Chapter 7 Entropy: A Measure of Disorder
Entropy and the Clausius Inequality The second law of thermodynamics leads to the definition of a new property called entropy, a quantitative measure of microscopic disorder for a system. Entropy is a measure of energy that is no longer available to perform useful work within the current environment. To obtain the working definition of entropy and, thus, the second law, let's derive the Clausius inequality. 2
Consider a heat reservoir giving up heat to a reversible heat engine, which in turn gives up heat to a piston-cylinder device as shown below. We apply the first law on an incremental basis to the combined system composed of the heat engine and the system. E E = ΔE δq ( δw + δw ) = de in out c R rev sys c where E c is the energy of the combined system. 3
Let W c be the work done by the combined system. Then the first law becomes δw = δw + δw c rev sys δq δw = de R c c If we assume that the engine is totally reversible, then δq T R R = δq T δ δq T Q R = R T Where T is in absolute units. The total net work done by the combined system becomes δ δw = T Q de T c R c 4
Now the total work done is found by taking the cyclic integral of the incremental work. If the system, like the heat engine, is required to undergo a cycle, then and the total net work becomes 5
If W c = W rev + W sys is positive, we have a cyclic device exchanging energy with a single heat reservoir and producing an equivalent amount of work; thus, the Kelvin-Planck statement of the second law is violated. But W c can be zero (no work done) or negative (work is done on the combined system) and not violate the Kelvin- Planck statement of the second law. Therefore, since T R > 0 (absolute temperature), we conclude 6
or Here Q is the net heat added to the system, Q net. This equation is called the Clausius inequality. The equality holds for the reversible process and the inequality holds for the irreversible process. 7
Example 7-1 For a particular power plant, the heat added and rejected both occur at constant temperature and no other processes experience any heat transfer. The heat is added in the amount of 3150 kj at 440 o C and is rejected in the amount of 1950 kj at 20 o C. Is the Clausius inequality satisfied and is the cycle reversible or irreversible? The Clausius inequality is satisfied. Since the inequality is less than zero, the cycle has at least one irreversible process and the cycle is irreversible. Calculate the net work, cycle efficiency, and Carnot efficiency based on T H and T L for this cycle. W net = Q in + Q out = (3150 1950) kj = 1200 kj 8
Example 7-2 For a particular power plant, the heat added and rejected both occur at constant temperature; no other processes experience any heat transfer. The heat is added in the amount of 3150 kj at 440 o C and is rejected in the amount of 1294.46 kj at 20 o C. Is the Clausius inequality satisfied and is the cycle reversible or irreversible? The Clausius inequality is satisfied. Since the cyclic integral is equal to zero, the cycle is made of reversible processes such as a Carnot cycle. The net work and cycle efficiency is given by 9
Definition of Entropy Let s take another look at the quantity If no irreversibilities occur within the system as well as the reversible cyclic device, then the cycle undergone by the combined system will be internally reversible. As such, it can be reversed. In the reversed cycle case, all the quantities will have the same magnitude but the opposite sign. Therefore, the work W C, which could not be a positive quantity in the regular case, cannot be a negative quantity in the reversed case. Then it follows that W C,int rev = 0 since it cannot be a positive or negative quantity, and therefore for internally reversible cycles. Thus we conclude that the equality in the Clausius inequality holds for totally or just internally reversible cycles and the inequality for the irreversible ones. 10
To develop a relation for the definition of entropy, let us examine this last equation more closely. Here we have a quantity whose cyclic integral is zero. Let us think for a moment what kind of quantities can have this characteristic. We know that the cyclic integral of work is not zero. (It is a good thing that it is not. Otherwise, heat engines that work on a cycle such as steam power plants would produce zero net work.) Neither is the cyclic integral of heat. Now consider the volume occupied by a gas in a piston-cylinder device undergoing a cycle, as shown to the right. When the piston returns to its initial position at the end of a cycle, the volume of the gas also returns to its initial value. Thus the net change in volume during a cycle is zero. This is also expressed as We see that the cyclic integral of a property is zero. A quantity whose cyclic integral is zero depends only on the state and not on the process path; thus it is a property. Therefore the quantity (δq net /T) int rev must be a property. 11
In-class Example Consider the cycle shown below composed of two reversible processes A and B. Apply the Clausius inequality for this cycle. We find that P 2 B A 1 Since the quantity (δq net /T) int rev is independent of the path and must be a property, we call this property the entropy S. A cycle composed of two reversible processes. V Apply the Clausius inequality for the cycle made of two internally reversible processes: 12
