what's thermodynamics system, what is thermodynamics and explain first law with example,what are thermodynamics properties, what are thermodynamics laws, what are the thermodynamics properties
JohenCorner Profile Pic
Published Date:02-08-2017
Your Website URL(Optional)
The Beginning 1.1 WHAT IS THERMODYNAMICS? Thermodynamics is the study of energy and the ways in which it can be used to improve the lives of people around the world. The efficient use of natural and renewable energy sources is one of the most important technical, political, and environmental issues of the 21st century. In mechanics courses, we study the concept of force andhowitcanbemadetodousefulthings.Inthermo- dynamics, we carry out a parallel study of energy and all its technological implications. The objects studied in mechanics are called bodies, and we analyze them through the use of free body diagrams. The objects studied in thermodynamics are called systems, and the free body diagrams of mechanics are replaced by system diagrams in thermodynamics. THERMO—WHAT? The word thermodynamics comes from the Greek words θερμη (therme, meaning “heat”)and δυναμις (dynamis, meaning “power”). Thermodynamics is the study of the various processes that change energy from one form into another (such as converting heat into work) and uses variables such as temperature, volume, and pressure.2 CHAPTER 1: The Beginning 1 Energy is one of the most useful concepts ever developed. Energy can be possessed by an object or a system, suchasacoiledspringorachemicalfuel,anditmaybetransmitted through empty space as electromagnetic radiation. The energy contained in a system is often only partially available for use. This, called the available energy of the system, is treated in detail later in this book. One of the basic laws of thermodynamics is that energy is conserved. This law is so important that it is called the first law of thermodynamics. It states that energy can be changed from one form to another, but it can- not be created or destroyed (that is, energy is “conserved”). Some of the more common forms of energy are: gravitational, kinetic, thermal, elastic, chemical, electrical, magnetic, and nuclear. Our ability to effi- ciently convert energy from one form into a more useful form has provided much of the technology we have today. 1.2 WHY IS THERMODYNAMICS IMPORTANT TODAY? The people of the world consume 1.06 cubic miles of oil each year as an energy source for a wide variety of uses 2 such as the engines shown in Figures 1.1 and 1.2. Coal, gas, and nuclear energy provide additional energy, 3 3 equivalent to another 1.57 mi of oil, making our total use of exhaustible energy sources equal to 2.63 mi of oil every year. We also use renewable energy from solar, biomass, wind (see Figure 1.3), and hydroelectric, in 3 amounts that are equivalent to an additional 0.37 mi of oil each year. This amounts to a total worldwide FIGURE 1.1 A cutaway of the Pratt & Whitney F-100 gas turbine engine. FIGURE 1.2 Corvette engine. 1 The word energy is the modern form of the ancient Greek term energeia, which literally means “in work” (en = in and ergon = work). 2 15 One cubic mile of oil is equal to 1.1 trillion gallons and contains 160 quadrillion (160 × 10 ) kilojoules of energy.1.2 Why Is Thermodynamics Important Today? 3 FIGURE 1.3 Sustainable wind energy technology. FIGURE 1.4 The Starship Enterprise in Star Trek. (Photo credit: Industrial Light & Magic, Copyright © 2008 by PARAMOUNT PICTURES. All Rights Reserved.) 3 energy use equivalent to 3.00 mi of oil each year. If the world energy demand continues at its present rate to create the technologies of the future (e.g., the Starships of Figure 1.4), we will need an energy supply equivalent 3 to consuming an astounding 270 mi of oil by 2050 (90 times more that we currently use). Where is all that energy going to come from? How are we going to use energy more efficiently so that we do not need to use so much? We address these and other questions in the study of thermodynamics. The study of energy is of fundamental importance to all fields of engineering. Energy, like momentum, is a unique subject and has a direct impact on virtually all technologies. In fact, things simply do not “work” with- out a flow of energy through them. In this text, we show how the subject touches all engineering fields through worked example problems and relevant homework problems at the end of the chapters.4 CHAPTER 1: The Beginning HOW IS THERMODYNAMICS USED IN ENGINEERING? ■ Mechanical engineers study the flow of energy in systems such as automotive engines (Figure 1.2), turbines, heat exchangers, bearings, gearboxes, air conditioners, refrigerators, nozzles, and diffusers. ■ Electrical engineers deal with electronic cooling problems, increasing the energy efficiency of large-scale electrical power generation, and the development of new electrical energy conversion technologies such as fuel cells. ■ Civil engineers deal with energy utilization in construction methods, solid waste disposal, geothermal power generation, transportation systems, and environmental impact analysis. ■ Materials engineers develop new energy-efficient metallurgical compounds, create high-temperature materials for engines, and utilize the unique properties of nanotechnology. ■ Industrial engineers minimize energy consumption and waste in manufacturing processes, develop new energy management methods, and improve safety conditions in the workplace. ■ Aerospace engineers develop energy management systems for air and space vehicles, space stations, and planetary habitation (Figure 1.4). ■ Biomedical engineers develop better energy conversion systems for the health care industry, design new diagnostic and treatment tools, and study the energy flows in living systems. All engineering fields utilize the conversion and use of energy to improve the human condition. 1.3 GETTING ANSWERS: A BASIC PROBLEM SOLVING TECHNIQUE Unlike mechanics, which deals with a relatively small range of applications, thermodynamics is truly global and can be applied to virtually any subject, technology, or object conceivable. You no longer can thumb through a book looking for the right equation to apply to your problem. You need a method or technique that guides you through the process of solving a problem in a prescribed way. In Chapter 4, we provide a more detailed technique for thermodynamics problem solving, but for the present, here are seven basic problem solving steps you should know and understand. 1. Read. Always begin by carefully reading the problem statement and try to visualize the “thing” about which the problem is written (a car, engine, rocket, etc.). The “thing” about which the problem is written is called the system in thermodynamics. This may seem simple, but it is key to understanding exactly what you are analyzing. 2. Sketch. Now draw a simple sketch of the system you visualized and add as much of the numerical information given in the problem statement as possible to the sketch. If you do not know what the “thing” in the problem statement looks like, just draw a blob and call it the system. You will not be able to remember all the numbers given in the problem statement, so write them in an appropriate spot on your sketch, so that they are easy to find when you need them. 3. Need. Write down exactly what you need to determine—what does the problem ask you to find? 4. Know. Make a list of the names, numerical values, and units of everything else given in the problem statement. For example, Initial velocity = 35 meters per second, mass = 5.5 kilograms. 