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Unformatted text preview: Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 741 Example 79 Air, initially at 17oC, is compressed in an isentropic process through a pressure ratio of 8:1. Find the final temperature assuming constant specific heats and variable specific heats, and using EES. a. Constant specific heats, isentropic process FG T IJ HT K
2 1 s = const . FPI =G J HPK
2 1 ( k −1)/ k For air, k = 1.4, and a pressure ratio of 8:1 means that P2/P1 = 8. FPI T =TG J HPK
2 1 2 1 ( k −1)/ k = (17 + 273) K 8 bg (1.4 −1)/1.4 = 525.3K = 252.3o C b. Variable specific heat method FG P IJ HPK
2 1 =
s = const . FP /P I GH P / P JK
2 ref 1 ref =
s = const . Pr 2 Pr 1 Using the air data from Table A17 for T1 = (17+273) K = 290 K, Pr1 = 1.2311. Pr 2 = Pr 1 P2 P1 = 12311(8) = 9.8488 .
Interpolating in the air table at this value of Pr2, gives T2 = 522.4 K = 249.4oC
Chapter 7  41 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 742 c. A second variable specific heat method. Using the air table, Table A17, for T1 = (17+273) K = 290 K, soT1 = 1.66802 kJ/kg⋅K. For the isentropic process
o s2 = s1o + R ln P2 P1 = 1.66802 kJ kJ + 0.287 ln 8 kg ⋅ K kg ⋅ K kJ = 2.26482 kg ⋅ K bg At this value of soT2, the air table gives T2 = 522.4 K= 249.4oC. This technique is based on the same information as the method shown in part b. d. Using the EES software with T in oC and P in kPa and assuming P1 = 100 kPa. s_1 = ENTROPY(Air, T=17, P=100) s_2 = s_1 T_2 = TEMPERATURE(Air, P=800, s=s_2) The solution is: s_1 = 5.668 kJ/kg⋅K s_2 = 5.668 kJ/kg⋅K T_2 = 249.6oC Chapter 7  42 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 743 Example 710 Air initially at 0.1 MPa, 27oC, is compressed reversibly to a final state. (a) Find the entropy change of the air when the final state is 0.5 MPa, 227oC. (b) Find the entropy change when the final state is 0.5 MPa, 180oC. (c) Find the temperature at 0.5 MPa that makes the entropy change zero. Assume air is an ideal gas with constant specific heats. Show the two processes on a Ts diagram. a. s2 − s1 = C p , av ln = 1.005 T2 P − R ln 2 T1 P1 0.5 MPa kJ ( 227 + 273) K kJ ln ln − 0.287 . 01 MPa kg ⋅ K ( 27 + 273) K kg ⋅ K kJ kg ⋅ K FG H IJ K FG H IJ K = +0.0507
b. s2 − s1 = C p , av ln = 1.005 T2 P − R ln 2 T1 P1 kJ (180 + 273) K kJ 0.5 MPa − 0.287 ln ln kg ⋅ K ( 27 + 273) K kg ⋅ K 0.1 MPa kJ kg ⋅ K FG H IJ K FG H IJ K = −0.0484 Chapter 7  43 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 744 c. FPI T =TG J HPK
2 1 2 1 ( k −1)/ k F 0.5 MPa IJ = ( 27 + 273) K G H 01MPa K .