Once again, since the quantity (δq net /T) int rev is independent of the path and must be a property, we call this property the entropy S. The entropy change occurring during a process is related to the heat transfer and the temperature of the system. The entropy is given the symbol S (kj/k), and the specific entropy is s (kj/kg K). The entropy change during a reversible process, sometimes called an internally reversible process, is defined as 13
Consider now the cycle 1-A-2-B-1, shown below, where process A is arbitrary that is, it can be either reversible or irreversible, and process B is internally reversible. A The integral along the internally reversible path, process B, is the entropy change S 1 S 2. Therefore, B or along A 14
Thus, in general the entropy change during a process satisfies where ds Q δ T net = holds for the internally reversible process > holds for the irreversible process Consider the effect of heat transfer on entropy for the internally reversible case. ds Q = δ T Where the temperature T is in absolute scale and hence always positive. Thus if: net δq δq δq net net net > 0, then ds > 0 = 0, then ds = 0 < 0, then ds < 0 This makes sense if you think about S being a measure of disorder, i.e., the more heat added to a system, the more disorder in the system, etc. 15
This last result shows why we have kept the subscript net on the heat transfer Q. It is important for you to recognize that Q has a sign depending on the direction of heat transfer. The net subscript is to remind us that Q is positive when added to a system and negative when leaving a system. Thus, the entropy change of the system will have the same sign as the heat transfer in a reversible process. From the above, we see that for a reversible, adiabatic process ds S = 0 = S 2 1 The reversible, adiabatic process is called an isentropic process. Entropy change is caused by heat transfer and irreversibilities. Heat transfer to a system increases the entropy; heat transfer from a system decreases it. The effect of irreversibilities is always to increase the entropy. In fact, a process in which the heat transfer is out of the system (which normally would lead to a decrease in entropy) may be so irreversible that the actual entropy change is positive. Friction is one source of irreversibilities in a system. 16
The entropy change during a process is obtained by integrating the ds equation over the process: Here, the inequality is to remind us that the entropy change of a system during an irreversible process is always greater than, called the entropy transfer. That is, some entropy is generated or created during an irreversible process, and this generation is due entirely to the presence of irreversibilities. The entropy generated during a process is called entropy generation and is denoted as S gen. We can remove the inequality by writing the following S gen is always a positive quantity or zero. Its value depends upon the process and thus it is not a property. S gen is zero for an internally reversible process. The integral is performed by applying the first law to the process to obtain the heat transfer as a function of the temperature. The integration is not easy to perform, in general. 17
Definition of Second Law of Thermodynamics Now consider an isolated system composed of several subsystems exchanging energy among themselves. Since the isolated system has no energy transfer across its system boundary, the heat transfer across the system boundary is zero. Applying the definition of entropy to the isolated system Thus, the total entropy change for the isolated system is Δ 0 S isolated 18
This equation is the working definition of the second law of thermodynamics. The second law, known as the principle of increase of entropy, is stated as The total entropy change of an isolated system during a process always increases or, in the limiting case of a reversible process, remains constant. Now consider a general system exchanging mass as well as energy with its surroundings. Sgen = ΔStotal = ΔSsys + ΔSsurr 0 where = holds for the totally reversible process > holds for the irreversible process 19
Thus, the entropy generated or the total entropy change (sometimes called the entropy change of the universe or net entropy change) due to the process of this isolated system is positive (for actual processes) or zero (for reversible processes). The total entropy change for a process is the amount of entropy generated during that process (S gen ), and it is equal to the sum of the entropy changes of the system and the surroundings. The entropy changes of the important system (closed system or control volume) and its surroundings do not both have to be positive. The entropy for a given system (important or surroundings) may decrease during a process, but the sum of the entropy changes of the system and its surroundings for an isolated system can never decrease. Entropy change is caused by heat transfer and irreversibilities. Heat transfer to a system increases the entropy, and heat transfer from a system decreases it. The effect of irreversibilities is always to increase the entropy. The increase in entropy principle can be summarized as follows: 20
