CHAPTER04 - Chapter 4 ENERGY ANALYSIS OF CLOSED SYSTEMS n...

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Unformatted text preview: Chapter 4 ENERGY ANALYSIS OF CLOSED SYSTEMS n Chap. 2, we considered various forms of energy and energy transfer, and we developed a general relation for the conservation of energy principle or energy balance. Then in Chap. 3, we learned how to determine the thermodynamics properties of substances. In this chapter, we apply the energy balance relation to systems that do not involve any mass flow across their boundaries; that is, closed systems. We start this chapter with a discussion of the moving boundary work or P dV work commonly encountered in reciprocating devices such as automotive engines and compressors. We continue by applying the general energy balance relation, which is simply expressed as Ein Eout Esystem, to systems that involve pure substance. Then we define specific heats, obtain relations for the internal energy and enthalpy of ideal gases in terms of specific heats and temperature changes, and perform energy balances on various systems that involve ideal gases. We repeat this for systems that involve solids and liquids, which are approximated as incompressible substances. I Objectives The objectives of Chapter 4 are to: Examine the moving boundary work or P dV work commonly encountered in reciprocating devices such as automotive engines and compressors. Identify the first law of thermodynamics as simply a statement of the conservation of energy principle for closed (fixed mass) systems. Develop the general energy balance applied to closed systems. Define the specific heat at constant volume and the specific heat at constant pressure. Relate the specific heats to the calculation of the changes in internal energy and enthalpy of ideal gases. Describe incompressible substances and determine the changes in their internal energy and enthalpy. Solve energy balance problems for closed (fixed mass) systems that involve heat and work interactions for general pure substances, ideal gases, and incompressible substances. | 165 166 | Thermodynamics INTERACTIVE TUTORIAL 41 MOVING BOUNDARY WORK SEE TUTORIAL CH. 4, SEC. 1 ON THE DVD. The moving boundary GAS FIGURE 41 The work associated with a moving boundary is called boundary work. F One form of mechanical work frequently encountered in practice is associated with the expansion or compression of a gas in a pistoncylinder device. During this process, part of the boundary (the inner face of the piston) moves back and forth. Therefore, the expansion and compression work is often called moving boundary work, or simply boundary work (Fig. 41). Some call it the P dV work for reasons explained later. Moving boundary work is the primary form of work involved in automobile engines. During their expansion, the combustion gases force the piston to move, which in turn forces the crankshaft to rotate. The moving boundary work associated with real engines or compressors cannot be determined exactly from a thermodynamic analysis alone because the piston usually moves at very high speeds, making it difficult for the gas inside to maintain equilibrium. Then the states through which the system passes during the process cannot be specified, and no process path can be drawn. Work, being a path function, cannot be determined analytically without a knowledge of the path. Therefore, the boundary work in real engines or compressors is determined by direct measurements. In this section, we analyze the moving boundary work for a quasiequilibrium process, a process during which the system remains nearly in equilibrium at all times. A quasi-equilibrium process, also called a quasistatic process, is closely approximated by real engines, especially when the piston moves at low velocities. Under identical conditions, the work output of the engines is found to be a maximum, and the work input to the compressors to be a minimum when quasi-equilibrium processes are used in place of nonquasi-equilibrium processes. Below, the work associated with a moving boundary is evaluated for a quasi-equilibrium process. Consider the gas enclosed in the pistoncylinder device shown in Fig. 42. The initial pressure of the gas is P, the total volume is V, and the crosssectional area of the piston is A. If the piston is allowed to move a distance ds in a quasi-equilibrium manner, the differential work done during this process is dWb F ds PA ds P dV (41) A P GAS ds FIGURE 42 A gas does a differential amount of work dWb as it forces the piston to move by a differential amount ds. That is, the boundary work in the differential form is equal to the product of the absolute pressure P and the differential change in the volume dV of the system. This expression also explains why the moving boundary work is sometimes called the P dV work. Note in Eq. 41 that P is the absolute pressure, which is always positive. However, the volume change dV is positive during an expansion process (volume increasing) and negative during a compression process (volume decreasing). Thus, the boundary work is positive during an expansion process and negative during a compression process. Therefore, Eq. 41 can be viewed as an expression for boundary work output, Wb,out. A negative result indicates boundary work input (compression). The total boundary work done during the entire process as the piston moves is obtained by adding all the differential works from the initial state to the final state: 2 Wb 1 P dV 1kJ2 (42) Chapter 4 This integral can be evaluated only if we know the functional relationship between P and V during the process. That is, P f (V) should be available. Note that P f (V) is simply the equation of the process path on a P-V diagram. The quasi-equilibrium expansion process described is shown on a P-V diagram in Fig. 4 3. On this diagram, the differential area dA is equal to P dV, which is the differential work. The total area A under the process curve 12 is obtained by adding these differential areas: 2 2 | 167 P 1 Process path 2 dA = P dV V1 dV V2 V Area A 1 dA 1 P dV (43) P A comparison of this equation with Eq. 42 reveals that the area under the process curve on a P-V diagram is equal, in magnitude, to the work done during a quasi-equilibrium expansion or compression process of a closed system. (On the P-v diagram, it represents the boundary work done per unit mass.) A gas can follow several different paths as it expands from state 1 to state 2. In general, each path will have a different area underneath it, and since this area represents the magnitude of the work, the work done will be different for each process (Fig. 4 4). This is expected, since work is a path function (i.e., it depends on the path followed as well as the end states). If work were not a path function, no cyclic devices (car engines, power plants) could operate as work-producing devices. The work produced by these devices during one part of the cycle would have to be consumed during another part, and there would be no net work output. The cycle shown in Fig. 45 produces a net work output because the work done by the system during the expansion process (area under path A) is greater than the work done on the system during the compression part of the cycle (area under path B), and the difference between these two is the net work done during the cycle (the colored area). If the relationship between P and V during an expansion or a compression process is given in terms of experimental data instead of in a functional form, obviously we cannot perform the integration analytically. But we can always plot the P-V diagram of the process, using these data points, and calculate the area underneath graphically to determine the work done. Strictly speaking, the pressure P in Eq. 42 is the pressure at the inner surface of the piston. It becomes equal to the pressure of the gas in the cylinder only if the process is quasi-equilibrium and thus the entire gas in the cylinder is at the same pressure at any given time. Equation 42 can also be used for nonquasi-equilibrium processes provided that the pressure at the inner face of the piston is used for P. (Besides, we cannot speak of the pressure of a system during a nonquasi-equilibrium process since properties are defined for equilibrium states only.) Therefore, we can generalize the boundary work relation by expressing it as 2 FIGURE 43 The area under the process curve on a P-V diagram represents the boundary work. P 1 WA = 10 kJ WB = 8 kJ WC = 5 kJ A B C 2 V1 V2 V FIGURE 44 The boundary work done during a process depends on the path followed as well as the end states. P 2 A Wnet B 1 V2 V1 V Wb 1 Pi dV (44) where Pi is the pressure at the inner face of the piston. Note that work is a mechanism for energy interaction between a system and its surroundings, and Wb represents the amount of energy transferred from the system during an expansion process (or to the system during a FIGURE 45 The net work done during a cycle is the difference between the work done by the system and the work done on the system. 168 | Thermodynamics compression process). Therefore, it has to appear somewhere else and we must be able to account for it since energy is conserved. In a car engine, for example, the boundary work done by the expanding hot gases is used to overcome friction between the piston and the cylinder, to push atmospheric air out of the way, and to rotate the crankshaft. Therefore, 2 Wb Wfriction Watm Wcrank 1 1Ffriction Patm A Fcrank 2 dx (45) Of course the work used to overcome friction appears as frictional heat and the energy transmitted through the crankshaft is transmitted to other components (such as the wheels) to perform certain functions. But note that the energy transferred by the system as work must equal the energy received by the crankshaft, the atmosphere, and the energy used to overcome friction. The use of the boundary work relation is not limited to the quasi-equilibrium processes of gases only. It can also be used for solids and liquids. EXAMPLE 41 Boundary Work for a Constant-Volume Process A rigid tank contains air at 500 kPa and 150C. As a result of heat transfer to the surroundings, the temperature and pressure inside the tank drop to 65C and 400 kPa, respectively. Determine the boundary work done during this process. Solution Air in a rigid tank is cooled, and both the pressure and temperature drop. The boundary work done is to be determined. Analysis A sketch of the system and the P-V diagram of the process are shown in Fig. 46. The boundary work can be determined from Eq. 42 to be 2 Wb 1 0 P dV 0 Discussion This is expected since a rigid tank has a constant volume and dV 0 in this equation. Therefore, there is no boundary work done during this process. That is, the boundary work done during a constant-volume process is always zero. This is also evident from the P-V diagram of the process (the area under the process curve is zero). P, kPa Heat 500 1 AIR P1 = 500 kPa T1 = 150C FIGURE 46 Schematic and P-V diagram for Example 41. P2 = 400 kPa T2 = 65C 400 2 V Chapter 4 EXAMPLE 42 Boundary Work for a Constant-Pressure Process | 169 A frictionless pistoncylinder device contains 10 lbm of steam at 60 psia and 320 F. Heat is now transferred to the steam until the temperature reaches 400 F. If the piston is not attached to a shaft and its mass is constant, determine the work done by the steam during this process. Solution Steam in a piston cylinder device is heated and the temperature rises at constant pressure. The boundary work done is to be determined. Analysis A sketch of the system and the P-v diagram of the process are shown in Fig. 47. Assumption The expansion process is quasi-equilibrium. Analysis Even though it is not explicitly stated, the pressure of the steam within the cylinder remains constant during this process since both the atmospheric pressure and the weight of the piston remain constant. Therefore, this is a constant-pressure process, and, from Eq. 42 2 2 Wb 1 P dV P0 1 dV P0 1V2 v1 2 V1 2 (46) or Wb mP0 1v2 since V mv. From the superheated vapor table (Table A6E), the specific volumes are determined to be v1 7.4863 ft3/lbm at state 1 (60 psia, 320 F) and v2 8.3548 ft3/lbm at state 2 (60 psia, 400 F). Substituting these values yields Wb 110 lbm2 160 psia 2 3 18.3548 96.4 Btu 7.48632 ft3>lbm4 a 1 Btu b 5.404 psia # ft3 Discussion The positive sign indicates that the work is done by the system. That is, the steam used 96.4 Btu of its energy to do this work. The magnitude of this work could also be determined by calculating the area under the process curve on the P-V diagram, which is simply P0 V for this case. P, psia 1 60 H2O m = 10 lbm P = 60 psia P0 = 60 psia 2 Heat Area = wb v1 = 7.4863 v2 = 8.3548 v, ft3/lbm FIGURE 47 Schematic and P-v diagram for Example 42. 170 | Thermodynamics EXAMPLE 43 Isothermal Compression of an Ideal Gas A pistoncylinder device initially contains 0.4 m3 of air at 100 kPa and 80C. The air is now compressed to 0.1 m3 in such a way that the temperature inside the cylinder remains constant. Determine the work done during this process. Solution Air in a pistoncylinder device is compressed isothermally. The boundary work done is to be determined. Analysis A sketch of the system and the P-V diagram of the process are shown in Fig. 48. Assumptions 1 The compression process is quasi-equilibrium. 2 At specified conditions, air can be considered to be an ideal gas since it is at a high temperature and low pressure relative to its critical-point values. Analysis For an ideal gas at constant temperature T0, PV mRT0 C or P C V where C is a constant. Substituting this into Eq. 42, we have 2 2 Wb 1 P dV 1 C dV V 2 C 1 dV V C ln V2 V1 P1V1 ln V2 V1 (47) In Eq. 47, P1V1 can be replaced by P2V2 or mRT0. Also, V2/V1 can be replaced by P1/P2 for this case since P1V1 P2V2. Substituting the numerical values into Eq. 47 yields Wb 1100 kPa2 10.4 m3 2 a ln 0.1 1 kJ ba b 0.4 1 kPa # m3 55.5 kJ Discussion The negative sign indicates that this work is done on the system (a work input), which is always the case for compression processes. P 2 T0 = 80C = const. AIR V1 = 0.4 m3 P1 = 100 kPa T0 = 80C = const. 0.1 1 0.4 V, m3 FIGURE 48 Schematic and P-V diagram for Example 43. Chapter 4 P 1 n n P1V1 = P2V 2 | 171 P1 PV n = const. GAS PV n = C = const. P2 2 V1 V2 V FIGURE 49 Schematic and P-V diagram for a polytropic process. Polytropic Process During actual expansion and compression processes of gases, pressure and C, where n and C are constants. A volume are often related by PV n process of this kind is called a polytropic process (Fig. 49). Below we develop a general expression for the work done during a polytropic process. The pressure for a polytropic process can be expressed as P CV n EXPERIMENT (48) Substituting this relation into Eq. 42, we obtain 2 2 Wb 1 P dV 1 CV n dV C V2 n 1 V1 n n 1 1 P2V2 1 P1V1 n (49) since C P1V1n P2V2n. For an ideal gas (PV also be written as Wb mR 1T2 1 n T1 2 n 1 mRT), this equation can 1kJ2 (410) For the special case of n 2 1 the boundary work becomes 2 Wb 1 P dV 1 CV 1 dV PV ln a V2 b V1 Use actual data from the experiment shown here to find the polytropic exponent for expanding air. See end-of-chapter problem 4174. Ronald Mullisen For an ideal gas this result is equivalent to the isothermal process discussed in the previous example. EXAMPLE 44 Expansion of a Gas against a Spring A pistoncylinder device contains 0.05 m3 of a gas initially at 200 kPa. At this state, a linear spring that has a spring constant of 150 kN/m is touching the piston but exerting no force on it. Now heat is transferred to the gas, causing the piston to rise and to compress the spring until the volume inside the cylinder doubles. If the cross-sectional area of the piston is 0.25 m2, determine (a) the final pressure inside the cylinder, (b) the total work done by 172 | Thermodynamics the gas, and (c) the fraction of this work done against the spring to compress it. Solution A gas in a pistoncylinder device equipped with a linear spring expands as a result of heating. The final gas pressure, the total work done, and the fraction of the work done to compress the spring are to be determined. Assumptions 1 The expansion process is quasi-equilibrium. 2 The spring is linear in the range of interest. Analysis A sketch of the system and the P-V diagram of the process are shown in Fig. 410. (a) The enclosed volume at the final state is V2 2V1 122 10.05 m3 2 10.1 0.052 m3 0.25 m2 0.1 m3 Then the displacement of the piston (and of the spring) becomes x V A 0.2 m The force applied by the linear spring at the final state is F kx 1150 kN>m2 10.2 m2 F A 30 kN 0.25 m2 30 kN The additional pressure applied by the spring on the gas at this state is P 120 kPa Without the spring, the pressure of the gas would remain constant at 200 kPa while the piston is rising. But under the effect of the spring, the pressure rises linearly from 200 kPa to 200 at the final state. 120 320 kPa (b) An easy way of finding the work done is to plot the process on a P-V diagram and find the area under the process curve. From Fig. 410 the area under the process curve (a trapezoid) is determined to be W area 1200 3202 kPa 2 3 10.1 0.052 m3 4 a 1 kJ b 1 kPa # m3 13 kJ k = 150 kN/m P, kPa 320 II 200 A = 0.25 m2 P1 = 200 kPa V1 = 0.05 m3 0.05 Heat I 0.1 V, m3 FIGURE 410 Schematic and P-V diagram for Example 44. Chapter 4 Note that the work is done by the system. (c) The work represented by the rectangular area (region I) is done against the piston and the atmosphere, and the work represented by the triangular area (region II) is done against the spring. Thus, | 173 Wspring Discussion 1 23 1320 x2 2 1 200 2 kPa4 10.05 m3 2 a 1 2 1150 1 kJ b 1 kPa # m3 02 4 a 3 kJ This result could also be obtained from 1 2 2 k 1x2 Wspring kN>m2 3 10.2 m2 2 1 kJ b 1 kN # m 3 kJ 42 ENERGY BALANCE FOR CLOSED SYSTEMS E in E out E system 1kJ2 Energy balance for any system undergoing any kind of process was expressed as (see Chap. 2) (411) Net energy transfer by heat, work, and mass INTERACTIVE TUTORIAL SEE TUTORIAL CH. 4, SEC. 2 ON THE DVD. or, in the rate form, as . E in . E out Rate of net energy transfer by heat, work, and mass For constant rates, the total quantities during a time interval t are related to the quantities per unit time as Q # Q t, W # W t, and E 1dE>dt2 t 1kJ>kg 2 1kJ2 (413) The energy balance can be expressed on a per unit mass basis as ein eout esystem (414) which is obtained by dividing all the quantities in Eq. 411 by the mass m of the system. Energy balance can also be expressed in the differential form as dE in dE out dE system or dein deout desystem (415) For a closed system undergoing a cycle, the initial and final states are identical, and thus Esystem E2 E1 0. Then the energy balance for a cycle simplifies to Ein Eout 0 or Ein Eout. Noting that a closed system does not involve any mass flow across its boundaries, the energy balance for a cycle can be expressed in terms of heat and work interactions as Wnet,out Q net,in or # Wnet,out # Q net,in 1for a cycle 2 (416) That is, the net work output during a cycle is equal to net heat input (Fig. 411). Change in internal, kinetic, potential, etc., energies dE system>dt 1kW2 (412) Rate of change in internal, kinetic, potential, etc., energies P Qnet = Wnet V FIGURE 411 For a cycle E 0, thus Q W. 174 | Thermodynamics The energy balance (or the first-law) relations already given are intuitive in nature and are easy to use when the magnitudes and directions of heat and work transfers are known. However, when performing a general analytical study or solving a problem that involves an unknown heat or work interaction, we need to assume a direction for the heat or work interactions. In such cases, it is common practice to use the classical thermodynamics sign convention and to assume heat to be transferred into the system (heat input) in the amount of Q and work to be done by the system (work output) in the amount of W, and then to solve the problem. The energy balance relation in that case for a closed system becomes Q net,in Wnet,out E system or Q W E (417) General Q W = E Stationary systems Q W = U Per unit mass q w = e Differential form q w = de FIGURE 412 Various forms of the first-law relation for closed systems. EXPERIMENT where Q Qnet,in Qin Qout is the net heat input and W Wnet,out Wout Win is the net work output. Obtaining a negative quantity for Q or W simply means that the assumed direction for that quantity is wrong and should be reversed. Various forms of this "traditional" first-law relation for closed systems are given in Fig. 412. The first law cannot be proven mathematically, but no process in nature is known to have violated the first law, and this should be taken as sufficient proof. Note that if it were possible to prove the first law on the basis of other physical principles, the first law then would be a consequence of those principles instead of being a fundamental physical law itself. As energy quantities, heat and work are not that different, and you probably wonder why we keep distinguishing them. After all, the change in the energy content of a system is equal to the amount of energy that crosses the system boundaries, and it makes no difference whether the energy crosses the boundary as heat or work. It seems as if the first-law relations would be much simpler if we had just one quantity that we could call energy interaction to represent both heat and work. Well, from the first-law point of view, heat and work are not different at all. From the second-law point of view, however, heat and work are very different, as is discussed in later chapters. EXAMPLE 45 Electric Heating of a Gas at Constant Pressure A pistoncylinder device contains 25 g of saturated water vapor that is maintained at a constant pressure of 300 kPa. A resistance heater within the cylinder is turned on and passes a current of 0.2 A for 5 min from a 120-V source. At the same time, a heat loss of 3.7 kJ occurs. (a) Show that for a closed system the boundary work Wb and the change in internal energy U in the first-law relation can be combined into one term, H, for a constantpressure process. (b) Determine the final temperature of the steam. Solution Saturated water vapor in a pistoncylinder device expands at constant pressure as a result of heating. It is to be shown that U Wb H, and the final temperature is to be determined. Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero, KE PE 0. Therefore, E U and internal energy is the only form of energy of the system that may change during this process. 2 Electrical wires constitute a very small part of the system, and thus the energy change of the wires can be neglected. Use actual data from the experiment shown here to verify the first law of thermodynamics. See end-of-chapter problem 4175. Ronald Mullisen Chapter 4 P, kPa | 175 H 2O m = 25 g P1 = P2 = 300 kPa Sat. vapor 0.2 A 120 V 5 min Qout = 3.7 kJ 300 1 2 EXPERIMENT v FIGURE 413 Schematic and P-v diagram for Example 45. Analysis We take the contents of the cylinder, including the resistance wires, as the system (Fig. 413). This is a closed system since no mass crosses the system boundary during the process. We observe that a pistoncylinder device typically involves a moving boundary and thus boundary work Wb. The pressure remains constant during the process and thus P2 P1. Also, heat is lost from the system and electrical work We is done on the system. (a) This part of the solution involves a general analysis for a closed system undergoing a quasi-equilibrium constant-pressure process, and thus we consider a general closed system. We take the direction of heat transfer Q to be to the system and the work W to be done by the system. We also express the work as the sum of boundary and other forms of work (such as electrical and shaft). Then the energy balance can be expressed as Use actual data from the experiment shown here to verify the first law of thermodynamics. See end-of-chapter problem 4176. Ronald Mullisen E in E out E system Change in internal, kinetic, potential, etc., energies Net energy transfer by heat, work, and mass Q Q Wother W Wb U U2 Q However, 0 0 KE PE U1 EXPERIMENT For a constant-pressure process, the boundary work is given as Wb P0(V2 V1). Substituting this into the preceding relation gives Wother S Q P0 1V2 V1 2 U2 P2V2 2 1kJ2 U1 1U1 P1V1 2 (418) P0 Also H U P2 P1 Q Wother H2 H1 1U2 PV, and thus Wother which is the desired relation (Fig. 414). This equation is very convenient to use in the analysis of closed systems undergoing a constant-pressure quasiequilibrium process since the boundary work is automatically taken care of by the enthalpy terms, and one no longer needs to determine it separately. Use actual data from the experiment shown here to verify the first law of thermodynamics. See end-of-chapter problem 4177. Ronald Mullisen 176 | Thermodynamics EXPERIMENT (b) The only other form of work in this case is the electrical work, which can be determined from We VI t 1120 V2 10.2 A2 1300 s2 a h1 h g @ 300 kPa 1 kJ>s 1000 VA b 7.2 kJ 1Table A52 State 1: P1 300 kPa f sat. vapor 2724.9 kJ>kg The enthalpy at the final state can be determined directly from Eq. 418 by expressing heat transfer from the system and work done on the system as negative quantities (since their directions are opposite to the assumed directions). Alternately, we can use the general energy balance relation with the simplification that the boundary work is considered automatically by replacing U by H for a constant-pressure expansion or compression process: Use actual data from the experiment shown here to verify the first law of thermodynamics. See end-of-chapter problem 4178. Ronald Mullisen E in E out E system Net energy transfer by heat, work, and mass U H Change in internal, kinetic, potential, etc., energies We,in Q out We,in Wb Q out 3.7 kJ h2 7.2 kJ 10.025 kg2 1h 2 m 1h 2 h1 2 2724.9 2 kJ>kg 1since P constant 2 2864.9 kJ>kg Now the final state is completely specified since we know both the pressure and the enthalpy. The temperature at this state is State 2: P = const. P2 h2 300 kPa f 2864.9 kJ>kg T2 200C 1Table A6 2 H Q W other Wb = U Q W other = H Therefore, the steam will be at 200C at the end of this process. Discussion Strictly speaking, the potential energy change of the steam is not zero for this process since the center of gravity of the steam rose somewhat. Assuming an elevation change of 1 m (which is rather unlikely), the change in the potential energy of the steam would be 0.0002 kJ, which is very small compared to the other terms in the first-law relation. Therefore, in problems of this kind, the potential energy term is always neglected. FIGURE 414 For a closed system undergoing a quasi-equilibrium, P constant process, U Wb H. EXAMPLE 46 Unrestrained Expansion of Water A rigid tank is divided into two equal parts by a partition. Initially, one side of the tank contains 5 kg of water at 200 kPa and 25C, and the other side is evacuated. The partition is then removed, and the water expands into the entire tank. The water is allowed to exchange heat with its surroundings until the temperature in the tank returns to the initial value of 25C. Determine (a) the volume of the tank, (b) the final pressure, and (c) the heat transfer for this process. Solution One half of a rigid tank is filled with liquid water while the other side is evacuated. The partition between the two parts is removed and water is allowed to expand and fill the entire tank while the temperature is maintained constant. The volume of tank, the final pressure, and the heat transfer are to be to determined. Chapter 4 Assumptions 1 The system is stationary and thus the kinetic and potential energy changes are zero, KE PE 0 and E U. 2 The direction of heat transfer is to the system (heat gain, Qin). A negative result for Qin indicates the assumed direction is wrong and thus it is a heat loss. 3 The volume of the rigid tank is constant, and thus there is no energy transfer as boundary work. 4 The water temperature remains constant during the process. 5 There is no electrical, shaft, or any other kind of work involved. Analysis We take the contents of the tank, including the evacuated space, as the system (Fig. 415). This is a closed system since no mass crosses the system boundary during the process. We observe that the water fills the entire tank when the partition is removed (possibly as a liquidvapor mixture). (a) Initially the water in the tank exists as a compressed liquid since its pressure (200 kPa) is greater than the saturation pressure at 25C (3.1698 kPa). Approximating the compressed liquid as a saturated liquid at the given temperature, we find | 177 v1 vf @ 25C V1 0.001003 m3>kg 0.001 m3>kg 1Table A4 2 Then the initial volume of the water is mv1 15 kg2 10.001 m3>kg2 12 2 10.005 m3 2 V2 m 0.01 m3 5 kg 0.005 m3 The total volume of the tank is twice this amount: Vtank 0.01 m3 (b) At the final state, the specific volume of the water is v2 0.002 m3>kg which is twice the initial value of the specific volume. This result is expected since the volume doubles while the amount of mass remains constant. At 25C: vf 0.001003 m3>kg and vg 43.340 m3>kg 1Table A4 2 1Table A4 2 Since vf v2 vg, the water is a saturated liquidvapor mixture at the final state, and thus the pressure is the saturation pressure at 25C: P2 Psat @ 25C 3.1698 kPa P, kPa System boundary Evacuated space H2 O m = 5 kg P1 = 200 kPa T1 = 25 C Partition 200 1 3.17 Qin 2 v FIGURE 415 Schematic and P-v diagram for Example 46. 178 | Thermodynamics Vacuum P=0 W=0 (c) Under stated assumptions and observations, the energy balance on the system can be expressed as E in E out m 1u2 E system Change in internal, kinetic, potential, etc., energies Net energy transfer by heat, work, and mass H 2O Heat Qin U u1 2 FIGURE 416 Expansion against a vacuum involves no work and thus no energy transfer. Notice that even though the water is expanding during this process, the system chosen involves fixed boundaries only (the dashed lines) and therefore the moving boundary work is zero (Fig. 416). Then W 0 since the system does not involve any other forms of work. (Can you reach the same conclusion by choosing the water as our system?) Initially, u1 uf @ 25C 104.83 kJ>kg The quality at the final state is determined from the specific volume information: 1 kg IRON 20 30C 1 kg WATER 20 30C x2 Then v2 vfg uf vf 0.002 43.34 0.001 0.001 2.3 10 5 u2 x2ufg 12.3 10 5 2 12304.3 kJ>kg2 4.5 kJ 41.8 kJ 104.83 kJ>kg 104.88 kJ>kg FIGURE 417 It takes different amounts of energy to raise the temperature of different substances by the same amount. Substituting yields Qin 15 kg2 3 1104.88 104.83 2 kJkg4 0.25 kJ Discussion The positive sign indicates that the assumed direction is correct, and heat is transferred to the water. m = 1 kg T = 1C Specific heat = 5 kJ/kg C 43 SPECIFIC HEATS 5 kJ FIGURE 418 Specific heat is the energy required to raise the temperature of a unit mass of a substance by one degree in a specified way. INTERACTIVE TUTORIAL SEE TUTORIAL CH. 4, SEC. 3 ON THE DVD. We know from experience that it takes different amounts of energy to raise the temperature of identical masses of different substances by one degree. For example, we need about 4.5 kJ of energy to raise the temperature of 1 kg of iron from 20 to 30C, whereas it takes about 9 times this energy (41.8 kJ to be exact) to raise the temperature of 1 kg of liquid water by the same amount (Fig. 417). Therefore, it is desirable to have a property that will enable us to compare the energy storage capabilities of various substances. This property is the specific heat. The specific heat is defined as the energy required to raise the temperature of a unit mass of a substance by one degree (Fig. 418). In general, this energy depends on how the process is executed. In thermodynamics, we are interested in two kinds of specific heats: specific heat at constant volume cv and specific heat at constant pressure cp. Physically, the specific heat at constant volume cv can be viewed as the energy required to raise the temperature of the unit mass of a substance by one degree as the volume is maintained constant. The energy required to Chapter 4 do the same as the pressure is maintained constant is the specific heat at constant pressure cp. This is illustrated in Fig. 419. The specific heat at constant pressure cp is always greater than cv because at constant pressure the system is allowed to expand and the energy for this expansion work must also be supplied to the system. Now we attempt to express the specific heats in terms of other thermodynamic properties. First, consider a fixed mass in a stationary closed system undergoing a constant-volume process (and thus no expansion or compression eout esystem work is involved). The conservation of energy principle ein for this process can be expressed in the differential form as dein deout du | 179 (2) (1) V = constant m = 1 kg T = 1C cv = 3.12 kJ kg. C P = constant m = 1 kg T = 1 C cp = 5.19 kJ kg.C 3.12 kJ 5.19 kJ The left-hand side of this equation represents the net amount of energy transferred to the system. From the definition of cv, this energy must be equal to cv dT, where dT is the differential change in temperature. Thus, cv d T du at constant volume FIGURE 419 Constant-volume and constantpressure specific heats cv and cp (values given are for helium gas). or cv a 0u b 0T v (419) Similarly, an expression for the specific heat at constant pressure cp can be obtained by considering a constant-pressure expansion or compression process. It yields cp a 0h b 0T p (420) cv = u Equations 4 19 and 4 20 are the defining equations for cv and cp, and their interpretation is given in Fig. 4 20. Note that cv and cp are expressed in terms of other properties; thus, they must be properties themselves. Like any other property, the specific heats of a substance depend on the state that, in general, is specified by two independent, intensive properties. That is, the energy required to raise the temperature of a substance by one degree is different at different temperatures and pressures (Fig. 421). But this difference is usually not very large. A few observations can be made from Eqs. 419 and 420. First, these equations are property relations and as such are independent of the type of processes. They are valid for any substance undergoing any process. The only relevance cv has to a constant-volume process is that cv happens to be the energy transferred to a system during a constant-volume process per unit mass per unit degree rise in temperature. This is how the values of cv are determined. This is also how the name specific heat at constant volume originated. Likewise, the energy transferred to a system per unit mass per unit temperature rise during a constant-pressure process happens to be equal to cp. This is how the values of cp can be determined and also explains the origin of the name specific heat at constant pressure. Another observation that can be made from Eqs. 419 and 420 is that cv is related to the changes in internal energy and cp to the changes in enthalpy. In fact, it would be more proper to define cv as the change in the internal energy of a substance per unit change in temperature at constant ( ( T v = the change in internal energy with temperature at constant volume cp = h T p = the change in enthalpy with temperature at constant pressure ( ( FIGURE 420 Formal definitions of cv and cp. 180 | AIR m = 1 kg Thermodynamics AIR m = 1 kg 1000 1001 K 300 301 K 0.718 kJ 0.855 kJ FIGURE 421 The specific heat of a substance changes with temperature. volume. Likewise, cp can be defined as the change in the enthalpy of a substance per unit change in temperature at constant pressure. In other words, cv is a measure of the variation of internal energy of a substance with temperature, and cp is a measure of the variation of enthalpy of a substance with temperature. Both the internal energy and enthalpy of a substance can be changed by the transfer of energy in any form, with heat being only one of them. Therefore, the term specific energy is probably more appropriate than the term specific heat, which implies that energy is transferred (and stored) in the form of heat. A common unit for specific heats is kJ/kg C or kJ/kg K. Notice that these two units are identical since T(C) T(K), and 1C change in temperature is equivalent to a change of 1 K. The specific heats are some times given on a molar basis. They are then denoted by cv and cp and have the unit kJ/kmol C or kJ/kmol K. INTERACTIVE TUTORIAL SEE TUTORIAL CH. 4, SEC. 4 ON THE DVD. 44 INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF IDEAL GASES Pv RT We defined an ideal gas as a gas whose temperature, pressure, and specific volume are related by It has been demonstrated mathematically (Chap. 12) and experimentally (Joule, 1843) that for an ideal gas the internal energy is a function of the temperature only. That is, u u 1T2 (421) Thermometer WATER AIR (high pressure) Evacuated FIGURE 422 Schematic of the experimental apparatus used by Joule. In his classical experiment, Joule submerged two tanks connected with a pipe and a valve in a water bath, as shown in Fig. 422. Initially, one tank contained air at a high pressure and the other tank was evacuated. When thermal equilibrium was attained, he opened the valve to let air pass from one tank to the other until the pressures equalized. Joule observed no change in the temperature of the water bath and assumed that no heat was transferred to or from the air. Since there was also no work done, he concluded that the internal energy of the air did not change even though the volume and the pressure changed. Therefore, he reasoned, the internal energy is a function of temperature only and not a function of pressure or specific volume. (Joule later showed that for gases that deviate significantly from ideal-gas behavior, the internal energy is not a function of temperature alone.) Using the definition of enthalpy and the equation of state of an ideal gas, we have h Pv u Pv f RT h u RT Since R is constant and u u(T), it follows that the enthalpy of an ideal gas is also a function of temperature only: h h 1T2 (422) Chapter 4 Since u and h depend only on temperature for an ideal gas, the specific heats cv and cp also depend, at most, on temperature only. Therefore, at a given temperature, u, h, cv, and cp of an ideal gas have fixed values regardless of the specific volume or pressure (Fig. 4 23). Thus, for ideal gases, the partial derivatives in Eqs. 4 19 and 420 can be replaced by ordinary derivatives. Then the differential changes in the internal energy and enthalpy of an ideal gas can be expressed as du cv 1T2 dT cp 1T2 dT (423) | 181 u= h= cv = cp = u(T ) h(T ) cv (T ) c p(T ) and dh (424) The change in internal energy or enthalpy for an ideal gas during a process from state 1 to state 2 is determined by integrating these equations: 2 FIGURE 423 For ideal gases, u, h, cv, and cp vary with temperature only. u u2 u1 1 cv 1T2 dT 1kJ>kg2 (425) and 2 h h2 h1 1 cp 1T2 dT 1kJ>kg2 (426) To carry out these integrations, we need to have relations for cv and cp as functions of temperature. At low pressures, all real gases approach ideal-gas behavior, and therefore their specific heats depend on temperature only. The specific heats of real gases at low pressures are called ideal-gas specific heats, or zero-pressure specific heats, and are often denoted cp0 and cv 0. Accurate analytical expressions for ideal-gas specific heats, based on direct measurements or calculations from statistical behavior of molecules, are available and are given as third-degree polynomials in the appendix (Table A2c) for several gases. A plot of cp0(T) data for some common gases is given in Fig. 4 24. The use of ideal-gas specific heat data is limited to low pressures, but these data can also be used at moderately high pressures with reasonable accuracy as long as the gas does not deviate from ideal-gas behavior significantly. The integrations in Eqs. 4 25 and 426 are straightforward but rather time-consuming and thus impractical. To avoid these laborious calculations, u and h data for a number of gases have been tabulated over small temperature intervals. These tables are obtained by choosing an arbitrary reference point and performing the integrations in Eqs. 425 and 426 by treating state 1 as the reference state. In the ideal-gas tables given in the appendix, zero kelvin is chosen as the reference state, and both the enthalpy and the internal energy are assigned zero values at that state (Fig. 425). The choice of the reference state has no effect on u or h calculations. The u and h data are given in kJ/kg for air (Table A17) and usually in kJ/kmol for other gases. The unit kJ/kmol is very convenient in the thermodynamic analysis of chemical reactions. Some observations can be made from Fig. 4 24. First, the specific heats of gases with complex molecules (molecules with two or more atoms) are higher and increase with temperature. Also, the variation of specific heats Cp0 kJ/kmol K 60 CO 2 H2 O 50 40 O2 H2 30 Air Ar, He, Ne, Kr, Xe, Rn 20 1000 2000 Temperature, K 3000 FIGURE 424 Ideal-gas constant-pressure specific heats for some gases (see Table A2c for cp equations). 