= 475.4 K = 202.4 o C
2 b P1 1 s P2 (1.4 −1)/1.4 The Ts plot is
T c a Give an explanation for the difference in the signs for the entropy changes. Chapter 7  44 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 745 Example 711 Nitrogen expands isentropically in a piston cylinder device from a temperature of 500 K while its volume doubles. What is the final temperature of the nitrogen, and how much work did the nitrogen do against the piston, in kJ/kg? System: The closed pistoncylinder device
T
Nitrogen system boundary 1 P1 W 2 P2 s P
P1 P2 1 s = const. 2 v Property Relation: Ideal gas equations, constant properties Process and Process Diagram: Isentropic expansion Conservation Principles: Second law: Chapter 7  45 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 746 Since we know T1 and the volume ratio, the isentropic process, ∆s = 0, allows us to find the final temperature. Assuming constant properties, the temperatures are related by FG v IJ T =T Hv K FG 1 IJ = 500 K H 2K
k −1 2 1 1 2 1.4 −1 = 378.9 K = 105.9 o C
Why did the temperature decrease? First law, closed system: Note, for the isentropic process (reversible, adiabatic); the heat transfer is zero. The conservation of energy for this closed system becomes Ein − E out = ∆E −W = ∆U W = − ∆U
Using the ideal gas relations, the work per unit mass is Chapter 7  46 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 747 W = − mCv (T2 − T1 ) W w = = − Cv (T2 − T1 ) m kJ = −0.743 (378.9 − 500) K kg ⋅ K kJ = 90.2 kg
Why is the work positive? Extra Assignment For the isentropic process Pvk = constant. Use the definition of boundary work to show that you get the same result as the last example. That is, determine the boundary work and show that you obtain the same expression as that for the polytropic boundary work. Example 712 A Carnot engine has 1 kg of air as the working fluid. Heat is supplied to the air at 800 K and rejected by the air at 300 K. At the beginning of the heat addition process, the pressure is 0.8 MPa and during heat addition the volume triples. (a) Calculate the net cycle work assuming air is an ideal gas with constant specific heats. (b) Calculate the amount of work done in the isentropic expansion process. Chapter 7  47 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 748 (c) Calculate the entropy change during the heat rejection process. System: The Carnot engine pistoncylinder device.
W Air system boundary Q Property Relation: Ideal gas equations, constant properties. Process and Process Diagram: Constant temperature heat addition.
T
TH 1 QH 2 TL 4 QL 3 s
Carnot Cycle Conservation Principles: a. Apply the first law, closed system, to the constant temperature heat addition process, 12. Chapter 7  48 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 749 Qnet ,12 − Wnet ,12 = ∆U12 Qnet ,12 = mCv (T2 − T1 ) = 0 = Wnet ,12 So for the ideal gas isothermal process, Wnet ,12 = Wother ,12 + Wb ,12 = PdV
1 dV 1 V V2 = mRT ln V1 =
2 z z 2 mRT FG IJ H K
kJ )(800 K ) ln(3) kg ⋅ K = 1kg (0.287 = 252.2 kJ
But Qnet ,12 = QH QH = 252.2 kJ Chapter 7  49 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 750 The cycle thermal efficiency is η th = Wnet , cycle QH For the Carnot cycle, the thermal efficiency is also given by TL 300 K ηth = 1 − = 1− TH 800 K = 0.625
The net work done by the cycle is Wnet , cycle = η th QH = 0.625(252.2 kJ ) = 157.6 kJ
b. Apply the first law, closed system, to the isentropic expansion process, 23. But the isentropic process is adiabatic, reversible; so, Q23 = 0. Chapter 7  50 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 751 Ein − E out = ∆E −W = ∆U W23 = − ∆U 23
Using the ideal gas relations, the work per unit mass is W23 = − mCv (T3 − T2 ) kJ )(300 − 800) K = − (1kg )(0.718 kg ⋅ K = 359.0 kJ
This is the work leaving the cycle in process 23. c. Using equation (634) ∆s34 = C p ln FG T IJ − R lnFG P IJ HT K H P K
4 4 3 3 But T4 = T3 = TL = 300 K, and we need to find P4 and P3. Consider process 12 where T1 = T2 = TH = 800 K, and, for ideal gases Chapter 7  51 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 752 m1 = m2 PV1 PV2 1 = 2 T1 T2 P2 = P 1 V1 3V1 1 3 = (800 kPa ) = 266.7 kPa
Consider process 23 where s3 = s2. T3 P3 = T2 P2
3 2 FG IJ H K FG T IJ P =P HT K
3 2 ( k −1)/ k k /( k −1) F 300K IJ = 266.7 kPa G H 800k K 1.4 /(1.4 −1) = 8.613 kPa Chapter 7  52 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 753 Now, consider process 41 where s4 = s1. T P4 = P 4 1 T1 k /( k −1) 300 K = 8000 kPa 800k = 25.834 kPa
Now, 1.4 /(1.4 −1) LMC lnF T I − R lnF P I OP ∆S = m GH T JK GH P JK Q N FG P IJ = − mR ln HPK kJ . F 25834kPa IJ = (1 kg )(0.287 ) lnG kg ⋅ K H 8.613kPa K
p 4 4 3 3 4 3 = −0.315