5. How. Because of the nature of thermodynamics, there are more equations than you are accustomed to working with. To be able to sort them all out, you need to get in the habit of listing the relevant equations and assumptions that you “might” be able to use to solve for the unknowns in the problem. Write down all of them. A BASIC PROBLEM SOLVING TECHNIQUE 1. Carefully read the problem statement and visualize what you are analyzing. 2. Draw a sketch of the object you visualized in step 1. 3. Now write down what you need to find, that is, make a list of the unknown(s). 4. List everything else you know about the problem (i.e., all the remaining information given in the problem statement). 5. Make a list of relevant equations to see how to solve the problem. 6. Solve these equations algebraically for the unknown(s). 7. Calculate the value(s) of the unknown(s), and check the units in each calculation. Read→ Sketch→ Need→ Know→ How→ Solve→ Calculate1.3 Getting Answers: A Basic Problem Solving Technique 5 6. Solve. Next, you need to algebraically solve the equations listed in step 5 for the unknowns. Because the number of variables in this subject can be large, the unknowns you need to determine may be inside one of your equations, and you need to solve for it algebraically. 7. Calculate. Finally, after you have successfully completed the first six steps, you compute the values of the unknowns, being careful to check the units in all your calculations for consistency. This technique requires discipline and patience on your part. However, if you follow these basic steps, you will be able solve the thermodynamics problems in the first three chapters of this textbook. The following example illustrates this problem solving technique. EXAMPLE 1.1 A new racecar with a JX-750 free-piston engine is traveling on a straight level test track at a velocity of 85.0 miles per hour. The driver accelerates at a constant rate for 5.00 seconds, at which point the car’s velocity has increased to 120. miph. Deter- 3 mine the acceleration of the car as it went from 85.0 to 120. mph. Solution 1. Readthe problem statement carefully. Sometimesyou may begiven miscellaneousinformationthatis not needed in the solution. For example,wedo notneed toknowwhatkind ofengineis usedinthe car,but wedoneed toknowthatthe car hasa constant acceleration for the 5.00s. 2. Draw a sketch of the problem, like the one in Figure 1.5. Transfer all the numerical information given in the problem statement onto your sketch so you need not search for it later. V = 85.0 mph V = 120. mph 1 2 FIGURE 1.5 Example 1.1, solution step 2. 3. What are we supposed to find? We need the acceleration of the car. 4. Weknowthefollowingthings:Theinitialvelocity=85.0mph,thefinalvelocity=120.mph,andthecaracceleratesfort=5.00s. 5. How are we going to find the car’s acceleration? In this case, the basic physics equation that defines acceleration is 2 2 a = dx /dt = dV/dt, and if the acceleration a is constant, then we can integrate this equation to get V = V + at. final initial Note that the acceleration must be constant to use this equation. Aha, that is why the acceleration was specified as constant in the problem statement. No additional equations are needed to solve this problem. 6. Now we can solve for the unknown acceleration, a: V −V final initial a = t 7. Now all we have to do is to insert the given numerical values and calculate the solution:  miles 120−85 miles hour a = = 7 5seconds hour/seconds Now check the units. Miles per hour times seconds makes no sense. Let us convert the car’s velocity from miles per hour 4 to feet per second before we calculate the acceleration :   5280feet/mile miles feet V = 85 = 125 initial hour 3600seconds/hour second and   5280feet/mile miles feet V = 120: = 176 final hour 3600seconds/hour second (Continued)6 CHAPTER 1: The Beginning EXAMPLE 1.1 (Continued) Then, the acceleration becomes feet 176−125 feet second 2 a = = 10:3 = 10:3ft/s 2 5seconds second Remember, the answer is not correct if the units are not correct. Following most of the Example problems in this text are a few Exercises, complete with answers, that are based on the Example. These exercises are designed to allow you to build your problem solving skills and develop self-confidence. The exercises are to be solved by following the solution structure of the preceding example problem. Here are typical exercise problems based on Example 1.1. Exercises 1. Determinetheacceleration of the race car in Example1.1if itsfinal velocity is 130.mph instead of 120.mph. 2 Answer: a =13.2feet/second . 2 2. If the racecar in Example 1.1 has a constant acceleration of 10.0 ft/s , determine its velocity after 6.00 s. Answer: V = 126 mph. 1 3. A dragster travels a straight level mile drag strip in 6.00 s from a standing start (i.e., X = V = 0). Determine the initial initial 4 1 2 average constant acceleration of the dragster. Hint: The basic physics equation you need here is X = V × t+( )at . final initial 2 2 Answer: a = 73.3 ft/s . 3 You may be wondering why there are decimal points and extra zeros added to some of these numbers. This is because we are indicating the number of significant figures represented by these values. The subject of significant figures is covered later in this chapter. 4 For future reference, there are “exactly” 5280 feet in one mile and “exactly” 3600 seconds in one hour. 1.4 UNITS AND DIMENSIONS In thermodynamics, you determine the energy of a system in its many forms and master the mechanisms by which the energy can be converted from one form to another. A key element in this process is the use of a con- sistent set of dimensions and units. A calculated engineering quantity always has two parts, the numerical value and the associated units. The result of any analysis must be correct in both categories: It must have the correct numerical value and it must have the correct units. Engineering students should understand the origins of and relationships among the several units systems currently in use within the profession. Earlier measurements were carried out with elementary and often incon- sistently defined units. In the material that follows, the development of measurement and units systems is pre- sented in some detail. The most important part of this material is that covering modern units systems. 1.5 HOW DO WE MEASURE THINGS? Metrology is the study of measurement,the source ofreproducible quantificationin scienceand engineering. It deals with the dimensions, units, and numbers necessary to make meaningful measurements and calculations. It does not dealwiththetechnology ofmeasurement, soit is not concerned with how measurements are actually made. We call each measurable characteristic of a quantity a dimension of that quantity. If the quantity exists in the material world, then it automatically has three spatial dimensions (length, width, and height), all of which are called length (L) dimensions. If the quantity changes in time, then it also has a temporal dimension called time (t). Some dimensions are not unique because they are made up of other dimensions. For example, an area (A)isa measurable characteristic of an object and therefore one of its dimensions. However, the area dimension is the 2 same as the length dimension squared (A = L ). On the other hand, we could say that the length dimension is the same as the square root of the area dimension. Even though there seems to be a lack of distinguishing characteristics that allow one dimension to be recognized as more fundamental than some other dimension, we easily recognize an apparent utilitarian hierarchy within a set of similar dimensions. We therefore choose to call some dimensions fundamental and all other dimensions related to the chosen fundamental dimensions secondary or derived. It is important to understand that not all systems of dimensional analysis have the same set of fundamental dimensions. Units provide us with a numerical scale whereby we can carry out a measurement of a quantity. They are estab- lished quite arbitrarily and are codified by civil law or cultural custom. How the dimension of length ends up1.5 How Do We Measure Things? 