182 | Thermodynamics with temperature is smooth and may be approximated as linear over small temperature intervals (a few hundred degrees or less). Therefore the specific heat functions in Eqs. 4 25 and 4 26 can be replaced by the constant average specific heat values. Then the integrations in these equations can be performed, yielding u2 u1 cv,avg 1T2 cp,avg 1T2 T1 2 T1 2 1kJ>kg2 1kJ>kg 2 (427) AIR T, K 0 . . 300 310 . . u, kJ/kg 0 . . 214.07 221.25 . . h, kJ/kg 0 . . 300.19 310.24 . . and h2 h1 (428) FIGURE 425 In the preparation of ideal-gas tables, 0 K is chosen as the reference temperature. cp Approximation Actual cp,avg 1 2 The specific heat values for some common gases are listed as a function of temperature in Table A2b. The average specific heats cp,avg and cv,avg are evaluated from this table at the average temperature (T1 + T2)/2, as shown in Fig. 426. If the final temperature T2 is not known, the specific heats may be evaluated at T1 or at the anticipated average temperature. Then T2 can be determined by using these specific heat values. The value of T2 can be refined, if necessary, by evaluating the specific heats at the new average temperature. Another way of determining the average specific heats is to evaluate them at T1 and T2 and then take their average. Usually both methods give reasonably good results, and one is not necessarily better than the other. Another observation that can be made from Fig. 4 24 is that the ideal-gas specific heats of monatomic gases such as argon, neon, and helium remain constant over the entire temperature range. Thus, u and h of monatomic gases can easily be evaluated from Eqs. 4 27 and 4 28. Note that the u and h relations given previously are not restricted to any kind of process. They are valid for all processes. The presence of the constant-volume specific heat cv in an equation should not lead one to believe that this equation is valid for a constant-volume process only. On the contrary, the relation u cv,avg T is valid for any ideal gas undergoing any process (Fig. 427). A similar argument can be given for cp and h. To summarize, there are three ways to determine the internal energy and enthalpy changes of ideal gases (Fig. 4 28): 1. By using the tabulated u and h data. This is the easiest and most accurate way when tables are readily available. 2. By using the cv or cp relations as a function of temperature and performing the integrations. This is very inconvenient for hand calculations but quite desirable for computerized calculations. The results obtained are very accurate. 3. By using average specific heats. This is very simple and certainly very convenient when property tables are not available. The results obtained are reasonably accurate if the temperature interval is not very large. T1 T avg T2 T FIGURE 426 For small temperature intervals, the specific heats may be assumed to vary linearly with temperature. Specific Heat Relations of Ideal Gases A special relationship between cp and cv for ideal gases can be obtained by differentiating the relation h u RT, which yields dh du R dT Chapter 4 Replacing dh by cp dT and du by cv dT and dividing the resulting expression by dT, we obtain cp cv R 1kJ>kg # K2 (429) AIR V = constant T1 = 20 C T2 = 30 C | 183 This is an important relationship for ideal gases since it enables us to determine cv from a knowledge of cp and the gas constant R. When the specific heats are given on a molar basis, R in the above equation should be replaced by the universal gas constant Ru (Fig. 429). cp cv Ru 1kJ>kmol # K2 (430) Q1 Q2 AIR P = constant T1 = 20 C T2 = 30 C u = cv T = 7.18 kJ/kg At this point, we introduce another ideal-gas property called the specific heat ratio k, defined as k cp cv (431) u = cv T = 7.18 kJ/kg The specific ratio also varies with temperature, but this variation is very mild. For monatomic gases, its value is essentially constant at 1.667. Many diatomic gases, including air, have a specific heat ratio of about 1.4 at room temperature. FIGURE 427 The relation u cv T is valid for any kind of process, constant-volume or not. EXAMPLE 47 Evaluation of the u of an Ideal Gas u = u 2 u1 (table) 2 Air at 300 K and 200 kPa is heated at constant pressure to 600 K. Determine the change in internal energy of air per unit mass, using (a) data from the air table (Table A17), (b) the functional form of the specific heat (Table A2c), and (c) the average specific heat value (Table A2b). u = 1 cv (T ) dT u cv,avg T Solution The internal energy change of air is to be determined in three different ways. Assumptions At specified conditions, air can be considered to be an ideal gas since it is at a high temperature and low pressure relative to its criticalpoint values. Analysis The internal energy change u of ideal gases depends on the initial and final temperatures only, and not on the type of process. Thus, the following solution is valid for any kind of process. (a) One way of determining the change in internal energy of air is to read the u values at T1 and T2 from Table A17 and take the difference: FIGURE 428 Three ways of calculating u. u1 u2 Thus, u @ 300 K u @ 600 K 1434.78 214.07 kJ>kg 434.78 kJ>kg u u2 u1 214.072 kJ>kg 220.71 kJ>kg (b) The c p(T) of air is given in Table A2c in the form of a third-degree polynomial expressed as cp 1T 2 a bT cT 2 dT 3 184 | Thermodynamics where a 28.11, b 0.1967 10 2, c d 1.966 10 9. From Eq. 430, 0.4802 10 5, and AIR at 300 K cv = 0.718 kJ/kg K R = 0.287 kJ/kg . K or cv = 20.80 kJ/kmol . K Ru = 8.314 kJ/kmol . K { cp = 1.005 kJ/kg . K cv 1T2 From Eq. 425, 2 cp Ru T2 1a 3 1a Ru 2 Ru 2 bT cT 2 dT 3 { cp = 29.114 kJ/kmol . K u 1 cv 1T2 dT bT cT 2 T1 dT 3 4 dT Performing the integration and substituting the values, we obtain FIGURE 429 The cp of an ideal gas can be determined from a knowledge of cv and R. u 6447 kJ>kmol The change in the internal energy on a unit-mass basis is determined by dividing this value by the molar mass of air (Table A1): u u M 6447 kJ>kmol 28.97 kg>kmol 222.5 kJ>kg which differs from the tabulated value by 0.8 percent. (c) The average value of the constant-volume specific heat cv,avg is determined from Table A2b at the average temperature of (T1 T2)/2 450 K to be cv,avg Thus, cv @ 450 K T1 2 0.733 kJ>kg # K 3002K4 u cv,avg 1T2 220 kJ>kg 10.733 kJ>kg # K2 3 1600 Discussion This answer differs from the tabulated value (220.71 kJ/kg) by only 0.4 percent. This close agreement is not surprising since the assumption that cv varies linearly with temperature is a reasonable one at temperature intervals of only a few hundred degrees. If we had used the cv value at T1 300 K instead of at Tavg, the result would be 215.4 kJ/kg, which is in error by about 2 percent. Errors of this magnitude are acceptable for most engineering purposes. EXAMPLE 48 Heating of a Gas in a Tank by Stirring An insulated rigid tank initially contains 1.5 lbm of helium at 80F and 50 psia. A paddle wheel with a power rating of 0.02 hp is operated within the tank for 30 min. Determine (a) the final temperature and (b) the final pressure of the helium gas. Solution Helium gas in an insulated rigid tank is stirred by a paddle wheel. The final temperature and pressure of helium are to be determined. Assumptions 1 Helium is an ideal gas since it is at a very high temperature relative to its critical-point value of 451F. 2 Constant specific heats can be used for helium. 3 The system is stationary and thus the kinetic and potential energy changes are zero, KE PE 0 and E U. 4 The volume of the tank is constant, and thus there is no boundary work. 5 The system is adiabatic and thus there is no heat transfer. Chapter 4 Analysis We take the contents of the tank as the system (Fig. 430). This is a closed system since no mass crosses the system boundary during the process. We observe that there is shaft work done on the system. (a) The amount of paddle-wheel work done on the system is | 185 Wsh # Wsh t 10.02 hp 2 10.5 h 2 a 2545 Btu>h 1 hp b 25.45 Btu Under the stated assumptions and observations, the energy balance on the system can be expressed as E in E out m 1u 2 u1 2 E system T1 2 Net energy transfer by heat, work, and mass Wsh,in U 25.45 Btu T2 102.5F He m = 1.5 lbm T1 = 80F P1 = 50 psia W sh Change in internal, kinetic, potential, etc., energies mcv,avg 1T2 As we pointed out earlier, the ideal-gas specific heats of monatomic gases (helium being one of them) are constant. The cv value of helium is determined from Table A2Ea to be cv 0.753 Btu/lbm F. Substituting this and other known quantities into the above equation, we obtain 11.5 lbm2 10.753 Btu>lbm # F 2 1T2 80F2 (b) The final pressure is determined from the ideal-gas relation P1V1 T1 P2V2 T2 where V1 and V2 are identical and cancel out. Then the final pressure becomes 50 psia 180 4602 R P2 1102.5 P2 460 2R 52.1 psia Discussion Note that the pressure in the ideal-gas relation is always the absolute pressure. P, psia P2 2 50 1 V2 = V1 V FIGURE 430 Schematic and P-V diagram for Example 48. 186 | Thermodynamics EXAMPLE 49 Heating of a Gas by a Resistance Heater A pistoncylinder device initially contains 0.5 m3 of nitrogen gas at 400 kPa and 27C. An electric heater within the device is turned on and is allowed to pass a current of 2 A for 5 min from a 120-V source. Nitrogen expands at constant pressure, and a heat loss of 2800 J occurs during the process. Determine the final temperature of nitrogen. Solution Nitrogen gas in a pistoncylinder device is heated by an electric resistance heater. Nitrogen expands at constant pressure while some heat is lost. The final temperature of nitrogen is to be determined. Assumptions 1 Nitrogen is an ideal gas since it is at a high temperature and low pressure relative to its critical-point values of 147C, and 3.39 MPa. 2 The system is stationary and thus the kinetic and potential energy changes are zero, KE PE 0 and E U. 3 The pressure remains constant during the process and thus P2 P1. 4 Nitrogen has constant specific heats at room temperature. Analysis We take the contents of the cylinder as the system (Fig. 431). This is a closed system since no mass crosses the system boundary during the process. We observe that a pistoncylinder device typically involves a moving boundary and thus boundary work, Wb. Also, heat is lost from the system and electrical work We is done on the system. First, let us determine the electrical work done on the nitrogen: We VI t 1120 V2 12 A 2 15 60 s2 a 1 kJ>s 1000 VA b 72 kJ The mass of nitrogen is determined from the ideal-gas relation: m P1V1 RT1 10.297 kPa # m3>kg # K 2 1300 K2 1400 kPa2 10.5 m3 2 2.245 kg Under the stated assumptions and observations, the energy balance on the system can be expressed as E in E out E system Change in internal, kinetic, potential, etc., energies Net energy transfer by heat, work, and mass We,in Q out We,in Wb,out Q out U H 72 kJ 2.8 kJ T2 56.7C m 1h 2 h1 2 mcp 1T2 T1 2 since U Wb H for a closed system undergoing a quasi-equilibrium expansion or compression process at constant pressure. From Table A2a, cp 1.039 kJ/kg K for nitrogen at room temperature. The only unknown quantity in the previous equation is T2, and it is found to be 12.245 kg2 11.039 kJ>kg # K 2 1T2 27C2 Discussion Note that we could also solve this problem by determining the boundary work and the internal energy change rather than the enthalpy change. Chapter 4 P, kPa | 187 2A 120 V N2 P = const. 2800 J 400 1 2 V 1 = 0.5 m 3 P 1 = 400 kPa T1 = 27C 0.5 V, m3 FIGURE 431 Schematic and P-V diagram for Example 49. EXAMPLE 410 Heating of a Gas at Constant Pressure A pistoncylinder device initially contains air at 150 kPa and 27C. At this state, the piston is resting on a pair of stops, as shown in Fig. 432, and the enclosed volume is 400 L. The mass of the piston is such that a 350-kPa pressure is required to move it. The air is now heated until its volume has doubled. Determine (a) the final temperature, (b) the work done by the air, and (c) the total heat transferred to the air. Solution Air in a pistoncylinder device with a set of stops is heated until its volume is doubled. The final temperature, work done, and the total heat transfer are to be determined. Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical-point values. 2 The system is stationary and thus the kinetic and potential energy changes are zero, KE PE 0 and E U. 3 The volume remains constant until the piston starts moving, and the pressure remains constant afterwards. 4 There are no electrical, shaft, or other forms of work involved. Analysis We take the contents of the cylinder as the system (Fig. 432). This is a closed system since no mass crosses the system boundary during the process. We observe that a piston-cylinder device typically involves a moving boundary and thus boundary work, Wb. Also, the boundary work is done by the system, and heat is transferred to the system. (a) The final temperature can be determined easily by using the ideal-gas relation between states 1 and 3 in the following form: P1V1 T1 1150 kPa2 1V1 2 P3V3 T3 300 K 1350 kPa2 12V1 2 T3 T3 1400 K 188 | Thermodynamics P, kPa 2 3 350 AIR V 1 = 400 L P 1 = 150 kPa A 150 Q 0.4 0.8 1 FIGURE 432 Schematic and P-V diagram for Example 410. T1 = 27C V, m3 EXPERIMENT (b) The work done could be determined by integration, but for this case it is much easier to find it from the area under the process curve on a P-V diagram, shown in Fig. 432: A Therefore, 1V2 V1 2P2 10.4 m3 2 1350 kPa2 W13 140 kJ 140 m3 # kPa The work is done by the system (to raise the piston and to push the atmospheric air out of the way), and thus it is work output. Use actual data from the experiment shown here to obtain the specific heat of aluminum. See end-of-chapter problem 4179. Ronald Mullisen (c) Under the stated assumptions and observations, the energy balance on the system between the initial and final states (process 13) can be expressed as E in E out E system m 1u 3 u1 2 Net energy transfer by heat, work, and mass EXPERIMENT The mass of the system can be determined from the ideal-gas relation: Q in P1V1 RT1 Change in internal, kinetic, potential, etc., energies Wb,out U m 10.287 kPa # m3>kg # K2 1300 K2 u1 u3 u @ 300 K u @ 1400 K 214.07 kJ>kg 1113.52 kJ>kg 1150 kPa2 10.4 m3 2 0.697 kg The internal energies are determined from the air table (Table A17) to be Thus, Use actual data from the experiment shown here to obtain the specific heat of aluminum. See end-of-chapter problem 4180. Ronald Mullisen Q in Discussion 140 kJ Q in 10.697 kg2 3 11113.52 767 kJ 214.072 kJ>kg4 The positive sign verifies that heat is transferred to the system. Chapter 4 | 189 45 INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF SOLIDS AND LIQUIDS INTERACTIVE TUTORIAL SEE TUTORIAL CH. 4, SEC. 5 ON THE DVD. A substance whose specific volume (or density) is constant is called an incompressible substance. The specific volumes of solids and liquids essentially remain constant during a process (Fig. 433). Therefore, liquids and solids can be approximated as incompressible substances without sacrificing much in accuracy. The constant-volume assumption should be taken to imply that the energy associated with the volume change is negligible compared with other forms of energy. Otherwise, this assumption would be ridiculous for studying the thermal stresses in solids (caused by volume change with temperature) or analyzing liquid-in-glass thermometers. It can be mathematically shown that (see Chap. 12) the constant-volume and constant-pressure specific heats are identical for incompressible substances (Fig. 434). Therefore, for solids and liquids, the subscripts on cp and cv can be dropped, and both specific heats can be represented by a single symbol c. That is, cp cv c (432) LIQUID vl = constant SOLID vs = constant FIGURE 433 The specific volumes of incompressible substances remain constant during a process. This result could also be deduced from the physical definitions of constantvolume and constant-pressure specific heats. Specific heat values for several common liquids and solids are given in Table A3. Internal Energy Changes Like those of ideal gases, the specific heats of incompressible substances depend on temperature only. Thus, the partial differentials in the defining equation of cv can be replaced by ordinary differentials, which yield du cv dT IRON 25C c = cv = cp = 0.45 kJ/kg . C c 1T 2 dT (433) The change in internal energy between states 1 and 2 is then obtained by integration: 2 u u2 u1 1 c 1T 2 dT 1kJ>kg 2 (434) FIGURE 434 The cv and cp values of incompressible substances are identical and are denoted by c. The variation of specific heat c with temperature should be known before this integration can be carried out. For small temperature intervals, a c value at the average temperature can be used and treated as a constant, yielding u cavg 1T2 T1 2 1kJ>kg2 (435) Enthalpy Changes Using the definition of enthalpy h u Pv and noting that v constant, the differential form of the enthalpy change of incompressible substances can be determined by differentiation to be 0 dh du v dP P dv du v dP 1kJ>kg2 (436) Integrating, h u v P cavg T v P (437) 190 | Thermodynamics For solids, the term v P is insignificant and thus h liquids, two special cases are commonly encountered: u cavg T. For 1. Constant-pressure processes, as in heaters ( P 0): h u cavg T 2. Constant-temperature processes, as in pumps ( T 0): h v P For a process between states 1 and 2, the last relation can be expressed as h2 h1 v(P2 P1). By taking state 2 to be the compressed liquid state at a given T and P and state 1 to be the saturated liquid state at the same temperature, the enthalpy of the compressed liquid can be expressed as h @P,T hf @ T vf @ T 1P Psat @ T 2 (438) as discussed in Chap. 3. This is an improvement over the assumption that the enthalpy of the compressed liquid could be taken as hf at the given temperature (that is, h@ P,T hf @ T). However, the contribution of the last term is often very small, and is neglected. (Note that at high temperature and pressures, Eq. 438 may overcorrect the enthalpy and result in a larger error than the approximation h hf @ T.) EXAMPLE 411 Enthalpy of Compressed Liquid Determine the enthalpy of liquid water at 100C and 15 MPa (a) by using compressed liquid tables, (b) by approximating it as a saturated liquid, and (c) by using the correction given by Eq. 438. Solution The enthalpy of liquid water is to be determined exactly and approximately. Analysis At 100C, the saturation pressure of water is 101.42 kPa, and since P Psat, the water exists as a compressed liquid at the specified state. (a) From compressed liquid tables, we read P T 15 MPa f 100C h 430.39 kJ>kg 1Table A7 2 This is the exact value. (b) Approximating the compressed liquid as a saturated liquid at 100C, as is commonly done, we obtain h (c) From Eq. 438, hf @ 100C 419.17 kJ>kg This value is in error by about 2.6 percent. h@P,T hf @ T vf @ T 1P Psat @ T 2 10.001 m3 kg2 3 115,000 101.422 kPa4 a 1 kJ b 1 kPa # m3 1419.17 kJ>kg2 434.07 kJ>kg Discussion Note that the correction term reduced the error from 2.6 to about 1 percent in this case. However, this improvement in accuracy is often not worth the extra effort involved. Chapter 4 EXAMPLE 412 Cooling of an Iron Block by Water | 191 A 50-kg iron block at 80C is dropped into an insulated tank that contains 0.5 m3 of liquid water at 25C. Determine the temperature when thermal equilibrium is reached. Solution An iron block is dropped into water in an insulated tank. The final temperature when thermal equilibrium is reached is to be determined. Assumptions 1 Both water and the iron block are incompressible substances. 