7. Development of the 2nd Law

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Geraldine Merritt
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7-1 7. Development of the 2nd Law 7.1 1st Law Limitations The 1 st Law describes energy accounting. Once we have a process (or string of processes) we can calculate the relevant energy interactions.

1 Development of the 2nd Law 7.1 1st Law Limitations The 1 st Law describes energy accounting. Once we have a process (or string of processes) we can calculate the relevant energy interactions. The 1 st Law has a major limitation though. It cannot tell us if a process (or string of processes) can occur in real life. As examples, consider the following two cases. A current I flows through a resistor having resistance R Ω while experiencing a voltage drop ΔV (Figure 7-1). We know for steady state operation that the resistor dissipates power I 2 R Ω. The 1 st Law tells us that there will be Q equal to I 2 R Ω. Figure 7-1. EFD of current flow through a resistor. Since the 1 st Law is an equality the energy accounting is still valid if we reverse the process and heat the resistor (Figure 7-2). In that case there should be a voltage increase ΔV as current flows through the resistor. Nobody has ever observed such behavior, nor do we expect it to occur. Figure 7-2. EFD of reversing current flow by heating the resistor.

2 7-2 Next consider a block moving along a surface. A force F is required to overcome the frictional opposition mgμ dynamic. Work supplied to the block is dissipated via heat transfer (assume the block is a perfect thermal insulator). Figure 7-3. EFD of a block moving along a surface. Again, you know from experience that reversing Q will not cause the block to move against the friction force and exert a force F on some other object (Figure 7-4). Figure 7-4. EFD of reversing a block moving by adding heat. We need an analytical tool to supplement the 1 st Law. This tool should indicate the possibility, or impossibility, of a process occurring in actuality. Mathematically this suggests an inequality because we have two possible outcomes and an equality can only give one. In terms of Thermodynamics we want this indicator to be a property because property values depend only on their state and because their differences do not depend on the path between the initial and final states. This brings us to the property entropy, S.

3 Entropy You may have a physical view for U, the internal energy of a system. You might picture a multitude of molecules that are translating, with nuclei rotating and vibrating, and electrons in various orbits about their corresponding nuclei. An increase in temperature causes increases in molecular energy stored in these modes. It s less likely that you have a similar depth of understanding for H, the enthalpy of a system. Recall that enthalpy is a combination of internal energy and the product pv, where pv is the amount of energy required to make a place for that system. Now think about the basic equations we ve discussed so far. The 1 st Law is based on energy conservation. No one has proven that energy is conserved, but you ve grown to accept it. Similarly, no one ever proved that mass is also conserved, but you ve also grown to accept it. At this point you need to accept certain additional truths that will eventually lead to the 2 nd Law of Thermodynamics: - there is a property called entropy, S - you won t have a very good physical understanding of it - it has its own basic equation, and that equation is an inequality - it is not a conserved quantity like mass or energy - it provides information on whether a process is possible or impossible in real life. The entropy basic equation for a fixed quantity of mass that can undergo only boundary work (this is termed a simple system) is:

4 7-4 of enthalpy: There is an analog that is obtained by using the definition Both can also be written in terms of specific properties. Inspection of the two entropy equations shows entropy must be a property since it is included in equations that contain nothing but properties. Let s compare the first entropy equation to the 1 st Law expression for a simple system: This comparison indicates that the change in entropy is somehow tied to Q. In fact, The change in S equation provided above should worry you a bit since the LHS is the change in a property, S, T is a property, but dq is not a property. This means the S equation as written can only hold for certain heat transfer paths. For reasons to be discussed later, we will call the acceptable paths reversible. Then, more accurately

5 7-5 Further consideration of this new equation shows that temperature is involved, and that T is associated with dq. It is reasonable to assume T corresponds to the system boundary temperature where dq is occurring, so Integrating along the path from the initial state (1) to the final state (2) gives: A perhaps surprising feature of the above expression is the absence of W. That demonstrates work interactions do not change entropy. This is an important point. The entropy change equation cannot be complete in the form given (i.e., as an equality) for three reasons. 1. An equality can t indicate which Thermodynamic paths are possible and which are impossible. 2. An equality suggests S is a conserved property, like m and E, and it s not. 3. An equality violates the fundamental Thermodynamic postulate: The entropy of the universe never decreases. If we use our entropy change equation to analyze the known universe, for which there are no known W or Q interactions at the boundaries, then our current version of the entropy change equation is

6 7-6 when it should be If we work backward from the above expression and consider the case with nonzero Q This inequality is due to Clausius. It demonstrates the change in system entropy is equal to that for a reversible process between the beginning and end states (the = sign), plus a supplement if the actual path is not reversible (the > sign). It can be written as an equality if we know the magnitude of the supplement due to the non-reversible aspect. If the supplement contribution is represented by S generation, If the process is reversible, S generation is zero by definition. If the process is both reversible and adiabatic (only W so no Q), then the entropy change is zero. DS = 0 is called an isentropic process. It is an idealization which engineers often use. Note that S generation can never be negative. The adiabatic and reversible (i.e., isentropic) process is most desirable because adiabaticity requires that only work interactions occur. This means the maximum amount of energy decrease between 1 and 2 goes into W for a work producing device, or the minimum amount of energy increase is required as