7 being measured in units of feet or meters has nothing to do with any physical law. It is solely dependent on the creativity and ingenuity of people. Therefore, whereas the basic concepts of dimensions are grounded in the fun- damental logic of descriptive analysis, the basic ideas behind the units systems are often grounded in the roots of past civilizations and cultures. ANCIENT UNITS SYSTEMS Intuition tells us that civilization should have evolved using the decimal system. People have ten fingers and ten toes, so the base 10 (decimal) number system would seem to be the most logical system to be adopted by prehistoric people. However, archaeological evidence has shown that the pre-Egyptian Sumerians used a base 60 (sexagesimal) number system, and ancient Egyptians and early American Indians used a base 5 number system. A base 12 (duodecimal) number system was developed and usedextensively duringthe Roman Empire. Today,mixed remains ofthese ancient number systemsaredeeply rootedin our culture. A fundamental element of a successful mercantile trade is that the basic units of commerce Cubit have easily understood subdivisions. Normally, the larger the base number of a particular number system, the more integer divisors it has. For example, 10 has only three divisors (1, 2, and 5), but 12 has five integer divisors (1, 2, 3, 4, and 6) and therefore makes a con- Hand siderably better fractional base. On the other hand, 60 has an advantage over 100 as a number base because the former it has 11 integer divisors whereas 100 has only 8. The measurements of length and time were undoubtedly the first to be of concern to prehistoric people. Perhaps the measurement of time came first, because people had to know the relationship of night to day and understand the passing of the seasons of the year. The most striking aspect of our current measure of time is that it is a mixture of three numerical bases; decimal (base 10) for counting days of the year, duodecimal (base 12) for dividing day and night into equal parts (hours), and sexagesimal (base 60) for dividing hours and minutes into equal parts. Nearly all early scales of length were initially based on the dimensions of parts of the adult Foot human body because people needed to carry their measurement scales with them (see Figure 1.6). Early units were usually related to each other in a binary (base 2) system. For Pace example, some of the early length units were: half-hand = 2 fingers; hand = 2 half-hands; FIGURE 1.6 span = 2 hands; forearm (cubit) = 2 spans; fathom = 2 forearms, and so forth. Measure- Egyptian man with measurements. ments of area and volume followed using such units as handful = 2 mouthfuls, jack = 2 handfuls, gill = 2 jacks, cup = 2 gills, and so forth. Weight was probably the third fundamental measure to be established, with the development of such units as the grain (i.e., the weight of a single grain of barley), the stone, and the talent (the maximum weight that could be comfortably carried continuously by an adult man). CRITICAL THINKING Where are Roman numerals still commonly used today? How would technology be different if we used Roman numerals for engineering calculations today? NURSERY RHYMES AND UNITS Many of the Mother Goose nursery rhymes were not originally written for children but in reality were British political poems or songs. For example, in 17th century England, the treasury of King Charles I (1625–1640) ran low, so he imposed a tax on the ancient unit of volume used for measuring honey and hard liquor, the jack (1 jack = 2 handfuls). The response of the people was to avoid the tax by consuming drink measured in units other than the jack. Eventually, the jack unit became so unpopular with the people that it was no longer used for anything. One of the few existing uses of the jack unit (Continued)8 CHAPTER 1: The Beginning NURSERY RHYMES AND UNITS Continued today is in the term jackpot. Coincidentally, the next larger unit size, the gill (1 gill = 2 jacks), also fell into disuse. The political meaning of the following popular Mother Goose rhyme should now become clear (Figure 1.7): Jack and Gill went up a hill to fetch a pail of water. Jack fell down and broke his crown and Gill came tumbling after. TheJackandGillinthisrhymearenotreallyalittleboyandgirl,theyaretheoldunitsofvolumemeasure.Jackfelldownreferstothe fallofthejackfrompopularusageasaresultofthetaximposedbythecrown,CharlesI.ThephraseandGillcametumblingafterrefers tothesubsequentdeclineintheusetothegillunitofvolumemeasure.The“real”jackandgillofthisrhymeareshowninFigure1.8. Jack Gill FIGURE 1.7 FIGURE 1.8 Jack and Jill. The real jack and gill. CRITICAL THINKING What other Mother Goose rhymes or children’s songs are not what they seem? 1.6 TEMPERATURE UNITS The development of a temperature unit of measure came late in the history of science. The problem with early temperature scales is that all of them were empirical, and their readings often depended on the material (usually a liquid or a gas) used to indicate the temperature change. In a liquid-in-glass thermometer, the difference between the coefficient of thermal expansion of the liquid and the glass causes the liquid to change height when the temperature changes. If the coefficient of thermal expansion depends in some way on temperature, then an accurate thermometer cannot be made simply by defining two fixed (calibration) points and subdivid- ing the difference between these two points into a uniform number of degrees. Unfortunately, the coefficients of thermal expansion of all liquids depend to some extent on temperature; consequently, the two-fixed-point method of defining a temperature scale is inherently prone to this type of measurement error. In 1848, William Thomson (1824–1907), later to become Lord Kelvin, developed a thermodynamic absolute temperature scale that was independent of the measuring material. He was further able to show that his thermo- dynamic absolute temperature scale was identical to the ideal gas absolute temperature scale developed earlier, and therefore an ideal gas thermometer could be calibrated to measure thermodynamic absolute temperatures. Thereafter, the absolute Celsius temperature scale was named the Kelvin scale in his honor. Because it was a real thermodynamic absolute temperature scale, it could be constructed from a single fixed calibration point once the degree size had been chosen. The triple point of water (0.01°C or 273.16 K) was selected as the fixed point.1.6 Temperature Units 9 THE DEVELOPMENT OF THERMOMETERS Thermometry is the technology of temperature measurement. Although people have always been able to experience the physiological sensations of hot and cold, the quantification and accurate measurement of these concepts did not occur until the 17th century. Ancient physicians judged the wellness of their patients by sensing fevers and chills with a touch of the hand (as we often do today). The Roman physician Galen (ca. 129–199) ascribed the fundamental differences in the health or “temperament” of a person to the proportions in which the four “humors” (phlegm, black bile, yellow bile, and 5 blood) were mixed within the body. Thus, both the term for wellness (temperament) and that for body heat (tempera- ture) were derived from the same Latin root temperamentum, meaning “a correct mixture of things.” Until the late 17th century, thermometers were graduated with arbitrary scales. However, it soon became clear that some form of temperature standardization was necessary, and by the early 18th century, 30 to 40 temperature scales were in use. These scales were usually based on the use of two fixed calibration points (standard temperatures) with the distance between them divided into arbitrarily chosen equally spaced degrees. The 100 division (i.e., base 10 or decimal) Celsius temperature scale became very popular during the 18th and 19th centu- ries and was commonly known as the centigrade (from the Latin centum for “100” and gradus for “step”) scale until 1948, when Celsius’s name was formally attached to it and the term centigrade was officially dropped. 