2 Constant specific heats at room temperature can be used for water and the iron. 3 The system is stationary and thus the kinetic and potential energy changes are zero, KE PE 0 and E U. 4 There are no electrical, shaft, or other forms of work involved. 5 The system is well-insulated and thus there is no heat transfer. Analysis We take the entire contents of the tank as the system (Fig. 435). This is a closed system since no mass crosses the system boundary during the process. We observe that the volume of a rigid tank is constant, and thus there is no boundary work. The energy balance on the system can be expressed as WATER 25C IRON m = 50 kg 80C 0.5 m 3 E in E out E system Change in internal, kinetic, potential, etc., energies Net energy transfer by heat, work, and mass FIGURE 435 Schematic for Example 412. The total internal energy U is an extensive property, and therefore it can be expressed as the sum of the internal energies of the parts of the system. Then the total internal energy change of the system becomes 3mc 1T2 The specific volume of liquid water at or about room temperature can be taken to be 0.001 m3/kg. Then the mass of the water is mwater The specific heats of iron and liquid water are determined from Table A3 to be ciron 0.45 kJ/kg C and cwater 4.18 kJ/kg C. Substituting these values into the energy equation, we obtain 150 kg 2 10.45 kJ>kg # C2 1T2 Therefore, when thermal equilibrium is established, both the water and iron will be at 25.6C. Discussion The small rise in water temperature is due to its large mass and large specific heat. T1 2 4 iron V v 0 U Usys 3mc 1T2 Uiron T1 2 4 water Uwater 0 0 0.5 m3 0.001 m3>kg 500 kg 80C2 T2 1500 kg2 14.18 kJ>kg # C2 1T2 25.6C 25C2 0 192 | Thermodynamics EXAMPLE 413 Temperature Rise due to Slapping If you ever slapped someone or got slapped yourself, you probably remember the burning sensation. Imagine you had the unfortunate occasion of being slapped by an angry person, which caused the temperature of the affected area of your face to rise by 1.8C (ouch!). Assuming the slapping hand has a mass of 1.2 kg and about 0.150 kg of the tissue on the face and the hand is affected by the incident, estimate the velocity of the hand just before impact. Take the specific heat of the tissue to be 3.8 kJ/kg C. Solution The face of a person is slapped. For the specified temperature rise of the affected part, the impact velocity of the hand is to be determined. Assumptions 1 The hand is brought to a complete stop after the impact. 2 The face takes the blow without significant movement. 3 No heat is transferred from the affected area to the surroundings, and thus the process is adiabatic. 4 No work is done on or by the system. 5 The potential energy change is zero, PE 0 and E U KE. Analysis We analyze this incident in a professional manner without involving any emotions. First, we identify the system, draw a sketch of it, and state our observations about the specifics of the problem. We take the hand and the affected portion of the face as the system (Fig. 436). This is a closed system since it involves a fixed amount of mass (no mass transfer). We observe that the kinetic energy of the hand decreases during the process, as evidenced by a decrease in velocity from initial value to zero, while the internal energy of the affected area increases, as evidenced by an increase in the temperature. There seems to be no significant energy transfer between the system and its surroundings during this process. Under the stated assumptions and observations, the energy balance on the system can be expressed as FIGURE 436 Schematic for Example 413. E in E out E system Change in internal, kinetic, potential, etc., energies Net energy transfer by heat, work, and mass That is, the decrease in the kinetic energy of the hand must be equal to the increase in the internal energy of the affected area. Solving for the velocity and substituting the given quantities, the impact velocity of the hand is determined to be Vhand Discussion Reconstruction of events such as this by making appropriate assumptions are commonly used in forensic engineering. 0 0 Uaffected tissue 1mcT2 affected tissue KEhand 3m 10 V 2 2>2 4 hand 2 1mcT2 affected tissue mhand B 2 10.15 kg2 13.8 kJ>kg # C2 11.8C2 1000 m2>s2 a b B 1.2 kg 1 kJ>kg 41.4 m>s 1or 149 km>h2 Chapter 4 TOPIC OF SPECIAL INTEREST* | 193 Thermodynamic Aspects of Biological Systems An important and exciting application area of thermodynamics is biological systems, which are the sites of rather complex and intriguing energy transfer and transformation processes. Biological systems are not in thermodynamic equilibrium, and thus they are not easy to analyze. Despite their complexity, biological systems are primarily made up of four simple elements: hydrogen, oxygen, carbon, and nitrogen. In the human body, hydrogen accounts for 63 percent, oxygen 25.5 percent, carbon 9.5 percent, and nitrogen 1.4 percent of all the atoms. The remaining 0.6 percent of the atoms comes from 20 other elements essential for life. By mass, about 72 percent of the human body is water. The building blocks of living organisms are cells, which resemble miniature factories performing functions that are vital for the survival of organisms. A biological system can be as simple as a single cell. The human body contains about 100 trillion cells with an average diameter of 0.01 mm. The membrane of the cell is a semipermeable wall that allows some substances to pass through it while excluding others. In a typical cell, thousands of chemical reactions occur every second during which some molecules are broken down and energy is released and some new molecules are formed. This high level of chemical activity in the cells, which maintains the human body at a temperature of 37C while performing the necessary bodily tasks, is called metabolism. In simple terms, metabolism refers to the burning of foods such as carbohydrates, fat, and protein. The rate of metabolism in the resting state is called the basal metabolic rate, which is the rate of metabolism required to keep a body performing the necessary functions (such as breathing and blood circulation) at zero external activity level. The metabolic rate can also be interpreted as the energy consumption rate for a body. For an average male (30 years old, 70 kg, 1.8-m2 body surface area), the basal metabolic rate is 84 W. That is, the body dissipates energy to the environment at a rate of 84 W, which means that the body is converting chemical energy of the food (or of the body fat if the person has not eaten) into thermal energy at a rate of 84 W (Fig. 437). The metabolic rate increases with the level of activity, and it may exceed 10 times the basal metabolic rate when a body is doing strenuous exercise. That is, two people doing heavy exercising in a room may be supplying more energy to the room than a 1-kW electrical resistance heater (Fig. 438). The fraction of sensible heat varies from about 40 percent in the case of heavy work to about 70 percent in the case of light work. The rest of the energy is rejected from the body by perspiration in the form of latent heat. The basal metabolic rate varies with sex, body size, general health conditions, and so forth, and decreases considerably with age. This is one of the reasons people tend to put on weight in their late twenties and thirties even though they do not increase their food intake. The brain and the liver are the major sites of metabolic activity. These two organs are responsible for almost 50 percent of the basal metabolic rate of an adult human body although they constitute only about 4 percent of the body mass. In small children, it is remarkable that about half of the basal metabolic activity occurs in the brain alone. *This section can be skipped without a loss in continuity. FIGURE 437 An average person dissipates energy to the surroundings at a rate of 84 W when resting. Vol. 124/PhotoDisc 1.2 kJ/s 1 kJ/s FIGURE 438 Two fast-dancing people supply more energy to a room than a 1-kW electric resistance heater. 194 | Thermodynamics A 300-W resistance heater A 300-W refrigerator A 300-W fan A 300-W TV Two people, each dissipating 150 W Three light bulbs, 100 W each A 100-W computer with a 200-W monitor Solar energy 300 W The biological reactions in cells occur essentially at constant temperature, pressure, and volume. The temperature of the cell tends to rise when some chemical energy is converted to heat, but this energy is quickly transferred to the circulatory system, which transports it to outer parts of the body and eventually to the environment through the skin. The muscle cells function very much like an engine, converting the chemical energy into mechanical energy (work) with a conversion efficiency of close to 20 percent. When the body does no net work on the environment (such as moving some furniture upstairs), the entire work is also converted to heat. In that case, the entire chemical energy in the food released during metabolism in the body is eventually transferred to the environment. A TV set that consumes electricity at a rate of 300 W must reject heat to its environment at a rate of 300 W in steady operation regardless of what goes on in the set. That is, turning on a 300-W TV set or three 100-W light bulbs will produce the same heating effect in a room as a 300-W resistance heater (Fig. 439). This is a consequence of the conservation of energy principle, which requires that the energy input into a system must equal the energy output when the total energy content of a system remains constant during a process. Food and Exercise The energy requirements of a body are met by the food we eat. The nutrients in the food are considered in three major groups: carbohydrates, proteins, and fats. Carbohydrates are characterized by having hydrogen and oxygen atoms in a 2:1 ratio in their molecules. The molecules of carbohydrates range from very simple (as in plain sugar) to very complex or large (as in starch). Bread and plain sugar are the major sources of carbohydrates. Proteins are very large molecules that contain carbon, hydrogen, oxygen, and nitrogen, and they are essential for the building and repairing of the body tissues. Proteins are made up of smaller building blocks called amino acids. Complete proteins such as meat, milk, and eggs have all the amino acids needed to build body tissues. Plant source proteins such as those in fruits, vegetables, and grains lack one or more amino acids, and are called incomplete proteins. Fats are relatively small molecules that consist of carbon, hydrogen, and oxygen. Vegetable oils and animal fats are major sources of fats. Most foods we eat contain all three nutrition groups at varying amounts. The typical average American diet consists of 45 percent carbohydrate, 40 percent fat, and 15 percent protein, although it is recommended that in a healthy diet less than 30 percent of the calories should come from fat. The energy content of a given food is determined by burning a small sample of the food in a device called a bomb calorimeter, which is basically a well-insulated rigid tank (Fig. 440). The tank contains a small combustion chamber surrounded by water. The food is ignited and burned in the combustion chamber in the presence of excess oxygen, and the energy released is transferred to the surrounding water. The energy content of the food is calculated on the basis of the conservation of energy principle by measuring the temperature rise of the water. The carbon in the food is converted into CO2 and hydrogen into H2O as the food burns. The same chemical reactions occur in the body, and thus the same amount of energy is released. Using dry (free of water) samples, the average energy contents of the three basic food groups are determined by bomb calorimeter measurements to be FIGURE 439 Some arrangements that supply a room the same amount of energy as a 300-W electric resistance heater. Mixer and motor Electrical switch Thermometer Water Bomb (combustion chamber) Insulation Food sample FIGURE 440 Schematic of a bomb calorimeter used to determine the energy content of food samples. Chapter 4 18.0 MJ/kg for carbohydrates, 22.2 MJ/kg for proteins, and 39.8 MJ/kg for fats. These food groups are not entirely metabolized in the human body, however. The fraction of metabolizable energy contents are 95.5 percent for carbohydrates, 77.5 percent for proteins, and 97.7 percent for fats. That is, the fats we eat are almost entirely metabolized in the body, but close to one quarter of the protein we eat is discarded from the body unburned. This corresponds to 4.1 Calories/g for proteins and carbohydrates and 9.3 Calories/g for fats (Fig. 441) commonly seen in nutrition books and on food labels. The energy contents of the foods we normally eat are much lower than the values above because of the large water content (water adds bulk to the food but it cannot be metabolized or burned, and thus it has no energy value). Most vegetables, fruits, and meats, for example, are mostly water. The average metabolizable energy contents of the three basic food groups are 4.2 MJ/kg for carbohydrates, 8.4 MJ/kg for proteins, and 33.1 MJ/kg for fats. Note that 1 kg of natural fat contains almost 8 times the metabolizable energy of 1 kg of natural carbohydrates. Thus, a person who fills his stomach with fatty foods is consuming much more energy than a person who fills his stomach with carbohydrates such as bread or rice. The metabolizable energy content of foods is usually expressed by nutritionists in terms of the capitalized Calories. One Calorie is equivalent to one kilocalorie (1000 calories), which is equivalent to 4.1868 kJ. That is, 1 Cal 1Calorie2 1000 calories 1 kcal 1kilocalorie2 4.1868 kJ | 195 FIGURE 441 Evaluating the calorie content of one serving of chocolate chip cookies (values are for Chips Ahoy cookies made by Nabisco). Vol. 30/PhotoDisc The calorie notation often causes confusion since it is not always followed in the tables or articles on nutrition. When the topic is food or fitness, a calorie normally means a kilocalorie whether it is capitalized or not. The daily calorie needs of people vary greatly with age, gender, the state of health, the activity level, the body weight, and the composition of the body as well as other factors. A small person needs fewer calories than a larger person of the same sex and age. An average man needs about 2400 to 2700 Calories a day. The daily need of an average woman varies from 1800 to 2200 Calories. The daily calorie needs are about 1600 for sedentary women and some older adults; 2000 for sedentary men and most older adults; 2200 for most children, teenage girls, and active women; 2800 for teenage boys, active men, and some very active women; and above 3000 for very active men. The average value of calorie intake is usually taken to be 2000 Calories per day. The daily calorie needs of a person can be determined by multiplying the body weight in pounds (which is 2.205 times the body weight in kg) by 11 for a sedentary person, 13 for a moderately active person, 15 for a moderate exerciser or physical laborer, and 18 for an extremely active exerciser or physical laborer. The extra calories a body consumes are usually stored as fat, which serves as the spare energy of the body for use when the energy intake of the body is less than the needed amount. Like other natural fat, 1 kg of human body fat contains about 33.1 MJ of metabolizable energy. Therefore, a starving person (zero energy intake) who uses up 2200 Calories (9211 kJ) a day can meet his daily energy intake requirements by burning only 9211/33,100 0.28 kg of body fat. So it is no surprise that people are known to survive over 100 days without eating. (They still need to drink water, however, to replenish the water lost through the lungs and the skin to avoid the dehydration that may occur in just a few 196 | Thermodynamics days.) Although the desire to get rid of the excess fat in a thin world may be overwhelming at times, starvation diets are not recommended because the body soon starts to consume its own muscle tissue in addition to fat. A healthy diet should involve regular exercise while allowing a reasonable amount of calorie intake. The average metabolizable energy contents of various foods and the energy consumption during various activities are given in Tables 41 and 42. Considering that no two hamburgers are alike, and that no two people walk exactly the same way, there is some uncertainty in these values, as you would expect. Therefore, you may encounter somewhat different values in other books or magazines for the same items. The rates of energy consumption listed in Table 42 during some activities are for a 68-kg adult. The energy consumed for smaller or larger adults can be determined using the proportionality of the metabolism rate and the body size. For example, the rate of energy consumption by a 68-kg bicyclist is listed in Table 42 to be 639 Calories/h. Then the rate of energy consumption by a 50-kg bicyclist is 150 kg2 639 Cal>h 68 kg 470 Cal>h For a 100-kg person, it would be 940 Cal/h. The thermodynamic analysis of the human body is rather complicated since it involves mass transfer (during breathing, perspiring, etc.) as well as energy transfer. As such, it should be treated as an open system. However, the energy transfer with mass is difficult to quantify. Therefore, the human body is often modeled as a closed system for simplicity by treating energy transported with mass as just energy transfer. For example, eating is modeled as the transfer of energy into the human body in the amount of the metabolizable energy content of the food. Dieting Most diets are based on calorie counting; that is, the conservation of energy principle: a person who consumes more calories than his or her body burns TABLE 41 Approximate metabolizable energy content of some common foods (1 Calorie 4.1868 kJ 3.968 Btu) Food Apple (one, medium) Baked potato (plain) Baked potato with cheese Bread (white, one slice) Butter (one teaspoon) Cheeseburger Chocolate candy bar (20 g) Cola (200 ml) Egg (one) Calories 70 250 550 70 35 325 105 87 80 Food Fish sandwich French fries (regular) Hamburger Hot dog Ice cream (100 ml, 10% fat) Lettuce salad with French dressing Calories 450 250 275 300 110 150 Food Milk (skim, 200 ml) Milk (whole, 200 ml) Peach (one, medium) Pie (one 1 slice, 23 cm 8 diameter) Pizza (large, cheese, one 1 slice) 8 Calories 76 136 65 300 350 Chapter 4 will gain weight whereas a person who consumes less calories than his or her body burns will lose weight. Yet, people who eat whatever they want whenever they want without gaining any weight are living proof that the caloriecounting technique alone does not work in dieting. Obviously there is more to dieting than keeping track of calories. It should be noted that the phrases weight gain and weight loss are misnomers. The correct phrases should be mass gain and mass