/*! 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

The temperature of a flowing gas is to be measured with a thermocouple junction and wire stretched between two legs of a sting, a wind tunnel test fixture. The junction is formed by butt-welding two wires of different material, as shown in the schematic. For wires of diameter \(D=125 \mu \mathrm{m}\) and a convection coefficient of \(h=700 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the minimum separation distance between the two legs of the sting, \(L=L_{1}+L_{2}\), to ensure that the sting temperature does not influence the junction temperature and, in turn, invalidate the gas temperature measurement. Consider two different types of thermocouple junctions consisting of (i) copper and constantan wires and (ii) chromel and alumel wires. Evaluate the thermal conductivity of copper and constantan at \(T=300 \mathrm{~K}\). Use \(k_{\mathrm{Ch}}=19 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{\mathrm{Al}}=29 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) for the thermal conductivities of the chromel and alumel wires, respectively.

An electrical current of 700 A flows through a stainless steel cable having a diameter of \(5 \mathrm{~mm}\) and an electrical resistance of \(6 \times 10^{-4} \mathrm{\Omega} / \mathrm{m}\) (i.e., per meter of cable length). The cable is in an environment having a temperature of \(30^{\circ} \mathrm{C}\), and the total coefficient associated with convection and radiation between the cable and the environment is approximately \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If the cable is bare, what is its surface temperature? (b) If a very thin coating of electrical insulation is applied to the cable, with a contact resistance of \(0.02 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), what are the insulation and cable surface temperatures? (c) There is some concern about the ability of the insulation to withstand elevated temperatures. What thickness of this insulation \((k=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) will yield the lowest value of the maximum insulation temperature? What is the value of the maximum temperature when this thickness is used?

A cylindrical shell of inner and outer radii, \(r_{i}\) and \(r_{o}\), respectively, is filled with a heat-generating material that provides a uniform volumetric generation rate \(\left(\mathrm{W} / \mathrm{m}^{3}\right)\) of \(\dot{q}\). The inner surface is insulated, while the outer surface of the shell is exposed to a fluid at \(T_{\infty}\) and a convection coefficient \(h\). (a) Obtain an expression for the steady-state temperature distribution \(T(r)\) in the shell, expressing your result in terms of \(r_{i}, r_{o}, \dot{q}, h, T_{\infty}\), and the thermal conductivity \(k\) of the shell material. (b) Determine an expression for the heat rate, \(q^{\prime}\left(r_{o}\right)\), at the outer radius of the shell in terms of \(\dot{q}\) and shell dimensions.

A new building to be located in a cold climate is being designed with a basement that has an \(L=200\)-mm-thick wall. Inner and outer basement wall temperatures are \(T_{i}=20^{\circ} \mathrm{C}\) and \(T_{o}=0^{\circ} \mathrm{C}\), respectively. The architect can specify the wall material to be either aerated concrete block with \(k_{\mathrm{ac}}=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), or stone mix concrete. To reduce the conduction heat flux through the stone mix wall to a level equivalent to that of the aerated concrete wall, what thickness of extruded polystyrene sheet must be applied onto the inner surface of the stone mix con-crete wall? Floor dimensions of the basement are \(20 \mathrm{~m} \times 30 \mathrm{~m}\), and the expected rental rate is \(\$ 50 / \mathrm{m}^{2} /\) month. What is the yearly cost, in terms of lost rental income, if the stone mix concrete wall with polystyrene insulation is specified?

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.

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