The First Law of Thermodynamics

The First Law of Thermodynamics
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The First Law of Thermodynamics and Energy Transport Mechanisms 4.1 INTRODUCCIÓN (INTRODUCTION) In this chapter, we begin the formal study of the first law of thermodynamics. The theory is presented first, and in subsequent chapters, it is applied to a variety of closed and open systems of engineering interest. In Chapter 4, the first law of thermodynamics and its associated energy balance are developed along with a detailed discussion of the energy transport mechanisms of work and heat. To understand the usefulness of the first law of thermo- dynamics, we need to study the energy transport modes and investigate the energy conversion efficiency of common technologies. In Chapter 5, the focus is on applying the theory presented in Chapter 4 to a series of steady state closed sys- tems, such as sealed, rigid containers; electrical apparatuses; and piston-cylinder devices. Chapter 5 ends with a brief discussion of the behavior of unsteady state closed systems. The first law of thermodynamics is expanded in Chapter 6 to cover open systems, and the conservation of mass law is introduced as a second independent basic equation. Then, appropriate applications are presented, dealing with a variety of common open system technologies of engineering interest, such as nozzles, diffusers, throttling devices, heat exchangers, and work-producing or work-absorbing machines. Chapter 6 ends with a brief discus- sion of the behavior of unsteady state open systems. 4.2 EMMY NOETHER AND THE CONSERVATION LAWS OF PHYSICS Throughout the long history of physics and engineering, we believed that the conservation laws of momentum, energy, and electric charge were unique laws of nature that had to be discovered and verified by physical experi- ments. And, in fact, these laws were discovered in this way. They are the heart and soul of mechanics, thermody- namics, and electronics, because they deal with things (momentum, energy, charge) that cannot be created nor destroyed and therefore are “conserved.” These conservation laws have broad application in engineering and physics and are considered to be the most fundamental laws in nature. We have never been able explain where these laws came from because they seem to have no logical source. They seemed to be part of the mystery that is nature. However, almost 100 years ago, the mathematician Emmy 1 Noether developed a theorem that uncovered their source, yet few seem to know of its existence. Emmy Noether’s theorem is fairly simple. It states that: For every symmetry exhibited by a system, there is a corresponding observable quantity that is conserved. The meaning of the word symmetry here is probably not what you think itis.Thesymmetrythateverybody thinks of is called bilateral symmetry, when two halves of a whole are each other’s mirror images (bilateral sym- metry is also called mirror symmetry). For example, a butterfly has bilateral symmetry. Emmy Noether was talk- ing about symmetry with respect to a mathematical operation. We say that something has mathematical symmetry if, when you perform some mathematical operation on it, it does not change in any way. For exam- ple, everyone knows that the equations of physics remain the same under a translation of the coordinate system. This really says that there are no absolute positions in space. What matters is not where an object is in absolute terms, but where it is relative to other objects, that is, its coordinate differences. The impact of Emmy Noether’s studies on symmetry and the behavior of the physical world is nothing less than astounding. Virtually every theory, including relativity and quantum physics, is based on symmetry principles. To quote just one expert, Dr. Lee Smolin, of the Perimeter Institute for Theoretical Physics, “The connection between symmetries and conservation laws is one of the great discoveries of twentieth century physics. But very few non-experts will have heard either of it or its maker—Emily Noether, a great German mathematician. But it is as essential to twentieth century physics as famous ideas like the impossibility of exceeding the speed of 2 light.” Noether’s theorem proving that symmetries imply conservation laws has been called the most important theo- rem in engineering and physics since the Pythagorean theorem. These symmetries define the limit of all possible conservation laws. Is it possible that, had Emmy Noether been a man, all the conservation laws of physics would be called Noether’s laws? 1 Noether,E.,1918. Invariantevariationsprobleme. Nachr. D. König. Gesellsch. D. Wiss. Zu Göttingen, Math-phys. Klasse 1918, pp. 235–257. An English translationcan be found at http://arxiv.org/PS_cache/physics/pdf/0503/0503066v1.pdf. 2 Dr. Lee Smolin was born in New York City in 1955. He held faculty positions at Yale, Syracuse, and Penn State Universities, where he helped to found the Center for Gravitational Physics and Geometry. In September 2001, he moved to Canada to be a founding member of the Perimeter Institute for Theoretical Physics.4.3 The First Law of Thermodynamics 101 AN EXAMPLE OF MATHEMATICAL SYMMETRY Here is a story about Carl Friedrich Gauss (1777–1855). When he was a young child, his teacher wanted to occupy him for a while, so he asked him to add up all the numbers from 1 to 100. That is, find X=1+2+3+ … + 100. To the teacher’s surprise, Gauss returned a few minutes later and said that the sum was 5050. Apparently Gauss noticed that the sum is the same regardless of whether the terms are added forward (from first to last) or backward (from last to first). In other words, X=1+2+3+ … + 100 = 100 + 99 + 98 + … + 1. If we then add these two ways together, we get X = 1+2+3+…+100 X = 100+99+98+…+1 2X = 101+101+…+101 So 2X = 100 × 101 and X = (100 × 101)/2 = 5050. Gauss had found a mathematical symmetry, and it tremendously simplified the problem. What is conserved here? It is the sum, X. It does not change no matter how you add the numbers. Table 4.1 Relation of Conservation Laws to Mathematical Symmetry Conservation Law Mathematical Symmetry The laws of physics are the same regardless of where we are in space. This positional symmetry implies Linear momentum that linear momentum is conserved. The laws of physics are the same if we rotate about an axis. This rotational symmetry implies that angular Angular momentum momentum is conserved. The laws of physics do not depend on what time it is. This temporal symmetry implies the conservation of Energy energy. The interactions of charged particles with an electromagnetic field remain the same if we multiply the fields Electric charge iφ by a complex number e . This implies the conservation of charge. In summary, Emmy Noether’s theorem shows us that (Table 4.1) ■ Symmetry under translation produces the conservation of linear momentum. ■ Symmetry under rotation produces the conservation of angular momentum. ■ Symmetry in time produces the conservation of energy. ■ Symmetry in magnetic fields produces the conservation of charge. 