loss. A man who goes to space loses practically all of his weight but none of his mass. When the topic is food and fitness, weight is understood to mean mass, and weight is expressed in mass units. Researchers on nutrition proposed several theories on dieting. One theory suggests that some people have very "food efficient" bodies. These people need fewer calories than other people do for the same activity, just like a fuel-efficient car needing less fuel for traveling a given distance. It is interesting that we want our cars to be fuel efficient but we do not want the same high efficiency for our bodies. One thing that frustrates the dieters is that the body interprets dieting as starvation and starts using the energy reserves of the body more stringently. Shifting from a normal 2000-Calorie daily diet to an 800-Calorie diet without exercise is observed to lower the basal metabolic rate by 10 to 20 percent. Although the metabolic rate returns to normal once the dieting stops, extended periods of low-calorie dieting without adequate exercise may result in the loss of considerable muscle tissue together with fat. With less muscle tissue to burn calories, the metabolic rate of the body declines and stays below normal even after a person starts eating normally. As a result, the person regains the weight he or she has lost in the form of fat, plus more. The basal metabolic rate remains about the same in people who exercise while dieting. Regular moderate exercise is part of any healthy dieting program for good reason: it builds or preserves muscle tissue that burns calories much faster than the fat tissue does. It is interesting that aerobic exercise continues burning calories for several hours after the workout, raising the overall metabolic rate considerably. Another theory suggests that people with too many fat cells developed during childhood or adolescence are much more likely to gain weight. Some people believe that the fat content of the bodies is controlled by the setting of a "fat control" mechanism, much like the temperature of a house is controlled by the thermostat setting. Some people put the blame for weight problems simply on the genes. Considering that 80 percent of the children of overweight parents are also overweight, heredity may indeed play an important role in the way a body stores fat. Researchers from the University of Washington and the Rockefeller University have identified a gene, called the RIIbeta, that seems to control the rate of metabolism. The body tries to keep the body fat at a particular level, called the set point, that differs from person to person (Fig. 442). This is done by speeding up the metabolism and thus burning extra calories much faster when a person tends to gain weight and by slowing down the metabolism and thus burning calories at a slower rate when a person tends to lose weight. Therefore, a person who just became slim burns fewer calories than does a person of the same size who has always been slim. Even exercise does not seem to change that. Then to keep the weight off, the newly slim | 197 TABLE 42 Approximate energy consumption of a 68-kg adult during some activities (1 Calorie 4.1868 kJ 3.968 Btu) Activity Basal metabolism Basketball Bicycling (21 km/h) Cross-country skiing (13 km/h) Driving a car Eating Fast dancing Fast running (13 km/h) Jogging (8 km/h) Swimming (fast) Swimming (slow) Tennis (advanced) Tennis (beginner) Walking (7.2 km/h) Watching TV Calories/h 72 550 639 936 180 99 600 936 540 860 288 480 288 432 72 Body fat level Set point New set point FIGURE 442 The body tends to keep the body fat level at a set point by speeding up metabolism when a person splurges and by slowing it down when the person starves. 198 | Thermodynamics person should consume no more calories than he or she can burn. Note that in people with high metabolic rates, the body dissipates the extra calories as body heat instead of storing them as fat, and thus there is no violation of the conservation of energy principle. In some people, a genetic flaw is believed to be responsible for the extremely low rates of metabolism. Several studies concluded that losing weight for such people is nearly impossible. That is, obesity is a biological phenomenon. However, even such people will not gain weight unless they eat more than their body can burn. They just must learn to be content with little food to remain slim, and forget about ever having a normal "eating" life. For most people, genetics determine the range of normal weights. A person may end up at the high or low end of that range, depending on eating and exercise habits. This also explains why some genetically identical twins are not so identical when it comes to body weight. Hormone imbalance is also believed to cause excessive weight gain or loss. Based on his experience, the first author of this book has also developed a diet called the "sensible diet." It consists of two simple rules: eat whatever you want whenever you want as much as you want provided that (1) you do not eat unless you are hungry and (2) you stop eating before you get stuffed. In other words, listen to your body and don't impose on it. Don't expect to see this unscientific diet advertised anywhere since there is nothing to be sold and thus no money to be made. Also, it is not as easy as it sounds since food is at the center stage of most leisure activities in social life, and eating and drinking have become synonymous with having a good time. However, it is comforting to know that the human body is quite forgiving of occasional impositions. Being overweight is associated with a long list of health risks from high blood pressure to some forms of cancer, especially for people who have a weight-related medical condition such as diabetes, hypertension, and heart disease. Therefore, people often wonder if their weight is in the proper range. Well, the answer to this question is not written in stone, but if you cannot see your toes or you can pinch your love handles more than an inch, you don't need an expert to tell you that you went over your range. On the other hand, some people who are obsessed with the weight issue try to lose more weight even though they are actually underweight. Therefore, it is useful to have a scientific criterion to determine physical fitness. The range of healthy weight for adults is usually expressed in terms of the body mass index (BMI), defined, in SI units, as BMI H 2 1m2 2 W 1kg2 with 19 BMI 6 19 BMI 25 BMI 7 25 underweight healthy weight overweight (439) TABLE 43 The range of healthy weight for adults of various heights (Source: National Institute of Health) English Units Healthy Height, weight, in. lbm* 58 60 62 64 66 68 70 72 74 76 91119 97127 103136 111146 118156 125165 133175 140185 148195 156205 SI Units Height, m 1.45 1.50 1.55 1.60 1.65 1.70 1.75 1.80 1.85 1.90 Healthy weight, kg* 4053 4356 4660 4964 5268 5572 5877 6281 6586 6990 *The upper and lower limits of healthy range correspond to mass body indexes of 19 and 25, respectively. where W is the weight (actually, the mass) of the person in kg and H is the height in m. Therefore, a BMI of 25 is the upper limit for the healthy weight and a person with a BMI of 27 is 8 percent overweight. It can be shown that the formula above is equivalent in English units to BMI 705 W/H 2 where W is in pounds and H is in inches. The proper range of weight for adults of various heights is given in Table 43 in both SI and English units. Chapter 4 EXAMPLE 414 Burning Off Lunch Calories | 199 A 90-kg man had two hamburgers, a regular serving of french fries, and a 200-ml Coke for lunch (Fig. 443). Determine how long it will take for him to burn the lunch calories off (a) by watching TV and (b) by fast swimming. What would your answers be for a 45-kg man? Solution A man had lunch at a restaurant. The times it will take for him to burn the lunch calories by watching TV and by fast swimming are to be determined. Assumptions The values in Tables 41 and 42 are applicable for food and exercise. Analysis (a) We take the human body as our system and treat it as a closed system whose energy content remains unchanged during the process. Then the conservation of energy principle requires that the energy input into the body must be equal to the energy output. The net energy input in this case is the metabolizable energy content of the food eaten. It is determined from Table 41 to be FIGURE 443 A typical lunch discussed in Example 414. Vol. 30/PhotoDisc Ein 2 Ehamburger 250 Efries 87 Ecola 2 275 887 Cal The rate of energy output for a 68-kg man watching TV is given in Table 42 to be 72 Calories/h. For a 90-kg man it becomes E out Therefore, it will take 190 kg2 t 72 Cal>h 68 kg 95.3 Cal>h 887 Cal 95.3 Cal>h 9.3 h to burn the lunch calories off by watching TV. (b) It can be shown in a similar manner that it takes only 47 min to burn the lunch calories off by fast swimming. Discussion The 45-kg man is half as large as the 90-kg man. Therefore, expending the same amount of energy takes twice as long in each case: 18.6 h by watching TV and 94 min by fast swimming. EXAMPLE 415 Losing Weight by Switching to Fat-Free Chips The fake fat olestra passes through the body undigested, and thus adds zero calorie to the diet. Although foods cooked with olestra taste pretty good, they may cause abdominal discomfort and the long-term effects are unknown. A 1-oz (28.3-g) serving of regular potato chips has 10 g of fat and 150 Calories, whereas 1 oz of the so-called fat-free chips fried in olestra has only 75 Calories. Consider a person who eats 1 oz of regular potato chips every day at lunch without gaining or losing any weight. Determine how much weight this person will lose in one year if he or she switches to fat-free chips (Fig. 444). FIGURE 444 Schematic for Example 415. 200 | Thermodynamics Solution A person switches from regular potato chips to fat-free ones. The weight the person loses in one year is to be determined. Assumptions Exercising and other eating habits remain the same. Analysis The person who switches to the fat-free chips consumes 75 fewer Calories a day. Then the annual reduction in calories consumed becomes Ereduced 175 Cal>day2 1365 day>year2 27,375 Cal>year The metabolizable energy content of 1 kg of body fat is 33,100 kJ. Therefore, assuming the deficit in the calorie intake is made up by burning body fat, the person who switches to fat-free chips will lose mfat Ereduced Energy content of fat 27,375 Cal 4.1868 kJ a b 33,100 kJ>kg 1 Cal 3.46 kg (about 7.6 pounds) of body fat that year. SUMMARY Work is the energy transferred as a force acts on a system through a distance. The most common form of mechanical work is the boundary work, which is the work associated with the expansion and compression of substances. On a P-V diagram, the area under the process curve represents the boundary work for a quasi-equilibrium process. Various forms of boundary work are expressed as follows: 2 E in E out E system Change in internal, kinetic, potential, etc., energies 1kJ2 Net energy transfer by heat, work, and mass It can also be expressed in the rate form as . . dE system>dt E in Eout Rate of net energy transfer by heat, work, and mass 1kW2 Rate of change in internal, kinetic, potential, etc., energies (1) General (2) Isobaric process Wb P0 1V2 Wb 1 P dV V1 2 Taking heat transfer to the system and work done by the system to be positive quantities, the energy balance for a closed system can also be expressed as Q where W U KE PE 1kJ2 (P1 P2 P0 constant) (3) Polytropic process P2V2 P1V1 Wb 1n 1 2 (PV n 1 n (4) Isothernal process of an ideal gas Wb P1V1 ln V2 V1 mRT0 ln V2 V1 (PV constant) W U Wother m 1u 2 1 2 2 m 1V 2 u1 2 Wb mRT0 constant) KE PE H mg 1z 2 z1 2 V22 1 U PE H. Thus, 1kJ2 The first law of thermodynamics is essentially an expression of the conservation of energy principle, also called the energy balance. The general energy balances for any system undergoing any process can be expressed as For a constant-pressure process, Wb Q Wother KE Chapter 4 The amount of energy needed to raise the temperature of a unit mass of a substance by one degree is called the specific heat at constant volume cv for a constant-volume process and the specific heat at constant pressure cp for a constantpressure process. They are defined as cv a 0u b 0T v and cp a 0h b 0T p | 201 where R is the gas constant. The specific heat ratio k is defined as k cp cv For incompressible substances (liquids and solids), both the constant-pressure and constant-volume specific heats are identical and denoted by c: cp 2 For ideal gases u, h, cv, and cp are functions of temperature alone. The u and h of ideal gases are expressed as 2 cv c 1kJ>kg # K2 T1 2 u h u2 h2 u1 1 2 cv 1T2 dT cp 1T2 dT cv,avg 1T2 cp,avg 1T2 T1 2 T1 2 The u and h of imcompressible substances are given by u 1 c 1T2 dT vP cavg 1T2 1kJ>kg2 h1 1 h u 1kJ>kg2 For ideal gases, cv and cp are related by cp cv R 1kJ>kg # K2 REFERENCES AND SUGGESTED READINGS 1. ASHRAE Handbook of Fundamentals. SI version. Atlanta, GA: American Society of Heating, Refrigerating, and Air-Conditioning Engineers, Inc., 1993. 2. ASHRAE Handbook of Refrigeration. SI version. Atlanta, GA: American Society of Heating, Refrigerating, and AirConditioning Engineers, Inc., 1994. PROBLEMS* Moving Boundary Work 4 1C On a P-v diagram, what does the area under the process curve represent? 4 2C Is the boundary work associated with constant-volume systems always zero? 4 3C An ideal gas at a given state expands to a fixed final volume first at constant pressure and then at constant temperature. For which case is the work done greater? 4 4C Show that 1 kPa m3 1 kJ. 45 A pistoncylinder device initially contains 0.07 m3 of nitrogen gas at 130 kPa and 120C. The nitrogen is now expanded polytropically to a state of 100 kPa and 100C. Determine the boundary work done during this process. 46 A pistoncylinder device with a set of stops initially contains 0.3 kg of steam at 1.0 MPa and 400C. The location of the stops corresponds to 60 percent of the initial volume. Now the steam is cooled. Determine the compression work if the final state is (a) 1.0 MPa and 250C and (b) 500 kPa. (c) Also determine the temperature at the final state in part (b). *Problems designated by a "C" are concept questions, and students are encouraged to answer them all. Problems designated by an "E" are in English units, and the SI users can ignore them. Problems with a CD-EES icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with a computer-EES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. Steam 0.3 kg 1 MPa 400C Q FIGURE P46 202 | Thermodynamics process, the pressure changes with volume according to the relation P aV b, where a 1200 kPa/m3 and b 600 kPa. Calculate the work done during this process (a) by plotting the process on a P-V diagram and finding the area under the process curve and (b) by performing the necessary integrations. 4 7 A pistoncylinder device initially contains 0.07 m3 of nitrogen gas at 130 kPa and 120C. The nitrogen is now expanded to a pressure of 100 kPa polytropically with a polytropic exponent whose value is equal to the specific heat ratio (called isentropic expansion). Determine the final temperature and the boundary work done during this process. 4 8 A mass of 5 kg of saturated water vapor at 300 kPa is heated at constant pressure until the temperature reaches 200C. Calculate the work done by the steam during this process. Answer: 165.9 kJ 4 9 A frictionless pistoncylinder device initially contains 200 L of saturated liquid refrigerant-134a. The piston is free to move, and its mass is such that it maintains a pressure of 900 kPa on the refrigerant. The refrigerant is now heated until its temperature rises to 70C. Calculate the work done during this process. Answer: 5571 kJ GAS P = aV + b FIGURE P414 415E During an expansion process, the pressure of a gas changes from 15 to 100 psia according to the relation P aV b, where a 5 psia/ft3 and b is a constant. If the initial volume of the gas is 7 ft3, calculate the work done during the process. Answer: 181 Btu During some actual expansion and compression processes in pistoncylinder devices, the gases C, have been observed to satisfy the relationship PV n where n and C are constants. Calculate the work done when a gas expands from 150 kPa and 0.03 m3 to a final volume of 0.2 m3 for the case of n 1.3. 417 Reconsider Prob. 416. Using the EES (or other) software, plot the process described in the problem on a P-V diagram, and investigate the effect of the polytropic exponent n on the boundary work. Let the polytropic exponent vary from 1.1 to 1.6. Plot the boundary work versus the polytropic exponent, and discuss the results. 416 R-134a P = const. FIGURE P49 4 10 Reconsider Prob. 4 9. Using EES (or other) software, investigate the effect of pressure on the work done. Let the pressure vary from 400 kPa to 1200 kPa. Plot the work done versus the pressure, and discuss the results. Explain why the plot is not linear. Also plot the process described in Prob. 4 9 on the P-v diagram. 4 11E A frictionless pistoncylinder device contains 16 lbm of superheated water vapor at 40 psia and 600F. Steam is now cooled at constant pressure until 70 percent of it, by mass, condenses. Determine the work done during this process. 4 12 A mass of 2.4 kg of air at 150 kPa and 12C is contained in a gas-tight, frictionless pistoncylinder device. The air is now compressed to a final pressure of 600 kPa. During the process, heat is transferred from the air such that the temperature inside the cylinder remains constant. Calculate the work input during this process. Answer: 272 kJ 4 13 Nitrogen at an initial state of 300 K, 150 kPa, and 0.2 m3 is compressed slowly in an isothermal process to a final pressure of 800 kPa. Determine the work done during this process. 4 14 A gas is compressed from an initial volume of 0.42 m3 to a final volume of 0.12 m3. During the quasi-equilibrium 418 A frictionless pistoncylinder device contains 2 kg of nitrogen at 100 kPa and 300 K. Nitrogen is now compressed constant until it slowly according to the relation PV1.4 reaches a final temperature of 360 K. Calculate the work input during this process. Answer: 89 kJ N2 PV1.4 = const. FIGURE P418 Chapter 4 4 19 The equation of state of a gas is given as v (P RuT, where the units of v and P are 10/ v 2) m3/kmol and kPa, respectively. Now 0.5 kmol of this gas is expanded in a quasi-equilibrium manner from 2 to 4 m3 at a constant temperature of 300 K. Determine (a) the unit of the quantity 10 in the equation and (b) the work done during this isothermal expansion process. Reconsider Prob. 4 19. Using the integration feature of the EES software, calculate the work done, and compare your result with the "hand-calculated" result obtained in Prob. 4 19. Plot the process described in the problem on a P-v diagram. 424 | 203 Reconsider Prob. 423. Using the EES software, investigate the effect of the spring constant on the final pressure in the cylinder and the boundary work done. Let the spring constant vary from 50 kN/m to 500 kN/m. Plot the final pressure and the boundary work against the spring constant, and discuss the results. 4 20 425 Determine the boundary work done by a gas during an expansion process if the pressure and volume values at various states are measured to be 300 kPa, 1 L; 290 kPa, 1.1 L; 270 kPa, 1.2 L; 250 kPa, 1.4 L; 220 kPa, 1.7 L; and 200 kPa, 2 L. 426 A pistoncylinder device initially contains 0.25 kg of nitrogen gas at 130 kPa and 120C. The nitrogen is now expanded isothermally to a pressure of 100 kPa. Determine the boundary work done during this process. Answer: 7.65 kJ 4 21 Carbon dioxide contained in a pistoncylinder device is compressed from 0.3 to 0.1 m3. During the process, the pressure and volume are related by P aV 2, where a 8 kPa m6. Calculate the work done on the carbon dioxide during this process. Answer: 53.3 kJ 4 22E Hydrogen is contained in a pistoncylinder device at 14.7 psia and 15 ft3. At this state, a linear spring (F x) with a spring constant of 15,000 lbf/ft is touching the piston but exerts no force on it. The cross-sectional area of the piston is 3 ft2. Heat is transferred to the hydrogen, causing it to expand until its volume doubles. Determine (a) the final pressure, (b) the total work done by the hydrogen, and (c) the fraction of this work done against the spring. Also, show the process on a P-V diagram. 