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Chapter 3: Problem 142

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/(m虏路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 m虏: \(A_c = 0.02 m^2\) \(A_s = 0.00448 m^2\) 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鈥攖he mechanism through which the majority of the heat is dissipated鈥攚hile 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鈥攚hether 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

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 \(\dot{q}\) ?

Rows of the thermoelectric modules of Example \(3.13\) are attached to the flat absorber plate of Problem 3.108. The rows of modules are separated by \(L_{\text {sep }}=0.5 \mathrm{~m}\) and the backs of the modules are cooled by water at a temperature of \(T_{w}=40^{\circ} \mathrm{C}\), with \(h=45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the electric power produced by one row of thermoelectric modules connected in series electrically with a load resistance of \(60 \Omega\). Calculate the heat transfer rate to the flowing water. Assume rows of 20 immediately adjacent modules, with the lengths of both the module rows and water tubing to be \(L_{\text {row }}=20 W\) where \(W=54 \mathrm{~mm}\) is the module dimension taken from Example 3.13. Neglect thermal contact resistances and the temperature drop across the tube wall, and assume that the high thermal conductivity tube wall creates a uniform temperature around the tube perimeter. Because of the thermal resistance provided by the thermoelectric modules, it is no longer appropriate to assume that the temperature of the absorber plate directly above a tube is equal to that of the water.

A thin metallic wire of thermal conductivity \(k\), diameter \(D\), and length \(2 L\) is annealed by passing an electrical current through the wire to induce a uniform volumetric heat generation \(\dot{q}\). The ambient air around the wire is at a temperature \(T_{\infty}\), while the ends of the wire at \(x=\pm L\) are also maintained at \(T_{\infty}\). Heat transfer from the wire to the air is characterized by the convection coefficient \(h\). Obtain expressions for the following: (a) The steady-state temperature distribution \(T(x)\) along the wire, (b) The maximum wire temperature. (c) The average wire temperature.

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 \(T_{o}\).

A device used to measure the surface temperature of an object to within a spatial resolution of approximately \(50 \mathrm{~nm}\) is shown in the schematic. It consists of an extremely sharp-tipped stylus and an extremely small cantilever that is scanned across the surface. The probe tip is of circular cross section and is fabricated of polycrystalline silicon dioxide. The ambient temperature is measured at the pivoted end of the cantilever as \(T_{\infty}=\) \(25^{\circ} \mathrm{C}\), and the device is equipped with a sensor to measure the temperature at the upper end of the sharp tip, \(T_{\text {sen. }}\). The thermal resistance between the sensing probe and the pivoted end is \(R_{t}=5 \times 10^{6} \mathrm{~K} / \mathrm{W}\). (a) Determine the thermal resistance between the surface temperature and the sensing temperature. (b) If the sensing temperature is \(T_{\text {sen }}=28.5^{\circ} \mathrm{C}\), determine the surface temperature. Hint: Although nanoscale heat transfer effects may be important, assume that the conduction occurring in the air adjacent to the probe tip can be described by Fourier's law and the thermal conductivity found in Table A. \(4 .\)
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