4.3 THE FIRST LAW OF THERMODYNAMICS In this chapter, we focus our attention on the detailed structure of the first law of thermodynamics. To completely understand this law, we need to study a variety of work and heat energy transport modes and to investigate the basic elements of energy conversion efficiency. An effective general technique for solving thermodynamics pro- blems is presented and illustrated. This technique is used in Chapters 5 and 6 and the remainder of the book. The simplest, most direct statement of the first law of thermodynamics is that energy is conserved. That is, energy can be neither created nor destroyed. The condition of zero energy production was expressed mathematically in Eq. (2.15): E = 0 (2.15) P By differentiating this with respect to time, we obtain an equation for the condition of a zero energy production rate: dE P _ = E = 0 (2.16) p dt Whereas Eqs. (2.15) and (2.16) are accurate and concise statements of the first law of thermodynamics, they are relatively useless by themselves, because they do not contain terms that can be used to calculate other variables. However, if these equations are substituted into the energy balance and energy rate balance equations, then the following equations result. For the energy balance, E = E +E ðas required by the first lawÞ G T P102 CHAPTER 4: The First Law of Thermodynamics and Energy Transport Mechanisms or E = E (4.1) G T The energy rate balance is _ _ _ E = E +E ðas required by the first lawÞ G T P or _ _ E = E (4.2) G T From now on, we frequently use the phrases energy balance and energy rate balance in identifying the proper equation to use in an analysis. So, for simplicity, we introduce the following abbreviations: EB = energy balance and ERB = energy rate balance In Chapter 3, we introduce the components of the total system energy E as the internal energy U, the kinetic 2 3 energy mV /2g , and the potential energy mgZ/g,or c c 2 mgZ mV E = U+ + (3.9) g 2g c c In this equation, V is the magnitude of the velocity of the center of mass of the entire system, Z is the height of the center of mass above a ground (or zero) potential datum, and g is the dimensional proportionality factor c (see Table 1.2 of Chapter 1). In Chapter 3, we also introduce the abbreviated form of this equation: E = U+KE+PE (3.10) and similarly for the specific energy e, 2 gZ E V e = = u+ + (3.12) g c m 2g c and e = u+ke+pe (3.13) In these equations, we continue the practice introduced in Chapter 2 of using uppercase letters to denote extensive properties and lowercase letters to denote intensive (specific) properties. The energy concepts described in these equations are illustrated in Figure 4.1. In equilibrium thermodynamics, the proper energy balance is given by Eq. (4.1), where the gain in energy E is to be interpreted as follows. The system is initially G in some equilibrium state (call it state 1), and after the application of some “pro- cess,” the system ends up in a different equilibrium state (call it state 2). If we Center of Velocity V now add a subscript to each symbol to denote the state at which the property is mass to be evaluated (E is the total energy of the system in state 1 and so forth), then 1 we can writethe energygain ofthesystem as E = Final total energy− Initial total energy (4.3) Internal System (either G energy open or closed) or U E = E −E (4.4) G 2 1 System boundary Height = Z and extendingthis toEq. (3.9),we obtain  mg m 2 2 E = U −U + ðV −V Þ+ ðZ −Z Þ (4.5) G 2 1 2 1 2 1 g 2g c Z = 0 c or  2 2 V −V g 2 1 FIGURE 4.1 E =mu −u + + ðZ −Z Þ (4.6) G 2 1 2 1 g 2g c c System energy components. 3 In this text, we use the symbol V to represent the magnitude of the average velocity V, and the symbol V to represent volume.4.3 The First Law of Thermodynamics 103 alternatively, E = U −U +KE −KE +PE −PE (4.7) G 2 1 2 1 2 1 and E = mðu −u +ke −ke +pe −pe Þ (4.8) G 2 1 2 1 2 1 In most of the engineering situations we encounter, either the system is not moving at all or it is moving without any changein velocityorheight.In these cases, E = U −U = mðu −u Þ = E G 2 1 2 1 T EXAMPLE 4.1 Figure 4.2 shows that 3.00 lbm of saturated water vapor at 10.0 psia is sealed in a rigid container aboard a spaceship traveling at 25,000. mph at an altitude of 200. mi. What energy transport is required to decelerate the water to zero velocity and bring it down to the surface of the Earth such that its final specific internal energy is 950.0 Btu/lbm? Neglect any change in the acceleration of gravity over this distance. p = 10.0 psia, x = 1.00 1 1 u = 950.0 Btu/1bm 2 Z = V =0 2 2 V = 2500. mph 1 Z = 200. miles 1 Sealed rigid container State 1 State 2 FIGURE 4.2 Example 4.1. Solution Let the system in this example be just the water in the container, then the process followed by the water is a constant volume process (the water is in a “rigid, sealed container”). Therefore, the problem statement can be outlined as follows: State 1 m = 3:0lbm, V = constant State 2  p = 10:0psia u = 950:0Btu/lbm 1 2 3 x = 1:00ðÞ saturated vapor v = v = 38:42ft /lbm 1 2 1 3 v = vðÞ at10:0psia = 38:42ft /lbm 1 g Notice how the process path gives us the value of a property (v ) in the final state. To determine the required energy 2 transport, we use the energy balance Eq. (4.1), along with the definition of the energy gain term E from Eq. (4.5): G EB: E = E +E ðas required by the first lawÞ G T P 0 and, assuming g is constant during this process,  mg m 2 2 E = E = U −U + V −V +ðÞ Z −Z G T 2 1 2 1 2 1 g c 2g c Here, V =Z =0, so 2 2 mg m 2 E = U −U − V − Z T 2 1 1 1 g c 2g c Table C.2a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics gives u = uðÞ 10:0psia = 1072:2Btu/lbm 1 g and the problem statement requires that u =950.0 Btu/lbm. Therefore, 2 U = mu =ðÞ 3:00lbmðÞ 1072:2Btu/lbm = 3216:6Btu 1 1 (Continued)104 CHAPTER 4: The First Law of Thermodynamics and Energy Transport Mechanisms EXAMPLE 4.1 (Continued) and = mu =ðÞ 3:00lbmðÞ 950:0Btu/lbm = 2850Btu U 2 2 so   2 5280 ft/mile 3:00lbm E =ð2850− 3216:6Þ Btu− ð25,000: mile/hÞ T 2 3600 s/h 1 Btu  2 3:00 lbmð32:174 ft/s Þ . 