4 23 A pistoncylinder device contains 50 kg of water at 250 kPa and 25C. The cross-sectional area of the piston is 0.1 m2. Heat is now transferred to the water, causing part of it to evaporate and expand. When the volume reaches 0.2 m3, the piston reaches a linear spring whose spring constant is 100 kN/m. More heat is transferred to the water until the piston rises 20 cm more. Determine (a) the final pressure and temperature and (b) the work done during this process. Also, show the process on a P-V diagram. Answers: (a) 450 kPa, 147.9C, (b) 44.5 kJ N2 130 kPa 120C FIGURE P426 427 A pistoncylinder device contains 0.15 kg of air initially at 2 MPa and 350C. The air is first expanded isothermally to 500 kPa, then compressed polytropically with a polytropic exponent of 1.2 to the initial pressure, and finally compressed at the constant pressure to the initial state. Determine the boundary work for each process and the net work of the cycle. Closed System Energy Analysis 428 A 0.5-m3 rigid tank contains refrigerant-134a initially at 160 kPa and 40 percent quality. Heat is now transferred to the refrigerant until the pressure reaches 700 kPa. Determine (a) the mass of the refrigerant in the tank and (b) the amount of heat transferred. Also, show the process on a P-v diagram with respect to saturation lines. A = 0.1 m H 2O m = 50 kg 2 FIGURE P423 429E A 20-ft3 rigid tank initially contains saturated refrigerant-134a vapor at 160 psia. As a result of heat transfer from the refrigerant, the pressure drops to 50 psia. Show the process on a P-v diagram with respect to saturation lines, and determine (a) the final temperature, (b) the amount of refrigerant that has condensed, and (c) the heat transfer. 204 | Thermodynamics 434 A pistoncylinder device contains 5 kg of refrigerant134a at 800 kPa and 70C. The refrigerant is now cooled at constant pressure until it exists as a liquid at 15C. Determine the amount of heat loss and show the process on a T-v diagram with respect to saturation lines. Answer: 1173 kJ 435E A pistoncylinder device contains 0.5 lbm of water initially at 120 psia and 2 ft3. Now 200 Btu of heat is transferred to the water while its pressure is held constant. Determine the final temperature of the water. Also, show the process on a T-v diagram with respect to saturation lines. 436 An insulated pistoncylinder device contains 5 L of saturated liquid water at a constant pressure of 175 kPa. Water is stirred by a paddle wheel while a current of 8 A flows for 45 min through a resistor placed in the water. If one-half of the liquid is evaporated during this constantpressure process and the paddle-wheel work amounts to 400 kJ, determine the voltage of the source. Also, show the process on a P-v diagram with respect to saturation lines. Answer: 224 V 4 30 A well-insulated rigid tank contains 5 kg of a saturated liquidvapor mixture of water at l00 kPa. Initially, three-quarters of the mass is in the liquid phase. An electric resistor placed in the tank is connected to a 110-V source, and a current of 8 A flows through the resistor when the switch is turned on. Determine how long it will take to vaporize all the liquid in the tank. Also, show the process on a T-v diagram with respect to saturation lines. H 2O V = constant We FIGURE P430 4 31 Reconsider Prob. 4 30. Using EES (or other) software, investigate the effect of the initial mass of water on the length of time required to completely vaporize the liquid. Let the initial mass vary from 1 to 10 kg. Plot the vaporization time against the initial mass, and discuss the results. H2O P = constant We Wsh 4 32 An insulated tank is divided into two parts by a partition. One part of the tank contains 2.5 kg of compressed liquid water at 60C and 600 kPa while the other part is evacuated. The partition is now removed, and the water expands to fill the entire tank. Determine the final temperature of the water and the volume of the tank for a final pressure of 10 kPa. FIGURE P436 437 A pistoncylinder device contains steam initially at 1 MPa, 450C, and 2.5 m3. Steam is allowed to cool at constant pressure until it first starts condensing. Show the process on a T-v diagram with respect to saturation lines and determine (a) the mass of the steam, (b) the final temperature, and (c) the amount of heat transfer. A pistoncylinder device initially contains steam at 200 kPa, 200C, and 0.5 m3. At this state, a linear spring (F x) is touching the piston but exerts no force on it. Heat is now slowly transferred to the steam, causing the pressure and the volume to rise to 500 kPa and 0.6 m3, respectively. Show the process on a P-v diagram with respect to saturation lines and determine (a) the final temperature, (b) the work done by the steam, and (c) the total heat transferred. Answers: (a) 1132C, (b) 35 kJ, (c) 808 kJ 438 Evacuated Partition H2O FIGURE P432 4 33 Reconsider Prob. 4 32. Using EES (or other) software, investigate the effect of the initial pressure of water on the final temperature in the tank. Let the initial pressure vary from 100 to 600 kPa. Plot the final temperature against the initial pressure, and discuss the results. Chapter 4 | 205 Q H2 O 200 kPa 200C 442 A 30-L electrical radiator containing heating oil is placed in a 50-m3 room. Both the room and the oil in the radiator are initially at 10C. The radiator with a rating of 1.8 kW is now turned on. At the same time, heat is lost from the room at an average rate of 0.35 kJ/s. After some time, the average temperature is measured to be 20C for the air in the room, and 50C for the oil in the radiator. Taking the density and the specific heat of the oil to be 950 kg/m3 and 2.2 kJ/kg C, respectively, determine how long the heater is kept on. Assume the room is well-sealed so that there are no air leaks. FIGURE P438 4 39 Reconsider Prob. 4 38. Using EES (or other) software, investigate the effect of the initial temperature of steam on the final temperature, the work done, and the total heat transfer. Let the initial temperature vary from 150 to 250C. Plot the final results against the initial temperature, and discuss the results. Room 10C Radiator Q 4 40 A pistoncylinder device initially contains 0.8 m3 of saturated water vapor at 250 kPa. At this state, the piston is resting on a set of stops, and the mass of the piston is such that a pressure of 300 kPa is required to move it. Heat is now slowly transferred to the steam until the volume doubles. Show the process on a P-v diagram with respect to saturation lines and determine (a) the final temperature, (b) the work done during this process, and (c) the total heat transfer. Answers: (a) 662C, (b) 240 kJ, (c) 1213 kJ FIGURE P442 Specific Heats, u, and h of Ideal Gases 443C Is the relation u mcv,avg T restricted to constantvolume processes only, or can it be used for any kind of process of an ideal gas? 444C Is the relation h mcp,avg T restricted to constantpressure processes only, or can it be used for any kind of process of an ideal gas? _ _ 445C Show that for an ideal gas cp cv Ru. 446C Is the energy required to heat air from 295 to 305 K the same as the energy required to heat it from 345 to 355 K? Assume the pressure remains constant in both cases. 447C In the relation u mcv T, what is the correct unit of cv -- kJ/kg C or kJ/kg K? 448C A fixed mass of an ideal gas is heated from 50 to 80C at a constant pressure of (a) 1 atm and (b) 3 atm. For which case do you think the energy required will be greater? Why? 449C A fixed mass of an ideal gas is heated from 50 to 80C at a constant volume of (a) 1 m3 and (b) 3 m3. For which case do you think the energy required will be greater? Why? 450C A fixed mass of an ideal gas is heated from 50 to 80C (a) at constant volume and (b) at constant pressure. For which case do you think the energy required will be greater? Why? 451 Determine the kJ/kg, as it is heated empirical specific heat (Table A2c), (b) the enthalpy change h of nitrogen, in from 600 to 1000 K, using (a) the equation as a function of temperature cp value at the average temperature 4 41 Two tanks (Tank A and Tank B) are separated by a partition. Initially Tank A contains 2-kg steam at 1 MPa and 300C while Tank B contains 3-kg saturated liquidvapor mixture with a vapor mass fraction of 50 percent. Now the partition is removed and the two sides are allowed to mix until the mechanical and thermal equilibrium are established. If the pressure at the final state is 300 kPa, determine (a) the temperature and quality of the steam (if mixture) at the final state and (b) the amount of heat lost from the tanks. TANK A 2 kg 1 MPa 300C TANK B 3 kg 150C x = 0.5 Q FIGURE P441 206 | Thermodynamics at a rate of 10,000 kJ/h, and a 100-W fan is used to distribute the warm air in the room. The rate of heat loss from the room is estimated to be about 5000 kJ/h. If the initial temperature of the room air is 10C, determine how long it will take for the air temperature to rise to 20C. Assume constant specific heats at room temperature. 459 A student living in a 4-m 6-m 6-m dormitory room turns on her 150-W fan before she leaves the room on a summer day, hoping that the room will be cooler when she comes back in the evening. Assuming all the doors and windows are tightly closed and disregarding any heat transfer through the walls and the windows, determine the temperature in the room when she comes back 10 h later. Use specific heat values at room temperature, and assume the room to be at 100 kPa and 15C in the morning when she leaves. Answer: 58.2C (Table A2b), and (c) the cp value at room temperature (Table A2a). Answers: (b) 447.8 kJ/kg, (b) 448.4 kJ/kg, (c) 415.6 kJ/kg 4 52E Determine the enthalpy change h of oxygen, in Btu/lbm, as it is heated from 800 to 1500 R, using (a) the empirical specific heat equation as a function of temperature (Table A2Ec), (b) the cp value at the average temperature (Table A2Eb), and (c) the cp value at room temperature (Table A2Ea). Answers: (a) 170.1 Btu/lbm, (b) 178.5 Btu/lbm, (c) 153.3 Btu/lbm 4 53 Determine the internal energy change u of hydrogen, in kJ/kg, as it is heated from 200 to 800 K, using (a) the empirical specific heat equation as a function of temperature (Table A2c), (b) the cv value at the average temperature (Table A2b), and (c) the cv value at room temperature (Table A2a). Closed-System Energy Analysis: Ideal Gases 4 54C Is it possible to compress an ideal gas isothermally in an adiabatic pistoncylinder device? Explain. 4 55E A rigid tank contains 20 lbm of air at 50 psia and 80F. The air is now heated until its pressure doubles. Determine (a) the volume of the tank and (b) the amount of heat transfer. Answers: (a) 80 ft3, (b) 1898 Btu 4 56 A 3-m3 rigid tank contains hydrogen at 250 kPa and 550 K. The gas is now cooled until its temperature drops to 350 K. Determine (a) the final pressure in the tank and (b) the amount of heat transfer. 4 57 A 4-m 5-m 6-m room is to be heated by a baseboard resistance heater. It is desired that the resistance heater be able to raise the air temperature in the room from 7 to 23C within 15 min. Assuming no heat losses from the room and an atmospheric pressure of 100 kPa, determine the required power of the resistance heater. Assume constant specific heats at room temperature. Answer: 1.91 kW 4 58 A 4-m 5-m 7-m room is heated by the radiator of a steam-heating system. The steam radiator transfers heat 5000 kJ/h ROOM 4m5m7m Steam Wpw IDEAL GAS 10,000 kJ/h 800 kPa 50C Evacuated ROOM 4m6m6m Fan FIGURE P459 460E A 10-ft3 tank contains oxygen initially at 14.7 psia and 80F. A paddle wheel within the tank is rotated until the pressure inside rises to 20 psia. During the process 20 Btu of heat is lost to the surroundings. Determine the paddle-wheel work done. Neglect the energy stored in the paddle wheel. 461 An insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains 4 kg of an ideal gas at 800 kPa and 50C, and the other part is evacuated. The partition is now removed, and the gas expands into the entire tank. Determine the final temperature and pressure in the tank. FIGURE P458 FIGURE P461 Chapter 4 4 62 A pistoncylinder device whose piston is resting on top of a set of stops initially contains 0.5 kg of helium gas at 100 kPa and 25C. The mass of the piston is such that 500 kPa of pressure is required to raise it. How much heat must be transferred to the helium before the piston starts rising? Answer: 1857 kJ | 207 transfer. Let the polytropic exponent vary from 1.1 to 1.6. Plot the boundary work and the heat transfer versus the polytropic exponent, and discuss the results. 469 A room is heated by a baseboard resistance heater. When the heat losses from the room on a winter day amount to 6500 kJ/h, the air temperature in the room remains constant even though the heater operates continuously. Determine the power rating of the heater, in kW. 4 63 An insulated pistoncylinder device contains 100 L of air at 400 kPa and 25C. A paddle wheel within the cylinder is rotated until 15 kJ of work is done on the air while the pressure is held constant. Determine the final temperature of the air. Neglect the energy stored in the paddle wheel. 4 64E A pistoncylinder device contains 25 ft3 of nitrogen at 40 psia and 700F. Nitrogen is now allowed to cool at constant pressure until the temperature drops to 200F. Using specific heats at the average temperature, determine the amount of heat loss. 4 65 A mass of 15 kg of air in a pistoncylinder device is heated from 25 to 77C by passing current through a resistance heater inside the cylinder. The pressure inside the cylinder is held constant at 300 kPa during the process, and a heat loss of 60 kJ occurs. Determine the electric energy supplied, in kWh. Answer: 0.235 kWh ROOM Q Tair = constant We FIGURE P469 470E A pistoncylinder device contains 3 ft3 of air at 60 psia and 150F. Heat is transferred to the air in the amount of 40 Btu as the air expands isothermally. Determine the amount of boundary work done during this process. AIR P = constant We Q 471 A pistoncylinder device contains 4 kg of argon at 250 kPa and 35C. During a quasi-equilibrium, isothermal expansion process, 15 kJ of boundary work is done by the system, and 3 kJ of paddle-wheel work is done on the system. Determine the heat transfer for this process. 472 A pistoncylinder device, whose piston is resting on a set of stops, initially contains 3 kg of air at 200 kPa and 27C. The mass of the piston is such that a pressure of 400 kPa is required to move it. Heat is now transferred to the air until its volume doubles. Determine the work done by the air and the total heat transferred to the air during this process. Also show the process on a P-v diagram. Answers: 516 kJ, 2674 kJ FIGURE P465 4 66 An insulated pistoncylinder device initially contains 0.3 m3 of carbon dioxide at 200 kPa and 27C. An electric switch is turned on, and a 110-V source supplies current to a resistance heater inside the cylinder for a period of 10 min. The pressure is held constant during the process, while the volume is doubled. Determine the current that passes through the resistance heater. 4 67 A pistoncylinder device contains 0.8 kg of nitrogen initially at 100 kPa and 27C. The nitrogen is now compressed slowly in a polytropic process during which PV1.3 constant until the volume is reduced by one-half. Determine the work done and the heat transfer for this process. 4 68 Reconsider Prob. 4 67. Using EES (or other) software, plot the process described in the problem on a P-V diagram, and investigate the effect of the polytropic exponent n on the boundary work and heat A pistoncylinder device, with a set of stops on the top, initially contains 3 kg of air at 200 kPa and 27C. Heat is now transferred to the air, and the piston rises until it hits the stops, at which point the volume is twice the initial volume. More heat is transferred until the pressure inside the cylinder also doubles. Determine the work done and the amount of heat transfer for this process. Also, show the process on a P-v diagram. 473 Closed-System Energy Analysis: Solids and Liquids 474 In a manufacturing facility, 5-cm-diameter brass balls (r 8522 kg/m3 and cp 0.385 kJ/kg C) initially at 120C are quenched in a water bath at 50C for a period of 2 min at 208 | Thermodynamics 478 Stainless steel ball bearings (r 8085 kg/m3 and cp 0.480 kJ/kg C) having a diameter of 1.2 cm are to be quenched in water at a rate of 800 per minute. The balls leave the oven at a uniform temperature of 900C and are exposed to air at 25C for a while before they are dropped into the water. If the temperature of the balls drops to 850C prior to quenching, determine the rate of heat transfer from the balls to the air. 479 Carbon steel balls (r 7833 kg/m3 and cp 0.465 kJ/kg C) 8 mm in diameter are annealed by heating them first to 900C in a furnace, and then allowing them to cool slowly to 100C in ambient air at 35C. If 2500 balls are to be annealed per hour, determine the total rate of heat transfer from the balls to the ambient air. Answer: 542 W a rate of 100 balls per minute. If the temperature of the balls after quenching is 74C, determine the rate at which heat needs to be removed from the water in order to keep its temperature constant at 50C. Brass balls 120C 50C Water bath FIGURE P474 4 75 Repeat Prob. 4 74 for aluminum balls. Furnace 900C Air, 35C Steel ball 100C 4 76E During a picnic on a hot summer day, all the cold drinks disappeared quickly, and the only available drinks were those at the ambient temperature of 75F. In an effort to cool a 12-fluid-oz drink in a can, a person grabs the can and starts shaking it in the iced water of the chest at 32F. Using the properties of water for the drink, determine the mass of ice that will melt by the time the canned drink cools to 45F. 4 77 Consider a 1000-W iron whose base plate is made of 0.5-cm-thick aluminum alloy 2024-T6 (r 2770 kg/m3 and 875 J/kg C). The base plate has a surface area of 0.03 cp m2. Initially, the iron is in thermal equilibrium with the ambient air at 22C. Assuming 85 percent of the heat generated in the resistance wires is transferred to the plate, determine the minimum time needed for the plate temperature to reach 140C. FIGURE P479 480 An electronic device dissipating 30 W has a mass of 20 g and a specific heat of 850 J/kg C. The device is lightly used, and it is on for 5 min and then off for several hours, during which it cools to the ambient temperature of 25C. Determine the highest possible temperature of the device at the end of the 5-min operating period. What would your answer be if the device were attached to a 0.2-kg aluminum heat sink? Assume the device and the heat sink to be nearly isothermal. 481 Reconsider Prob. 480. Using EES (or other) software, investigate the effect of the mass of the heat sink on the maximum device temperature. Let the mass of heat sink vary from 0 to 1 kg. Plot the maximum temperature against the mass of heat sink, and discuss the results. 482 An ordinary egg can be approximated as a 5.5-cmdiameter sphere. The egg is initially at a uniform temperature of 8C and is dropped into boiling water at 97C. Taking the 3.32 properties of the egg to be r 1020 kg/m3 and cp kJ/kg C, determine how much heat is transferred to the egg by the time the average temperature of the egg rises to 80C. 483E ln a production facility, 1.2-in-thick 2-ft 2-ft square brass plates (r 532.5 lbm/ft3 and cp 0.091 Btu/lbm F) that are initially at a uniform temperature of 75F are heated by passing them through an oven at 1300F at a rate of 300 per minute. If the plates remain in the oven until their average temperature rises to 1000F, determine the rate of heat transfer to the plates in the furnace. FIGURE P477 Vol. 58/PhotoDisc Chapter 4 Furnace, 1300F | 209 494 A 55-kg man gives in to temptation and eats an entire 1-L box of ice cream. How long does this man need to jog to burn off the calories he consumed from the ice cream? Answer: 2.52 h 1.2 in. 495 Consider a man who has 20 kg of body fat when he goes on a hunger strike. Determine how long he can survive on his body fat alone. 