Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 45

The Biot number during a heat transfer process between a sphere and its surroundings is determined to be \(0.02\). Would you use lumped system analysis or the transient temperature charts when determining the midpoint temperature of the sphere? Why?

Short Answer

Expert verified
Answer: Given the Biot number of 0.02, which is less than 0.1, we should use lumped system analysis for the heat transfer process. This is because the rate of heat conduction within the sphere is much greater than the rate of heat transfer on the surface, making the temperature gradients within the sphere negligible.

Step by step solution

01

Understanding the Biot number

The Biot number (Bi) is a dimensionless parameter used in heat transfer problems to characterize the relative importance of heat conduction within an object as compared to heat transfer across the object's surface. It is defined as the ratio of the thermal resistance of heat conduction within an object to the thermal resistance of heat convection on the object's surface. Mathematically, the Biot number is given by: Bi = \(hL_c/k\), where \(h\) is the convective heat transfer coefficient, \(L_c\) is the characteristic length (one common choice for spheres is the radius), and \(k\) is the thermal conductivity of the material.
02

Criteria for lumped system analysis

Lumped system analysis assumes that the temperature inside the object is uniform at any given time, and that the heat transfer takes place only at the object's surface. This simplifies the analysis by reducing the problem to a single time-dependent variable (the temperature of the object) rather than a time- and space-dependent variable. In general, lumped system analysis can be used when the Biot number is small (Bi \(<\) 0.1), which implies that the rate of heat conduction within the object is much greater than the rate of heat transfer on the surface. In other words, the temperature gradients inside the object are insignificant compared to the gradients at the surface.
03

Criteria for transient temperature charts

Transient temperature charts, also known as Heisler charts, are used to characterize heat transfer in objects where the temperature does vary meaningfully within the object at a given time. These charts are used to estimate the temperature profiles inside an object during the heat transfer process. Transient temperature charts are typically used when the Biot number is greater than 0.1, which indicates that the temperature gradients within the object are not negligible and must be considered for an accurate analysis.
04

Solution based on the Biot number

In this exercise, the given Biot number during the heat transfer process between the sphere and its surroundings is 0.02. Since the Biot number is less than 0.1, we would use lumped system analysis to determine the midpoint temperature of the sphere. This is because the rate of heat conduction within the sphere is much greater than the rate of heat transfer on the surface, and therefore the temperature gradients within the sphere are negligible.

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

A 5-cm-high rectangular ice block \((k=2.22 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=0.124 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\) initially at \(-20^{\circ} \mathrm{C}\) is placed on a table on its square base \(4 \mathrm{~cm} \times 4 \mathrm{~cm}\) in size in a room at \(18^{\circ} \mathrm{C}\). The heat transfer coefficient on the exposed surfaces of the ice block is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Disregarding any heat transfer from the base to the table, determine how long it will be before the ice block starts melting. Where on the ice block will the first liquid droplets appear?

What is a semi-infinite medium? Give examples of solid bodies that can be treated as semi-infinite mediums for heat transfer purposes.

During a fire, the trunks of some dry oak trees (es) \(\left(k=0.17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\) and \(\left.\alpha=1.28 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\) that are initially at a uniform temperature of \(30^{\circ} \mathrm{C}\) are exposed to hot gases at \(520^{\circ} \mathrm{C}\) for a period of \(5 \mathrm{~h}\), with a heat transfer coefficient of \(65 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the surface. The ignition temperature of the trees is \(410^{\circ} \mathrm{C}\). Treating the trunks of the trees as long cylindrical rods of diameter \(20 \mathrm{~cm}\), determine if these dry trees will ignite as the fire sweeps through them. Solve this problem using analytical one-term approximation method (not the Heisler charts).

A small chicken \(\left(k=0.45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=0.15 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) can be approximated as an \(11.25\)-cm-diameter solid sphere. The chicken is initially at a uniform temperature of \(8^{\circ} \mathrm{C}\) and is to be cooked in an oven maintained at \(220^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). With this idealization, the temperature at the center of the chicken after a 90 -min period is (a) \(25^{\circ} \mathrm{C}\) (b) \(61^{\circ} \mathrm{C}\) (c) \(89^{\circ} \mathrm{C}\) (d) \(122^{\circ} \mathrm{C}\) (e) \(168^{\circ} \mathrm{C}\)

Long cylindrical AISI stainless steel rods \((k=\) \(7.74 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\) and \(\left.\alpha=0.135 \mathrm{ft}^{2} / \mathrm{h}\right)\) of 4 -in-diameter are heat treated by drawing them at a velocity of \(7 \mathrm{ft} / \mathrm{min}\) through a 21 -ft-long oven maintained at \(1700^{\circ} \mathrm{F}\). The heat transfer coefficient in the oven is \(20 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\). If the rods enter the oven at \(70^{\circ} \mathrm{F}\), determine their centerline temperature when they leave. Solve this problem using analytical one-term approximation method (not the Heisler charts).
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Electrostatics

Read Explanation

Radiation

Read Explanation

Space Physics

Read Explanation

Nuclear Physics

Read Explanation

Particle Model of Matter

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