778:16ft lbf 1 Btu × − ð200: milesÞð5280 ft/mileÞ . . . lbm ft lbm ft 778:16ft lbf 32:174 32:174 2 2 . . lbf s lbf s =−366:6−80,550−4071 = −85,000Btu ðto three significant figuresÞ Therefore, 85,000 Btu of energy must be transferred out of the water (E is negative here) by some mechanism. This can be T done, for example, by having the spaceship (and the water) do work on the atmosphere by aerodynamic drag as it lands. Exercises 1. What would be the value of u in Example 4.1 if E were zero? Answer: u =29,300 Btu/lbm. (What is the physical state 2 T 2 of the water now?) 2. Which causes the larger change in E : G a. A velocity increase from 0 to 1 ft/s or an increase in height from 0 to 1 ft? b. A velocity increase from 0 to 100 ft/s or a height increase from 0 to 100 ft? Answers: (a) height, (b) velocity. 3. Determine the value of E that must occur when you stop a 1300. kg automobile traveling at 100. km/h on a level road T with no change in internal energy. Answer: E =502 kJ. T _ In nonequilibrium systems, we use the energy rate balance equation with E defined as G  mg d m 2 _ _ E = U+ V + Z = E (4.9) G T g dt 2g c c system Equation (4.9) can become quite complicated for open systems whose total mass is rapidly changing (such as with rockets), because it expands as follows (using U=mu):   2 g gZ V V _ _ _ _ E = m u_ + ðVÞ+ ðZÞ + u+ + m _ = E (4.10) G T g g g c c 2g c c _ _ Notice that, in this equation, V = dV/dt is the magnitude of the instantaneous acceleration, and Z is the magnitude of the instantaneous vertical velocity. The equilibrium thermodynamics energy balance and the nonequilibrium energy rate balance are fairly simple concepts; however, their implementation can be quite complex. Each of the gain, transport, and production terms may expand into many separate terms, all of which must be evaluated in an analysis. Next, we investigate the structure of the energy transport and energy transport rate terms. 4.4 ENERGY TRANSPORT MECHANISMS There are three energy transport mechanisms, any or all of which may be operating in any given system: (1) heat, 4 (2) work, and (3) mass flow. These three mechanisms and their sign conventions are illustrated in Figure 4.3. Note that the sign conventions for heat and work shown in Figure 4.3 are not the same. Heat transfer into a system is taken as positive, whereas work must be produced by or come out of a system to be positive. This is the conventional mechanical engineering sign convention and reflects the traditional view that heat coming out 4 The types of work transports of energy included here are only those due to dissipative or nonconservative forces. For example, the work associated with gravitational or electrostatic forces is not considered a work mode because it is conservative (i.e., it is representable by the gradient of a scalar quantity) and is consequently nondissipative. Energy transports resulting from the actions of conservative forces have their own individual terms in the energy balance equation (such as mgZ/g for the gravitational potential c energy).4.4 Energy Transport Mechanisms 105 System System boundary boundary −W −W +W +W +Q E T E T +Q −Q −Q +E Mass flow −E (a) Closed system (b) Open system FIGURE 4.3 Energy transport mechanisms. WHAT ARE HEAT AND WORK ANYWAY? Heat is usually defined as energy transport to or from a system due to a temperature difference between the system and its surroundings. This can occur by only three modes: conduction, convection, and radiation. Work is more difficult to define. It is often defined as a force moving through a distance, but this is only one type of work; there are many other work modes as well. Since the only energy transport modes for moving energy across a system’s boundary are heat, mass flow, and work, the simplest definition of work is that it is any energy transport mode that is 5 neither heat nor mass flow. 5 Work can also be defined using the concept of a “generalized” force moving through a “generalized” displacement, see Table 4.2 later in this chapter. of a system is “lost” (i.e.,negative),whilework producedbya system(such asanengine)should beassigneda posi- tive value. By definition, a closed system has no mass crossing its system boundary, so it can experience only work and heat transport mechanisms. Also, since the gain, transport, and production terms in the balance equation are defined to be net values (see Eq. (2.10)), we define 1. The net heat transport of energy into a system=∑Q =Q and the net heat transport rate of energy into a i i _ _ system=∑Q = Q: i i 2. The net work transport of energy out of a system=∑W =W and the net work transport rate of energy out of i i _ _ a system=∑W = W : i i 3. The net mass transport of energy into the system=∑ E = ∑E and the net mass transport rate of i i mass flow _ _ energy into the system=∑E =∑E : i i mass flow Thus, for a closed system, the total energy transport becomes E = Q−W (4.11) T and the total energy transport rate is _ _ _ E = Q−W (4.12) T For open systems, the same quantities are E = Q−W+∑Emass (4.13) T flow and _ _ _ _ mass E = Q−W +∑E (4.14) T flow In Eqs. (4.13) and (4.14), note that we write the summation signs on the net mass transport of energy terms, but for simplicity, we do not write the summation signsontheworkorheattransportterms.Thisisbecause you often have open systems with more than one mass flow stream, but seldom do you have more than one106 CHAPTER 4: The First Law of Thermodynamics and Energy Transport Mechanisms _ _ type of work or heat transport present. However, you must always remember that W, W, Q,and Q are also net terms and represent a summation of all the different types of work and heat transports of energy present. This is illustrated in the following example. EXAMPLE 4.2 Determine the energy transport rate for the system shown in Figure 4.4. Fuel flow Top heat loss Exhaust flow E = 15,000 Btu/min 180,000 Btu/h E = 500. Btu/min fuel exhaust System 200. hp Engine−generator set boundary Bottom heat loss 54,000 Btu/h 50.0 hp Electrical workout FIGURE 4.4 Example 4.2 Solution From Eq. (4.14), the total energy transport rate is _ _ _ _ E = Q−W +∑Emass T flow where _ Q = net heat transferintothe system 3 3 3 = −180:×10 Btu/h−54:0×10 Btu/h = −234×10 Btu/h and _ W = net work rateoutof the system = 200:hp+50:0hp = 250:hp while _ ∑Emass = net mass flow of energyintothe system flow 3 3 = 15:0×10 Btu/min−500:Btu/min = 14:5×10 Btu/min So _ . E =ð‒234×103Btu/hÞ½ 1h/ðÞ 60min −ðÞ 250:hp½ 42:4Btu/ðÞ hp min +14:5×103Btu/min = 0:00Btu/min T Exercises 4. Determine the energy transport rate that occurs in Example 4.2 when the work mode directions are reversed. 