Extra Problem kJ K Use a second approach to find ∆S34 by noting that the temperature of process 34 is constant and applying the basic definition of entropy for an internally reversible process, dS = δQ/T. Chapter 7  53 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 754 Reversible SteadyFlow Work Isentropic, Steady Flow through Turbines, Pumps, and Compressors Consider a turbine, pump, compressor, or other steadyflow control volume, workproducing device. The general first law for the steadyflow control volume is Ein = Eout Vi 2 Ve2 Qnet + ∑ mi (hi + + gzi ) = Wnet + ∑ me (he + + gze ) 2 2 inlets exits
For a oneentrance, oneexit device undergoing an internally reversible process, this general equation of the conservation of energy reduces to, on a unit of mass basis δ wrev = δ qrev − dh − dke − dpe But δ qrev = T ds δ wrev = T ds − dh − dke − dp
Using the Gibb’s second equation, this becomes dh = T ds + v dP δ wrev = −v dP − dke − dpe
Integrating over the process, this becomes Chapter 7  54 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 755 wrev = − v dP − ∆ke − ∆pe
1 z 2 FG kJ IJ H kg K Neglecting changes in kinetic and potential energies, reversible work becomes wrev = − v dP
1 z 2 FG kJ IJ H kg K Based on the classical sign convention, this is the work done by the control volume. When work is done on the control volume such as compressors or pumps, the reversible work going into the control volume is wrev , in = v dP + ∆ke + ∆pe
1 z 2 FG kJ IJ H kg K Turbine Since the fluid pressure drops as the fluid flows through the turbine, dP < 0, and the specific volume is always greater than zero, wrev, turbine > 0. To perform the integral, the pressurevolume relation must be known for the process. Compressor and Pump Since the fluid pressure rises as the fluid flows through the compressor or pump, dP > 0, and the specific volume is always greater than zero, wrev, in > 0, or work is supplied to the compressor or pump. To perform the integral, the pressurevolume relation must be known for the process. The term compressor is usually applied to the compression of a gas. The term pump is usually applied when increasing the pressure of a liquid.
Chapter 7  55 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 756 Pumping an incompressible liquid For an incompressible liquid, the specific volume is approximately constant. Taking v approximately equal to v1, the specific volume of the liquid entering the pump, the work can be expressed as wrev , in = v dP + ∆ke + ∆pe
1 z 2 FG kJ IJ H kg K = v∆P + ∆ke + ∆pe
For the steadyflow of an incompressible fluid through a device that involves no work interactions (such as nozzles or a pipe section), the work term is zero, and the equation above can be expressed as the wellknow Bernoulli equation in fluid mechanics. v ( P2 − P1 ) + ∆ke + ∆pe = 0
Extra Assignment Using the above discussion, find the turbine and compressor work per unit mass flow for an ideal gas undergoing an isentropic process, where the pressurevolume relation is Pvk = constant, between two temperatures, T1 and T2. Compare your results with the first law analysis of Chapter 5 for control volumes. Chapter 7  56 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 757 Example 713 Saturated liquid water at 10 kPa leaves the condenser of a steam power plant and is pumped to the boiler pressure of 5 MPa. Calculate the work for an isentropic pumping process. a. From the above analysis, the work for the reversible process can be applied to the isentropic process (it is left for the student to show this is true) as WC = mv1 ( P2 − P1 )
Here at 10 kPa, v1 = vf = 0.001010 m3/kg. The work per unit mass flow is wC = WC = v1 ( P2 − P1 ) m m3 kJ = 0.001010 (5000 − 10) kPa 3 m kPa kg kJ = 5.04 kg
−Wnet = m(h2 − h1 ) −(0 − WC ) = m(h2 − h1 ) b. Using the steam table data for the isentropic process, we have Chapter 7  57 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 758 From the saturation pressure table, kJ h1 = 191.81 P = 10 kPa kg 1 kJ Sat. Liquid s1 = 0.6492 kg ⋅ K Since the process is isentropic, s2 = s1. Interpolation in the compressed liquid tables gives kJ kJ h2 = 197.42 s2 = s1 = 0.6492 kg kg ⋅ K The work per unit mass flow is P2 = 5 MPa wC = WC = (h2 − h1 ) m kJ kg = (197.42 − 191.81) = 5.61 kJ kg The first method for finding the pump work is adequate for this case. Chapter 7  58 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 759 Turbine, Compressor (Pump), and Nozzle Efficiencies Most steadyflow devices operate under adiabatic conditions, and the ideal process for these devices is the isentropic process. The parameter that describes how a device approximates a corresponding isentropic device is called the isentropic or adiabatic efficiency. It is defined for turbines, compressors, and nozzles as follows: Turbine: The isentropic work is the maximum possible work output that the adiabatic turbine can produce; therefore, the actual work is less than the isentropic work. Since efficiencies are defined to be less than 1, the turbine isentropic efficiency is defined as ηT = Actual turbine work w = a Isentropic turbine work ws ηT ≅ h1 − h2 a h1 − h2 s Welldesigned large turbines may have isentropic efficiencies above 90 percent. Small turbines may have isentropic efficiencies below 70 percent.
