/*! 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 157 Aluminum fins of triangular prof... [FREE SOLUTION] | 91Ó°ÊÓ

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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

A long cylindrical rod of diameter \(200 \mathrm{~mm}\) with thermal conductivity of \(0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) experiences uniform volumetric heat generation of \(24,000 \mathrm{~W} / \mathrm{m}^{3}\). The rod is encapsulated by a circular sleeve having an outer diameter of \(400 \mathrm{~mm}\) and a thermal conductivity of \(4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The outer surface of the sleeve is exposed to cross flow of air at \(27^{\circ} \mathrm{C}\) with a convection coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Find the temperature at the interface between the rod and sleeve and on the outer surface. (b) What is the temperature at the center of the rod?

An annular aluminum fin of rectangular profile is attached to a circular tube having an outside diameter of \(25 \mathrm{~mm}\) and a surface temperature of \(250^{\circ} \mathrm{C}\). The fin is \(1 \mathrm{~mm}\) thick and \(10 \mathrm{~mm}\) long, and the temperature and the convection coefficient associated with the adjoining fluid are \(25^{\circ} \mathrm{C}\) and \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (a) What is the heat loss per fin? (b) If 200 such fins are spaced at \(5-\mathrm{mm}\) increments along the tube length, what is the heat loss per meter of tube length?

Two stainless steel plates \(10 \mathrm{~mm}\) thick are subjected to a contact pressure of 1 bar under vacuum conditions for which there is an overall temperature drop of \(100^{\circ} \mathrm{C}\) across the plates. What is the heat flux through the plates? What is the temperature drop across the contact plane?

Consider the composite wall of Example 3.7. In the Comments section, temperature distributions in the wall were determined assuming negligible contact resistance between materials A and B. Compute and plot the temperature distributions if the thermal contact resistance is \(R_{t, c}^{\prime \prime}=10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\).

To maximize production and minimize pumping costs, crude oil is heated to reduce its viscosity during transportation from a production field. (a) Consider a pipe-in-pipe configuration consisting of concentric steel tubes with an intervening insulating material. The inner tube is used to transport warm crude oil through cold ocean water. The inner steel pipe \(\left(k_{s}=35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) has an inside diameter of \(D_{i, 1}=150 \mathrm{~mm}\) and wall thickness \(t_{i}=10 \mathrm{~mm}\) while the outer steel pipe has an inside diameter of \(D_{i, 2}=250 \mathrm{~mm}\) and wall thickness \(t_{o}=t_{i}\). Determine the maximum allowable crude oil temperature to ensure the polyurethane foam insulation \(\left(k_{p}=\right.\) \(0.075 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) between the two pipes does not exceed its maximum service temperature of \(T_{p, \max }=\) \(70^{\circ} \mathrm{C}\). The ocean water is at \(T_{\infty, o}=-5^{\circ} \mathrm{C}\) and provides an external convection heat transfer coefficient of \(h_{o}=500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The convection coefficient associated with the flowing crude oil is \(h_{i}=450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) It is proposed to enhance the performance of the pipe-in-pipe device by replacing a thin \(\left(t_{a}=5 \mathrm{~mm}\right)\) section of polyurethane located at the outside of the inner pipe with an aerogel insulation material \(\left(k_{a}=0.012 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\). Determine the maximum allowable crude oil temperature to ensure maximum polyurethane temperatures are below \(T_{p, \max }=70^{\circ} \mathrm{C}\).

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