3 _ Answer: E = 21:2×10 Btu/min: T 5. Determine the net rate of energy gain of a closed system that receives heat at a rate of 4500. kJ/s and produces work at a _ rate of 1500. kJ/s. Answer: E = 3000:KJ/s: G 6. An insulated open system has a net gain of 700. Btu of energy while producing 500. Btu of work. Determine the mass 3 flow energy transport. Answer:E =1.20 × 10 Btu. mass flow The system of Example 4.2 has no net energy transport rate, even though it has six energy transport rates. Note _ _ that the energy rate balance (Eq. (4.2)) for this system is E = E ; therefore, this system also has no net gain of G T energy. That is, the total energy E of this system is constant in time.4.5 Point and Path Functions 107 4.5 POINT AND PATH FUNCTIONS A quantity, say y, that has a value at every point within its range is called a point function. Its derivative is written as dy, and its integral from state 1 to state 2 is Z 2 dy = y −y 2 1 1 Thus, the value of the integral depends only on the values of y at the end points of the integration path and is independent of the actual path taken between these end points. This is a fundamental characteristic of point functions. All intensive and extensive thermodynamic properties are point functions. Therefore, we can write Z Z Z 2 2 2 dE = E −E ; du = u −u ; dm = m −m 2 1 2 1 2 1 1 1 1 and so forth. Aquantity,say x, whose value depends on the path taken between two points within its range is called a path function. Since path functions do not differentiate or integrate in the same manner as point functions, we cannot use the same differential and integral notation for both path and point functions. Instead, we let dx denote the differential of the path function x, and we define its integral over the path from state 1 to state 2 as  Z Z 2 2 dx = x Note: dx≠ ðx −x Þ (4.15) 1 2 2 1 1 1 A path function does not have a value at a point. It has a value only for a path of points, and this value is directly determined by all the points on the path, not just its end points. For example, the area A under the curve of the point function w=f(y) is a path function because dA =wdy = fðyÞdy and Z Z 2 Y2 dA = A = fðyÞdy = area underfyðÞbetween the points y and y 1 2 1 2 1 Y1 Clearly,if the path f(y) is changed, thenthe area A is alsochanged.Consequently,wesaythat A is a path function. 1 2 1 2 We see in the next sections that both the work and heat transports of energy are path functions. Therefore, we write the differentials of these quantities as dW and dQ, and their integrals as Z 2 dW = W (4.16) 1 2 1 and Z 2 dQ = Q (4.17) 1 2 1 Since the associated rate equations contain the time differential, we define power as the work rate, or _ W = dW/dt (4.18) and, similarly, the heat transfer rate is _ Q = dQ/dt (4.19) Each of the different types of work or heat transport of energy is called a mode. A system that has no operating work modes is said to be aergonic. Similarly, a system that changes its state without any work transport of energy having NOTE Z 2 Since work and heat are not thermodynamic properties and therefore not point functions, dW≠W −W and 2 1 Z Z Z 1 2 2 2 dW≠ΔW: Similarly, dQ≠Q −Q , and dQ≠ΔQ: Equations (4.16) and (4.17) are the only correct ways to write these 2 1 1 1 1 path function integrals.108 CHAPTER 4: The First Law of Thermodynamics and Energy Transport Mechanisms WHAT IS AERGONIC ANYWAY? The term aergonic comes from the Greek roots a meaning “not” and ergon meaning “work,” and it should be interpreted to mean “no work has occurred.” It is the analog of the word adiabatic, meaning no heat transfer has occurred, introduced later in this chapter. Substituting Eqs. (4.8) and (4.11) into Eq. (4.1) and rearranging gives the general closed system energy balance equation for a system undergoing a process from state 1 to state 2 as General closed system energy balance: Q − W =ðE −E Þ 1 2 1 2 2 1 (4.20) system 2 2 = m½ðu −u Þ+ðV −V Þ/ð2g Þ+ðZ −Z Þg/g  2 1 c 2 1 c 2 1 system andsubstituting Eq.(4.10)withm= constant andEq. (4.12)intoEq.(4.2) givesthegeneral closed systemenergyratebalance as General closed system energy rate balance: (4.21) _ _ _ _ Q−W =ðdE/dtÞ =ðmu_ +mVV/g +mg Z/g Þ c c system system Similarly, substitutingEqs.(4.9)and(4.14) intoEq.(4.2)gives the generalopen system energy ratebalanceas General open system energy rate balance: (4.22) 2 _ _ _ g g c Q−W +∑Emass =ðÞ d/dtðÞ mu+mV /2g +mZ / c system flow where themass of thesystemis no longer required tobe constant. occurred is said to have undergone an aergonic process. While there are only three modes of heat transport, there are many modes of work transport. In the following segments, four mechanical work modes and five nonmecha- nical work modes are studied in detail. 4.6 MECHANICAL WORK MODES OF ENERGY TRANSPORT In mechanics, we recognize that work is done whenever a force moves through a distance. When this force is a mechanical force F , we call this work mode mechanical work and define it as  . . dW =ðF Þ dx −ðF Þ dx (4.23) applied by the system applied on the system mechanical CAN YOU ANSWER THIS QUESTION FROM 1936? On page 66 of the October 1936 issue of Modern Mechanix is a discussion of the oddities of science that reads: “Modern science states that energy cannot be destroyed. Scientists are now wondering what happens to the energy contained in a compressed spring destroyed in acid.” How would you answer this question more than 70 years later? The person who wrote this in 1936 did not understand the concept of internal energy. Then, neglecting any changes in kinetic or potential energy, an energy balance on the system gives Q − W =ðE −E Þ =ðU −U Þ 1 2 1 2 2 1 2 1 system system where U = U + U =(m u + m u ). Now, U = F(ΔX), the work done in compressing the spring. Finally, 1 acid spring acid acid spring spring spring U = U =(m + m )u . If we make the reasonable assumption that the spring dissolved without any heat 2 acid+spring acid spring acid+spring transfer ( Q = 0) and aergonically ( W = 0), then the energy balance equation gives U = U , and solving it for the final 1 2 1 2 2 1 specific internal energy of the acid-spring solution, we find that m u +m u acid acid spring spring u = acid+spring m +m spring acid So the answer to the 1936 question is this:Theenergycontainedinthecompressedspring ends up as part of the energy of the combined acid-spring solution. That is, since the mechanical work that went into compressing the spring ended up as part of the spring’s internal energy, when the spring was dissolved in the acid, the internal energy in the spring became part of the internal energy of the acid-spring solution. Also, if we assume that the acid-spring solution is a simple incompressible liquid with an internal energy that depends only on temperature, then we can write u = cT, and we see that the energy contained in the compressed spring reappears acid+spring as an increase in the temperature of the resulting acid-spring solution.4.6 Mechanical Work Modes of Energy Transport 109 Area A L p F ± F x ± dx Area A x dx (a) Moving system boundary (c) Elastic work Wire frame T ± F b Film Moving slider dΘ x ± dx (b) Shaft work (d) Surface tension work FIGURE 4.5 Four classical types of mechanical work. or Z Z x2 x2 . . ð W Þ = ðF Þ dx − ðF Þ dx (4.24) 2 1 mechanical applied by the system applied on the system x1 x1 Note that our sign convention requires that work done by the system be positive, while work done on the system be negative. In thermodynamics, the four classical types of mechanical work (Figure 4.5) are 1. Moving system boundary work. 2. Rotating shaft work. 3. Elastic work. 4. Surface tension work. These are very important work modes in engineering analysis and the following material provides a detailed discussion of their major characteristics. 