dot{q}, L\), and \(T_{s, 1}\). (b) Compare the heat flux found in part (a) with the heat flux associated with a plane wall without energy generation whose surface temperatures are \(T_{s, 1}\) and \(T_{s, 2}\).

The wall of a spherical tank of \(1-m\) diameter contains an exothermic chemical reaction and is at \(200^{\circ} \mathrm{C}\) when the ambient air temperature is \(25^{\circ} \mathrm{C}\). What thickness of urethane foam is required to reduce the exterior temperature to \(40^{\circ} \mathrm{C}\), assuming the convection coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for both situations? What is the percentage reduction in heat rate achieved by using the insulation?

A spherical tank of \(3-\mathrm{m}\) diameter contains a liquifiedpetroleum gas at \(-60^{\circ} \mathrm{C}\). Insulation with a thermal conductivity of \(0.06 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thickness \(250 \mathrm{~mm}\) is applied to the tank to reduce the heat gain. (a) Determine the radial position in the insulation layer at which the temperature is \(0^{\circ} \mathrm{C}\) when the ambient air temperature is \(20^{\circ} \mathrm{C}\) and the convection coefficient on the outer surface is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) If the insulation is pervious to moisture from the atmospheric air, what conclusions can you reach about the formation of ice in the insulation? What effect will ice formation have on heat gain to the LP gas? How could this situation be avoided?

A composite wall separates combustion gases at \(2600^{\circ} \mathrm{C}\) from a liquid coolant at \(100^{\circ} \mathrm{C}\), with gas- and liquid-side convection coefficients of 50 and 1000 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The wall is composed of a \(10-\mathrm{mm}\)-thick layer of beryllium oxide on the gas side and a 20 -mm-thick slab of stainless steel (AISI 304) on the liquid side. The contact resistance between the oxide and the steel is \(0.05 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). What is the heat loss per unit surface area of the composite? Sketch the temperature distribution from the gas to the liquid.

The energy transferred from the anterior chamber of the eye through the cornea varies considerably depending on whether a contact lens is worn. Treat the eye as a spherical system and assume the system to be at steady state. The convection coefficient \(h_{o}\) is unchanged with and without the contact lens in place. The cornea and the lens cover one-third of the spherical surface area. Values of the parameters representing this situation are as follows: \(\begin{array}{ll}r_{1}=10.2 \mathrm{~mm} & r_{2}=12.7 \mathrm{~mm} \\\ r_{3}=16.5 \mathrm{~mm} & T_{\infty, o}=21^{\circ} \mathrm{C} \\ T_{\infty \infty, i}=37^{\circ} \mathrm{C} & k_{2}=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ k_{1}=0.35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} & h_{o}=6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \\ h_{i}=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} & \end{array}\) (a) Construct the thermal circuits, labeling all potentials and flows for the systems excluding the contact lens and including the contact lens. Write resistance elements in terms of appropriate parameters. (b) Determine the heat loss from the anterior chamber with and without the contact lens in place. (c) Discuss the implication of your results.

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