Chapter 7  59 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 760 Compressor and Pump: The isentropic work is the minimum possible work that the adiabatic compressor requires; therefore, the actual work is greater than the isentropic work. Since efficiencies are defined to be less than 1, the compressor isentropic efficiency is defined as T1 P1 WC Compressor or pump T2 P2 Isentropic compressor work ws = Actual compressor work wa h −h ηC ≅ 2s 1 h2 a − h1 ηC = Welldesigned compressors have isentropic efficiencies in the range from 75 to 85 percent. Review the efficiency of a pump and an isothermal compressor on your own. Chapter 7  60 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 761 Nozzle: The isentropic kinetic energy at the nozzle exit is the maximum possible kinetic energy at the nozzle exit; therefore, the actual kinetic energy at the nozzle exit is less than the isentropic value. Since efficiencies are defined to be less than 1, the nozzle isentropic efficiency is defined as T1 P1 V1 Nozzle T2 P2 V2 V22a 2 V22s 2 Actual KE at nozzle exit V22a / 2 ηN = = Isentropic KE at nozzle exit V22s / 2
For steadyflow, no work, neglecting potential energies, and neglecting the inlet kinetic energy, the conservation of energy for the nozzle is V22a h1 = h2 a + 2 Chapter 7  61 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 762 The nozzle efficiency is written as ηN ≅ h1 − h2 a h1 − h2 s Nozzle efficiencies are typically above 90 percent, and nozzle efficiencies above 95 percent are not uncommon. Example 714 The isentropic work of the turbine in Example 76 is 1152.2 kJ/kg. If the isentropic efficiency of the turbine is 90 percent, calculate the actual work. Find the actual turbine exit temperature or quality of the steam. ηT = w Actual turbine work = a Isentropic turbine work ws kJ kJ ) = 1037.7 kg kg wa = ηT ws = (0.9)(1153.0 h1 − h2 a ηT ≅ h1 − h2 s
Now to find the actual exit state for the steam. From Example 76, steam enters the turbine at 1 MPa, 600oC, and expands to 0.01 MPa. Chapter 7  62 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 763 From the steam tables at state 1 kJ h1 = 3698.6 P = 1 MPa kg 1 kJ T1 = 600o C s1 = 8.0311 kg ⋅ K At the end of the isentropic expansion process, see Example 76. kJ h2 s = 2545.6 kg kJ s2 s = s1 = 8.0311 kg ⋅ K x2 s = 0.984 The actual turbine work per unit mass flow is (see Example 76) P2 = 0.01 MPa wa = h1 − h2 a h2 a = h1 − wa kJ = (3698.6 − 1037.7) kg kJ = 2660.9 kg
For the actual turbine exit state 2a, the computer software gives P2 = 0.01 MPa h2 a URT = 86.85 C  kJ VS = 2660.9 TSuperheated kg W
o 2a
Chapter 7  63 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 764 A second method for finding the actual state 2 comes directly from the expression for the turbine isentropic efficiency. Solve for h2a. h2 a = h1 − ηT ( h1 − h2 s ) kJ kJ = 3698.6 − (0.9)(3698.6 − 2545.6) kg kg kJ = 2660.9 kg
Then P2 and h2a give T2a = 86.85oC. Example 715 Air enters a compressor and is compressed adiabatically from 0.1 MPa, 27oC, to a final state of 0.5 MPa. Find the work done on the air for a compressor isentropic efficiency of 80 percent. System: The compressor control volume
T1 P1 WC Compressor or pump T2 P2 Property Relation: Ideal gas equations, assume constant properties.