5 It was thought that illness occurred when these four humors were not in balance, and that their balance could be restored by draining off one of them (i.e., by “bleeding” the patient). Table 1.1 Early Temperature Scales Inventor and Date Fixed Points Isaac Newton (1701) Freezing water (0°N) and human body temperature (12°N) a Daniel Fahrenheit (1724) Old: Freezing saltwater mixture (0°F) and human body temperature (96°F) New: Freezing water (32°F) and boiling water (212°F) René Reaumur (1730) Freezing water (0°Re) and boiling water (80°Re) b Anders Celsius (1742) Freezing water (0°C) and boiling water (100°C) a The modern Fahrenheit scale uses the freezing point of water (32°F) and the boiling point of water (212°F) as its fixed points. This change to more stable fixed points resulted in changing the average body temperature reading from 96°F on the old Fahrenheit scale to 98.6°Fonthe new Fahrenheit scale. b Initially,Celsiuschosethefreezingpointofwatertobe100°andtheboilingpointofwatertobe0°,butthisscalewassooninvertedtoitspresentform. The difference between the boiling and freezing points of water at atmospheric pressure then became 100 K or, alternatively, 100°C, making the Kelvin and Celsius degree size the same. Soon thereafter, an absolute temperature scale based on the Fahrenheit scale was developed, named after the Scottish engineer William Rankine (1820–1872). Some early temperature scales with fixed calibration points are shown in Table 1.1. Note that both the Newton and the Fahrenheit scales are duodecimal (i.e., base 12). EXAMPLE 1.2 Convert 55 degrees on the modern Fahrenheit scale (Figure 1.9) into (a) degrees Newton, (b) degrees Reaumer, and (c) Kelvin. ?N 55°F ?Re ?K (a) Fahrenheit (b) Newton (c) Reaumer (d) Kelvin FIGURE 1.9 Example 1.2. (Continued)10 CHAPTER 1: The Beginning EXAMPLE 1.2 (Continued) Solution (a) From Table 1.1, we find that both 0°N and 32°F correspond to the freezing point of water, and body heat (temperature) corresponds to 12°N and 98.6°F (on the modern Fahrenheit scale) on these scales. Since both these scales are linear temperature scales, we can construct a simple proportional relation between the two scales as 98:6−55 12−x = 98:6−32 12−0 where x is the temperature on the Newton scale that corresponds to 55°F. Solving for x gives  98:6−55 x = 12 1− = 4:14°N⁡ 98:6−32 (b) Since the Reaumur scale is also a linear scale with 0°Re and 80°Re corresponding to 32°F and 212°F, respectively, we can establish the following proportion for the Reaumur temperature y that corresponds to 55°F: 80−y 212−55 = 212−32 80−0 from which we can solve for  212−55 y = 80 1− = 10:2°Re⁡ 212−32 (c) Here we have 273.15 K and 373.15 K corresponding to 32°F and 212°F, respectively. The proportionality between these scales is then 212−55 373:15−z = 212−32 373:15−273:15 from which we can compute the Kelvin temperature z that corresponds to 55°Fas  212−55 z = 373:15−ð373:15−273:15Þ = 285:9K⁡ 212−32 Notice that we do not use the degree symbol (°) with either the Kelvin or the Rankine absolute temperature scale symbols. The reason for this is by international agreement as explained later in this chapter. Exercises 4. Convert 20.0°C into Kelvin and Rankine. Answer: 293.2 K and 527.7 R. 5. Convert 30°C into degrees Newton and degrees Reaumur. Answer: 9.7°N and 24°Re. 6. Convert 500. K into Rankine, degrees Celsius, and degrees Fahrenheit. Answer: 900 R, 226.9°C, and 440.3°F. 1.7 CLASSICAL MECHANICAL AND ELECTRICAL UNITS SYSTEMS The establishment of a stable system of units requires the identification of certain measures that must be taken as absolutely fundamental and indefinable. For example, one cannot define length, time, or mass in terms of more fundamental dimensions. They all seem to be fundamental quantities. Since we have so many quantities that can be taken as fundamental, we have no single unique system of units. Instead, there are many equivalent units systems, built on different fundamental dimensions. However, all the existing units systems today have one thing in common—they have all been developed from the same set of fundamental equations of physics, equations more or less arbitrarily chosen for this task. It turns out that all the equations of physics are mere proportionalities into which one must always introduce a “constant of proportionality” to obtain an equality. These proportionality constants are intimately related to the system of units used in producing the numerical calculations. Consequently, three basic decisions must be made in establishing a consistent system of units: 1. The choice of the fundamental quantities on which the system of units is to be based. 2. The choice of the fundamental equations that serve to define the secondary quantities of the system of units. 3. The choice of the magnitude and dimensions of the inherent constants of proportionality that appear in the fundamental equations. With this degree of flexibility, it is easy to see why such a large number of measurement units systems have evolved throughout history.1.7 Classical Mechanical and Electrical Units Systems 11 Table 1.2 Five Units Systems in Use Today System Name TypeFM L t g =1/k C 1 MKS (SI) MLt newton (N) kilogram (kg) meter (m) second (s) 1 (dimensionless) CGS MLt dyne (d) gram (g) centimeter (cm) second (s) 1 (dimensionless) Absolute English MLt poundal (pd) pound mass (lbm) foot (ft) second (s) 1 (dimensionless) Technical English FLt pound force (lbf) slug (sg) foot (ft) second (s) 1 (dimensionless) 2 Engineering English FMLt pound force (lbf) pound mass (lbm) foot (ft) second (s) 32.174 lbm·ft/lbf·s The classical mechanical units system uses Newton’s second law as the fundamental equation. This law is a proportionality defined as F = k ma (1.1) 1 The wide variety of choices available for the fundamental quantities that can be used in this system has pro- duced a large number of units systems. Over a period of time, three systems, based on different sets of funda- mental quantities, have become popular: ■ MLt system, which considers mass (M), length (L), and time (t) as independent fundamental quantities. ■ FLt system, which considers force (F), length (L), and time (t) as independent fundamental quantities. ■ FLMt system, which considers all four as independent fundamental quantities. Table 1.2 shows the various popular mechanical units systems that have evolved along these lines. Also listed are the names arbitrarily given to the various derived units and the value and units of the constant of propor- tionality, k , which appears in Newton’s second law, Eq. (1.1). 