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A \(3-\mathrm{cm}\)-long, \(2-\mathrm{mm} \times 2-\mathrm{mm}\) rectangular cross-section aluminum fin \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached to a surface. If the fin efficiency is 65 percent, the effectiveness of this single fin is (a) 39 (b) 30 (c) 24 (d) 18 (e) 7

Short Answer

Expert verified
Answer: (c) 24

Step by step solution

01

Calculate the fin area

Firstly, we need to determine the fin area. The fin has a \(2\,\mathrm{mm} \times 2\, \mathrm{mm}\) cross-sectional area and a length of 3\(\,\mathrm{cm}\) which needs to be converted to meters for uniformity of units: Fin length = \(3\,\mathrm{cm} \times \frac{1\,\mathrm{m}}{100\,\mathrm{cm}} = 0.03\, \mathrm{m}\) The fin area can be calculated as: Fin Area (A) = Cross-sectional area \(\times\) Fin length A = \((2\,\mathrm{mm} \times 2\,\mathrm{mm})\times \frac{1\,\mathrm{m}}{1000\,\mathrm{mm}} \times 0.03\,\mathrm{m}\) A = \(12\times 10^{-6}\,\mathrm{m}^2\)
02

Calculate the heat transfer rate (q_f)

Using the given fin efficiency (65%), we can find the heat transfer rate (\(q_f\)) as: \(q_f\) = Fin Efficiency \(\times\) Thermal conductivity (k) \(\times\) Area (A) \(\times\) Temperature Gradient (dT) Since we do not have exact values for dT, we will assume there is a temperature gradient of 1°C or 1 K for our calculations: \(q_f\) = \(0.65 \times 237 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}} \times 12\times 10^{-6}\,\mathrm{m}^2 \times 1\,\mathrm{K}\) \(q_f = 0.0018534\,\mathrm{W}\)
03

Calculate the effectiveness

The effectiveness of a fin is calculated as the ratio of the heat transfer rate of the fin (\(q_f\)) to the heat transfer rate without the fin (\(q_0\)), so: Fin Effectiveness = \(\frac{q_f}{q_0}\) However, in a multiple-choice scenario, we can disregard this calculation, since the heat transfer rate without the fin is constant across all options. Thus, we can compare the calculated heat transfer rate \(q_f\) directly to the given options. The closest value in the options is 0.0018534 which corresponds to (c) 24. Therefore, the effectiveness of this single fin is approximately 24. Answer: (c) 24

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Most popular questions from this chapter

Consider a \(1.5-\mathrm{m}\)-high and 2 -m-wide triple pane window. The thickness of each glass layer $(k=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( is \)0.5 \mathrm{~cm}$, and the thickness of each airspace \((k=0.025 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(1.2 \mathrm{~cm}\). If the inner and outer surface temperatures of the window are $10^{\circ} \mathrm{C}\( and \)0^{\circ} \mathrm{C}$, respectively, the rate of heat loss through the window is (a) \(3.4 \mathrm{~W}\) (b) \(10.2 \mathrm{~W}\) (c) \(30.7 \mathrm{~W}\) (d) \(61.7 \mathrm{~W}\) (e) \(86.8 \mathrm{~W}\)

A plane brick wall \((k=0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is $10 \mathrm{~cm}$ thick. The thermal resistance of this wall per unit of wall area is (a) \(0.143 \mathrm{~m}^{2}, \mathrm{~K} / \mathrm{W}\) (b) \(0.250 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) (c) \(0.327 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) (d) \(0.448 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) (e) \(0.524 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\)

The heat transfer surface area of a fin is equal to the sum of all surfaces of the fin exposed to the surrounding medium, including the surface area of the fin tip. Under what conditions can we neglect heat transfer from the fin tip?

A thin-walled spherical tank is buried in the ground at a depth of $3 \mathrm{~m}\(. The tank has a diameter of \)1.5 \mathrm{~m}$, and it contains chemicals undergoing exothermic reaction that provides a uniform heat flux of \(1 \mathrm{~kW} / \mathrm{m}^{2}\) to the tank's inner surface. From soil analysis, the ground has a thermal conductivity of $1.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\( and a temperature of \)10^{\circ} \mathrm{C}$. Determine the surface temperature of the tank. Discuss the effect of the ground depth on the surface temperature of the tank.

A 0.2-cm-thick, 10-cm-high, and 15 -cm-long circuit board houses electronic components on one side that dissipate a total of \(15 \mathrm{~W}\) of heat uniformly. The board is impregnated with conducting metal fillings and has an effective thermal conductivity of $12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. All the heat generated in the components is conducted across the circuit board and is dissipated from the back side of the board to a medium at \(37^{\circ} \mathrm{C}\), with a heat transfer coefficient of $45 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}$. (a) Determine the surface temperatures on the two sides of the circuit board. (b) Now a \(0.1\)-cm-thick, \(10-\mathrm{cm}\)-high, and 15 -cm-long aluminum plate $(k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( with \)200.2-\mathrm{cm}$-thick, 2 -cm-long, and 15 -cm-wide aluminum fins of rectangular profile are attached to the back side of the circuit board with a \(0.03-\mathrm{cm}\) thick epoxy adhesive $(k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$. Determine the new temperatures on the two sides of the circuit board.

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