Chapter 7  64 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 765 T 2s 2a P2 1 P1 s Process and Process Diagram: First, assume isentropic, steadyflow and then apply the compressor isentropic efficiency to find the actual work. Conservation Principles: For the isentropic case, Qnet = 0. Assuming steadystate, steadyflow, and neglecting changes in kinetic and potential energies for one entrance, one exit, the first law is Ein = E out m1h1 + WCs = m2 h2 s
The conservation of mass gives m1 = m2 = m
The conservation of energy reduces to Chapter 7  65 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 766 WCs = m(h2 s − h1 ) wCs = WCs = (h2 s − h1 ) m Using the ideal gas assumption with constant specific heats, the isentropic work per unit mass flow is wCs = C p (T2 s − T1 )
The isentropic temperature at state 2 is found from the isentropic relation T2 s F 0.5 MPa IJ = (27 + 273) K G H 01MPa K .
= 475.4 K
The conservation of energy becomes FG P IJ =T HPK
1 2 1 ( k −1)/ k (1.4 −1)/1.4 wCs = C p (T2 s − T1 ) kJ (475.4 − 300) K kg ⋅ K kJ = 176.0 kg . = 1005
Chapter 7  66 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 767 The compressor isentropic efficiency is defined as ηC =
wCa = ws wa ηC wcs kJ kJ kg = 220 kg 0.8 176 =
Example 716 Nitrogen expands in a nozzle from a temperature of 500 K while its pressure decreases by factor of two. What is the exit velocity of the nitrogen when the nozzle isentropic efficiency is 95 percent? System: The nozzle control volume. T
T1 P1 V1 Nozzle T2 P2 V2 1 P1 2s 2a P2 s Property Relation: The ideal gas equations, assuming constant specific heats Chapter 7  67 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 768 Process and Process Diagram: First assume an isentropic process and then apply the nozzle isentropic efficiency to find the actual exit velocity. Conservation Principles: For the isentropic case, Qnet = 0. Assume steadystate, steadyflow, no work is done. Neglect the inlet kinetic energy and changes in potential energies. Then for one entrance, one exit, the first law reduces to Ein = E out V22s m1h1 = m2 (h2 s + ) 2
The conservation of mass gives m1 = m2 = m
The conservation of energy reduces to V2 s = 2(h1 − h2 s )
Using the ideal gas assumption with constant specific heats, the isentropic exit velocity is V2 s = 2C p (T1 − T2 s )
The isentropic temperature at state 2 is found from the isentropic relation Chapter 7  68 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 769 T2 s FG P IJ =T HPK
1 2 1 ( k −1)/ k = (500) K = 410.0 K FG 0.5P IJ H PK
1 1 (1.4 −1)/1.4 V2 s = 2C p (T1 − T2 s ) = F . kJ IJ (500 − 410.0) K 10 m / s 2G 1005 kJ / kg H kg ⋅ K K
3 2 2 = 442.8 m s The nozzle exit velocity is obtained from the nozzle isentropic efficiency as V22a / 2 ηN = 2 V2 s / 2 V2 a = V2 s m m 0.95 = 421.8 η N = 442.8 s s Chapter 7  69
Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 770 Entropy Balance The principle of increase of entropy for any system is expressed as an entropy balance given by Ein Sin System ∆Esystem ∆Ssystem ∆Sgen≥0 Eout Sout F Total I F Total I F Total I F Change in theI GG entropy JJ − GG entropyJJ + GG entropy JJ = GG total entropy JJ H enteringK H leaving K H generated K H of the system K
or Sin − Sout + S gen = ∆S system
The entropy balance relation can be stated as: the entropy change of a system during a process is equal to the net entropy transfer through the system boundary and the entropy generated within the system as a result of irreversibilities. Entropy change of a system The entropy change of a system is the result of the process occurring within the system. Chapter 7  70 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 771 Entropy change = Entropy at final state – Entropy at initial state ∆Ssystem = S final − Sinitial = S2 − S1
Mechanisms of Entropy Transfer, Sin and Sout Entropy can be transferred to or from a system by two mechanisms: heat transfer and mass flow. Entropy transfer occurs at the system boundary as it crosses the boundary, and it represents the entropy gained or lost by a system during the process. The only form of entropy interaction associated with a closed system is heat transfer, and thus the entropy transfer for an adiabatic closed system is zero. Heat transfer The ratio of the heat transfer Q at a location to the absolute temperature T at that location is called the entropy flow or entropy transfer and is given as Entropy transfer by heat transfer: Sheat = Q (T = constant ) T Q/T represents the entropy transfer accompanied by heat transfer, and the direction of entropy transfer is the same as the direction of heat transfer since the absolute temperature T is always a positive quantity. When the temperature is not constant, the entropy transfer for process 12 can be determined by integration (or by summation if appropriate) as Sheat = z 2 1 δQ Qk ≅∑ T Tk Chapter 7  71 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 772 Work Work is entropyfree, and no entropy is transferred by work. Energy is transferred by both work and heat, whereas entropy is transferred only by heat and mass. Entropy transfer by work :