1 In Table 1.2, the four units in boldface type have the following definitions: 2 . 1newton =1kg m/s (1.2) 2 . 1dyne =1g cm/s (1.3) 2 . 1poundal = 1lbm ft/s (1.4) 2 . /ft (1.5) 1slug = 1lbf s These definitions are arrived at from Newton’s second law using the fact that k has been arbitrarily chosen to be 1 unity and dimensionless in each of these units systems. Because of the form of k in the Engineering English system, engineering texts have evolved a rather strange and 1 unfortunate convention regarding its use. It is common to let g = 1/k,where g in the Engineering English c 1 c units system is simply . 1 lbm ft EngineeringEnglishunits: g = = 32:174 (1.6) c . 2 k lbf s 1 and in all the other units systems described in Table 1.2, it is 1 Allotherunitssystems: g = = 1ðdimensionlessÞ (1.7) c k 1 This symbolism was originally chosen apparently because the value (but not the dimensions)of g happens to be c the same as that of standard gravity in the Engineering English units system. However, this symbolism is awk- ward because it tends to make you think that g is the same as local gravity, which it definitely is not. Like k , g is c 1 c 2 nothing more than a proportionality constant with dimensions of ML/(Ft ). Because the use of g is so wide- c spread today and it is important that you are able to recognize the meaning of g when you see it elsewhere, it c is used in all the relevant equations in this text. For example, we now write Newton’s second law as ma F = (1.8) g c Until the mid-20th century, most English speaking countries used the Engineering English units system. But, because of world trade pressures and the worldwide acceptance of the SI system, most engineering thermo- dynamics texts today (including this one) present example and homework problems in both the old Engineering English and the new SI units systems. The dimensions of energy are the same as the dimensions of work, which are force × distance,andthedimen- sions of power are the same as the dimensions of work divided by time, or force × distance ÷ time. The corre- sponding units and their secondary names (when they exist) are shown in Table 1.3.12 CHAPTER 1: The Beginning WHICH WEIGHS MORE—A POUND OF FEATHERS OR A POUND OF GOLD? The avoirdupois (from the French meaning “to have weight”) pound contains 7000 barleycorns and is divided into 16 ounces. It was used primarily for weighing ordinary commodities, such as wood, bricks, feathers, and so forth. The troy pound was named after the French city Troyes and was used to weigh only precious metals (gold, silver, etc.), gems, and drugs. The English troy pound contains only 5760 barleycorns and is subdivided into 12 ounces, as was the original Roman pound. The English word ounce is also derived from the Latin word uncia, meaning “the twelfth part of.” Consequently, the avoirdupois pound is considerably larger (by a factor of 7000/5760 = 1.215) than the troy pound and the coexistence of both pound units produced considerable confusion over the years. So a pound of feathers actually does weigh more than a pound of gold, because the weight of the feathers is measured with the avoirdupois pound, whereas the weight of the gold is measured with the troy pound. Today, all engineering calculations done in an English units system are done with the 16 ounce, 7000 grain, avoirdupois pound. Table 1.3 Units of Energy and Power System Name Energy Power 2 2 2 3 MKS (SI) N·m=kg·m /s = joule (J) N·m/s = kg·m /s = J/s = watt (W) 2 2 2 3 CGS dyn·cm = g·cm /s = erg dyn·cm/s = g·cm /s = erg/s Absolute English foot·poundal (ft·pdl) ft·pdl/s Technical English ft·lbf ft·lbf/s Engineering English ft·lbf (1 Btu = 778.17 ft·lbf) ft·lbf/s (1 hp = 550 ft·lbf/s) –5 –7 Note: 1 dyn = 10 N and 1 erg = 10 J. EXAMPLE 1.3 In Table 1.2, the Technical English units system uses force (F), length (L), and time (t)asthe 1 chunk= ? feet= ? meters fundamental dimensions. Then, the mass unit “slug” was defined such that k and g came 1 c out to be unity (1) and dimensionless. Define a new units system in which the force, mass, and time dimensions are taken to be fundamental with units of lbf, lbm, and s, and the length unit is defined such that k is unity (1) and dimensionless. Call this new length unit 1 the chunk and find its conversion factor into the Engineering English and SI units systems FIGURE 1.10 (Figure 1.10). Example 1.3. Solution From Eq. (1.1), we see that the length unit must be defined via Newton’ssecondlaw, F = k ma.Sincewewant k to be 1 1 unity and dimensionless, we set F = k = 1ðdimensionlessÞ 1 ma In our new system, we arbitrarily require 1 lbf to be the force calculated from Newton’s second law when 1 lbm is acceler- 2 ated at a rate of 1 chunk/s . Then, from the preceding k equation, we get 1 1lbf = 1ðdimensionlessÞ 2 1lbmð1chunk/s Þ so that 2 . lbf s 1chunk = 1 lbm In the Engineering English units system, 1 lbf accelerates 1 lbm at a rate of  . F 1lbf lbm ft ft a = ðg Þ = 32:174 = 32:174 c 2 2 . m 1lbm lbf s s1.7 Classical Mechanical and Electrical Units Systems 13 Since the lbf, lbm, and s have the same meaning in both the new system and the traditional Engineering English units system, it follows that chunk ft 1 = 32:174 2 2 s s and that  1m 1chunk = 32:174ft =ð32:174ftÞ = 9:806m 3:281ft Exercises 7. Determine the weight at standard gravity of an object whose mass is 1.0 slug. Answer: Since force and weight are the . From Table 1.2, we find that, in the Absolute English units system, g = 1 same, Eq. (1.8) gives F = W = mg/g c c 2 2 (dimensionless). So the weight of 1.0 slug is W = (1.0 slug)(32.174 ft/s )/1 = 32.174 slug (ft/s ). But, from Eq. (1.8), 2 2 2 we see that 1.0 slug = 1.0 lbf·s /ft, so the weight of 1 slug is then W = 32.174 (lbf·s /ft)(ft/s ) = 32.174 lbf. 8. Determine the mass of an object whose weight at standard gravity is 1 poundal. Answer: Using the same technique as in 2 2 Exercise 7, show that the mass of 1 poundal is m = Fg/g = Wg/g = (1 poundal)(1)/32.174 ft/s = 0.03108 pdl·s /ft = c c 0.03108 lbm. 6 9. W. H. Snedegar whimsically suggested the following new names for some of the SI units : 2 1 far = 1 meter (m); 1 jog = 1 m/s; 1 pant = 1 m/s 1 shove = 1 newton (N); 1 grunt = 1 joule (J); 1 varoom = 1 watt (W) 2 1 lump = 1 kilogram (kg); 1 gasp = 1 pascal (Pa); 1 flab =1kg·m and so forth. Of course the Snedegar units would use the same unit prefixes as SI (see Table 1.5 later). For example, a km would be a kilofar, a kJ would be a kilogrunt, a MPa would be a megagasp, and an incremental length (incremental far) would probably be called a near. In this system the fundamental mass, length, and time (M, L, t) units are the lump, far, and second. All other Snedegar units are secondary, being defined by some basic equation. For example, the secondary unit for velocity, the jog, is defined from the definition of the dimensions of velocity as length per unit time (L/t), or 1 jog =1far/s.Thiscan,however,producesomeproblemsinusage.Inmechanics,theunitsofmicrostrain would be microfar/far. Since a microfar is closer to a near than a far, microstrain units would probably become a near/ far. Such logistical inconsistency often adds confusion to an otherwise well-defined system of units. Determine the relation between the primary and secondary Snedegar units for (a) force, (b) momentum (ML/t), (c) accelera- 2 2 2 tion, (d) work, (e) power, and (f) stress (F/L ). Answers:(a)1shove =1lump·far/s;(b)1lump·jog = 1 lump·far/s ; 2 2 2 2 3 (c) 1 pant =1far/s;(d)1grunt = 1 shove·far = lump·far /s ; (e) 1 varoom =1grunt/s = shove·far/s = lump·far /s ; 2 2 (f) 1 gasp = shove/far = 1 lump/far·s . 