Introduction to Agricultural Engineering: A Problem Solving Approach

351 Pages·1992·12.75 MB·English

by  Dr. Lawrence O. Roth Ph.D.

AN INTRODUCTION TO
AGRICULTURAL ENGINEERING
A PROBLEM SOLVING APPROACH
Second Edition
Dr. Lawrence O. Roth, Ph.D., PEe
and
Dr. Harry L. Field, Ed.D.
A Chapman & Hall Food Science Book
AN ASPEN PUBLICATION®
Aspen Publishers, Inc.
Gaithersburg, Maryland
1999
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Library of Congress Cataloging-in-Publication Data
Roth, Lawrence 0.
Introduction to agricultural engineering: a problem solving approach
Lawrence O. Roth and Harry L. Field-2nd ed
p. em.
"An AV I Book"
Originally published: New Yor!< : Chapman & Hall, 1991.
Includes bibliographical references and index.
(Formerlypublisbed by Chapman & Hall, ISBN 0-412-06991-1) ISBN 0-8342-1308-7
1. Agricultural engineering. I. Field, Harry L. , 1949 II. Title
S675.R62 1991
63O-dc20
91-31458
CIP
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Library of Congress Catalog Card Number 91-31458
ISBN: 0-8342-1308-7
2 3 4 5
CONTENTS
Preface v
1. Problem Solving 1
2. Significant Figures and Standard Form 10
3. Simple Machines 17
4. Work. Power. Torque. and Horsepower 28
5. Internal Combustion Engines 38
6. Power Trains 48
7. Tractors and Power Units 63
8. Equipment Efficiency and Capacity 77
9. Machinery Calibration 91
10. Economics of Agricultural Machinery III
11. Sound and Noise 135
12. Measuring Distance 142
13. Angles and Areas 154
14. Land Description 178
15. Differential and Profile Leveling 186
16. Weather 203
17. Water Runoff 211
18. Erosion and Erosion Control 218
19. Irrigation 226
20. Handling. Moisture Management. and Storage of
Biological Products 242
21. Animal Waste Management 258
22. Insulation and Heat Flow 265
23. Heating. Ventilation and Air-conditioning 272
24. Selection of Structural Members 291
25. Principles of Electricity 297
26. Series and Parallel Circuits 304
27. Sizing Conductors 314
28. Electric Motors 323
Appendix I 331
Appendix II 332
Appendix III 333
Appendix IV 335
Appendix V 336
Appendix VI 337
Appendix VII 338
Appendix VIII 339
Appendix IX 340
Appendix X 341
Appendix XI 342
Index 343
iii
Preface
This book is for use in introductory courses in colleges of
agriculture and in other applications requiring a problematic
an
approach to agriculture. It is intended as a replacement for
Introduction to Agricultural Engineering by Roth, Crow, and
Mahoney. Parts of the previous book have been revised and
included, but some sections have been removed and new ones
added. Problem solving has been expanded to include a chapter
on techniques, and suggestions are incorporated throughout the
example problems.
The topics and treatment were selected for three reasons: (1) to
acquaint students with a wide range of applications of
engineering principles to agriculture, (2) to present a selection of
independent but related, topics, and (3) to develop and enhance
the problem solving ability of the students.
Each chapter contains educational objectives, introductory
material, example problems (where appropriate), and sample
problems, with answers, that can be used for self-assessment.
Most chapters are self-contained and can be used independently
of the others. Those that are sequential are organiZed in a logical
order to ensure that the knowledge and skills needed are
presented in a previous chapter.
As principal author I wish to express my gratitude to Dr.
Lawrence O. Roth for his contributions of subject matter and
gUidance. I also wish to thank Professor Earl E. Baugher for his
expertise as technical editor, and my wife Marsha for her help and
patience.
HARRY FIELD
v
1
Problem Solving
OBJECTIVES
1. Be able to define problem solving.
2. Be able to describe the common problem-solving
methods.
3. Be able to select the appropriate method for solving a
problem.
INTRODUCTION
We are faced with a host of problems on a daily basis. Some
problems involve people and human relations, whereas others
require a mathematical solution. In this chapter, we will deal
with problems involVing mathematical solutions and several ways
in which these problems can be approached.