Some Remarks about Entropy 1. Processes can occur in a certain direction only, not in just any direction, such that S gen 0. 2. Entropy is a nonconserved property, and there is no such thing as the conservation of entropy principle. The entropy of the universe is continuously increasing. 3. The performance of engineering systems is degraded by the presence of irreversibilities, and entropy generation is a measure of the magnitudes of the irreversibilities present during that process. 21
Heat Transfer as the Area under a T-S Curve For the reversible process, the equation for ds implies that δq ds net = = δq T TdS net or the incremental heat transfer in a process is the product of the temperature and the differential of the entropy, the differential area under the process curve plotted on the T-S diagram. In this figure, the heat transfer in an internally reversible process is shown as the area under the process curve plotted on the T-S diagram. 22
Isothermal and Reversible Process For an isothermal, reversible process, the temperature is constant and the integral to find the entropy change is readily performed. If the system has a constant temperature, T 0, the entropy change becomes For a process occurring over a varying temperature, the entropy change must be found by integration over the process. 23
Adiabatic and Reversible (Isentropic) Process For an adiabatic process, one in which there is no heat transfer, the entropy change is If the process is adiabatic and reversible, the equality holds and the entropy change is ΔS = S2 S1 = S = S 2 1 0 s or on a per unit mass basis, we have s = = S m s 2 1 The adiabatic, reversible process is a constant entropy process and is called isentropic. We shall see later for an ideal gas that an isentropic process is the same as the polytropic process where the polytropic exponent n = k = C p /C v. 24
The principle of increase of entropy for a closed system exchanging heat with its surroundings at a constant temperature T surr is found by using the equation for the entropy generated for an isolated system. A general closed system (a cup of coffee) exchanging heat with its surroundings Surroundings T surr Q out, sys System Boundary S = ΔS = ΔS + ΔS ΔS = ( S S ) ΔS gen total sys surr sys surr = Q 2 1 net, surr T surr sys 0 or where Qnet, surr Sgen = Δ Stotal = m( s s ) sys + T 2 1 0 surr Q = Q = ( 0 Q ) = Q net, surr net, sys out, sys out, sys 25
Effect of Heat Transfer on Entropy Let's apply the second law to the following situation. Consider the transfer of heat from a heat reservoir at temperature T to a heat reservoir at temperature T - ΔT > 0 where ΔT > 0, as shown below. Two heat reservoirs exchanging heat over a finite temperature difference The second law for the isolated system composed of the two heat reservoirs is Sgen = ΔStotal = ΔSsys + ΔSsurr 0 S = ΔS = ΔS @ + ΔS @ Δ gen total HR T HR T T 26
In general, if the heat reservoirs are internally reversible Q going out of HR@T Q going in to HR@(T-ΔT) S gen Q Q = ΔSTotal = + + T T ΔT or Now as ΔT 0, S gen 0 and the process becomes totally reversible. Therefore, for reversible heat transfer ΔT must be small. As ΔT gets large, S gen increases and the process becomes irreversible. 27
Example 7-3 Find the total entropy change, or entropy generation, for the transfer of 1000 kj of heat energy from a heat reservoir at 1000 K to a heat reservoir at 500 K. HR at T=1000 K HR at T-ΔT = 500K Q=1000 kj 1000 K T 500 K Areas = 1000 kj 0 1 2 S, kj/k The second law for the isolated system is What happens when the low-temperature reservoir is at 750 K? The effect of decreasing the ΔT for heat transfer is to reduce the entropy generation or total entropy change of the universe due to the isolated system and the irreversibilities associated with the heat transfer process. 28
Third Law of Thermodynamics The third law of thermodynamics states that the entropy of a pure crystalline substance at absolute zero temperature is zero. This law provides an absolute reference point for the determination of entropy. The entropy determined relative to this point is called absolute entropy. Entropy as a Property: The Tds Relations Entropy is a property, and it can be expressed in terms of more familiar properties (P,v,T) through the Tds relations. These relations come from the analysis of a reversible closed system that does boundary work and has heat added. Writing the first law for the closed system in differential form on a per unit mass basis δq int rev δq δw = du int rev int rev, out δw int rev, out δq int rev = TdS System used to find expressions for ds δw int rev, out = PdV TdS P dv = du 29
On a unit mass basis we obtain the first Tds equation, or Gibbs equation, as Tds = du + Pdv Recall that the enthalpy is related to the internal energy by h = u + Pv. Using this relation in the above equation, the second Tds equation is Tds= dh vdp These last two relations have many uses in thermodynamics and serve as the starting point in developing entropy-change relations for processes. The successful use of Tds relations depends on the availability of property relations. Such relations do not exist in an easily used form for a general pure substance but are available for incompressible substances (liquids, solids) and ideal gases. 30
So, for the general pure substance, such as water and the refrigerants, we must resort to property tables (and/or graphs) to find values of entropy and entropy changes. 31