6 Snedegar, W. H., “Letter to the Editor,” 1983. Am. J. Phys. 51, 684. EXAMPLE 1.4 Time passes. You graduate from college and go on to become a famous NASA design engineer. You have sole responsibility for the design and launch of the famous Bubble-II space telescope system. The telescope weighs exactly 25,000 lbf on the surface of the Earth andis tobeinstalled in anasynchronousEarth orbit withan orbital velocity ofexactly 5000mph(Figure 1.11). 2 a. What is the value of g (in lbm·ft/lbf·s ) in this orbit? c 2 b. How much will the telescope weigh (in lbf) in Earth orbit where the local acceleration of gravity is only 2.50 ft/s . Weight = 25,000. lbf V = 5000. mph FIGURE 1.11 Example 1.4. (Continued)14 CHAPTER 1: The Beginning EXAMPLE 1.4 (Continued) Solution a. From the text, we see that g is always a constant. It does not depend on the local acceleration of gravity. From Eq. (1.6) c 2 and Table 1.2, we find that in the Engineering English units system g = 32.174 lbm·ft/lbf·s . c b. Since weight is force due to gravity, we have W = F = mg/g , and the mass can be computed from m = Wg/g,or c c  . lbm ft ð25,000lbfÞ 32:174 2 . lbf s m = = 25,000lbm ft 32:174 2 s Then the weight in Earth orbit is 2 mg ð25,000lbmÞð2:50ft/s Þ orbit W = = = 1940lbf orbit . g lbm ft c 32:174 2 . lbf s Exercises 10. Suppose you use the SI units system in Example 1.4. What is the value of g in the orbit? Answer: 1.0 and dimensionless c (see Table 1.2). 2 2 11. Suppose the orbit in Example 1.4 changes so that the local acceleration of gravity is decreased from 2.50 ft/s to 1.75 ft/s . Determine the new weight of the telescope in orbit. Answer: W = 1360 lbf. orbit 12. If the telescope in Example 1.4 weighs 112 kN on the surface of the Earth, how much does it weigh on the surface of 2 the Moon, where the local gravity is only 1.60m/s ? Answer: W = 18.3 kN. moon 1.8 CHEMICAL UNITS A good deal of energy conversion technology comes from converting the chemical energy of fuels into thermal energy. Therefore, we need to be aware of the nature of units used in chemical reactions. A chemical reaction equation is essentially a molecular mass balance equation. For example, the equation A + B = C tells us that one molecule of A reacts with one molecule of B to yield one molecule of C.Since 23 the molecular mass of substance A, M , contains the same number of molecules (6.022 × 10 , Avogadro’s A constant) as the molecular masses M and M of substances B and C, the coefficients in their chemical B C reaction equation are also equal to the number of molecular masses involved in the reaction as well as the number of molecules. Chemists find it convenient to use a mass unit that is proportional to the molecular masses of the substances involved in a reaction. Since chemists use only small amounts of chemicals in laboratory experiments, the centimeter-gram-second (CGS) units system has proven to be ideal for their work. Therefore, chemists defined their molecular mass unit as the amount of any chemical substance that has a mass in grams numerically equal to the molecular mass of the substance and gave it the name mole. However, the chemists’ mole unit is problematic, in that most of the other physical sciences do not use the CGS units system and the actual size of the molar mass unit depends on the size of the mass unit in the units system being used. Strictly speaking, the molar mass unit used by chemists should be called a gram mole,becausetheword mole by itself does not convey the type of mass unit used in the units system. Conse- quently, we call the molar mass of a substance in the SI system a kilogram mole; in the Absolute and Engi- neering English systems it is a pound mole; and in the Technical English system it is a slug mole.Inthistext, we abbreviate gram mole as gmole, kilogram mole as kgmole, and pound mole as lbmole. Clearly, these are all different amounts of mass, since 1 gmole ≠1kgmole ≠ 1 lbmole ≠ 1 slug mole. For example, 1 pound mole of water would have a mass of 18 lbm, whereas 1 gram mole would have a mass of only 18 g (0.04 lbm), so that there is an enormous difference in the molar masses of a substance depending on the units system being used. Since the molar amount n of a substance having a mass m is given by m n = (1.9) M1.9 Modern Units Systems 15 7 where M is the molecular mass of the substance, it is clear that the molecular mass must have units of mass/ mass-mole. Therefore, we can write the molecular mass of water as M = 18g/gmole = 18lbm/lbmole = 18kg/kgmole = ::: H O 2 The numerical value of the molecular mass is constant, but it has units that must be taken into account when- ever it is used in an equation. EXAMPLE 1.5 A cylindrical drinking glass, 0.07 m in diameter and 0.15 m high, is three-quarters full of water (Figure 1.12). Determine the number of kilogram moles of water in the glass. The density of 3 . liquid water is exactly 1000 kg/m Solution The mass of water in the glass is equal to the volume of water present multiplied by the density of 3/4 full 0.15 m water, or   2 2 3 m = πR L ×ρ =πðÞ 0:035mðÞ 0:75×0:15m 1000kg/m = 0:433kg The molecular mass of water is 18 kg/kgmole, and Eq. (1.9) gives the number of moles present as 0.07 m 0:433kg m n = = = 0:024kgmole M kg 18 FIGURE 1.12 kgmole Example 1.5. Exercises 13. Determine the number of lbmole in a cubic foot of air whose mass is 0.075 lbm. The molecular mass of air is 28.97 lbm/lbmole. Answer: n = 0.00259 lbmole. 14. How many kilograms are contained in 1 kgmole of a polymer with a molecular mass of 6 6 2.5 × 10 kg/kgmole? Answer: m = 2.5 × 10 kg. 15. Exactly 2 kgmole of xenon has a mass of 262.6 kg. What is the molecular mass of xenon? Answer: M = 131.3 kg/kgmole. 1.9 MODERN UNITS SYSTEMS The units systems commonly used in thermodynamics today are the traditional Engineering English system and the metric SI system. Table 1.4 lists various common derived secondary units of the SI system, and Table 1.5 shows the approved SI prefixes, along with their names and symbols. You need to understand the difference between the units of absolute pressure and gauge pressure. In the Engi- neering English units system, we add the letter a or g to the psi (pounds per square inch) pressure units to make this distinction. Thus, atmospheric pressure can be written as 14.7 psia or as 0 psig. In the SI units system, we add the word that applies (and not the letter a or g) immediately after the unit name or symbol. For example, atmospheric pressure in the SI system is 101,325 Pa absolute or 0 Pa gauge. When the words absolute or gauge do not appear on a pressure unit, assume it is absolute pressure. In 1967, the degree symbol (°) was officially dropped from the absolute temperature unit, and the notational scheme was introduced wherein all unit names were to be written without capitalization (unless, of course, they HOW DO I KNOW WHETHER IT IS ABSOLUTE OR GAUGE PRESSURE? When the clarifying term absolute or gauge is not present in a pressure unit in the textbook, assume that pressure unit is absolute. For example, the pressure 15.2 kPa is interpreted to mean 15.2 kPa absolute. 