/*! 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} Q33E An important concern in the stud... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Q33E (page 1)

An important concern in the study of heat transfer is to determine the steady-state temperature distribution of a thin plate when the temperature around the boundary is known. Assume the plate shown in the figure represents a cross section of a metal beam, with negligible heat flow in the direction perpendicular to the plate. Let \({T_1},...,{T_4}\) denote the temperatures at the four interior nodes of the mesh in the figure. The temperature at a node is approximately equal to the average of the four nearest nodes鈥攖o the left, above, to the right, and below. For instance,

\({T_1} = \left( {10 + 20 + {T_2} + {T_4}} \right)/4\), or \(4{T_1} - {T_2} - {T_4} = 30\)

33. Write a system of four equations whose solution gives estimates

for the temperatures \({T_1},...,{T_4}\).

Short Answer

Expert verified

The system of equations whose solution gives the estimates for the temperatures \({T_1},...,{T_4}\)is shown below:

\(\begin{array}{c}4{T_1} - {T_2} - {T_4} = 30\\ - {T_1} + 4{T_2} - {T_3} = 60\\ - {T_2} + 4{T_3} - {T_4} = 70\\ - {T_1} - {T_3} + 4{T_4} = 40\end{array}\)

Step by step solution

01

Write the second equation for temperature 2

From the figure, it is observed that for temperature 2, i.e., \({T_2}\) (lies at the interior nodes of the mesh), the nearest temperatures are \({T_1}\), \(20^\circ \), \(40^\circ \), and \({T_3}\).

It is given thatthe temperature at a node is approximately equal to the average of the four nearest nodes. So, temperature \({T_2}\) at a node is approximately equal to the average of the four temperatures at the nearest nodes \({T_1}\), \(20^\circ \), \(40^\circ \), and \({T_3}\) as shown below:

\({T_2} = \left( {{T_1} + 20 + 40 + {T_3}} \right)/4\)

Simplify the above expression.

\(4{T_2} - {T_1} - {T_3} = 60\)

02

Write the second equation for temperature 3

From the figure, it is observed that for temperature 3, i.e., \({T_3}\) (lies at the interior nodes of the mesh), the nearest temperatures are \({T_4}\), \(30^\circ \), \(40^\circ \), and \({T_2}\).

It is given thatthe temperature at a node is approximately equal to the average of the four nearest nodes. So, the temperature \({T_3}\) at a node is approximately equal to the average of the four temperatures at the nearest nodes \({T_4}\), \(30^\circ \), \(40^\circ \), and \({T_2}\) as shown below:

\({T_3} = \left( {{T_4} + 30 + 40 + {T_2}} \right)/4\)

Simplify the above expression.

\(4{T_3} - {T_4} - {T_2} = 70\)

03

Write the second equation for temperature 4

From the figure, it is observed that for temperature 4, i.e., \({T_4}\) (lies at the interior nodes of the mesh), the nearest temperatures are \({T_1}\), \(10^\circ \), \(30^\circ \), and \({T_3}\).

It is given thatthe temperature at a node is approximately equal to the average of the four nearest nodes. So, temperature \({T_4}\) at a node is approximately equal to the average of the four temperatures at the nearest nodes \({T_1}\), \(10^\circ \), \(30^\circ \), and \({T_3}\) as shown below:

\({T_4} = \left( {{T_1} + 10 + 30 + {T_3}} \right)/4\)

Simplify the above expression.

\(4{T_4} - {T_1} - {T_3} = 40\)

04

Rearrange the obtained system of equations

The obtained system of equations is \(4{T_2} - {T_1} - {T_3} = 60\), \(4{T_3} - {T_4} - {T_2} = 70\),and \(4{T_4} - {T_1} - {T_3} = 40\). Also, the given equation is \(4{T_1} - {T_2} - {T_4} = 30\).

Rearrange all the equations as shown below:

\(\begin{array}{c}4{T_1} - {T_2} - {T_4} = 30\\ - {T_1} + 4{T_2} - {T_3} = 60\\ - {T_2} + 4{T_3} - {T_4} = 70\\ - {T_1} - {T_3} + 4{T_4} = 40\end{array}\)

Thus, the system of four equations whose solution gives the estimates for the temperatures \({T_1},...,{T_4}\)is shown below:

\(\begin{array}{c}4{T_1} - {T_2} - {T_4} = 30\\ - {T_1} + 4{T_2} - {T_3} = 60\\ - {T_2} + 4{T_3} - {T_4} = 70\\ - {T_1} - {T_3} + 4{T_4} = 40\end{array}\)

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 91影视!

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

In Exercises 32, find the elementary row operation that transforms the first matrix into the second, and then find the reverse row operation that transforms the second matrix into the first.

32. \(\left[ {\begin{array}{*{20}{c}}1&2&{ - 5}&0\\0&1&{ - 3}&{ - 2}\\0&{ - 3}&9&5\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}1&2&{ - 5}&0\\0&1&{ - 3}&{ - 2}\\0&0&0&{ - 1}\end{array}} \right]\)

A Givens rotation is a linear transformation from \({\mathbb{R}^{\bf{n}}}\) to \({\mathbb{R}^{\bf{n}}}\) used in computer programs to create a zero entry in a vector (usually a column of matrix). The standard matrix of a given rotations in \({\mathbb{R}^{\bf{2}}}\) has the form

\(\left( {\begin{aligned}{*{20}{c}}a&{ - b}\\b&a\end{aligned}} \right)\), \({a^2} + {b^2} = 1\)

Find \(a\) and \(b\) such that \(\left( {\begin{aligned}{*{20}{c}}4\\3\end{aligned}} \right)\) is rotated into \(\left( {\begin{aligned}{*{20}{c}}5\\0\end{aligned}} \right)\).

Determine h and k such that the solution set of the system (i) is empty, (ii) contains a unique solution, and (iii) contains infinitely many solutions.

a. \({x_1} + 3{x_2} = k\)

\(4{x_1} + h{x_2} = 8\)

b. \( - 2{x_1} + h{x_2} = 1\)

\(6{x_1} + k{x_2} = - 2\)

In Exercises 9, write a vector equation that is equivalent to

the given system of equations.

9. \({x_2} + 5{x_3} = 0\)

\(\begin{array}{c}4{x_1} + 6{x_2} - {x_3} = 0\\ - {x_1} + 3{x_2} - 8{x_3} = 0\end{array}\)

Find the general solutions of the systems whose augmented matrices are given

11. \(\left[ {\begin{array}{*{20}{c}}3&{ - 4}&2&0\\{ - 9}&{12}&{ - 6}&0\\{ - 6}&8&{ - 4}&0\end{array}} \right]\).
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Logic and Functions

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.