Mass flow S work = 0 Mass contains entropy as well as energy, and the entropy and energy contents of a system are proportional to the mass. When a mass in the amount m enters or leaves a system, entropy in the amount of ms enters or leaves, where s is the specific entropy of the mass. Entropy transfer by mass: Smass = ms Entropy Generation, Sgen Irreversibilities such as friction, mixing, chemical reactions, heat transfer through a finite temperature difference, unrestrained expansion, nonquasiequilibrium expansion, or compression always cause the entropy of a system to increase, and entropy generation is a measure of the entropy created by such effects during a process. Chapter 7  72 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 773 For a reversible process, the entropy generation is zero and the entropy change of a system is equal to the entropy transfer. The entropy transfer by heat is zero for an adiabatic system and the entropy transfer by mass is zero for a closed system. The entropy balance for any system undergoing any process can be expressed in the general form as Sin − Sout + S gen = ∆Ssystem
Net entropy transfer by heat and mass Entropy generation Change in entropy 3 ( kJ / K ) The entropy balance for any system undergoing any process can be expressed in the general rate form, as Sin − Sout
Rate of net entropy transfer by heat and mass + S gen Rate of entropy generation 3 = ∆S system
Rate of change of entropy ( kW / K ) where the rates of entropy transfer by heat transferred at a rate of mass flowing at a rate of m are Sheat = Q / T and Smass = ms . Q and The entropy balance can also be expressed on a unitmass basis as ( sin − sout ) + sgen = ∆ssystem ( kJ / kg ⋅ K ) The term Sgen is the entropy generation within the system boundary only, and not the entropy generation that may occur outside the system boundary during the process as a result of external irreversibilities. Sgen = 0 for the internally reversible process, but not necessarily zero for the totally reversible process. The total entropy generated during any process is
Chapter 7  73 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 774 obtained by applying the entropy balance to an Isolated System that contains the system itself and its immediate surroundings. ∆ Closed Systems Taking the positive direction of heat transfer to the system to be positive, the general entropy balance for the closed system is Qk ∑ T + S gen = ∆Ssystem = S2 − S1 k
For an adiabatic process (Q = 0), this reduces to ( kJ / K ) Adiabatic closed system: S gen = ∆Sadiabatic system
Chapter 7  74 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 775 A general closed system and its surroundings (an isolated system) can be treated as an adiabatic system, and the entropy balance becomes System + surroundings: S gen = ∑ ∆S = ∆S system + ∆S surroundings Control Volumes The entropy balance for control volumes differs from that for closed systems in that the entropy exchange due to mass flow must be included. ∑ T + ∑m s − ∑m s
i i k Qk e e + S gen = ( S2 − S1 ) CV ( kJ / K ) In the rate form we have Qk ∑ T + ∑ mi si − ∑ me se + S gen = ∆SCV ( kW / K ) k
This entropy balance relation is stated as: the rate of entropy change within the control volume during a process is equal to the sum of the rate of entropy transfer through the control volume boundary by heat transfer, the net rate of entropy transfer into the control volume by mass flow, and the rate of entropy generation within the boundaries of the control volume as a result of irreversibilities. Chapter 7  75 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 776 For a general steadyflow process, by setting ∆SCV = 0 the entropy balance simplifies to S gen = ∑ me se − ∑ mi si − ∑ Qk Tk For a singlestream (one inlet and one exit), steadyflow device, the entropy balance becomes S gen Qk = m( se − si ) − ∑ Tk For an adiabatic singlestream device, the entropy balance becomes S gen = m( se − si ) This states that the specific entropy of the fluid must increase as it flows through an adiabatic device since S gen ≥ 0 . If the flow through the device is reversible and adiabatic, then the entropy will remain constant regardless of the changes in other properties. Chapter 7  76 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 777 Therefore, for steadyflow, singlestream, adiabatic and reversible process: se = si
Example 717 An inventor claims to have developed a water mixing device in which 10 kg/s of water at 25oC and 0.1 MPa and 0.5 kg/s of water at 100oC, 0.1 MPa, are mixed to produce 10.5 kg/s of water as a saturated liquid at 0.1 MPa. If the surroundings to this device are at 20oC, is this process possible? If not, what temperature must the surroundings have for the process to be possible? System: The mixing chamber control volume.
