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Chapter 3: Problem 157

Aluminum fins of triangular profile are attached to a plane wall whose surface temperature is \(250^{\circ} \mathrm{C}\). The fin base thickness is \(2 \mathrm{~mm}\), and its length is \(6 \mathrm{~mm}\). The system is in ambient air at a temperature of \(20^{\circ} \mathrm{C}\), and the surface convection coefficient is \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What are the fin efficiency and effectiveness? (b) What is the heat dissipated per unit width by a single fin?

Short Answer

Expert verified
The short answer to the given problem is as follows: (a) Fin efficiency and effectiveness can be calculated using the formulas: Fin efficiency: \(\eta_{fin} = \frac{tanh(mL)}{mL}\) Fin effectiveness: \(Fin \; effectiveness = \frac{fin \; efficiency \times A}{A_c}\) After calculating the area, mL, and other parameters, we find the fin efficiency and effectiveness. (b) The heat dissipated per unit width by a single fin can be calculated using the formula: \(q = fin \; efficiency * h * A * (T_{base} - T_{ambient})\) Plug in the fin efficiency, area, convection coefficient, and temperatures to find the heat dissipated per unit width.

Step by step solution

01

Calculate the fin area

The fin is of triangular shape with base thickness = 2 mm and length = 6 mm. Convert the dimensions to meters for consistency. \(base thickness = 2 * 10^{-3} meter\) \(length = 6 * 10^{-3} meter\) The area of the triangle fin can be found using the formula for the area of a triangle. \(Area = 0.5 * base_{thickness} * length\) \(A = 0.5 * (2 * 10^{-3}) * (6 * 10^{-3})\)
02

Calculate the parameter (m*L) and determine the fin efficiency

In order to calculate the fin efficiency, we need to find the value of mL. \(m = \sqrt{\frac{hP}{kA_c}}\) Where: - h is the convection coefficient - P is the perimeter of the fin base - k is the thermal conductivity of aluminum - \(A_c\) is the cross-sectional area We are given \(h = 40 \frac{W}{m^2 \cdot K}\). Using the thermal conductivity of aluminum, \(k = 237 \frac{W}{m \cdot K}\). For an equilateral triangular fin: \(P = base_{thickness} + 2 * length\) \(A_c = 0.5 * base_{thickness} * length\) Calculating mL: \(m * L = L * \sqrt{\frac{hP}{kA_c}}\)
03

Determine the fin effectiveness

Fin effectiveness is defined as the fin heat transfer rate divided by the heat transfer rate for the same situation without a fin (from the base of the fin to ambient air). Fin effectiveness can be calculated using the fin efficiency. \(Fin \; effectiveness = \frac{fin \; efficiency \times A}{A_c}\)
04

Calculate the heat dissipated per unit width

To calculate the heat dissipated per unit width by a single fin, we can use the fin efficiency and the base temperature difference between the wall and ambient air. \(q = fin \; efficiency * h * A * (T_{base} - T_{ambient})\) We were given: \(T_{base} = 250^\circ C\) \(T_{ambient} = 20^\circ C\) Plug in these values to find the heat dissipated.