The temperature-entropy and enthalpy-entropy diagrams for water are shown below. The h-s diagram, called the Mollier diagram, is a useful aid in solving steam power plant problems. 32
Example 7-4 Find the entropy and/or temperature of steam at the following states: P T Region s kj/(kg K) 5 MPa 120 o C 1 MPa 50 o C 1.8 MPa 400 o C 40 kpa Quality, x = 0.9 40 kpa 7.1794 The Student should work this out as practice, just like in Chapter 3. 33
Answer to Example 7-4 Find the entropy and/or temperature of steam at the following states: P T Region s kj/kg K 5 MPa 120 o C Compressed Liquid and in the table 1.5233 1 MPa 50 o C Compressed liquid but not in the table s = s f at 50 o C = 0.7038 1.8 MPa 400 o C Superheated 7.1794 40 kpa T=T sat =75.87 o C Quality, x = 0.9 Saturated mixture s = s f + x s fg = 7.0056 40 kpa T=T sat =75.87 o C s f <s<s g at P Saturated mixture X = (s-s f )/s fg = 0.9262 7.1794 34
Example 7-5 Determine the entropy change of water contained in a closed system as it changes phase from saturated liquid to saturated vapor when the pressure is 0.1 MPa and constant. Why is the entropy change positive for this process? System: The water contained in the system (a piston-cylinder device) Property Relation: Steam tables Process and Process Diagram: Constant pressure (sketch the process relative to the saturation lines) Conservation Principles: Using the definition of entropy change, the entropy change of the water per mass is Δ s= s s = s s = s 2 1 = 6.0562 g f fg kj kg K The entropy change is positive because: (Heat is added to the water.) 35
Example 7-6 Steam at 1 MPa, 600 o C, expands in a turbine to 0.01 MPa. If the process is isentropic, find the final temperature, the final enthalpy of the steam, and the turbine work. System: The control volume formed by the turbine 1 W out Property Relation: Steam tables Control surface 2 Process and Process Diagram: Isentropic (sketch the process relative to the saturation lines on the T-s diagram) Conservation Principles: Assume: steady-state, steady-flow, one entrance, one exit, neglect KE and PE Conservation of mass: m& = m& = m& 1 2 36
Example 7-6 (Continued) First Law or conservation of energy: The process is isentropic and thus adiabatic and reversible; therefore Q = 0. The conservation of energy becomes E& in = E& out mh & = mh & + W& 1 1 2 2 Since the mass flow rates in and out are equal, solve for the work done per unit mass out E& & in = Eout W& = mh & m& h out 1 1 2 2 = mh & ( 1 h2) & w W out = = h h m& 1 2 37
Example 7-6 (Continued) Now, let s go to the steam tables to find the h s. kj h = 3698.6 = 1 = 600 kj = 1 P1 MPa kg o T1 C s 1 8.0311 kg K The process is isentropic, therefore; s 2 = s 1 = 8.0311 kj/(kg K ) At P 2 = 0.01 MPa, s f = 0.6492 kj/kg K, and s g = 8.1488 kj/(kg K); thus, s f < s 2 < s g. 38
Example 7-6 (Continued) State 2 is in the saturation region, and the quality is needed to specify the state. s = s + x s x 2 f 2 2 = s 2 s s fg f fg 8.0311 0.6492 = = 0.984 7.4996 h2 = hf + x2hfg = 191.8 + (0.984)(2392.1) kj = 2545.6 kg Since state 2 is in the two-phase region, T 2 = T sat at P2 = 45.81 o C. w= h h 1 2 = (3698.6 2545.6) kj = 1153 kg kj kg 39
Entropy Change and Isentropic Processes The entropy-change and isentropic relations for a process can be summarized as follows: 1.Pure substances: Any process: Δs= s2 s1 (kj/kg K) Isentropic process: s = s 2 1 2.Incompressible substances (Liquids and Solids): ds = du T + P T dv The change in internal energy and volume for an incompressible substance is du dv = C dt 0 The entropy change now becomes 40
If the specific heat for the incompressible substance is constant, equal to C av, then the entropy change (kj/kg K) is Any process: Isentropic process: T s2 s1 = Cav ln T T = T 2 1 2 1 3. Ideal gases: a. Constant specific heats (approximate treatment): Any process: (can you fill in the steps?) or T s s = C ln 2 + R ln 2 1 vav, T T s2 s1 = Cpav, ln Rln T 1 v v P P 2 2 1 1 2 1 41
Or, on a unit-mole basis (kj/kmol K), T2 s2 s1 = Cvav, ln + Ruln T 1 v v 2 1 and T s2 s1 = Cpav, ln R T ln P 2 2 u 1 P1 Isentropic process in which Δs = 0: (Can you fill in the steps here?) P 2 v 1 = P v 1 s = const. 2 k For an isentropic process this last result looks like Pv k = constant which is the polytropic process equation Pv n = constant with n = k = C p /C v. 42
43
b.variable specific heats (exact treatment): From Tds = dh - vdp, we obtain 2 Cp ( T) P2 Δ s = dt Rln 1 T P The first term can be integrated relative to a reference state at temperature T ref. 1 The integrals on the right-hand side of the above equation are called the standard state entropies, s o, at state 1, T 1, and state 2, T 2 ; so is a function of temperature only. 44
Therefore, for any process: o o s s = s s Rln 2 1 2 1 P P 2 1 in kj/kg K, or (in kj/kmol K) o o s s = s s R ln 2 1 2 1 u P P 2 1 The standard state entropies are found in Tables A-17 for air on a mass basis and Tables A-18 through A-25 for other gases on a mole basis. When using this variable specific heat approach to finding the entropy change for an ideal gas, remember to include the pressure term along with the standard state entropy terms--the tables don t warn you to do this. 45
Standard-State Entropies For Air (per mass) 46
Standard-State Entropies For Nitrogen (per mole) 47
Isentropic process: Δs = 0 o o s s = s s Rln 2 1 2 1 P P 2 1 Leads to o o P2 s2 = s1 + Rln (kj/kg K) P 1 If we are given T 1, P 1, and P 2, we find s o 1 at T 1, calculate s o 2, and then determine from the tables T 2, u 2, and h 2. 48