10 kg/s P1 = 0.1 MPa T1 = 25oC 0.5 kg/s P1 = 0.1 MPa T1 = 100oC 10.5 kg/s P3 = 0.1 MPa Saturated liquid Mixing chamber Qsys ? Surroundings To = 20oC Property Relation: The steam tables Process and Process Diagram: Assume steadyflow Chapter 7  77 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 778 Conservation Principles: First let’s determine if there is a heat transfer from the surroundings to the mixing chamber. Assume there is no work done during the mixing process, and neglect kinetic and potential energy changes. Then for two entrances and one exit, the first law becomes Qnet F V + gz IJ = W + ∑ m Gh + H 2 K
2 i i i i inlets net F V + gz IJ + ∑ m Gh + H 2 K
e e 2 e e exits Qnet + m1h1 + m2 h2 = m3h3
kJ h1 ≅ h f @ T1 = 104.83 P = 0.1 MPa kg 1 kJ T1 = 25o C s1 ≅ s f @ T1 = 0.3672 kg ⋅ K kJ h2 = 2675.8 P2 = 0.1 MPa kg kJ T2 = 100o C s2 = 7.3611 kg ⋅ K kJ h3 = 417.51 P3 = 0.1 MPa kg kJ Sat. liquid s3 = 1.3028 kg ⋅ K Chapter 7  78 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 779 Qnet = m3 h3 − m1h1 − m2 h2 kg kJ kg kJ kg kJ 417.51 − 10 104.83 − 0.5 2675.8 s kg s kg s kg kJ = +1997.7 s = 10.5
So, 1996.33 kJ/s of heat energy must be transferred from the surroundings to this mixing process, or Qnet , surr = −Qnet , CV . For the process to be possible, the second law must be satisfied. Write the second law for the isolated system, Qk ∑ T + ∑ mi si − ∑ me se + S gen = ∆SCV k
For steadyflow
∆SCV = 0 . Solving for entropy generation, we have S gen = ∑ me se − ∑ mi si − ∑ = m3 s3 − m1s1 − m2 s2 − = 10.5 Qk Tk Qcv Tsurr kg kJ kg kJ − 10 0.3672 1.3028 s kg ⋅ K s kg ⋅ K kg kJ 1997.7 kJ / s − 0.5 7.3611 − s kg ⋅ K (20 + 273) K kJ = −0.491 K ⋅s
Chapter 7  79 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 780 Since S gen must be ≥ 0 to satisfy the second law, this process is impossible, and the inventor's claim is false. To find the minimum value of the surrounding temperature to make this mixing process possible, set S gen = 0 and solve for Tsurr. S gen = ∑ me se − ∑ mi si − ∑ Tsurr = = Qcv m3 s3 − m1s1 − m2 s2 Qk =0 Tk 1997.7 kJ / s kg kJ kg kJ kg kJ − 10 0.3672 − 0.5 7.3611 10.5 1.3028 s kg ⋅ K s kg ⋅ K s kg ⋅ K = 315.75K
One way to think about this process is as follows: Heat is transferred from the surroundings at 315.75 K (42.75oC) in the amount of 1997.7 kJ/s to increase the water temperature to approximately 42.75oC before the water is mixed with the superheated steam. Recall that the surroundings must be at a temperature greater than the water for the heat transfer to take place from the surroundings to the water. Chapter 7  80 Student Study Guide for 5th edition of Thermodynamics by Y. A. Çengel & M. A. Boles 781 Answer to Example 74 Find the entropy and/or temperature of steam at the following states: P 5 MPa 1 MPa T 120oC 50oC Region Compressed Liquid and in the table Compressed liquid but not in the table Superheated Quality, x = 0.9 Saturated mixture sf<s<sg at P Saturated mixture X = (ssf)/sfg = 0.9262 s kJ/kg⋅K 1.5233 s = sf at 50oC = 0.7038 7.1794 s = sf + x sfg = 7.0056 7.1794 1.8 400oC MPa 40 kPa T=Tsat =75.87o C 40 kPa T=Tsat =75.87o C Chapter 7  81 ...
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This note was uploaded on 09/17/2009 for the course MAE 301 taught by Professor Hassan during the Fall '08 term at N.C. State.
 Fall '08
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