4.6.1 Moving System Boundary Work Whenever a system boundary moves such that the total volume of the system changes, moving system boundary work occurs. This is sometimes called expansion or compression work, and it has wide application in mechanical power technology. In this case, the force is applied by the system through the pressure p (see Figure 4.5a), so . . F = pA and F dx =pA dx = pdV, where p is the pressure acting on the system boundary, A is the area vector (defined to be normal to the system boundary and pointing outward), dx is the differential boundary move- . ment, and dV is the differential volume A dx : Consequently, ðdWÞ =pdV (4.25) moving boundary and for moving boundary work, Moving boundary work: Z 2 ð W Þ = pdV 2 1 (4.26) moving 1 boundary EXAMPLE 4.3 The sealed, rigid tank shown in Figure 4.6 contains air at 0.100 MPa and 20.0°C. The tank is then heated until the pressure in the tank reaches 0.800 MPa. Determine the mechanical moving boundary work produced in this process. (Continued)110 CHAPTER 4: The First Law of Thermodynamics and Energy Transport Mechanisms EXAMPLE 4.3 (Continued) Heated sealed Solution rigid container Let the system be the material inside the tank. The process of heating the tank is one of constant volume (the tank is “rigid”). p = 0.100 MPa 1 Work = ? Therefore, since the system volume, V, is constant, dV = 0 and p = 0.800 MPa 2 T = 20°C 1 the moving boundary work is: Z 2 State 1 State 2 ð W Þ = pdV = 0 2 1 moving 1 boundary FIGURE 4.6 Example 4.3. Therefore, no moving boundaryworkoccursduring this process. Since a “rigid” container cannot change its volume, its moving boundary work is always zero regardless of the process it undergoes. EXAMPLE 4.4 The weather balloon in Figure 4.7 is inflated from a constant pressure, compressed gas source at 20.0 psia. Determine the moving system boundary work as the balloon expands from a diameter of 1.00 ft to 10.0 ft. Solution   3 3 Assume the balloon is a sphere, then V = 4 πR = 1 πD . The process here p = 20.0 psia 2 3 6 is one of constant pressure, so p=constant, and p = 20.0 psia 1 Z Z 2 2 ð W Þ = pdV = p dV = pðV −V Þ 2 1 moving 2 1 D = 10.0 ft dia 1 1 2 boundary D = 1.00 ft dia 1    2 lbf 144in π 3 3 3 = 20:0 ½ð10:0 −1:00 Þft  2 2 in 6 ft 6 . = 1:51×10 ft lbf State 1 State 2 The work is positive because the balloon does work on the atmosphere as it expands and pushes the atmosphere out of the way. FIGURE 4.7 Example 4.4. Exercises 7. In Example 4.3, is the moving boundary work always zero for a sealed, rigid container? Are any other work modes always zero for this type of system? Could a piston-cylinder apparatus be modeled as a sealed, rigid system? Answers: Yes, no, no. (It is sealed and the components, the piston and the cylinder, are rigid, but the piston can move, producing a change in the enclosed volume.) 8. Determine the moving boundary work for the balloon in Example 4.4 as it deflates from a diameter of 10. ft to a dia- meterof5.0ftataconstantpressureof20.psia.Whatdoestheworkontheballoon?Answer:( W ) =–1.3× 1 2 moving boundary 6 10 ft·lbf. The surrounding atmosphere does work on the balloon as it deflates, that is why the work is negative. 9. If the pressure inside a system depends on volume according to the relation p = K +K V+K /V, where K , K , and K 1 2 3 1 2 3 are constants, determine the appropriate equation for the moving boundary work done as the volume changes from 2 2 V to V : Answer:ðÞ W = K ðV −V Þ+K ðV −V Þ/2+K lnðV /V Þ: 1 2 1 2 3 moving boundary 1 2 2 1 2 1 2 1 To carry out the integration indicated in Eq. (4.26), the exact p = pðVÞ pressure volume function must be known. This function is usually given in the process path specification of a problem statement. For example, in Example 4.3, the process is one of constant volume (the container is rigid), so dV = 0; and in Example 4.4, the filling process is isobaric (p=constant), so the integral of Eq. (4.26) is very easy. In general, outside of these two cases, the integration of Eq. (4.26) is not trivial and must be determined with great care. As an example of a nontrivial integration of Eq. (4.26), consider a process that obeys the relation n pV = constant (4.27)4.6 Mechanical Work Modes of Energy Transport 111 or n n p V = p V 1 2 1 2 6 where the exponent n is a constant. Such processes are called polytropic processes. The moving system boundary work of any substance undergoing a polytropic process is Z Z 2 2 constant ð W Þ = pdV = dV 2 polytropic 1 n V 1 1 moving boundary For n=1, this integral becomes V V 2 2 ð W Þ = p V ln = p V ln (4.28) 2 1 2 1 polytropic ðn=1Þ 1 2 V V 1 1 moving boundary and for n≠ 1, it becomes p V −p V 2 1 2 1 ð W Þ = (4.29) 2 polytropicðn≠1Þ, 1 1−n moving boundary If the material undergoing a polytropic process is an ideal gas, then it must simultaneously satisfy both of the following equations: 1. The ideal gas equation of state, pV = mRT: n V = constant: 2. The polytropic process equation, p Combining these two equations by eliminating the pressure p gives n−1 mRTV = constant or, for a fixed mass system, n−1 n−1 T V = T V 1 2 1 2 or 1−n  1−n V 2 T v 2 2 = = (4.30) T V v 1 1 1 Similarly, eliminating V in these two equations (for a fixed mass system) gives the polytropic process equations for an ideal gas: Polytropic process equations for an ideal gas   ðn−1Þ/n 1−n p (4.31) T 2 v 2 2 = = T p v 1 1 1 Finally, if we have an ideal gas undergoing a polytropic process with n≠ 1, then its moving system boundary work is given by Eq. (4.29), with p V −p V = mRðÞ T −T as the polytropic work equation for an ideal gas (n≠ 1): 2 1 2 1 2 1 Polytropic work equation for an ideal gasðÞ n≠1 mR (4.32) ð W Þ = ðT −T Þ 2 2 1 1 polytropicðn≠1Þ 1−n ideal gas moving boundary 6 The term polytropic comes from the Greek roots poly meaning “many” and trope meaning “turns” or “paths.”112 CHAPTER 4: The First Law of Thermodynamics and Energy Transport Mechanisms EXAMPLE 4.5 Figure 4.8 shows a new process in which 0.0100 kg of methane (an ideal gas) is compressed from a pressure of 0.100 MPa and a temperature of 20.0°C to a pressure of 10.0 MPa in a polytropic process with n=1.35. Determine the moving bound- ary work required. Polytropic process with n = 1.35 Methane Methane m = 0.0100 kg m = 0.0100 kg 2 1 p = 10.0 MPa p = 0.100 MPa 2 1 W =? 