MATHEMATICAL PROBLEM SOLVING
Mathematical problem solving is the process by which an
individual uses previously acquired knowledge, skills, and
understanding to satisfy the demands of an unfamiliar situation.
The essence of the process is the ability to use information and
facts to arrive at a solution.
Problems can be solved in different ways. One of the objectives
of this chapter is to increase the reader's knowledge of problem
solving methods. Six different approaches to solving
mathematical problems will be discussed: diagrams and sketches,
patterns, equations, units cancellation, use of formulas, and
intuitive reasoning.
DIAGRAMS AND SKETCHES
Some problems involve the determination of a quantity of items,
such as the number of nails per sheet of plywood or the number
of studs in a wall. In solving these types of problems, it usually is
helpful to draw a sketch or a diagram.
Problem: How many posts are needed to build a fence 100 feet
long with posts 10 feet apart?
1
2 AN INTRODUCTION TO AGRICULTURAL ENGINEERING
Solution: For many people the first response would be 10:
100ft
Posts= ft
10 post
= 10 posts
but a diagram (Figure 1.1) shows that the correct number of posts
is 11.
1234567891011
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I. ... I I~ I
10 ft ..
100ft
Figure 1.1. Number of posts.
This is an example of a situation where a wrong answer is
possible if you do not interpret the problem correctly. In this
example, 10 represents the number of spaces, not the number of
posts.
PATTERNS
The solution to some problems may depend upon one's being able
to discover a pattern in an array of numbers or values.
Frequently it is convenient to examine the patterns in a sample
rather than the entire population. Once a pattern is discovered
and shown to be consistent for the sample. it can be used to
predict the solution for the entire population.
Problem: A dairy farmer has five children. Each child is
responsible for one part of the daily ration for the family's 100
dairy cows. The oldest is responsible for the grain. the second for
the minerals. the third for the hay. the fourth for the silage. and
the fifth for water. Instead of feeding each cow. the first child
deCides she will not feed the cows at all that day. The second
child decides just to feed every other cow. the third child feeds
every third cow, and so on. Dad soon discovers how the cows
were fed. and needs to know which cows did not receive any feed
or water.
PROBLEM SOLVING 3
Solution: When one is faced with this type of problem, it is
usually helpful to set up a table. In this case it would be very
time-consuming to set up a table for all 100 cows. Instead, select
a sample of the cows. If a pattern is true for the sample, there is
a high probability that the pattern will be true for a large group.
Determining the size of a sample is not always easy. Pick one,
and if a clear pattern does not appear, increase the size until a
pattern develops. We will start with the fIrst 10 cows (Table 1-1).
Table 1-1. Patterns in numbers, fIrst sample.
Cow Number
Ration Child 1 2 3 4 5 6 7 8 9 10
Grain 1 N N N N N N N N N N
Mineral 2 Y Y Y Y Y
Hay 3 Y Y Y
SUage 4 Y Y
Water 5 Y Y
In this sample, cows # 1 and #7 did not receive any grain,
mineral, hay, silage, or water. Is this enough information to
establish a pattern? We Will predict that the next cow that did
not receive any feed or water is # 11. Why? To test this
prediction. the sample size must be extended to include a larger
number of cows.
Table 1-2. Patterns in numbers, second sample.
Cow Number
Ration Child 11 12 13 14 15 16 17 18 19 20
Grain 1 N N N N N N N N N N
Mineral 2 Y Y Y Y Y
Hay 3 Y Y Y
SUage 4 Y Y Y
Water 5 Y Y
Table 1-2 shows that the prediction was right; cow # 11, along
with #13. #17, and #19, did not receive any grain. minerals, hay,
silage, or water. It is now safe to conSider that the prediction
could be used to identify all of the animals within the herd that
4 AN INTRODUCTION TO AGRICULTURAL ENGINEERING
did not receive any grain, minerals, hay, silage, or water (those
animals represented by prime numbers, that is, a number
divisible only by itself and one).
EQUATIONS
The solution to some problems requires the development of a
mathematical equation based on a pattern or another type of
relationship between the numbers. Many equations already have
been developed and have been adopted for everyday use. For
example, the area of a Circle is:
=
A xr2 (1-1)
Occasionally it is necessary to develop an equation before the
problem can be solved.