When air undergoes an isentropic process when variable specific heat data are required, there is another approach to finding the properties at the end of the isentropic process. Consider the entropy change written as 2 Cp ( T) P Δ s = dt Rln 1 T P 2 1 Letting T 1 = T ref, P 1 = P ref = 1atm, T 2 = T, P 2 = P, and setting the entropy change equal to zero yields 49
Relative Pressure We define the relative pressure P r as the pressure ratio necessary to have an isentropic process between the reference temperature and the actual temperature and is a function of the actual temperature. This parameter is a function of temperature only and is found in the air tables, Table A-17. The relative pressure is not available for other gases in this text. The ratio of pressures in an isentropic process is related to the ratio of relative pressures. P P / P P = = P P / P P 2 2 ref r 2 1 s = const. 1 ref s = const. r1 50
Relative Pressures For Air 51
There is a second approach to finding data at the end of an ideal gas isentropic process when variable specific heat data are required. Consider the following entropy change equation set equal to zero. From Tds = du + Pdv, we obtain for ideal gases Letting T 1 = T ref, v 1 = v ref, T 2 = T, v 2 = v, and setting the entropy change equal to zero yield 52
Relative Volume We define the relative volume v r as the above volume ratio. v r is the volume ratio necessary to have an isentropic process between the reference temperature and the actual temperature and is a function of the actual temperature. This parameter is a function of temperature only and is found in the air tables, Table A-17. The relative volume is not available for other gases in this text. 53
Relative Volumes For Air 54
Extra Assignment For an ideal gas having constant specific heats and undergoing a polytropic process in a closed system, Pv n = constant, with n = k, find the heat transfer by applying the first law. Based on the above discussion of isentropic processes, explain your answer. Compare your results to this problem to a similar extra assignment problem in Chapter 4. Example 7-7 Aluminum at 100 o C is placed in a large, insulated tank having 10 kg of water at a temperature of 30 o C. If the mass of the aluminum is 0.5 kg, find the final equilibrium temperature of the aluminum and water, the entropy change of the aluminum and the water, and the total entropy change of the universe because of this process. Before we work the problem, what do you think the answers ought to be? Are entropy changes going to be positive or negative? What about the entropy generated as the process takes place? 55
Example 7-7 (Continued) System: Closed system including the aluminum and water. Water Tank insulated boundary AL Property Relation:? Process: Constant volume, adiabatic, no work energy exchange between the aluminum and water. Conservation Principles: Apply the first law, closed system to the aluminum-water system. Q W = ΔUsystem 0 0= ΔU + water ΔU Using the solid and incompressible liquid relations, we have AL m C ( T T) + m C ( T T) = water water 2 1 water AL AL 2 1 AL 0 56
Example 7-7 (Continued) But at equilibrium, T 2,AL = T 2,water = T 2 mwatercwater ( T1) water + malcal ( T1) AL T2 = mwatercwater + malcal 10kgwater ( 418. kj / kgwater K)( 303K) + 05. kg AL( 0941. kj / kg AL K)( 373K) = 10kgwater ( 418. kj / kgwater K) + 05. kg AL( 0. 941kJ / kg AL K) = 3038. K The second law gives the entropy production, or total entropy change of the universe, as S = ΔS = ΔS + ΔS 0 gen total water AL Using the entropy change equation for solids and liquids, Why is S gen or ΔS Total positive? 57
Example 7-7 (Continued) Why is ΔS AL negative? Why is ΔS water positive? S = ΔS = ΔS + ΔS gen total water AL = (. 01101 0. 0966) =+ 00135. kj K kj K 58
Example 7-8 Carbon dioxide initially at 50 kpa, 400 K, undergoes a process in a closed system until its pressure and temperature are 2 MPa and 800 K, respectively. Assuming ideal gas behavior, find the entropy change of the carbon dioxide by first assuming constant specific heats and then assuming variable specific heats. Compare your results with the real gas data obtained from the EES software. (a) Assume the Table A-2(a) data at 300 K are adequate; then C p = 0.846 kj/kg K and R = 0.1889 kj/kg-k. 59
Example 7-8 (b) For variable specific heat data, use the carbon dioxide data from Table A-20. (c) Using EES for carbon dioxide as a real gas: Deltas=ENTROPY(CarbonDioxide,T=800,P=2000)- ENTROPY(CarbonDioxide,T=400,P=50) = +0.03452 kj/kg K (d) Repeat the constant specific heat calculation assuming C p is a constant at the average of the specific heats for the temperatures. Then C p = 1.054 kj/kg K (see Table A-2(b)). 60
Example 7-8 (Continued) It looks like the 300 K data give completely incorrect results here. If the compression process is adiabatic, why is Δs positive for this process? 61
Example 7-9 Air, initially at 17 o C, 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 For air, k = 1.4, and a pressure ratio of 8:1 means that P 2 /P 1 = 8 b.variable specific heat method 62
Example 7-9 (Continued) Using the air data from Table A-17 for T 1 = (17+273) K = 290 K, P r1 = 1.2311. P = P P 2 r P r2 1 1 = 12311. (8) = 9. 8488 Interpolating in the air table at this value of P r2, gives T 2 = 522.4 K = 249.4 o C c.a second variable specific heat method. Using the air table, Table A-17, for T 1 = (17+273) K = 290 K, s o T1 = 1.66802 kj/kg K. For the isentropic process 63