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Key Concepts

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

Convection Coefficient
The convection coefficient, often denoted by the letter \( h \), is a critical factor in understanding how well heat travels between a solid surface and the surrounding fluid, such as air. In the context of this problem, the convection coefficient is given as \( 40 \, \text{W/m}^2 \cdot \text{K} \). This tells us the rate of heat transfer per unit area per degree of temperature difference between the surface and the fluid. Understanding this coefficient is important because:
  • It determines how much heat is transferred from the wall to the air.
  • A higher convection coefficient means better heat transfer.
  • It's influenced by factors like fluid velocity and properties, surface roughness, and temperature difference.
In design, choosing the right material and conditions to optimize the convection coefficient can significantly impact the efficiency and performance of a cooling or heating system.
Fin Efficiency
Fin efficiency is an important measure that helps us understand how effective a fin is in transferring heat. It is defined as the ratio of the actual heat transferred by the fin to the heat that would be transferred if the entire fin were at the base temperature. To calculate fin efficiency, we use the formula:\[\text{Efficiency} = \frac{\text{actual heat transfer}}{\text{ideal heat transfer}}\]Achieving high fin efficiency is crucial because:
  • It indicates more effective use of material to provide cooling or heating.
  • High fin efficiency reduces waste and improves system performance.
  • It informs decisions about whether to use more or fewer fins.
Factors like fin shape, size, material, and placement in the fluid influence efficiency. In this exercise, calculating fin efficiency also required finding the parameter \( mL \), which incorporates thermal properties and geometric dimensions.
Thermal Conductivity
Thermal conductivity, represented by \( k \), is a material's ability to conduct heat. In this example, aluminum, with a thermal conductivity of \( 237 \, \text{W/m} \cdot \text{K} \), is used for the fins. This high thermal conductivity means aluminum efficiently transfers heat from its hotter region to cooler parts, aiding in rapid thermal equilibration. Key aspects of thermal conductivity include:
  • It's a constant for each material, crucial in choosing the right material for thermal management.
  • Higher values mean better heat conduction, saving energy and improving efficiency.
  • Affecting how quickly a fin can remove heat from its base to the tip.
In applications requiring quick heat dissipation, materials like aluminum are ideal due to their high thermal conductivity, ensuring that designs are efficient and effective.
Triangular Fin
A triangular fin is a specific shape used in engineering to enhance heat transfer from a surface. This fin type is often chosen due to:
  • Its reduced weight compared to rectangular or other fin profiles.
  • Potential for increased surface area optimizing material usage and heat dissipation in a compact form.
  • Performance in applications requiring detailed thermal analyses and custom shapes.
The fin's triangular shape provides a tapered surface area from base to tip, ensuring effective thermal paths and helping to reduce material and manufacturing costs. In this exercise:- The triangular fin has specific dimensions: base thickness of \( 2 \text{mm} \) and length of \( 6 \text{mm} \).- These dimensions were crucial for calculating the perimeter and cross-sectional area necessary for solving heat transfer scenarios.- Utilizing a triangular design can result in improved heat dissipation due to an increased perimeter, providing more efficient heat transfer paths.

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Most popular questions from this chapter

Electric current flows through a long rod generating thermal energy at a uniform volumetric rate of \(\dot{q}=\) \(2 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). The rod is concentric with a hollow ceramic cylinder, creating an enclosure that is filled with air. The thermal resistance per unit length due to radiation between the enclosure surfaces is \(R_{\mathrm{rad}}^{\prime}=0.30 \mathrm{~m} \cdot \mathrm{K} / \mathrm{W}\), and the coefficient associated with free convection in the enclosure is \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Construct a thermal circuit that can be used to calculate the surface temperature of the rod, \(T_{r}\). Label all temperatures, heat rates, and thermal resistances, and evaluate each thermal resistance. (b) Calculate the surface temperature of the rod for the prescribed conditions.

Circular copper rods of diameter \(D=1 \mathrm{~mm}\) and length \(L=25 \mathrm{~mm}\) are used to enhance heat transfer from a surface that is maintained at \(T_{s, 1}=100^{\circ} \mathrm{C}\). One end of the rod is attached to this surface (at \(x=0\) ), while the other end \((x=25 \mathrm{~mm})\) is joined to a second surface, which is maintained at \(T_{s, 2}=0^{\circ} \mathrm{C}\). Air flowing between the surfaces (and over the rods) is also at a temperature of \(T_{\infty}=0^{\circ} \mathrm{C}\), and a convection coefficient of \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained. (a) What is the rate of heat transfer by convection from a single copper rod to the air? (b) What is the total rate of heat transfer from a \(1 \mathrm{~m} \times 1 \mathrm{~m}\) section of the surface at \(100^{\circ} \mathrm{C}\), if a bundle of the rods is installed on 4 -mm centers?