7 7-7 W for a work consuming device. In addition, reversible means there are no dissipation processes, such as friction in mechanical systems or resistance in electrical ones, so all the work is usable via a rotating shaft, a reciprocating piston, etc. We will regularly make use of the isentropic (adiabatic + reversible) model. The equation for entropy change provided above is appropriate when no mass crosses a system boundary. When mass transfer is present we must form a rate version and add terms representing the entropy entering and exiting the system Here Q j is the rate heat transfer across the system boundary where the temperature is T boundary,j. The same assumptions apply as for Conservation of Mass and the 1 st Law: if the process is steady flow/steady state the accumulation term disappears, and if there is no mass flow in or out the summation terms disappear. 7.3 Non-Reversible and Reversible Processes We have seen two examples of processes that are not reversible a block sliding along a surface where friction is present and a current flowing through a resistor. Are there other examples? The answer is yes, and many will be obvious. In addition, from our entropy change expression above we see that all non-reversible processes have non-zero (positive) S generation. All mixing processes are non-reversible. This includes mixing two chemically identical species initially at different pressures or temperatures, or mixing two chemically different species at the same initial temperatures and pressures (or any

8 7-8 combination). Your intuition should guide you here because you have never seen two mixed gases separate. [Think of the implications if the oxygen and nitrogen in air were to suddenly move to different portions of a room.] Any Q through a finite DT is also non-reversible. Again, your intuition should guide you since you know energy transfer due to a temperature gradient is always from the high temperature material to the lower temperature one. All unrestrained expansions are yet another example of non-reversible processes. The flow of a gas into an initially evacuated chamber is an obvious case. However, any rapid gas expansion, or expansion through a finite pressure difference caused by fluid viscosity, is non-reversible. This includes shock waves. Although we won t cover them in this course, chemical reactions are almost always non-reversible. Again, have you ever seen the products of hydrocarbon combustion reform themselves into the original fuel and oxidizer? An important realization is that S generation is positive for all of these examples, and for all other non-reversible paths. As a reminder, it is the value of S generation which indicates if a process is reversible, non-reversible or impossible. When S generation = 0 for a substance, the substance undergoes a reversible process. If S generation > 0 for a substance, the substance undergoes a nonreversible process. Finally, if S generation < 0 you have either made a mistake or the process the substance undergoes is impossible. 7.4 Thermal Reservoirs The entropy change for any system undergoing a reversible, isothermal process is easy to calculate:

9 7-9 A good question is Which, if any, practical systems behave in a reversible fashion? In other words, are there real systems that keep their temperature (nearly) constant while also transferring energy via Q? One thing that s clear is the internal energy of such a system must be very large compared to the energy transferred: E >> Q. If we substitute m c DT for E, as is the case of a liquid, either m, c or both must be large. Water has one of the largest liquid c values so a huge mass of water is a candidate. The Great Lakes are a case in point. In practice, we can use something smaller a liquid temperature change no more than 1% of the average value might suffice. This is <3 C for a substance that starts at room temperature. A two-phase liquid slowly undergoing a phase change is another example. As long as the pressure doesn t vary, the temperature will remain constant. A substance that can provide energy transfer Q while maintaining its temperature constant is called a thermal reservoir. The state of any thermal reservoir is uniquely specified by its temperature. Thus, if we have the temperature of a thermal reservoir we have all of its properties. In addition, since the temperature of a thermal reservoir doesn t change its properties (including its entropy) do not change, even while heat transfer is occurring. As an example of a thermal reservoir, consider the energy transfer between the outdoors (at temperature T high ) through a wall

10 7-10 into a living space (at a temperature T low ). We assume that Q is such that neither T high nor T low vary. The temperature within the wall is also at steady state. The appropriate EFD is shown in Figure 7-5. Figure 7-5. EFD for energy transfer through a wall. With the wall material as the system, and assuming that it can be modeled as a thermal reservoir, the entropy equation gives with DS = 0 because the state of material inside the wall doesn t change. The result is a S generation that is positive, indicating a non-reversible process (as we expect). 7.5 The role thermal reservoirs play in cyclic processes and systems Recall that IC engines operate cyclically, as do steam power plants, and gas turbine engines for aircraft and ground use. Vapor compression refrigerators are another common example of a cyclic system, and gas (Brayton cycle) refrigerators are also exist. In each case, the hardware contains at least one heat exchanger which receives heat from, or sends heat to, an external source or