496 Consider two identical 50-kg women, Candy and Wendy, who are doing identical things and eating identical food except that Candy eats her baked potato with four teaspoons of butter while Wendy eats hers plain every evening. Determine the difference in the weights of Candy and Wendy after one year. Answer: 6.5 kg 497 A woman who used to drink about one liter of regular cola every day switches to diet cola (zero calorie) and starts eating two slices of apple pie every day. Is she now consuming fewer or more calories? 498 A 60-kg man used to have an apple every day after dinner without losing or gaining any weight. He now eats a 200-ml serving of ice cream instead of an apple and walks 20 min every day. On this new diet, how much weight will he lose or gain per month? Answer: 0.087-kg gain 499 The average specific heat of the human body is 3.6 kJ/kg C. If the body temperature of an 80-kg man rises from 37C to 39C during strenuous exercise, determine the increase in the thermal energy of the body as a result of this rise in body temperature. 4100E Alcohol provides 7 Calories per gram, but it provides no essential nutrients. A 1.5 ounce serving of 80-proof liquor contains 100 Calories in alcohol alone. Sweet wines and beer provide additional calories since they also contain carbohydrates. About 75 percent of American adults drink some sort of alcoholic beverage, which adds an average of 210 Calories a day to their diet. Determine how many pounds less an average American adult will weigh per year if he or she quit drinking alcoholic beverages and started drinking diet soda. 4101 A 12-oz serving of a regular beer contains 13 g of alcohol and 13 g of carbohydrates, and thus 150 Calories. A 12-oz serving of a light beer contains 11 g of alcohol and 5 g Regular beer Light beer Brass plate, 75F FIGURE P483E 4 84 Long cylindrical steel rods (r 7833 kg/m3 and cp 0.465 kJ/kg C) of 10-cm diameter are heat-treated by drawing them at a velocity of 3 m/min through an oven maintained at 900C. If the rods enter the oven at 30C and leave at a mean temperature of 700C, determine the rate of heat transfer to the rods in the oven. Special Topic: Biological Systems 4 85C What is metabolism? What is basal metabolic rate? What is the value of basal metabolic rate for an average man? 4 86C For what is the energy released during metabolism in humans used? 4 87C Is the metabolizable energy content of a food the same as the energy released when it is burned in a bomb calorimeter? If not, how does it differ? 4 88C Is the number of prospective occupants an important consideration in the design of heating and cooling systems of classrooms? Explain. 4 89C What do you think of a diet program that allows for generous amounts of bread and rice provided that no butter or margarine is added? 4 90 Consider two identical rooms, one with a 2-kW electric resistance heater and the other with three couples fast dancing. In which room will the air temperature rise faster? 4 91 Consider two identical 80-kg men who are eating identical meals and doing identical things except that one of them jogs for 30 min every day while the other watches TV. Determine the weight difference between the two in a month. Answer: 1.045 kg 4 92 Consider a classroom that is losing heat to the outdoors at a rate of 20,000 kJ/h. If there are 30 students in class, each dissipating sensible heat at a rate of 100 W, determine if it is necessary to turn the heater in the classroom on to prevent the room temperature from dropping. 4 93 A 68-kg woman is planning to bicycle for an hour. If she is to meet her entire energy needs while bicycling by eating 30-g chocolate candy bars, determine how many candy bars she needs to take with her. 12 oz. 150 cal 12 oz. 100 cal FIGURE P4101 210 | Thermodynamics where W is the weight (actually, the mass) of the person in kg and H is the height in m, and the range of healthy weight is 19 BMI 25. Convert the previous formula to English units such that the weight is in pounds and the height in inches. Also, calculate your own BMI, and if it is not in the healthy range, determine how many pounds (or kg) you need to gain or lose to be fit. 4109 The body mass index (BMI) of a 1.7-m tall woman who normally has 3 large slices of cheese pizza and a 400-ml Coke for lunch is 30. She now decides to change her lunch to 2 slices of pizza and a 200-ml Coke. Assuming that the deficit in the calorie intake is made up by burning body fat, determine how long it will take for the BMI of this person to drop to 25. Use the data in the text for calories and take the metabolizable energy content of 1 kg of body fat to be 33,100 kJ. Answer: 262 days of carbohydrates, and thus 100 Calories. An average person burns 700 Calories per hour while exercising on a treadmill. Determine how long it will take to burn the calories from a 12-oz can of (a) regular beer and (b) light beer on a treadmill. 4 102 A 5-oz serving of a Bloody Mary contains 14 g of alcohol and 5 g of carbohydrates, and thus 116 Calories. A 2.5-oz serving of a martini contains 22 g of alcohol and a negligible amount of carbohydrates, and thus 156 Calories. An average person burns 600 Calories per hour while exercising on a cross-country ski machine. Determine how long it will take to burn the calories from one serving of (a) a Bloody Mary and (b) a martini on this cross-country ski machine. 4 103E A 176-pound man and a 132-pound woman went to Burger King for lunch. The man had a BK Big Fish sandwich (720 Cal), medium french fries (400 Cal), and a large Coke (225 Cal). The woman had a basic hamburger (330 Cal), medium french fries (400 Cal), and a diet Coke (0 Cal). After lunch, they start shoveling snow and burn calories at a rate of 360 Cal/h for the woman and 480 Cal/h for the man. Determine how long each one of them needs to shovel snow to burn off the lunch calories. 4 104 Consider two friends who go to Burger King every day for lunch. One of them orders a Double Whopper sandwich, large fries, and a large Coke (total Calories 1600) while the other orders a Whopper Junior, small fries, and a small Coke (total Calories 800) every day. If these two friends are very much alike otherwise and they have the same metabolic rate, determine the weight difference between these two friends in a year. 4 105E A 150-pound person goes to Hardee's for dinner and orders a regular roast beef (270 Cal) and a big roast beef (410 Cal) sandwich together with a 12-oz can of Pepsi (150 Cal). A 150-pound person burns 400 Calories per hour while climbing stairs. Determine how long this person needs to climb stairs to burn off the dinner calories. 4 106 A person eats a McDonald's Big Mac sandwich (530 Cal), a second person eats a Burger King Whopper sandwich (640 Cal), and a third person eats 50 olives with regular french fries (350 Cal) for lunch. Determine who consumes the most calories. An olive contains about 5 Calories. 4 107 A 100-kg man decides to lose 5 kg without cutting down his intake of 3000 Calories a day. Instead, he starts fast swimming, fast dancing, jogging, and biking each for an hour every day. He sleeps or relaxes the rest of the day. Determine how long it will take him to lose 5 kg. 4 108E The range of healthy weight for adults is usually expressed in terms of the body mass index (BMI), defined, in SI units, as BMI H 2 1m2 2 W 1kg2 Review Problems 4110 Consider a pistoncylinder device that contains 0.5 kg air. Now, heat is transferred to the air at constant pressure and the air temperature increases by 5C. Determine the expansion work done during this process. 4111 In solar-heated buildings, energy is often stored as sensible heat in rocks, concrete, or water during the day for use at night. To minimize the storage space, it is desirable to use a material that can store a large amount of heat while experiencing a small temperature change. A large amount of heat can be stored essentially at constant temperature during a phase change process, and thus materials that change phase at about room temperature such as glaubers salt (sodium sulfate decahydrate), which has a melting point of 32C and a heat of fusion of 329 kJ/L, are very suitable for this purpose. Determine how much heat can be stored in a 5-m3 storage space using (a) glaubers salt undergoing a phase change, (b) granite rocks with a heat capacity of 2.32 kJ/kg C and a temperature change of 20C, and (c) water with a heat capacity of 4.00 kJ/kg C and a temperature change of 20C. 4112 A pistoncylinder device contains 0.8 kg of an ideal gas. Now, the gas is cooled at constant pressure until its temperature decreases by 10C. If 16.6 kJ of compression work Ideal gas 0.8 kg T = 10C Q FIGURE P4112 Chapter 4 is done during this process, determine the gas constant and the molar mass of the gas. Also, determine the constantvolume and constant-pressure specific heats of the gas if its specific heat ratio is 1.667. 4 113 The temperature of air changes from 0 to 10C while its velocity changes from zero to a final velocity, and its elevation changes from zero to a final elevation. At which values of final air velocity and final elevation will the internal, kinetic, and potential energy changes be equal? Answers: 119.8 m/s, 731.9 m | 211 the vapor phase. Heat is now transferred to the water, and the piston, which is resting on a set of stops, starts moving when the pressure inside reaches 300 kPa. Heat transfer continues until the total volume increases by 20 percent. Determine (a) the initial and final temperatures, (b) the mass of liquid water when the piston first starts moving, and (c) the work done during this process. Also, show the process on a P-v diagram. 4116E A spherical balloon contains 10 lbm of air at 30 psia and 800 R. The balloon material is such that the pressure inside is always proportional to the square of the diameter. Determine the work done when the volume of the balloon doubles as a result of heat transfer. Answer: 715 Btu 4117E Reconsider Prob. 4116E. Using the integration feature of the EES software, determine the work done. Compare the result with your "handcalculated" result. 4 114 A frictionless pistoncylinder device initially contains air at 200 kPa and 0.2 m3. At this state, a linear spring (F x) is touching the piston but exerts no force on it. The air is now heated to a final state of 0.5 m3 and 800 kPa. Determine (a) the total work done by the air and (b) the work done against the spring. Also, show the process on a P-v diagram. Answers: (a) 150 kJ, (b) 90 kJ 4118 A mass of 12 kg of saturated refrigerant-134a vapor is contained in a pistoncylinder device at 240 kPa. Now 300 kJ of heat is transferred to the refrigerant at constant pressure while a 110-V source supplies current to a resistor within the cylinder for 6 min. Determine the current supplied if the final temperature is 70C. Also, show the process on a T-v diagram with respect to the saturation lines. Answer: 12.8 A AIR P = 200 kPa 1 V1 = 0.2 m3 R-134a P = constant FIGURE P4114 We 4 115 A mass of 5 kg of saturated liquidvapor mixture of water is contained in a pistoncylinder device at 125 kPa. Initially, 2 kg of the water is in the liquid phase and the rest is in Q FIGURE P4118 4119 A mass of 0.2 kg of saturated refrigerant-134a is contained in a pistoncylinder device at 200 kPa. Initially, 75 percent of the mass is in the liquid phase. Now heat is transferred to the refrigerant at constant pressure until the cylinder contains vapor only. Show the process on a P-v diagram with respect to saturation lines. Determine (a) the volume occupied by the refrigerant initially, (b) the work done, and (c) the total heat transfer. 4120 A pistoncylinder device contains helium gas initially at 150 kPa, 20C, and 0.5 m3. The helium is now compressed in a polytropic process (PV n constant) to 400 kPa and 140C. Determine the heat loss or gain during this process. Answer: 11.2 kJ loss H 2O m = 5 kg FIGURE P4115 212 | Thermodynamics 4124 One ton (1000 kg) of liquid water at 80C is brought into a well-insulated and well-sealed 4-m 5-m 6-m room initially at 22C and 100 kPa. Assuming constant specific heats for both air and water at room temperature, determine the final equilibrium temperature in the room. Answer: 78.6C He PV n = constant Q FIGURE P4120 4 121 A frictionless pistoncylinder device and a rigid tank initially contain 12 kg of an ideal gas each at the same temperature, pressure, and volume. It is desired to raise the temperatures of both systems by 15C. Determine the amount of extra heat that must be supplied to the gas in the cylinder which is maintained at constant pressure to achieve this result. Assume the molar mass of the gas is 25. 4 122 A passive solar house that is losing heat to the outdoors at an average rate of 50,000 kJ/h is maintained at 22C at all times during a winter night for 10 h. The house is to be heated by 50 glass containers each containing 20 L of water that is heated to 80C during the day by absorbing solar energy. A thermostat-controlled 15-kW back-up electric resistance heater turns on whenever necessary to keep the house at 22C. (a) How long did the electric heating system run that night? (b) How long would the electric heater run that night if the house incorporated no solar heating? Answers: (a) 4.77 h, (b) 9.26 h 4125 A 4-m 5-m 6-m room is to be heated by one ton (1000 kg) of liquid water contained in a tank that is placed in the room. The room is losing heat to the outside at an average rate of 8000 kJ/h. The room is initially at 20C and 100 kPa and is maintained at an average temperature of 20C at all times. If the hot water is to meet the heating requirements of this room for a 24-h period, determine the minimum temperature of the water when it is first brought into the room. Assume constant specific heats for both air and water at room temperature. 4126 The energy content of a certain food is to be determined in a bomb calorimeter that contains 3 kg of water by burning a 2-g sample of it in the presence of 100 g of air in the reaction chamber. If the water temperature rises by 3.2C when equilibrium is established, determine the energy content of the food, in kJ/kg, by neglecting the thermal energy stored in the reaction chamber and the energy supplied by the mixer. What is a rough estimate of the error involved in neglecting the thermal energy stored in the reaction chamber? Answer: 20,060 kJ/kg Reaction chamber Food T = 3.2C FIGURE P4126 22C Water 80C Pump 4127 A 68-kg man whose average body temperature is 39C drinks 1 L of cold water at 3C in an effort to cool down. Taking the average specific heat of the human body to be 3.6 kJ/kg C, determine the drop in the average body temperature of this person under the influence of this cold water. 4128 A 0.2-L glass of water at 20C is to be cooled with ice to 5C. Determine how much ice needs to be added to the water, in grams, if the ice is at (a) 0C and (b) 8C. Also determine how much water would be needed if the cooling is to be done with cold water at 0C. The melting temperature and the heat of fusion of ice at atmospheric pressure are FIGURE P4122 4 123 An 1800-W electric resistance heating element is immersed in 40 kg of water initially at 20C. Determine how long it will take for this heater to raise the water temperature to 80C. Chapter 4 0C and 333.7 kJ/kg, respectively, and the density of water is 1 kg/L. 4 129 Reconsider Prob. 4 128. Using EES (or other) software, investigate the effect of the initial temperature of the ice on the final mass required. Let the ice temperature vary from 20 to 0C. Plot the mass of ice against the initial temperature of ice, and discuss the results. | 213 evaporates in 25 min. Determine the power rating of the electric heating element immersed in water. Also, determine how long it will take for this heater to raise the temperature of 1 L of cold water from 18C to the boiling temperature. 1 atm 4 130 In order to cool 1 ton of water at 20C in an insulated tank, a person pours 80 kg of ice at 5C into the water. Determine the final equilibrium temperature in the tank. The melting temperature and the heat of fusion of ice at atmospheric pressure are 0C and 333.7 kJ/kg, respectively. Answer: 12.4C Coffee maker 1L 4 131 An insulated pistoncylinder device initially contains 0.01 m3 of saturated liquidvapor mixture with a quality of 0.2 at 120C. Now some ice at 0C is added to the cylinder. If the cylinder contains saturated liquid at 120C when thermal equilibrium is established, determine the amount of ice added. The melting temperature and the heat of fusion of ice at atmospheric pressure are 0C and 333.7 kJ/kg, respectively. 4 132 The early steam engines were driven by the atmospheric pressure acting on the piston fitted into a cylinder filled with saturated steam. A vacuum was created in the cylinder by cooling the cylinder externally with cold water, and thus condensing the steam. Consider a pistoncylinder device with a piston surface area of 0.1 m2 initially filled with 0.05 m3 of saturated water vapor at the atmospheric pressure of 100 kPa. Now cold water is poured outside the cylinder, and the steam inside starts condensing as a result of heat transfer to the cooling water outside. If the piston is stuck at its initial position, determine the friction force acting on the piston and the amount of heat transfer when the temperature inside the cylinder drops to 30C. Cold water FIGURE P4133 4134 Two rigid tanks are connected by a valve. Tank A contains 0.2 m3 of water at 400 kPa and 80 percent quality. Tank B contains 0.5 m3 of water at 200 kPa and 250C. The valve is now opened, and the two tanks eventually come to the same state. Determine the pressure and the amount of heat transfer when the system reaches thermal equilibrium with the surroundings at 25C. Answers: 3.17 kPa, 2170 kJ H 2O 400 kPa Q A B H2 O 200 kPa FIGURE P4134 4135 Reconsider Prob. 4134. Using EES (or other) software, investigate the effect of the environment temperature on the final pressure and the heat transfer. Let the environment temperature vary from 0 to 50C. Plot the final results against the environment temperature, and discuss the results. 0.05 m 3 100 kPa Steam FIGURE P4132 4 133 Water is boiled at sea level in a coffee maker equipped with an immersion-type electric heating element. The coffee maker contains 1 L of water when full. Once boiling starts, it is observed that half of the water in the coffee maker 4136 A rigid tank containing 0.4 m3 of air at 400 kPa and 30C is connected by a valve to a pistoncylinder device with zero clearance. The mass of the piston is such that a pressure of 200 kPa is required to raise the piston. The valve is now opened slightly, and air is allowed to flow into the cylinder until the pressure in the tank drops to 200 kPa. During this process, heat is exchanged with the surroundings such that 214 | Thermodynamics N2 1 m3 500 kPa 80C He 1 m3 500 kPa 25C the entire air remains at 30C at all times. Determine the heat transfer for this process. AIR T = const. Q FIGURE P4138 4139 Repeat Prob. 4138 by assuming the piston is made of 5 kg of copper initially at the average temperature of the two gases on both sides. Answer: 56C FIGURE P4136 4 137 A well-insulated 4-m 4-m 5-m room initially at 10C is heated by the radiator of a steam heating system. The radiator has a volume of 15 L and is filled with superheated vapor at 200 kPa and 200C. At this moment both the inlet and the exit valves to the radiator are closed. A 120-W fan is used to distribute the air in the room. The pressure of the steam is observed to drop to 100 kPa after 30 min as a result of heat transfer to the room. Assuming constant specific heats for air at room temperature, determine the average temperature of air in 30 min. Assume the air pressure in the room remains constant at 100 kPa. 4140 Reconsider Prob. 4139. Using EES (or other) software, investigate the effect of the mass of the copper piston on the final equilibrium temperature. Let the mass of piston vary from 1 to 10 kg. Plot the final temperature against the mass of piston, and discuss the results. 