7 Most texts call M the molecular weight, probably out of historical tradition. However, M clearly has units of mass, not weight, and therefore is more appropriately named molecular mass.16 CHAPTER 1: The Beginning Table 1.4 Some Common Derived SI Units Dimension Name Symbol Formula Expression in Terms of SI Fundamental Units –1 Frequency hertz Hz 1/s s 2 –2 Force newton N kg·m/s m·kg·s 2 –2 Energy joule J N·mm ·kg·s 2 –3 Power watt W J/s m ·kg·s Electric charge coulomb C A·sA·s 2 –3 –1 Electric potential volt V W/A m ·kg·s ·A 2 –3 –2 Electric resistance ohm Ω V/A m ·kg·s ·A –2 –1 4 2 Electric capacitance farad F C/V m ·kg ·s ·A 2 –2 –1 Magnetic flux weber Wb V·sm ·kg·s ·A 2 –1 –2 Pressure or stress pascal Pa N/m m ·kg·s –2 –1 3 2 Conductance siemens S A/V m ·kg ·s ·A 2 –2 –1 Magnetic flux density tesla T Wb/m kg·s ·A 2 –2 –2 Inductance henry H Wb/A m ·kg·s ·A Luminous flux lumen lm cd·sr cd·sr 2 –2 Illuminance lux lx lm/m m ·cd·sr Source: Adapted from the American Society for Testing and Materials, 1980. Standard for Metric Practice, ASTM 380-79. Copyright ASTM. Reprinted with permission. ASTM International, 100 Barr Harbor Drive, PO Box C700, West Conshohocken, PA 19428. Table 1.5 SI Unit Prefixes Multiples Prefixes Symbols 18 10 exa E 15 10 peta P 12 10 tera T 9 10 giga G 6 10 mega M 3 10 kilo k 2 10 hecto h 0 10 —— 1 10 deka da –1 10 deci d –2 10 centi c –3 10 milli m –6 10 micro μ –9 10 nano n –12 10 pico p –15 10 femto f –18 10 atto a Source: Adapted with permission from the American Society for Testing and Materials, 1980. Standard for Metric Practice, ASTM 380-79. ASTM International, 100 Barr Harbor Drive, PO Box C700, West Conshohocken, PA 19428. appear at the beginning of a sentence) regardless of whether they were derived from proper names or not. There- fore the name of the SI absolute temperature unit was reduced from degree Kelvin to simply kelvin even though the unit was named after Lord Kelvin. However, when the name of a unit is to be abbreviated, it was decided that the name abbreviation was to be capitalized if the unit was derived from a proper name. Therefore, the kelvin abso- lute temperature unit is abbreviated as K (not °K, k, or °k). Similarly, the SI unit of force, the newton, named after Sir Isaac Newton (1642–1727), is abbreviated N. The following list illustrates a variety of units from the SI and other systems, all of which were derived from proper names: ampere(A),becquerel(Bq),celsius(°C),coulomb(C),farad(F),fahrenheit(°F),gauss(G),gray(Gy),henry(H), hertz(Hz),joule(J),kelvin(K),newton(N),ohm(Ω),pascal(Pa),poiseuille(P),rankine(R),siemens(S),stoke(St), tesla(T),volt(V),watt(W),weber(Wb).1.10 Significant Figures 17 CRITICAL THINKING Suppose someone wanted to name a new unit of measure after you. What name would you choose and how would it be abbreviated so that your unit would not be confused with other existing unit abbreviations? Note that we still use the degree symbol (°) with the celsius and fahrenheit temperature units. This is due partly to tradition and partly to distinguish their abbreviations from those of the coulomb and farad. In this text, we also drop the degree symbol on the rankine absolute temperature unit, even though it is not part of the SI system. This is done simply to be consistent with the SI notation scheme and because the rankine abbreviation, R, does not conflict with that of any other popular unit. Note that abbreviations use two letters only when necessary to prevent 8 them from being confused with other established unit abbreviations or to express prefixes (e.g., kg for kilogram). All other units whose names were not derived from the names of historically important people are both written and abbreviated with lowercase letters; for example, meter (m), kilogram (kg), and second (s). Obvious viola- tions of this rule occur when any unit name appears at the beginning of a sentence or when its abbreviation is part of a capitalized title, such as in the MKSA System of Units. Also, a unit abbreviation is never pluralized, whereas the unit’s name may be pluralized. For example, kilograms is abbreviated as kg and not kgs, and newtons as N and not Ns. Finally, unit name abbreviations are never written with a terminal period unless they appear at the end of a sentence. For example, the correct abbreviation of sec- onds is s, not sec. or secs. 1.10 SIGNIFICANT FIGURES Using theproper number of significant figures in calculationsis an important part of carrying out credible engineer- ing work. Twotypes of numbers are used in engineering calculations: exact values, such asan integer number used in counting(e.g.,5 ingotsof steel) or numbersfixedbydefinition (e.g.,3600 seconds=1hour);and inexact values,such asnumbersproducedbyphysicalmeasurements(e.g.,the diameterofapipe or the velocityorheightof anobject). Every physical measurement is inexact to some degree. The number of significant figures used to record a measurement is used as an indication of the accuracy of the measurement itself. For example, if you measure the diameter of a shaft with a ruler that could be read to two significant figures, the result might be 3.5 inches, but if it were measured with a micrometer that could bereadtofoursignificant figures it might be 3.512 inches. So, when you are given a value for some variable as, say, the number 4, you see that it is measured with a precision of only one significant figure. But if the value you are given is 4.0, you see that it is measured with two significant figures, and 4.00 indicates it is measured with three significant figures. “EXACT” NUMBERS HAVE AN INFINITE NUMBER OF SIGNIFICANT FIGURES Exact numbers, such as the number of people in a room, have an infinite number of significant figures. Exact numbers are not measurements made with instruments. For example, there are defined numbers, such as 1 foot = 12 inches, so there are “exactly” 12 inches in 1 foot. If a number is “exact,” it does not affect the accuracy of a calculation. Some other examples are 100 years in a century, 2 molecules of hydrogen react with 1 molecule of oxygen to form 1 molecule of water, 500 sheets of paperin 1 ream,60seconds in 1 minute, and 1000grams in 1 kilogram. WHAT IS A “SIGNIFICANT FIGURE”? Asignificantfigureisanyoneofthedigits1,2,3,4,5,6,7,8,and9.Zeroisalsoasignificantfigureexceptwhenused simply to fix the decimal point or to fill the places of unknown or discarded digits. 8 Non-SI units systems do not generally follow this simple rule. For example, the English length unit, foot, could be abbreviated f rather than ft. However, the latter abbreviation is well established within society and changing it at this time would only cause confusion.18 CHAPTER 1: The Beginning A number reported as 0.000452 has only three significant figures (4, 5, and 2), since the leading zeros are used simply to fix the decimal point. But the number 7305 has four significant figures. The number 2300 may have two, three, or four significant figures. To convey which ending zeros of a number are significant, it should be 3 3 3 written as 2.3×10 if it has only two significant figures, 2.30×10 if it has three, and 2.300×10 if it has four. Remember that the identification of the number of significant figures associated with a measurement comes only through a detailed knowledge of how the measurement is carried out. Computations often deal with numbers having unequal numbers of significant figures. A number of rules have been developed for various computations. The rule for addition and subtraction of figures follows. Next comes the rule for multiplicationand divisionof figures. The operation ofroundingvalues upordownalso followsspecific rules. Do you need to maintain the correct number of significant figures in all the steps of a calculation? No, just keep one or two more digits in intermediate results than you need in your final answer. These rules are summarized in Table 1.6. RULE FOR