A cylindrical pin fin of diameter \(0.6 \mathrm{~cm}\) and length of \(3 \mathrm{~cm}\) with negligible heat loss from the tip has an efficiency of 0.7. The effectiveness of this fin is (a) \(0.3\) (b) \(0.7\) (c) 2 (d) 8 (e) 14

Short Answer

Expert verified
Answer: The fin effectiveness of the given cylindrical pin fin is 0.7.

Step by step solution

01

Understand Given Information and Required Formula

We are given the following information: - Diameter \((D) = 0.6\,\text{cm} = 0.006\,\text{m}\) - Length \((L) = 3\,\text{cm} = 0.03\,\text{m}\) - Fin Efficiency \((\eta_f)= 0.7\) We're tasked to find the fin effectiveness. The formula for fin effectiveness in terms of fin efficiency, heat transfer rate from an unfinned area of the same base dimensions \((q_s)\), and actual heat transfer from the fin \((q_f)\) is as follows: Fin Effectiveness \((\varepsilon) = \frac{q_f}{q_s} = \frac{\eta_f \cdot q_s}{q_s}\)
02

Substitute Given Values into the Formula

Now, let's substitute the given values of the fin efficiency into the formula: \(\varepsilon = \frac{0.7 \cdot q_s}{q_s}\)
03

Simplify and Calculate the Fin Effectiveness

We can simplify the equation for fin effectiveness by canceling out the \(q_s\) term: \(\varepsilon = 0.7\) Fin Effectiveness \((\varepsilon) = 0.7\) The correct answer is (b) 0.7.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Heat Transfer
Heat transfer is a fundamental concept in thermal engineering that involves the movement of thermal energy from one place to another. This process occurs due to a temperature difference within a body or between different bodies, and it can occur through three main mechanisms: conduction, convection, and radiation.

Conduction is the transfer of heat through a solid material when there is a temperature gradient. In our example with the cylindrical pin fin, heat conducts from the base where the fin is attached to the cooler surroundings at the tip, even though in this exercise the tip is considered to have negligible heat loss.

Convection occurs when a fluid (either liquid or gas) is heated and the less dense portion rises, while the denser and cooler part sinks. Fins increase the surface area exposed to the fluid, which enhances heat transfer by convection.

Radiation is the transfer of heat in the form of electromagnetic waves without the need for a physical medium. All objects with a temperature above absolute zero (-273.15°C) emit thermal radiation.

Understanding the principles of heat transfer is crucial when analyzing the functionality and efficiency of a fin since fins are designed to maximize heat dissipation through conduction and convection.
Fin Efficiency
Fin efficiency, denoted as \(\eta_f\), is a measure of how well a fin conducts heat relative to its ideal capacity. The ideal, or maximum, heat transfer would occur if the entire fin were at the base temperature, which is highly unlikely due to the drop in temperature along the fin's length. The efficiency is thus defined as the actual heat transfer rate from the fin to the air divided by the heat transfer rate if the entire fin were at the base temperature.

An efficiency of 0.7, as in the given exercise, indicates that the fin is transferring 70% of the heat compared to the ideal case. It's also important to note that the efficiency of a fin is affected by its geometric characteristics, material properties, and the surrounding thermal environment. For instance, a fin's performance can change based on factors like its shape and size, the conductivity of the material it's made from, and the convective heat transfer coefficient of the fluid in contact with the fin.
Cylindrical Pin Fin
A cylindrical pin fin is a particular type of heat transfer enhancement device commonly used to increase the heat dissipation from a surface. These fins are shaped like rods or pins and extend out from the surface to be cooled. They work on the principle of increasing the surface area exposed to cooling air or fluid, thus facilitating greater heat transfer from the surface to the cooling medium through conduction and convection.

In the given exercise, we have a cylindrical pin fin with a diameter of 0.006 m and a length of 0.03 m. The simplicity of a cylindrical pin fin's geometry often makes it easy to manufacture and analyze, but its performance can vary depending on the fin's aspect ratio and the conditions of the cooling medium. This type of fin is particularly effective in forced convection scenarios where air or fluid is moved over the fin by mechanical means.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A total of 10 rectangular aluminum fins \((k=\) \(203 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) are placed on the outside flat surface of an electronic device. Each fin is \(100 \mathrm{~mm}\) wide, \(20 \mathrm{~mm}\) high and \(4 \mathrm{~mm}\) thick. The fins are located parallel to each other at a center- tocenter distance of \(8 \mathrm{~mm}\). The temperature at the outside surface of the electronic device is \(60^{\circ} \mathrm{C}\). The air is at \(20^{\circ} \mathrm{C}\), and the heat transfer coefficient is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine \((a)\) the rate of heat loss from the electronic device to the surrounding air and \((b)\) the fin effectiveness.

A 1-m-inner-diameter liquid-oxygen storage tank at a hospital keeps the liquid oxygen at \(90 \mathrm{~K}\). The tank consists of a \(0.5-\mathrm{cm}\)-thick aluminum ( \(k=170 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) shell whose exterior is covered with a 10 -cm-thick layer of insulation \((k=\) \(0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The insulation is exposed to the ambient air at \(20^{\circ} \mathrm{C}\) and the heat transfer coefficient on the exterior side of the insulation is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The temperature of the exterior surface of the insulation is (a) \(13^{\circ} \mathrm{C}\) (b) \(9^{\circ} \mathrm{C}\) (c) \(2^{\circ} \mathrm{C}\) (d) \(-3^{\circ} \mathrm{C}\) (e) \(-12^{\circ} \mathrm{C}\)

Consider a tube for transporting steam that is not centered properly in a cylindrical insulation material \((k=\) \(0.73 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The tube diameter is \(D_{1}=20 \mathrm{~cm}\) and the insulation diameter is \(D_{2}=40 \mathrm{~cm}\). The distance between the center of the tube and the center of the insulation is \(z=5 \mathrm{~mm}\). If the surface of the tube maintains a temperature of \(100^{\circ} \mathrm{C}\) and the outer surface temperature of the insulation is constant at \(30^{\circ} \mathrm{C}\), determine the rate of heat transfer per unit length of the tube through the insulation.

The plumbing system of a house involves a \(0.5-\mathrm{m}\) section of a plastic pipe \((k=0.16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of inner diameter \(2 \mathrm{~cm}\) and outer diameter \(2.4 \mathrm{~cm}\) exposed to the ambient air. During a cold and windy night, the ambient air temperature remains at about \(-5^{\circ} \mathrm{C}\) for a period of \(14 \mathrm{~h}\). The combined convection and radiation heat transfer coefficient on the outer surface of the pipe is estimated to be \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the heat of fusion of water is \(333.7 \mathrm{~kJ} / \mathrm{kg}\). Assuming the pipe to contain stationary water initially at \(0^{\circ} \mathrm{C}\), determine if the water in that section of the pipe will completely freeze that night.

Hot water at an average temperature of \(70^{\circ} \mathrm{C}\) is flowing through a \(15-\mathrm{m}\) section of a cast iron pipe \((k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose inner and outer diameters are \(4 \mathrm{~cm}\) and \(4.6 \mathrm{~cm}\), respectively. The outer surface of the pipe, whose emissivity is \(0.7\), is exposed to the cold air at \(10^{\circ} \mathrm{C}\) in the basement, with a heat transfer coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat transfer coefficient at the inner surface of the pipe is \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the walls of the basement to be at \(10^{\circ} \mathrm{C}\) also, determine the rate of heat loss from the hot water. Also, determine the average velocity of the water in the pipe if the temperature of the water drops by \(3^{\circ} \mathrm{C}\) as it passes through the basement.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free