Example 7-9 (Continued) At this value of s o T2, the air table gives T 2 = 522.4 K= 249.4o C. This technique is based on the same information as the method shown in part b. d.using the EES software with T in o C and P in kpa and assuming P 1 = 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 = 249.6 o C 64
Example 7-10 Air initially at 0.1 MPa, 27 o C, is compressed reversibly to a final state. (a) Find the entropy change of the air when the final state is 0.5 MPa, 227 o C. (b) Find the entropy change when the final state is 0.5 MPa, 180 o C. (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 T-s diagram. a. 65
Example 7-10 (Continued) b. c. 66
Example 7-10 (Continued) The T-s plot is T 2 c a P 2 b P 1 1 s Give an explanation for the difference in the signs for the entropy changes. 67
Example 7-11 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 piston-cylinder device 68
Example 7-11 (Continued) Property Relation: Ideal gas equations, constant properties Process and Process Diagram: Isentropic expansion Conservation Principles: Second law: Since we know T 1 and the volume ratio, the isentropic process, Δs = 0, allows us to find the final temperature. Assuming constant properties, the temperatures are related by Why did the temperature decrease? 69
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 Eout = ΔE W = ΔU W = ΔU Using the ideal gas relations, the work per unit mass is W = mc ( T T) v 2 1 w = W = Cv ( T2 T1) m Why is the work positive? kj = 0743. ( 3789. 500) K kg K kj = 90. 2 kg 70
Extra Assignment For the isentropic process Pv k = 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 7-12 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. (c) Calculate the entropy change during the heat rejection process. System: The Carnot engine piston-cylinder device. 71
Example 7-12 (Continued) Property Relation: Ideal gas equations, constant properties. Process and Process Diagram: Constant temperature heat addition. 72
Example 7-12 (Continued) Conservation Principles: a. Apply the first law, closed system, to the constant temperature heat addition process, 1-2. Q W = ΔU net, 12 net, 12 12 Q = mc ( T T ) = = W v net, 12 net, 12 2 1 0 So for the ideal gas isothermal process, 73
Example 7-12 (Continued) But Q Q net,12 H = Q H = 252.2kJ The cycle thermal efficiency is η th = W net, cycle Q H For the Carnot cycle, the thermal efficiency is also given by TL 300K η th = 1 = 1 T 800 K H = 0.625 The net work done by the cycle is W = η Q net, cycle th H = 0. 625( 252. 2 kj) = 157. 6kJ 74
Example 7-12 (Continued) b. Apply the first law, closed system, to the isentropic expansion process, 2-3. But the isentropic process is adiabatic, reversible; so, Q 23 = 0. E E = ΔE in W = W out ΔU = ΔU 23 23 Using the ideal gas relations, the work per unit mass is W = mc ( T T ) 23 v 3 2 kj = ( 1kg)( 0. 718 )( 300 800) K kg K = 359. 0kJ This is the work leaving the cycle in process 2-3. 75
Example 7-12 (Continued) c. Using equation (6-34) But T 4 = T 3 = T L = 300 K, and we need to find P 4 and P 3. Consider process 1-2 where T 1 = T 2 = T H = 800 K, and, for ideal gases m m 1 2 PV T P = = 1 1 2 2 1 2 2 1 PV T V1 = P 3 V 1 1 = (800 kpa) 3 = 266.7 kpa 76
Example 7-12 (Continued) Consider process 2-3 where s 3 = s 2. Now, consider process 4-1 where s 4 = s 1. T P P T 4 4 = 1 1 k/( k 1) 300K = 8000 kpa 800k = 25.834kPa 1.4/(1.4 1) 77
Example 7-12 (Continued) Now, Extra Problem Use a second approach to find ΔS 34 by noting that the temperature of process 3-4 is constant and applying the basic definition of entropy for an internally reversible process, ds = δq/t. 78
Reversible Steady-Flow Work Isentropic, Steady Flow through Turbines, Pumps, and Compressors Consider a turbine, pump, compressor, or other steady-flow control volume, workproducing device. The general first law for the steady-flow control volume is E& & in = Eout r r 2 2 Q& Vi m& ( h gz ) W& Ve net + i i + + i = net + m& e( he + + gze) 2 2 inlets For a one-entrance, one-exit device undergoing an internally reversible process, this general equation of the conservation of energy reduces to, on a unit of mass basis exits δ w = δ q dh dke dpe rev rev rev rev But δ q = T ds δ w = T ds dh dke dp 79
Using the Gibb s second equation, this becomes dh = T ds + v dp δ w = vdp dke dpe rev Integrating over the process, this becomes Neglecting changes in kinetic and potential energies, reversible work becomes 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 80
Turbine Since the fluid pressure drops as the fluid flows through the turbine, dp < 0, and the specific volume is always greater than zero, w rev, turbine > 0. To perform the integral, the pressure-volume 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, w rev, in > 0, or work is supplied to the compressor or pump. To perform the integral, the pressure-volume 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. Pumping an incompressible liquid For an incompressible liquid, the specific volume is approximately constant. Taking v approximately equal to v 1, the specific volume of the liquid (at the given temperature) entering the pump, the work can be expressed as 81