ADDITION AND SUBTRACTION The sum or difference of two numbers should contain no more significant figures farther to the right of the decimal point than occur in the least accurate number used in the operation. For example, 114.2+1.31=115.51, which must be rounded to 115.5, since the least precise number in this operation is 114.2 (having only one place to the right of the decimal point). Similarly, 114.2–1.31=112.89, which must now be rounded to 112.9. This rule is vitally important when subtracting two numbers of similar magnitudes, since their difference may be much less significant than the two numbers that were subtracted. For example, 114.212−114.0=0.212, which must be rounded to 0.2 since 114.0 has only one significant figure to the right of the decimal point. In this case, the result has only one signifi- cant figure even though the “measured” numbers each had four or more significant figures. RULE FOR MULTIPLICATION AND DIVISION The product or quotient should contain no more significant figures than are contained in the term with the least number of sig- nificant figures used in the operation. For example, 114.2 × 1.31=149.602, which must be rounded to 150, since the term 1.31containsonly three significant figures. Also, 114.2/1.31=87.1756, whichmust berounded to87.2 for the samereason. RULES FOR ROUNDING 1. When the discarded value is less than 5, the next remaining value should not be changed. For example, if we round 114.2 to three significant figures it becomes 114; if we rounded it to two significant figures it becomes 110; and rounding it to one significant figure produces 100. 2. When the discarded value is greater than 5 (or is 5 followed by at least one digit other than 0), the next remaining value should be increased by 1. For example, 117.879 rounded to five significant figures is 117.88; rounded to four significant figures, it becomes 117.9; and rounding it to three significant figures produces 118. 3. When the discarded value is exactly equal to 5 followed only by zeros, then the next remaining value should be rounded up if it is an odd number, but remain unchanged if it is an even number. For example, 1.55 rounds to two significant figures as 1.6, and 1.65 also rounds to two significant figures as 1.6. Table 1.6 Significant Figures Number of Significant Figures Represented Written Form of a Number by These Numbers –5 4 3 or 0.1 or 0.01 or 0.001 or 3×10 or 5×10 One significant figure 3 3.1 or 50. or 0.010 or 0.00036 or 7.0×10 Two significant figures 4 3.14 or 500. or 0.0155 or 0.00106 or 7.51×10 Three significant figures 8 3.142 or 1,000. or 0.1050 or 0.0004570 or 3.540×10 Four significant figures 4 3.1416 or 10,000. or 0.0030078 or 1.2500×10 Five significant figures 3.14159 or 100,000. or 186,285 Six significant figures1.10 Significant Figures 19 WHAT ABOUT INTERMEDIATE CALCULATIONS? When doing multi-step calculations, keep one or two more digits in intermediate results than needed in your final answer. If you round-off all your intermediate answers to the correct number of significant figures, you discard the information con- tained in the next digit, and the last digit in your final answer might be incorrect. For example, the calculation 12 × 12 × 1.5 has an answer with two significant figures. But you should use the intermediate results without rounding because 12 × 12 = 144, and 144 × 1.5 = 216→ 220. But, if you round 144 to 140, you obtain 140 × 1.5 = 210, which is pretty far off. It is best to wait until the end of a calculation to round to the correct number of significant figures. Never round in the middle of a multi-step calculation, round only the final answer. NOW TEST YOURSELF a. The number 106.750 has ___ significant figures. b The number 0.0003507 has ___ significant figures. 4 c. The number 3.7 × 10 has ___ significant figures. d. The number 2.7182818 has ___ significant figures. (Answers: a. six, b. four, c. two, d. eight) In a textbook, it is often awkward to write each value with the proper number of significant figures. However, the examples and problems in this textbook have a specific number of significant figures indicated in the mea- sured values. For example, a mass that has been measured to three significant figures is given as, say, 10.0 kg, and a temperature measured to three significant figures is given as, say, 200.ºC (note the decimal point). This is followed throughout the remainder of this textbook. EXAMPLE 1.6 Theinsidediameterofacircularwaterpipeismeasuredtotwo significant figures with a ruler measure and found to be 2.5 inches (Figure 1.13). Determine the cross-sectional area of the pipe to the correct number of significant figures. D = 2.5 inches FIGURE 1.13 Example 1.6. Solution 2 2 2 From elementary algebra, the cross-sectional area of a circle is A = πD /4, so A = π(2.5 inches) /4 = 4.9087 in , which pipe 2 must be rounded to 4.9 in , since the least accurate value in this calculation is the pipe diameter, with only two significant figures. Exercises 16. Determine the cross-sectional area of a circular metal rod measured to two significant figures with a tape measure and 2 found to be 0.025 m. Answer: A = 0.00049 m to two significant figures. bar (Continued)20 CHAPTER 1: The Beginning EXAMPLE 1.6 (Continued) 17. A cubical box is measured to three significant figures with a ruler and found to be 1.21 ft on one side, 1.22 ft on another side, and 1.20 ft on the third side. Determine the volume of the box to the proper number of significant figures. 3 Answer: 1.77 ft . 18. A shaft is measured with a micrometer and found to have a diameter of 1.735 inches (to four significant figures). Determine the circumference of the shaft to the proper number of significant figures. Answer: 5.451 inches. 1.11 POTENTIAL AND KINETIC ENERGIES In classical physics, the term potential energy usually refers to gravitational potential energy and represents the work done against the local gravitational force in changing the position of an object. It depends on the mass m of the object and its height Z above a reference level, written as mgZ Potentialenergy¼PE¼k mgZ = (1.10) 1 g c where k is defined by Eqs. (1.6) and (1.7) as 1/g (see Table 1.2). 1 c Kinetic energy represents the work associated with changing the motion of an object and can occur in two forms: translational and rotational. The total kinetic energy of an object is the sum of both forms of its kinetic energy. The translational kinetic energy of an object is the kinetic energy resulting from a translation velocity V, written as 2 2 mV mV Translationalkineticenergy =ðKEÞ = k = (1.11) 1 trans 2 2g c The rotational kinetic energy of an object is the kinetic energy resulting from a rotation about some axis with an angular velocity ω, written as 2 2 Iω Iω Rotationalkineticenergy =ðKEÞ = k = (1.12) 1 rot 2 2g c where I is the mass moment of inertia of the object about the axis of rotation. The mass moment of inertia of an object is the integral of a mass element dm located at a radial distance r from the axis of rotation: Z 2 dm (1.13) I = r Table 1.7 provides equations for the mass moment of inertia of various common geometrical shapes with a total mass m. Table 1.7 Mass Moments of Inertia of Various Common Shapes x x Thin rectangular plate Slender circular rod 2 2 I = m(a + b )/12 2 x = mL /2 y I x 2 I = ma /12 y 2 I = mb /12 z L z x x Solid rectangular prism Thin disk 2 2 I = m(a + b )/12 x 2 R I = mR /2 2 2 x z I = m(a + L )/12 y 2 z y I = I = mR /4 2 2 y z I = m(b + L )/12 z y

Advise: Why You Wasting Money in Costly SEO Tools, Use World's Best Free SEO Tool Ubersuggest.