Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 4

How do rating problems in heat transfer differ from the sizing problems?

Short Answer

Expert verified
Answer: The primary difference between rating and sizing problems in heat transfer is that rating problems involve determining the performance of a given heat exchanger with known design and dimensions, while sizing problems involve designing a heat exchanger or finding the necessary dimensions and specifications to achieve a specified performance.

Step by step solution

01

Definition of Rating Problems

Rating problems in heat transfer involve determining the performance of a given heat exchanger. In this type of problem, the design is already given, and you are asked to calculate the heat transfer rate or efficiency of the heat exchanger using specified inlet and outlet temperatures, fluid flow rates, and the physical properties of the fluids.
02

Definition of Sizing Problems

Sizing problems in heat transfer involve designing a heat exchanger or finding the necessary dimensions and specifications. In this type of problem, the desired heat transfer rate is given, and you are required to determine the size, length, or other design characteristics of a heat exchanger that will achieve this specified performance.
03

Main Differences

The primary difference between rating and sizing problems is the information given and the desired outcome: 1. In rating problems, the design, dimensions, and specifications of the heat exchanger are known, and the goal is to determine the performance (heat transfer rate, efficiency, etc.). 2. In sizing problems, the desired performance is given, and the goal is to design a heat exchanger or determine its necessary dimensions and specifications to achieve that performance. Another key difference is the focus of the calculations: 1. Rating problems involve heat transfer coefficients, temperature differences, and heat transfer rates based on the known design details. 2. Sizing problems necessitate a deeper understanding of heat transfer processes and mechanisms, as well as correlations and equations to design a heat exchanger that meets the desired performance criteria.

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.

Start your free trial

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

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

Heat is lost through a brick wall \((k=0.72 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), which is \(4 \mathrm{~m}\) long, \(3 \mathrm{~m}\) wide, and \(25 \mathrm{~cm}\) thick at a rate of \(500 \mathrm{~W}\). If the inner surface of the wall is at \(22^{\circ} \mathrm{C}\), the temperature at the midplane of the wall is (a) \(0^{\circ} \mathrm{C}\) (b) \(7.5^{\circ} \mathrm{C}\) (c) \(11.0^{\circ} \mathrm{C}\) (d) \(14.8^{\circ} \mathrm{C}\) (e) \(22^{\circ} \mathrm{C}\)

Consider a flat-plate solar collector placed on the roof of a house. The temperatures at the inner and outer surfaces of the glass cover are measured to be \(33^{\circ} \mathrm{C}\) and \(31^{\circ} \mathrm{C}\), respectively. The glass cover has a surface area of \(2.5 \mathrm{~m}^{2}\), a thickness of \(0.6 \mathrm{~cm}\), and a thermal conductivity of \(0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Heat is lost from the outer surface of the cover by convection and radiation with a convection heat transfer coefficient of \(10 \mathrm{~W} /\) \(\mathrm{m}^{2} \cdot \mathrm{K}\) and an ambient temperature of \(15^{\circ} \mathrm{C}\). Determine the fraction of heat lost from the glass cover by radiation.

A transistor with a height of \(0.4 \mathrm{~cm}\) and a diameter of \(0.6 \mathrm{~cm}\) is mounted on a circuit board. The transistor is cooled by air flowing over it with an average heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the air temperature is \(55^{\circ} \mathrm{C}\) and the transistor case temperature is not to exceed \(70^{\circ} \mathrm{C}\), determine the amount of power this transistor can dissipate safely. Disregard any heat transfer from the transistor base.

Consider a sealed 20-cm-high electronic box whose base dimensions are \(50 \mathrm{~cm} \times 50 \mathrm{~cm}\) placed in a vacuum chamber. The emissivity of the outer surface of the box is \(0.95\). If the electronic components in the box dissipate a total of \(120 \mathrm{~W}\) of power and the outer surface temperature of the box is not to exceed \(55^{\circ} \mathrm{C}\), determine the temperature at which the surrounding surfaces must be kept if this box is to be cooled by radiation alone. Assume the heat transfer from the bottom surface of the box to the stand to be negligible.

Consider a house in Atlanta, Georgia, that is maintained at \(22^{\circ} \mathrm{C}\) and has a total of \(20 \mathrm{~m}^{2}\) of window area. The windows are double-door type with wood frames and metal spacers and have a \(U\)-factor of \(2.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (see Prob. 1-125 for the definition of \(U\)-factor). The winter average temperature of Atlanta is \(11.3^{\circ} \mathrm{C}\). Determine the average rate of heat loss through the windows in winter.
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Nuclear Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Space Physics

Read Explanation

Electricity

Read Explanation

Energy Physics

Read Explanation
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