1 2 = 20.0°C T State 1 1 State 2 FIGURE 4.8 Example 4.5. Solution Since the methane behaves as an ideal gas and n≠ 1, we can find the work required from Eq. (4.32): mR ð W Þ = ðT −T Þ 2 2 1 1 polytropicðn≠1Þ 1−n ideal gas moving boundary where the value of T can be found from Eq. (4.31): 2  ðn−1Þ/n  ð1:35−1Þ/1:35 p 2 10:0MPa T = T =ð20:0+273:15KÞ = 967K = 694°C 2 1 0:100MPa p 1 Using Table C.13b of Thermodynamic Tables to accompany Modern Engineering Thermodynamics to find the value of the gas constant for methane, R =0.518 kJ/kg·K, Eq. (4.32) then gives methane . ð0:0100kgÞð0:518kJ=kg KÞ ð W Þ = ð967−293:15Þ = −9:98kJ 1 2 polytropic ðn≠1Þ 1−1:35 ideal gas moving boundary The work comes out negative, because it is being done on the system. Exercises 10. Determine the work required in Example 4.5 if the final pressure of the methane is 0.500 MPa. Answer:−2.25 kJ. 11. If the work required in Example 4.5 is−5.00 kJ, determine the final temperature and pressure of the methane. Answer: T =631 K, p =1.92 MPa. 2 2 12. If the gas used in Example 4.5 were air, determine the work required to compress it polytropically from 14.7 psia, 70.0°F to 150.°F with n=1.33. Answer: W =−285.1 ft·lbf 1 2 4.6.2 Rotating Shaft Work Whenever a rotating shaft carrying a torque load crosses a system boundary, rotating shaft work is done. In this case (see Figure 4.5b), . ðdWÞ = T dθ (4.33) rotating shaft and, for rotating shaft work, Rotating shaft work Z 2 . = T dθ (4.34) ð W Þ 2 1 rotating 1 shaft where T is the torque vector produced by the system on the shaft and dθ is its angular displacement vector. These two vectors are in the direction of the shaft axis. Normally, thermodynamic problem statements do not require rotating shaft work to be calculated from Eq. (4.34). The rotating shaft work is usually openly given4.6 Mechanical Work Modes of Energy Transport 113 as part of the problem statement. For example, if you are analyzing an automobile internal combustion engine producing 150. ft·lbf of work at the crankshaft, you must be able to recognize that ( W ) = 1 2 rotating shaft 150. ft·lbf. WHEN IS SHAFT WORK NOT SHAFT WORK? Suppose you have a system that contains a fluid, and this fluid is in contact with a mixing blade or an impeller driven by a shaft passing through the system boundary (see Figure 4.9). This would constitute an example of shaft work. Shaft work crosses the system boundary The shaft and the blade or impeller are inside the system and their physical and System thermodynamic properties are part of the system’s properties. You have a hetero- boundary genous system made up of the fluid and the solid shaft and blade. If the mass of the fluid is large enough and the size of the shaft and blade is small enough, then their impact on the system’s properties can be neglected and the system can be Fluid considered to consist of the fluid alone. However, this is not always the case. Sup- pose now you exclude the shaft and the blade or impeller from the system by restricting the system to be only the fluid and redraw the system boundaries so FIGURE 4.9 that they pass along the surface of the shaftandblade(seeFigure4.10).Now, Shaft work in a system containing a fluid. your system consists of a pure substance (the fluid), but what kind of work mode do you now have? No shaft work crosses Since the onlyworkmodes wecananalyze are “reversible,” the fluid mediumcannot the system boundary possess viscosity (fluid friction), and consequently, there can be no shear forces on the blade. The only force a viscousless fluid can exert on the blades is a pressure System force, p. Asthe blade moves, the system boundary must moveaccordingly tokeep up boundary withit,andthepressureforceontheblademust alsomove.Thisisjustthedefinition ofthe movingboundaryworkmode.Consequently, thistype ofshaft workisnotreally Fluid shaft workatall,it is reallymoving boundarywork. Anotherexampleis the shaftwork fromaninternalcombustionengine. It isproduced FIGURE 4.10 inside the engine by moving boundary piston-cylinder work, and in a frictionless A new system boundary that omits the reversible engine, these two work modes are equivalent. However, in a real engine, shaft and the blade. where friction and other losses are present, these two work modes are not equivalent (see Figure 4.11). Not all shaft work can be viewed as moving boundary work. The shaft work from an electric motor or a mechanical gearbox is not equivalent to moving boundary work (see Figure 4.12). Shaft work Shaft work Real Reversible (irreversible) W W W irr rev rev engine engine FIGURE 4.11 Reversible and irreversible work in an IC engine. W elect W shaft−out W W shaft shaft−in Electric motor Mechanical gearbox FIGURE 4.12 Shaft work from systems without internal moving boundaries.114 CHAPTER 4: The First Law of Thermodynamics and Energy Transport Mechanisms 4.6.3 Elastic Work Whenever we compress or extend an elastic solid (like a spring), we perform elastic work. Consider a force ±F applied on the end of an elastic rod (see Figure 4.5c). The normal stress σ in the rod is j F j σ=± (4.35) A where j F j is the magnitude of the force and A is the cross-sectional area of the rod. Since the force F and its . corresponding displacement dx are always in the same direction, the vector dot product F dx always reduces to Fdx,where F = j F j and dx = jdx j, and when the force is applied on the system from the surroundings rather than being produced by the system, the work is negative and its increment is . dW = −F dx = −Fdx = −σAdx (4.36) The strain ε in the rod is defined as dV dx Adx Adx = = = (4.37) dε = L AL V V where L is the length of the rod and AL is its volume V: Then, Adx = dV =Vdε (4.38) and Eq. (4.36) becomes dW = −σAdx = −σVdε (4.39) Therefore, for elastic work, Elastic work Z 2 (4.40) ðÞ W = − σVdε 1 2 elastic 1 EXAMPLE 4.6 Determine an expression for the work involved in deforming a constant volume elastic solid that obeys Hooke’s law of elasti- city (see Figure 4.13). W = ? 1 2 L L+ΔL F = EA(ΔL/L) State 1 State 2 FIGURE 4.13 Example 4.6. Solution Here we have V =constant. Also, from strength of materials we can write Hooke’s law as σ=Eε, where E is Young’s modulus of elasticity. Then, Eq. (4.40) becomes Z Z Z 2 2 2 ð W Þ = − σVdε = − EVεdε = −EV εdε 2 1 elastic 1 1 1  2 2 V ε −ε 2 1 2 2 = −EV = − ðσ −σ Þ 2 1 2 2E 2 2 2 2 Thus, if ε ε , then ð W Þ is negative and work is being put into the system; and if ε ε , then ð W Þ is positive 2 2 2 1 1 elastic 2 1 1 elastic and work is being produced by the system. Note that both tensile strains (ε 0) and compressive strains (ε 0) are possi- 2 ble here. But, the resulting work formula deals only with ε and consequently gives the correct result regardless of the strain direction.4.6 Mechanical Work Modes of Energy Transport 115 Exercises 13. What type of rigid system has zero elastic work regardless of the loading? Answer: A perfectly rigid system (E=∞). 