Problem: How much wire is needed to build a single wire fence
around a rectangular field measuring 450 feet long and 350 feet
wide?
Solution: In this example there are three quantities: length,
width, and perimeter. It should be obviouS that the perimeter is a
function of the other two. Begin by assigning the variables L to
represent the length, W to represent the width, and Pr to
represent the perimeter. Then, because a rectangle has two
lengths and two widths, the perimeter can be found as follows:
Pr= (2 x L) + (2 x W) (1-2)
=
(2 x 450 ft) + (2 x 350 ft)
= 900 + 700
=
1600 ft
UNITS CANCELLATION
Some problems are more complex than the examples we have
used, and many do not have patterns or previously developed
equations. Equations can be developed for some of these
problems, but an alternative approach is units cancellation.
Problems of this type will usually involve several quantities. All of
these quantities, except x, will have a unit such as feet, pounds,
PROBLEM SOLVING 5
gallons, and so on. Units cancellation follows two mathematical
principles: (1) the units of measure associated with the numbers
(feet, gallons, minutes, etc.) follow the same mathematical rules
as the numbers; (2) the units of the numbers behave according to
the rules of fractions. For example:
2 x 2 = 4 (or 22)
With units of feet the same equation is:
2 ft x 2 ft = 4 ft2
To review the rules of fractions study the following example:
3 4 3x4 3
"4x 5=4x5=5
In this example the 4's in the numerator and denominator
cancel out (4/4 = 1). If units of measure are included, they behave
in the same way:
3 ton 4 hr 3 ton ton
4 hr x 5 day = 5 day = 0.6 day
In this example the 4's and the units associated with them
cancel out. The uncancelled units become the units for the
answer. The following example shows another variation of this
principle (where gal = gallon and hr = hour):
5 ~ x 3 hr = 15 gal (1-3)
In this example the units for hours cancel each other, leaving
the units of the answer in gallons. Now we will examine a
problem where units cancellation is used to find the correct
answer.
Problem: What is the weight (lb) of one pint of water?
Solution: If a scale and a one pint measure were available, it
would be a simple task to weight one pint of water. An alternative
is to use the conversion factors found in a table of weights and
measures (Appendix 1) and units cancellation.
Note, in this example two types of measure are used, volume
and weight. The real nature of the problem is to find the
conversion value(s) that will convert from volume (pints) to weight
6 AN INTRODUCTION TO AGRICULTURAL ENGINEERING
(pounds). To begin. refer to appendix I and identify conversion
factors that use both volume and weight. You should find that 1
cubic foot contains 7.48 gallons. and 1 gallon contains 8 pints.
This is a start. but you need something more. If you also know
that water weighs 62.4 pounds per cubic foot. the problem can be
= = = =
solved with (lb pounds. gal gallons, pt pints. and ft3 cubic
feet):
lb 62.4 lb 1 ft3 ~
(1-4)
pC Ift3 x 7.48 gal x 8 pt
lb
= 1.04 pt
The units of pints. gallons. and cubic feet are arranged so they
cancel each other. leaving in the answer the desired units of
pounds and pints.
This example illustrates several principles of units cancellation.
First. it is very important to begin by writing down the correct
units for the answer. Second. write down the equals sign (=). and
then begin entering the values and their units. The first value
entered should have one of the desired units in the correct
position. even if it is a units conversion value from the table.
Starting with one of the units of measure in the correct position
will eliminate the possibility of having the problem inverted. Next
enter a value that will cancel out the unwanted units. if any. in
the first value entered. Continue to add variables and cancel
units until the units of the answer are all that remain. If all of
the units cancel except those that are desired for the answer. and
the units are in the correct position. then the only possible
mistake is a math error.
The process of units cancellation is also useful for problems
requiring the development of a new unit. For example. a very
common quantity in agriculture is power. Power can have
different units. depending on whether it is electrical or
mechanical. The units of mechanical power are foot-pounds per
minute. The solution to a problem in which a 24-ounce weight
=
was moved 15 feet in 5 seconds would look like this (with OZ
ounces, sec = seconds. lb = pounds. ft = feet):