For the steady-flow 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 well-know Bernoulli equation in fluid mechanics. Extra Assignment vp ( P) + Δke+ Δpe= 2 1 0 Using the above discussion, find the turbine and compressor work per unit mass flow for an ideal gas undergoing an isentropic process, where the pressure-volume relation is Pv k = constant, between two temperatures, T 1 and T 2. Compare your results with the first law analysis of Chapter 5 for control volumes. 82
Example 7-13 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 W& mv & ( P P) C = Here at 10 kpa, v 1 = v f = 0.001010 m3/kg. 1 2 1 The work per unit mass flow is W& C w m v P P C = = 1( 2 1) & 3 m = 0. 001010 (5000 10) kpa kg kj = 504. kg kj 3 mkpa 83
Example 7-13 (Continued) b. Using the steam table data for the isentropic process, we have From the saturation pressure table, W& = m& ( h h ) net ( 0 W& ) = m & ( h h ) C 2 1 2 1 kj h1 = 191.81 P1 = 10 kpa kg Sat. Liquid kj s = 1 0.6492 kg K Since the process is isentropic, s 2 = s 1. Interpolation in the compressed liquid tables gives The work per unit mass flow is P 2 = 5 MPa kj kj h = 197.42 = = kg K 2 s2 s1 0.6492 kg W& C wc = = ( h2 h1) m& = (197.42 191.81) kj = 5.61 kg kj kg The first method for finding the pump work is adequate for this case. 84
Turbine, Compressor (Pump), and Nozzle Efficiencies Most steady-flow 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 in the next few slides 85
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 Actual turbine work w h1 h2a a η T = = or ηt Isentropic turbine work w h1 h2s s Well-designed large turbines may have isentropic efficiencies above 90 percent. Small turbines may have isentropic efficiencies below 70 percent. 86
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 T 1 P 1 Compressor or pump &W C η C Isentropic compressor work Actual compressor work s 2s 1 = = or η C w w a h h h h 2a 1 Well-designed compressors have isentropic efficiencies in the range from 75 to 85 percent. 87
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 T 1 P 1 T 2 P 2 η N Nozzle Actual KE at nozzle exit = = Isentropic KE at nozzle exit r V 2 a r 2 2 2s V / 2 / 2 For steady-flow, no work, neglecting potential energies, and neglecting the inlet kinetic energy, the conservation of energy for the nozzle is h r V2 = h a + 2 1 2 2 a 1 2a The nozzle efficiency is written as η N Nozzle efficiencies are typically above 90 percent, and nozzle efficiencies above 95 percent are not uncommon. h h h h 1 2s 88
Example 7-14 The isentropic work of the turbine in Example 7-6 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. η T w Actual turbine work = = Isentropic turbine work kj = η w = (0.9)(1153.0 ) = 1037.7 kg a T s w w a s kj kg η T h h h h 1 2a 1 2s Now to find the actual exit state for the steam. From Example 7-6, steam enters the turbine at 1 MPa, 600 o C, and expands to 0.01 MPa. 89
Example 7-14 (Continued) From the steam tables at state 1 kj h = 3698.6 = = 600 kj = 1 P1 1MPa kg o T1 C s 1 8.0311 kg K At the end of the isentropic expansion process, see Example 7-6. P s 2 = 0.01MPa kj h2s = 2545.6 kj kg = s = 8.0311 kg K x2s = 0.984 2s 1 The actual turbine work per unit mass flow is (see Example 7-6) wa = h h h = h w 2a 1 1 2a a = (3698.6 1037.7) = 2660.9 kj kg kj kg 90
Example 7-14 (Continued) For the actual turbine exit state 2a, the computer software gives A second method for finding the actual state 2 comes directly from the expression for the turbine isentropic efficiency. Solve for h 2 a. Then P 2 and h 2 a give T 2 a = 86.85 o C. h2a = h1 η T( h1 h2s) kj kj = 3698.6 (0.9)(3698.6 2545.6) kg kg kj = 2660.9 kg 91
Example 7-15 Air enters a compressor and is compressed adiabatically from 0.1 MPa, 27 o C, 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 Property Relation: Ideal gas equations, assume constant properties. 2 a T P 2 2 s 1 P 1 s 92
Example 7-15 (Continued) Process and Process Diagram: First, assume isentropic, steady-flow and then apply the compressor isentropic efficiency to find the actual work. Conservation Principles: For the isentropic case, Q net = 0. Assuming steady-state, steady-flow, and neglecting changes in kinetic and potential energies for one entrance, one exit, the first law is The conservation of mass gives The conservation of energy reduces to E& & in = Eout mh & + W& = mh & 1 1 Cs 2 2s m& = m& = m& 1 2 W& Cs = m& ( h2s h1) W& Cs wcs = = ( h2s h1) m& 93
Example 7-15 (Continued) Using the ideal gas assumption with constant specific heats, the isentropic work per unit mass flow is w = C ( T T) Cs p 2s 1 The isentropic temperature at state 2 is found from the isentropic relation The conservation of energy becomes w = C ( T T) Cs p 2s 1 kj = 1005. ( 475. 4 300) K kg K kj = 176. 0 kg 94
Example 7-15 (Continued) The compressor isentropic efficiency is defined as Example 7-16 η w C Ca ws = w a wcs = η C kj 176 kg = = 0.8 kj 220 kg 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. 95
Example 7-16 (Continued) Property Relation: The ideal gas equations, assuming constant specific heats 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, Q net = 0. Assume steady-state, steady-flow, 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 E& & in = Eout r 2 V2 s mh & 1 1= m& 2( h2s + ) 2 96