14. If the system analyzed in Example 4.6 was a rectangular steel bar, 1.0 inch square by 12 inches long, determine the elastic 3 2 6 2 work required to stress it from 0.0 to 10. × 10 lbf/in . Use E =30.×10 lbf/in . Answer:( W ) =−1.7 ft·lbf. steel 1 2 elastic 15. Ten joules of elastic work is applied to a circular brass rod 0.0100 m in diameter and 1.00 m long. Determine the 11 resulting stress and strain in the bar if it is initially unloaded. Use E =1.05 × 10 Pa. Answer: σ=164 MPa and brass −3 ε=1.56 × 10 m/m. 4.6.4 Surface Tension Work Surface tension work is the two-dimensional analog of the elastic work just considered. Figure 4.5d shows a soap film on a wire loop. One side of the loop has a movable wire slider that can either compress or extend the film. As in the case of the elastic solid, the force and deflection are always in the same direction and the force is applied to the system, so we can modify Eq. (4.36) to read . dW = − F dx = −Fdx = −ð2σ bÞdx (4.41) s where σ is the surface tension of the film, and b is the length of the moving part of the film. The factor of 2 s appears because the film normally has two surfaces (top and bottom) in contact with air. Now, 2b·dx=dA= change in the film’s surface area, so Eq. (4.41) becomes dW = −σ dA (4.42) s and, for the surface tension work, Surface tension work Z 2 (4.43) ð W Þ 2 = − σ dA 1 s surface tension 1 EXAMPLE 4.7 Determinetheamountofsurfacetensionworkrequiredtoinflatethesoap D = 0 1 W = ? surf. tension D = 0.0500 m bubbleshowninFigure4.14fromadiameterofzeroto0.0500m.Thesurface 2 tension of the soap film can be taken to be a constant 0.0400 N/m. State 1 State 2 Solution FIGURE 4.14 Here, σ =constant=0.0400 N/m. Note that we are not calculating the s Example 4.7. surface area of the bubble here from its geometric elements, but wish only to find the change in area between states 1 and 2. Consequently, the area integral in this instance can be treated as a point function rather than as a path function. So Eq. (4.43) becomes Z 2 ð W Þ = −σ dA = −σ ðA −A Þ 2 s s 2 1 1 surface tension 1 =0. Now, since a soap bubble has two surfaces (the outside and inside films), where A 1  2 0:0500m 2 2 A = 2ð4πR Þ = 2ð4πÞ = 0:0157m 2 2 and 2 ð W Þ = −ð0:0400N/mÞð0:0157−0m Þ 2 1 surface tension −4 −4 . = −6:28 × 10 N m = −6:28 × 10 J −4 −7 = −ð6:28 × 10 JÞð1Btu/1055 JÞ = −5:96 × 10 Btu116 CHAPTER 4: The First Law of Thermodynamics and Energy Transport Mechanisms Table 4.2 Generalized Forces and Generalized Displacements Work Mode Generalized Force F Generalized Displacement dχ Moving system boundary p (pressure) dV (volume) Shaft T (torque) dθ (angular displacement) Elastic −σ (stress) Vdε (volume) Surface tension −σ (surface tension) dA (surface area) s Example 4.7 shows that it would take all of the surface tension energy stored in nearly 2 million 5 cm diameter soap bubbles to raise the temperature of one pound-mass of water by one degree Fahrenheit. Notice that, in each of the four cases of classical mechanical work, the work differential dW was given by the product of what we can call a generalized force F and a generalized displacement dχ; that is, dW = Fdχ (4.44) where F and dχ for each of the four classical mechanical work modes are identified in Table 4.2. In Eq. (4.44), the scalar or dot product is implied if F and dχ are vectors. The application of these work modes may change the thermodynamic state of the system and thus may produce a change in the system’s thermodynamic properties. Finally, note that the generalized forces are all intensive properties, whereas the generalized displacements are all extensive properties. We can generalize the work concept to nonmechanical systems by including any work mode given by Eq. (4.44) when the generalized force F is an intensive property forcing function and the generalized displacement dχ is an extensive property response function. We are now in a position to analyze the remaining work mode energy transport mechanisms. 4.7 NONMECHANICAL WORK MODES OF ENERGY TRANSPORT Of the wide variety of nonmechanical work modes available, the following five are of significant engineering value: 1. Electrical current flow. 2. Electrical polarization. 3. Magnetic. 4. Chemical. 5. Mechanochemical. Materialsareelectricallyclassifiedasconductors,nonconductors(dielectricsorinsulators),andsemiconductors.Apure conductorisasubstancethathasmobilecharges(electrons)freetomoveinanappliedelectricfield.Theyconstitutethe flow of electricalcurrent. Pure nonconductors havenofree electrons whatsoever,and a semiconductor isa material that behavesasadielectric(nonconductor)atlowtemperaturesbutbecomesconductingathighertemperatures. As an electric field E is applied to a pure conductor, the free electrons migrate to the conductor’s outer surface, where they create their own electric field, which opposes the applied field. As more and more electrons reach the outer surface, the electric field inside the object grows weaker and weaker, eventually vanishing altogether. At equilibrium, there is no electric field within a pure conductor. A pure nonconductor has no free electrons with which to neutralize the applied electric field. The externally applied field therefore acts on the internal molecules, and normally nonpolar molecules become polar and develop electric dipoles. Some molecules are naturally polar in the absence of an electric field (e.g., water). The applied electric field rotates and aligns the newly created or naturally polar molecules. Complete alignment is normally prevented by molecular vibrations. But, when the applied field is strong enough to overcome the vibration randomizing effects and further increases in field strength have no effect on the material, the material is said to be saturated by the applied field. The process of electric dipole creation, rotation, and alignment in an applied electric field is known as dielectric polarization. Therefore, two work modes arise from the application of an electric field to a material. The first is the work asso- ciated with the free electron (current) flow, and the second is the work associated with dielectric polarization. For a pure conductor, the polarization work is always zero; and for a pure nonconductor, the current flow work is always zero. We always treat these as separate work modes.4.7 Nonmechanical Work Modes of Energy Transport 117 4.7.1 Electrical Current Flow Work Electrical current flow work occurs whenever current-carrying wires (pure conductors) cross the system boundary. This is the most common type of nonmechanical work mode encountered in thermodynamic system analysis. The generalized force