A steam pipe of \(0.12-\mathrm{m}\) outside diameter is insulated with a layer of calcium silicate. (a) If the insulation is \(20 \mathrm{~mm}\) thick and its inner and outer surfaces are maintained at \(T_{s, 1}=800 \mathrm{~K}\) and \(T_{s, 2}=490 \mathrm{~K}\), respectively, what is the heat loss per unit length \(\left(q^{\prime}\right)\) of the pipe? (b) We wish to explore the effect of insulation thickness on the heat loss \(q^{\prime}\) and outer surface temperature \(T_{s, 2}\), with the inner surface temperature fixed at \(T_{s, 1}=\) \(800 \mathrm{~K}\). The outer surface is exposed to an airflow \(\left(T_{\infty}=25^{\circ} \mathrm{C}\right)\) that maintains a convection coefficient of \(h=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and to large surroundings for which \(T_{\text {sur }}=T_{\infty}=25^{\circ} \mathrm{C}\). The surface emissivity of calcium silicate is approximately \(0.8\). Compute and plot the temperature distribution in the insulation as a function of the dimensionless radial coordinate, \(\left(r-r_{1}\right) /\left(r_{2}-r_{1}\right)\), where \(r_{1}=0.06 \mathrm{~m}\) and \(r_{2}\) is a variable \(\left(0.06

In a test to determine the friction coefficient \(\mu\) associated with a disk brake, one disk and its shaft are rotated at a constant angular velocity \(\omega\), while an equivalent disk/shaft assembly is stationary. Each disk has an outer radius of \(r_{2}=180 \mathrm{~mm}\), a shaft radius of \(r_{1}=20 \mathrm{~mm}\), a thickness of \(t=12 \mathrm{~mm}\), and a thermal conductivity of \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). A known force \(F\) is applied to the system, and the corresponding torque \(\tau\) required to maintain rotation is measured. The disk contact pressure may be assumed to be uniform (i.e., independent of location on the interface), and the disks may be assumed to be well insulated from the surroundings. (a) Obtain an expression that may be used to evaluate \(\mu\) from known quantities. (b) For the region \(r_{1} \leq r \leq r_{2}\), determine the radial temperature distribution \(T(r)\) in the disk, where \(T\left(r_{1}\right)=T_{1}\) is presumed to be known. (c) Consider test conditions for which \(F=200 \mathrm{~N}\), \(\omega=40 \mathrm{rad} / \mathrm{s}, \tau=8 \mathrm{~N} \cdot \mathrm{m}\), and \(T_{1}=80^{\circ} \mathrm{C}\). Evaluate the friction coefficient and the maximum disk temperature.

Approximately \(10^{6}\) discrete electrical components can be placed on a single integrated circuit (chip), with electrical heat dissipation as high as \(30,000 \mathrm{~W} / \mathrm{m}^{2}\). The chip, which is very thin, is exposed to a dielectric liquid at its outer surface, with \(h_{o}=1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty, 0}=20^{\circ} \mathrm{C}\), and is joined to a circuit board at its inner surface. The thermal contact resistance between the chip and the board is \(10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), and the board thickness and thermal conductivity are \(L_{b}=5 \mathrm{~mm}\) and \(k_{b}=1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively. The other surface of the board is exposed to ambient air for which \(h_{i}=40\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty, i}=20^{\circ} \mathrm{C}\). (a) Sketch the equivalent thermal circuit corresponding to steady-state conditions. In variable form, label appropriate resistances, temperatures, and heat fluxes. (b) Under steady-state conditions for which the chip heat dissipation is \(q_{c}^{\prime \prime}=30,000 \mathrm{~W} / \mathrm{m}^{2}\), what is the chip temperature? (c) The maximum allowable heat flux, \(q_{c, m}^{\prime \prime}\), is determined by the constraint that the chip temperature must not exceed \(85^{\circ} \mathrm{C}\). Determine \(q_{c, m}^{\prime \prime}\) for the foregoing conditions. If air is used in lieu of the dielectric liquid, the convection coefficient is reduced by approximately an order of magnitude. What is the value of \(q_{c, m}^{\prime \prime}\) for \(h_{o}=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) ? With air cooling, can significant improvements be realized by using an aluminum oxide circuit board and/or by using a conductive paste at the chip/board interface for which \(R_{t, c}^{n}=10^{-5} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) ?
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