Example 7-16 (Continued) The conservation of mass gives m& = m& = m& 1 2 The conservation of energy reduces to r V = 2( h h ) 2s 1 2s Using the ideal gas assumption with constant specific heats, the isentropic exit velocity is r V = 2C ( T T ) 2s p 1 2s The isentropic temperature at state 2 is found from the isentropic relation 97
Example 7-16 (Continued) The nozzle exit velocity is obtained from the nozzle isentropic efficiency as η r V N r 2 V2 a /2 = r 2 V2 s /2 r m m = V η = 442.8 0.95 = 421.8 s s 2a 2s N 98
Entropy Balance The principle of increase of entropy for any system is expressed as an entropy balance given by or Sin Sout + Sgen = ΔSsystem 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. 99
Entropy change of a system The entropy change of a system is the result of the process occurring within the system. Entropy change = Entropy at final state Entropy at initial state Mechanisms of Entropy Transfer, S in and S out 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 100
Q Entropy transfer by heat transfer: S T T heat = ( =constant) 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 1-2 can be determined by integration (or by summation if appropriate) as Work Work is entropy-free, 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: = 0 S work 101
Mass flow 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, S gen Irreversibilities such as friction, mixing, chemical reactions, heat transfer through a finite temperature difference, unrestrained expansion, non-quasi-equilibrium 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. 102
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 The entropy balance for any system undergoing any process can be expressed in the general rate form, as where the rates of entropy transfer by heat transferred at a rate of and mass flowing at a rate of &m are S & & / & heat = Q T and Smass = ms & &Q 103
The entropy balance can also be expressed on a unit-mass basis as ( s s ) + s = Δs ( kj / kg K) in out gen system The term S gen 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. S gen = 0 for the internally reversible process, but not necessarily zero for the totally reversible process. The total entropy generated during any process is obtained by applying the entropy balance to an Isolated System that contains the system itself and its immediate surroundings. 104
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 + Sgen = ΔSsystem = S2 S1 ( kj / K) T For an adiabatic process (Q = 0), this reduces to k Adiabatic closed system: S = ΔS gen adiabatic system A general closed system and its surroundings (an isolated system) can be treated as an adiabatic system, and the entropy balance becomes Control Volumes System + surroundings: S = ΔS = ΔS + ΔS gen system surroundings The entropy balance for control volumes differs from that for closed systems in that the entropy exchange due to mass flow must be included. Q T k i i e e gen 2 1 CV k + ms ms + S = ( S S ) ( kj/ K) 105
In the rate form we have Q& T k i i e e gen CV k + ms & ms & + S& = ΔS& ( kw/ 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. For a general steady-flow process, by setting Δ S & CV = 0 the entropy balance simplifies to S& = m& s ms & gen e e i i For a single-stream (one inlet and one exit), steady-flow device, the entropy balance becomes & Q& Sgen = m& ( se si ) T k k Q& T k k 106
For an adiabatic single-stream device, the entropy balance becomes S& = m& ( s s ) gen e i 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. Therefore, for steady-flow, single-stream, adiabatic and reversible process: s e = s i 107
Example 7-17 An inventor claims to have developed a water mixing device in which 10 kg/s of water at 25 o C and 0.1 MPa and 0.5 kg/s of water at 100 o C, 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 20 o C, is this process possible? If not, what temperature must the surroundings have for the process to be possible? System: The mixing chamber control volume. Property Relation: The steam tables Process and Process Diagram: Assume steady-flow 108
Example 7-17 (Continued) 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 1 f @ T1 P MPa kg T 1 1 kj h h = 104.83 = 0.1 o = 25 C kj s1 sf @ T 0.3672 1 = kg K 2 P2 MPa kg T 2 kj h = 2675.8 = 0.1 o = 100 C kj s2 = 7.3611 kg K 109
Example 7-17 (Continued) kj h3 = 417.51 P3 = 0.1MPa kg Sat. liquid kj s3 1.3028 = kg K Q& = m& h mh & m& h net 3 3 1 1 2 2 kg kj kg kj kg kj = 10.5 417.51 10 104.83 0.5 2675.8 s kg s kg s kg kj =+ 1997.7 s So, 1996.33 kj/s of heat energy must be transferred from the surroundings to this mixing process, or Q & = Q & net, surr net, CV For the process to be possible, the second law must be satisfied. Write the second law for the isolated system, Q& k + ms & ms & S& S& i i e e+ gen= Δ CV T k For steady-flow Δ &S CV = 0. Solving for entropy generation, we have 110
Example 7-17 (Continued) Q S & gen = m & ese ms & i i T Q& cv = ms & 3 3 ms & 1 1 ms & 2 2 T & surr kg kj kg kj = 10.5 1.3028 10 0.3672 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 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 = 0 and solve for T surr. &S gen k k 111
Example 7-17 (Continued) Qk S& gen = m& ese ms & i i = 0 Tk Q& cv Tsurr = ms & ms & ms & & 3 3 1 1 2 2 1997.7 kj / s = kg kj kg kj kg kj 10.5 1.3028 10 0.3672 0.5 7.3611 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 (42.75 o C) in the amount of 1997.7 kj/s to increase the water temperature to approximately 42.75 o C 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. 112