here is the intensive property voltage (the electric potential) ϕ, and the extensive property 7 generalized displacement is the charge q. Then, assuming the voltage is applied to the system, ðdWÞ = −ϕdq electrical current and Z 2 ð W Þ = − ϕdq (4.45) 2 electrical 1 1 current Electrical current i is defined as dq i = dt so dq=idt, and ðdWÞ = −ϕidt (4.46) electrical current Then, electric current work is Electrical current work Z 2 (4.47) ð W Þ = ϕidt 2 1 electrical 1 current From Ohm’s law, the instantaneous voltage ϕ across a pure resistance R carrying an alternating current, described by i=i sin(2πft), is max ϕ = Ri = Ri sinð2πftÞ max where f is the frequency and ϕ =Ri . Thus, Eq. (4.47) gives the electrical current work of n cycles of an max max alternating electrical current applied to a pure resistance from time 0 to time t=n/f as Z t = n/f 2 ð W Þ = −ϕ i sin ð2πftÞdt 2 max max 1 electrical 0 current (4.48) i ðt/2Þ = −ϕ max max 2 2 = −ϕ i t = −ϕ ðt/RÞ = −i Rt e e e e pffiffiffi pffiffiffi where ϕ and i are the effective voltage and current defined by ϕ =ϕ / 2 and i = i / 2: e e e max e max Electrical work can exist in either open or closed systems (we do not consider the flow of electrons across a sys- tem boundary to be a mass flow term). When the electron supply is going into a finite system, such as a battery or a capacitor, Eq. (4.45) or (4.47) is convenient to use. But, when an essentially infinite supply of voltage and current is used, it is more convenient to use the instantaneous rate at which electrical work is done, or the elec- trical power, defined as dW _ ðWÞ = = −ϕi (4.49) electrical dt current OHM’SLAW This law was discovered experimentally by George Simon Ohm (1787–1854) in 1826. Basically, it states that, for a given conductor, the current is directly proportional to the potential difference, usually written as ϕ = Ri, where R is the electrical resistance in units of ohms, where 1 ohm = 1 volt/ampere. 7 The electrical potential ϕ and the electric field strength vector E are related by E=−∇(φ), where∇( ) is the gradient operator.118 CHAPTER 4: The First Law of Thermodynamics and Energy Transport Mechanisms The instantaneous electrical power −ϕi of an alternating current circuit varies in time with the excitation frequency f. However, it is common to report the electrical power of an ac device as the instantaneous power averaged over one cycle of oscillation, or Z Z 1/f 1/f 2 _ ðWÞ = −f ϕidt = −fϕ i sin ð2πftÞdt max max electrical 0 0 (4.50) ðpure resistanceÞ 2 2 = −ϕ i /2 = −ϕ i = −ϕ /R = −i R max e max e e e where ϕ and i are the effective voltage and current defined earlier. e e EXAMPLE 4.8 Consider the 120. V, 144 Ω (ohm), alternating current incandescent lightbulb shown in Figure 4.15 to be a pure resistance. Determine a. The electrical current work when the bulb is operated for 1.50 h. b. Its electrical power consumption. 144 ohm Solution a. Since thevoltageand currentratingsofac devicesarealways givenintermsoftheir a) W = ? for 1.50 hour elect effective values,φ =120. V and, from Ohm’slaw, i =ϕ /R=120./144=0.833 A. e e e b) W= ? Then, from Eq. (4.48), 120 .V ð W Þ = −ϕ i t = −ð120:VÞð0:833AÞð1:50hÞ 2 e 1 e electrical current FIGURE 4.15 . . . = −150:V A h = −150:W h Example 4.8. b. From Eq. (4.50), _ . ðWÞ = −ϕ i = −ð120:VÞð0:833AÞ = −100:V A = −100:W e e electrical current The minus signs appear because electrical work and power go into the system. Exercises 16. Determine the work and power consumption in Example 4.8 when the bulb is operated for 8.00 h instead of 1.50 h. _ Answer:( W ) =−800. W·h, and W =−100. W. 1 2 electrical electrical 17. Determine the effective current drawn by a 1.00 hp ac electric motor operating on a standard 120. V effective power line. Answer: i =6.22 A. e 18. Determine the electrical power dissipated by an 8-bit microprocessor computer chip that draws 90.0 mA at 5.00 V dc. _ =−450. mW. Answer: W electrical 4.7.2 Electrical Polarization Work The electric dipole formation, rotation, and alignment that occur when an electric field is applied to a noncon- ductor or a semiconductor constitutes an electric polarization work mode. The generalized force is the intensive property E (in V/m), the electric field strength vector, and the generalized displacement is the extensive property 2 P (in A·s/m ), the polarization vector of the medium (defined to be the sum of the electric dipole rotation moments of all the molecules in the system). Then, assuming the electric field is applied to the system, . ðdWÞ = − E dP (4.51) electrical polarization and Z 2 . ð W Þ = − E dP (4.52) 2 1 electrical 1 polarization4.7 Nonmechanical Work Modes of Energy Transport 119 Table 4.3 The Electric Susceptibility of Various Materials Material Temperature (°C/°F) χ (dimensionless) e −4 Air (14.7 psia) 20/68 5.36 × 10 Plexiglass 27/81 2.40 Neoprene rubber 24/75 5.7 Glycerine 25/77 41.5 Water 25/77 77.5 Source: Reprinted by permission of the publisher from Zemansky, M. W., Abbott, M. M., Van Ness, H. C., 1975. Basic Engineering Thermodynamics, second ed. McGraw-Hill, New York. Since the effect of the electric field is to orient the dipoles coincident with the field, then E and P are always parallel and point in the same direction. Therefore, if we let the magnitude of E be E and the magnitude of P be P, then Eqs. (4.51) and (4.52) reduce to ðdWÞ = −EdP (4.53) electrical polarization and Z 2 ð W Þ = − EdP (4.54) 2 1 electrical 1 polarization Many substances (particularly gases) correlate well with the following dielectric equation of state: P =ε χ VE (4.55) 0 e −12 2 where V is the volume of the dielectric substance,ε is the electric permittivity of vacuum (8.85419 × 10 N/V ), 0 andχ is the electric susceptibility (a dimensionless number) of the material. Table 4.3 gives values ofχ for various e e materials. EXAMPLE 4.9 120. volts The parallel plate capacitor shown in Figure 4.16 is charged to a potential difference of 120. V at 25.0°C. The plates are square with a side length of 0.100 m and are separated by0.0100 m. If the gap between the plates is filled with water, determine the polarization work required in the charging of the Water at capacitor. 25.0°C 0.100 m square Solution Here,wecanusethedielectricequationofstate,Eq.(4.55).Then,Eq.(4.54) becomes Z Z 2 2  0.0100 m 2 2 ðÞ W = − EdP = − ðε χ VEÞdE = −ε χVE −E /2 1 2 0 0 e e 2 1 electric 1 1 po1arization FIGURE 4.16 Example 4.9. Fromthe problemstatement,wehave 2 −4 3 V = AL =ðÞ 0:100mðÞ 0:0100m = 1:00 × 10 m If we assume that the electrical potential ϕ varies linearly between the plates, then we can write E = j−∇ðϕÞj =ðvoltage differenceÞ/ðplate gapÞwithE = 0ðuncharged platesÞ 1 and 120:V 4 E = = 1:20 × 10 V/mðÞ charged plates 2 0:0100m (Continued)

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