4141 An insulated rigid tank initially contains 1.4-kg saturated liquid water and water vapor at 200C. At this state, 25 percent of the volume is occupied by liquid water and the rest by vapor. Now an electric resistor placed in the tank is turned on, and the tank is observed to contain saturated water vapor after 20 min. Determine (a) the volume of the tank, (b) the final temperature, and (c) the electric power rating of the resistor. Answers: (a ) 0.00648 m3, (b) 371C, (c) 1.58 kW 10C 4m4m5m Fan Steam radiator We Water 1.4 kg, 200C FIGURE P4141 4142 A vertical 12-cm diameter pistoncylinder device contains an ideal gas at the ambient conditons of 1 bar and 24C. Initially, the inner face of the piston is 20 cm from the base of the cylinder. Now an external shaft connected to the piston exerts a force corresponding to a boundary work input of 0.1 kJ. The temperature of the gas remains constant during the process. Determine (a) the amount of heat transfer, (b) the final pressure in the cylinder, and (c) the distance that the piston is displaced. 4143 A pistoncylinder device initially contains 0.15-kg steam at 3.5 MPa, superheated by 5C. Now the steam loses heat to the surroundings and the piston moves down, hitting a set of stops at which point the cylinder contains saturated liquid water. The cooling continues until the cylinder contains water at 200C. Determine (a) the final pressure and the quality (if mix- FIGURE P4137 4 138 Consider a well-insulated horizontal rigid cylinder that is divided into two compartments by a piston that is free to move but does not allow either gas to leak into the other side. Initially, one side of the piston contains 1 m3 of N2 gas at 500 kPa and 80C while the other side contains 1 m3 of He gas at 500 kPa and 25C. Now thermal equilibrium is established in the cylinder as a result of heat transfer through the piston. Using constant specific heats at room temperature, determine the final equilibrium temperature in the cylinder. What would your answer be if the piston were not free to move? Chapter 4 ture), (b) the boundary work, (c) the amount of heat transfer when the piston first hits the stops, (d) and the total heat transfer. | 215 explosion energy eexp is usually expressed per unit volume, and it is obtained by dividing the quantity above by the total V of the vessel: eexp u1 v1 u2 Steam 0.15 kg 3.5 MPa Q where v1 is the specific volume of the fluid before the explosion. Show that the specific explosion energy of an ideal gas with constant specific heat is eexp P1 k 1 a1 T2 b T1 FIGURE P4143 4 144 An insulated rigid tank is divided into two compartments of different volumes. Initially, each compartment contains the same ideal gas at identical pressure but at different temperatures and masses. The wall separating the two compartments is removed and the two gases are allowed to mix. Assuming constant specific heats, find the simplest expression for the mixture temperature written in the form T3 fa m1 m2 , ,T,T b m3 m3 1 2 Also, determine the total explosion energy of 20 m3 of air at 5 MPa and 100C when the surroundings are at 20C. P2 T2 Steam boiler P1 T1 where m3 and T3 are the mass and temperature of the final mixture, respectively. FIGURE P4145 4146 Using the relations in Prob. 4145, determine the explosive energy of 20 m3 of steam at 10 MPa and 500C assuming the steam condenses and becomes a liquid at 25C after the explosion. To how many kilograms of TNT is this explosive energy equivalent? The explosive energy of TNT is about 3250 kJ/kg. SIDE 1 Mass = m1 Temperature = T1 SIDE 2 Mass = m 2 Temperature = T2 FIGURE P4144 4 145 Catastrophic explosions of steam boilers in the 1800s and early 1900s resulted in hundreds of deaths, which prompted the development of the ASME Boiler and Pressure Vessel Code in 1915. Considering that the pressurized fluid in a vessel eventually reaches equilibrium with its surroundings shortly after the explosion, the work that a pressurized fluid would do if allowed to expand adiabatically to the state of the surroundings can be viewed as the explosive energy of the pressurized fluid. Because of the very short time period of the explosion and the apparent stability afterward, the explosion process can be considered to be adiabatic with no changes in kinetic and potential energies. The closed-system conservation of energy relation in this case reduces to Wout m(u1 u2). Then the explosive energy Eexp becomes Eexp m 1u1 u2 2 Fundamentals of Engineering (FE) Exam Problems 4147 A room is filled with saturated steam at 100C. Now a 5-kg bowling ball at 25C is brought to the room. Heat is transferred to the ball from the steam, and the temperature of the ball rises to 100C while some steam condenses on the ball as it loses heat (but it still remains at 100C). The specific heat of the ball can be taken to be 1.8 kJ/kg C. The mass of steam that condensed during this process is (a) 80 g (b) 128 g (c) 299 g (d) 351 g (e) 405 g 4148 A frictionless pistoncylinder device and a rigid tank contain 2 kmol of an ideal gas at the same temperature, pressure, and volume. Now heat is transferred, and the temperature of both systems is raised by 10C. The amount of extra heat that must be supplied to the gas in the cylinder that is maintained at constant pressure is (a) 0 kJ (b) 42 kJ (c) 83 kJ (d ) 102 kJ (e) 166 kJ where the subscripts 1 and 2 refer to the state of the fluid before and after the explosion, respectively. The specific 216 | Thermodynamics and 3 kJ of paddle-wheel work is done on the system. The heat transfer during this process is (a) 12 kJ (b) 18 kJ (c) 2.4 kJ (d) 3.5 kJ (e) 60 kJ 4 149 The specific heat of a material is given in a strange unit to be c 3.60 kJ/kg F. The specific heat of this material in the SI units of kJ/kg C is (a) 2.00 kJ/kg C (b) 3.20 kJ/kg C (c) 3.60 kJ/kg C (d ) 4.80 kJ/kg C (e) 6.48 kJ/kg C 4 150 A 3-m3 rigid tank contains nitrogen gas at 500 kPa and 300 K. Now heat is transferred to the nitrogen in the tank and the pressure of nitrogen rises to 800 kPa. The work done during this process is (a) 500 kJ (b) 1500 kJ (c) 0 kJ 0.8-m3 (d ) 900 kJ (e) 2400 kJ 4156 A container equipped with a resistance heater and a mixer is initially filled with 3.6 kg of saturated water vapor at 120C. Now the heater and the mixer are turned on; the steam is compressed, and there is heat loss to the surrounding air. At the end of the process, the temperature and pressure of steam in the container are measured to be 300C and 0.5 MPa. The net energy transfer to the steam during this process is (a) 274 kJ (b) 914 kJ (c) 1213 kJ (d) 988 kJ (e) 1291 kJ 4 151 A rigid tank contains nitrogen gas at 600 kPa and 300 K. Now the gas is compressed isothermally to a volume of 0.1 m3. The work done on the gas during this compression process is (a) 746 kJ (b) 0 kJ (c) 420 kJ (d ) 998 kJ (e) 1890 kJ 4157 A 6-pack canned drink is to be cooled from 25C to 3C. The mass of each canned drink is 0.355 kg. The drinks can be treated as water, and the energy stored in the aluminum can itself is negligible. The amount of heat transfer from the 6 canned drinks is (a) 33 kJ (b) 37 kJ (c) 47 kJ (d) 196 kJ (e) 223 kJ 4 152 A well-sealed room contains 60 kg of air at 200 kPa and 25C. Now solar energy enters the room at an average rate of 0.8 kJ/s while a 120-W fan is turned on to circulate the air in the room. If heat transfer through the walls is negligible, the air temperature in the room in 30 min will be (a) 25.6C (b) 49.8C (c) 53.4C (d ) 52.5C (e) 63.4C 4158 A glass of water with a mass of 0.45 kg at 20C is to be cooled to 0C by dropping ice cubes at 0C into it. The latent heat of fusion of ice is 334 kJ/kg, and the specific heat of water is 4.18 kJ/kg C. The amount of ice that needs to be added is (a) 56 g (b) 113 g (c) 124 g (d) 224 g (e) 450 g 4 153 A 2-kW baseboard electric resistance heater in a vacant room is turned on and kept on for 15 min. The mass of the air in the room is 75 kg, and the room is tightly sealed so that no air can leak in or out. The temperature rise of air at the end of 15 min is (a) 8.5C (b) 12.4C (c) 24.0C (d) 33.4C (e) 54.8C 4159 A 2-kW electric resistance heater submerged in 5-kg water is turned on and kept on for 10 min. During the process, 300 kJ of heat is lost from the water. The temperature rise of water is (a) 0.4C (b) 43.1C (c) 57.4C (d ) 71.8C (e) 180.0C 4 154 A room contains 60 kg of air at 100 kPa and 15C. The room has a 250-W refrigerator (the refrigerator consumes 250 W of electricity when running), a 120-W TV, a 1kW electric resistance heater, and a 50-W fan. During a cold winter day, it is observed that the refrigerator, the TV, the fan, and the electric resistance heater are running continuously but the air temperature in the room remains constant. The rate of heat loss from the room that day is (a) 3312 kJ/h (b) 4752 kJ/h (c) 5112 kJ/h (d) 2952 kJ/h (e) 4680 kJ/h 4160 3 kg of liquid water initially at 12C is to be heated at 95C in a teapot equipped with a 1200-W electric heating element inside. The specific heat of water can be taken to be 4.18 kJ/kg C, and the heat loss from the water during heating can be neglected. The time it takes to heat water to the desired temperature is (a) 4.8 min (b) 14.5 min (c) 6.7 min (d ) 9.0 min (e) 18.6 min 4 155 A pistoncylinder device contains 5 kg of air at 400 kPa and 30C. During a quasi-equilibium isothermal expansion process, 15 kJ of boundary work is done by the system, 4161 An ordinary egg with a mass of 0.1 kg and a specific heat of 3.32 kJ/kg C is dropped into boiling water at 95C. Chapter 4 If the initial temperature of the egg is 5C, the maximum amount of heat transfer to the egg is (a) 12 kJ (b) 30 kJ (c) 24 kJ (d ) 18 kJ (e) infinity | 217 until the temperature rises to 1200C, and the piston moves to maintain a constant pressure. 1. The initial pressure of the steam is most nearly (a) 250 kPa (d ) 1000 kPa (b) 500 kPa (e) 1250 kPa (c) 750 kPa 2. The work done by the steam on the piston is most nearly (a) 230 kJ/kg (d ) 2340 kJ/kg (b) 1100 kJ/kg (e) 840 kJ/kg (c) 2140 kJ/kg 3. The total heat transferred to the steam is most nearly (a) 230 kJ/kg (d ) 2340 kJ/kg (b) 1100 kJ/kg (e) 840 kJ/kg (c) 2140 kJ/kg 4 162 An apple with an average mass of 0.18 kg and average specific heat of 3.65 kJ/kg C is cooled from 22C to 5C. The amount of heat transferred from the apple is (a) 0.85 kJ (b) 62.1 kJ (c) 17.7 kJ (d ) 11.2 kJ (e) 7.1 kJ 4 163 The specific heat at constant pressure for an ideal gas is given by cp 0.9 (2.7 10 4)T (kJ/kg K) where T is in kelvin. The change in the enthalpy for this ideal gas undergoing a process in which the temperature changes from 27 to 127C is most nearly (a) 90 kJ/kg (d ) 108.9 kJ/kg (b) 92.1 kJ/kg (e) 105.2 kJ/kg (c) 99.5 kJ/kg 4 164 The specific heat at constant volume for an ideal gas is given by cv 0.7 (2.7 10 4)T (kJ/kg K) where T is in kelvin. The change in the internal energy for this ideal gas undergoing a process in which the temperature changes from 27 to 127C is most nearly (a) 70 kJ/kg (d ) 82.1 kJ/kg (b) 72.1 kJ/kg (e) 84.0 kJ/kg (c) 79.5 kJ/kg 4 165 A pistoncylinder device contains an ideal gas. The gas undergoes two successive cooling processes by rejecting heat to the surroundings. First the gas is cooled at constant 3 T1. Then the piston is held stationary pressure until T2 4 1 while the gas is further cooled to T3 T1, where all temper2 atures are in K. 1. The ratio of the final volume to the initial volume of the gas is (a) 0.25 (d ) 0.75 (b) 0.50 (e) 1.0 (c) 0.67 2. The work done on the gas by the piston is (a) RT1/4 (d ) (cv cp)T1/4 (e) cv (T1 T2)/2 (b) cvT1/2 (c) cpT1/2 3. The total heat transferred from the gas is (a) RT1/4 (d ) (cv cp)T1/4 (e) cv (T1 T3)/2 (b) cvT1/2 (c) cpT1/2 4 166 Saturated steam vapor is contained in a pistoncylinder device. While heat is added to the steam, the piston is held stationary, and the pressure and temperature become 1.2 MPa and 700C, respectively. Additional heat is added to the steam Design, Essay, and Experiment Problems 4167 Using a thermometer, measure the boiling temperature of water and calculate the corresponding saturation pressure. From this information, estimate the altitude of your town and compare it with the actual altitude value. 4168 Find out how the specific heats of gases, liquids, and solids are determined in national laboratories. Describe the experimental apparatus and the procedures used. 4169 Design an experiment complete with instrumentation to determine the specific heats of a gas using a resistance heater. Discuss how the experiment will be conducted, what measurements need to be taken, and how the specific heats will be determined. What are the sources of error in your system? How can you minimize the experimental error? 4170 Design an experiment complete with instrumentation to determine the specific heats of a liquid using a resistance heater. Discuss how the experiment will be conducted, what measurements need to be taken, and how the specific heats will be determined. What are the sources of error in your system? How can you minimize the experimental error? How would you modify this system to determine the specific heat of a solid? 4171 You are asked to design a heating system for a swimming pool that is 2 m deep, 25 m long, and 25 m wide. Your client desires that the heating system be large enough to raise the water temperature from 20 to 30C in 3 h. The rate of heat loss from the water to the air at the outdoor design conditions is determined to be 960 W/m2, and the heater must also be able to maintain the pool at 30C at those conditions. Heat losses to the ground are expected to be small and can be disregarded. The heater considered is a natural gas furnace whose efficiency is 80 percent. What heater size (in kW input) would you recommend to your client? 4172 It is claimed that fruits and vegetables are cooled by 6C for each percentage point of weight loss as moisture 218 | Thermodynamics the heat unavoidably transferred into the hinge clamps. Verify the first law of thermodynamics using the video clip, the complete write-up, and the data provided on the DVD accompanying this book. 4178 First Law of Thermodynamics--Bicycle Braking Experiment The first law of thermodynamics is verified yet again--this time with a bicycle. A bicycle front caliper brake is removed and replaced with a lever-mounted, copper calorimeter friction pad. The calorimeter friction pad rubs on the front tire, heats up, brings the bicycle to a stop, and verifies the first law of thermodynamics. Used in the data reduction analysis are aerodynamics drag and rolling friction, which are obtained using bicycle coast-down data read into a cassette audio recorder by the bicycle rider. Verify the first law of thermodynamics using the video clip, the complete write-up, and the data provided on the DVD accompanying this book. 4179 Specific Heat of Aluminum--Electric Calorimeter Experiment The specific heat of aluminum is obtained with an electric calorimeter. The design consists of two individual calorimeters--each an assembly of 13 aluminum plates with electric resistance heater wires laced in-between the plates. The exterior surfaces of both calorimeters and the surrounding insulation are identical. However, the interior plates are different--one calorimeter has solid interior plates and the other has perforated interior plates. By initially adjusting the electrical power into each calorimeter the temperature-versustime curves for each calorimeter are matched. This curve match allows cancellation of the unknown heat loss from each calorimeter and cancellation of the unknown heater thermal capacity to deliver an accurate specific heat value. Obtain the specific heat of aluminum using the video clip, the complete write-up, and the data provided on the DVD accompanying this book. 4180 Specific Heat of Aluminum--Transient Cooling Experiment The specific heat of aluminum is obtained with an entirely different experiment than the one described in Prob. 4179. In the present experiment a hollow, aluminum cylinder calorimeter is fitted with a plug forming a watertight cavity. The calorimeter is heated with a hair drier and then allowed to cool in still air. Two tests are performed: one with water in the cavity and one without water in the cavity. Transient temperature measurements from the two tests give different cooling rates characterized with Trendlines in EXCEL. These Trendlines are used to compute the aluminum specific heat. Obtain the specific heat of aluminum using the video clip, the complete write-up, and the data provided on the DVD accompanying this book. during vacuum cooling. Using calculations, demonstrate if this claim is reasonable. 4 173 A 1982 U.S. Department of Energy article (FS #204) states that a leak of one drip of hot water per second can cost $1.00 per month. Making reasonable assumptions about the drop size and the unit cost of energy, determine if this claim is reasonable. 4 174 Polytropic Expansion of Air Experiment The expansion on compression of a gas can be described by the polyc, where p is pressure, v is specific tropic relation pv n volume, c is a constant and the exponent n depends on the thermodynamic process. In our experiment compressed air in a steel pressure vessel is discharged to the atmosphere while temperature and pressure measurements of the air inside the vessel are recorded. There measurements, along with the first law of thermodynamics, are used to produce the polytropic exponent n for the process. Obtain the polytropic exponent n for the process using the video clip, the complete write-up, and the data provided on the DVD accompanying this book. 4 175 First Law of Thermodynamics--Lead Smashing Experiment The first law of thermodynamics is verified with a lead smashing experiment. A small piece of lead, instrumented with a thermocouple, is smashed with two steel cylinders. The cylinders are suspended by nylon chords and swing as pendulums from opposite directions, simultaneously striking the lead. The loss in gravitational potential energy of the cylinders is equated to the rise in internal energy of the lead. Verify the first law of thermodynamics using the video clip, the complete write-up, and the data provided on the DVD accompanying this book. 4 176 First Law of Thermodynamics--Friction Bearing Experiment The first law of thermodynamics is verified with a friction bearing experiment. A copper friction bearing is attached to one end of a wood shaft that is driven in rotation with a falling weight turning a pulley attached to the shaft. Friction causes the bearing to heat up. Data reduction analysis accounts for gravitational potential energy, elastic potential energy, translational and rotational kinetic energy, internal energy, and heat loss from the bearing. Verify the first law of thermodynamics using the video clip, the complete write-up, and the data provided on the DVD accompanying this book. 4 177 First Law of Thermodynamics--Copper Cold Working Experiment The first law of thermodynamics is verified again, but this time with a copper hinge calorimeter that is "worked" by a swinging pendulum, which causes a rise in the hinge temperature. The loss in potential energy of the pendulum is equated to the rise in internal energy of the hinge, plus ...
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