11 7-11 sink. Because they are easy to analyze using the 1 st and 2 nd Laws, we will assume all the system external heat interactions involve a thermal reservoir. In the case of an IC engine cycle, the EFD might look like Figure 7-6. Figure 7-6. EFD for IC engine cycle operating between two thermal reservoirs. The system would be the air (or other fluid) enclosed by the piston-cylinder. Energy, Q high, is supplied from thermal reservoir T high. Some fraction of Q high is transformed into work, W, with the remainder, Q low, being wasted. The fact that some of Q high must be rejected (wasted) is a consequence of the 2 nd Law. The practical implication is that no device that uses a thermal energy source can have an energy efficiency of 100%. 7.6 Evaluating Entropy and its Changes Evaluating entropy for any substance is done just as for h, u, n, T, and u. Tables are available for compressed liquids, saturated liquids, two-phase mixtures, saturated vapors, and superheated vapors. The entropy entry is always at the far right of each table or sub-table. Charts are also available, although they have different ordinates and abscissas Charts for Entropy We have seen that p-v diagrams are most useful when identifying the state of a substance, and also when graphically

12 7-12 computing boundary work. p-h diagrams are more appropriate for systems where mass flows into and out of a system since Dh is often quickly related to W and Dp=0 when idealized heat exchangers are present. Charts with s on one of the axes ought to serve similar functions: provide insight into Ds and at least a quantitative indication of the change for an important energy quantity. Both objectives can be met by having T as the ordinate and s as the abscissa. Temperature on the ordinate gives a measure of the internal energy (or enthalpy) of a substance. The integral of the T- s path gives the reversible heat transfer. T-s diagrams find their greatest use when analyzing cyclic systems such as heat engines, refrigerators, and heat pumps. The areas under the heat exchanger T-s diagrams provide the high temperature and low temperature Q values. The vertical distances are proportional to the compression and expansion work interactions. Finally, the area enclosed by the succession of paths equals the net heat transfer for the cycle, which must also equal the net work. Figure 7-7 is an example of a cycle having four processes. Two of the processes are reversible and adiabatic (therefore isentropic) W production and consumption. The other two are isothermal (therefore potentially reversible) Qs. Just as for p-v and p-h charts, the phase regions from left to right are: compressed liquid, saturated liquid, two-phase mixture, saturated vapor, and superheated vapor. In Figure 7-7, the lower left state is saturated liquid, the upper left state is compressed liquid, the upper right state is superheated vapor, and the lower right state is saturated vapor.

13 Ammonia m3/kg 150 T [ C] kpa 5000 kpa 2500 kpa 1000 kpa 500 kpa 200 kpa Figure 7-7. T-s diagram for a cycle having four processes Entropy changes for incompressible substances The entropy change for an incompressible substance can be calculated from one of the Tds equations: s [kj/kg-k] du = c V dt, so Recall that dn = 0 for an incompressible substance, and that

14 7-14 Since there is effectively only one specific heat for an incompressible substance, c, and since it is quite often constant for practical situations, See Chapter 4 for a table of specific heats for incompressible substances Entropy changes for ideal gases An expression for ideal gas entropy change is developed from the other Tds equation Recall that dh= c p dt so The ideal gas law is used to recast n in terms of p, R and T The fact that the gas constant for a substance has a fixed value makes the second term on the RHS easy to evaluate. However, the first term on the RHS is complicated because c p is dependent on T. Fortunately, the argument of the integral depends only on T so it can be split into two as shown below, and integrated knowing only T 0 and the integral upper bound

15 7-15 where T 0 is the temperature of some reference state, taken to be 0 K in this case. The integrals on the RHS have been given the symbols s 0 and are tabulated in the ideal gas tables Consequently, Note that two properties must be known to compute ideal gas Ds some pair from the trio p, n and T. 7.7 Isentropic Processes Mechanical work producing ( active ) devices require the least energy input and provide the greatest energy output if the mass within them undergoes an adiabatic process. In addition, reversible processes are optimal. A combination of reversible and adiabatic should lead to optimum active device performance. The 2 nd Law indicates that an adiabatic and reversible process is an isentropic one so isentropic processes become our target when designing work producing/consuming systems. The same is true for nozzles and diffusers, along with piping networks. Isentropic processes for compressed liquids, saturated liquids, two-phase mixtures, saturated vapors, and superheated vapors are determined using property charts and tables. One simply finds the initial entropy and equates the final entropy to it.

16 7-16 Once the final entropy magnitude is known, another property is typically needed to specify the state so that h, p, n, T and u can be determined. An isentropic process for an incompressible substance is particularly easy to analyze. Since Ds depends only on T for this instance, Ds=0 requires DT=0. This is easy to remember if you recall that when two of the three I s, incompressible/isentropic/isothermal, are true the third one must also be true. Isentropic processes for ideal gases are analyzed using the ideal gas tables. When Ds=0 we have Some tables have tabulated values for the exponential. If available, it s written as 7.8 The Carnot Cycle Carnot s formulation of the cycle bearing his name is one of the greatest breakthroughs in Thermodynamics. His cycle is a powerful tool for Thermodynamic analysis because it represents the highest performing cyclic device. The formulation of the Carnot power cycle uses the concept of a reversible path providing maximum performance for a single process. In that case heat transfer must occur via an infinitesimal temperature difference. Realizing that it would be difficult to

17 7-17 maintain an infinitesimal DT if either the system or surroundings temperatures were changing, all heat transfers are required to take place during isothermal processes. The optimal work interactions will occur in the absence of irreversibilities. Since the system temperature will change during the work paths the processes must also be adiabatic. This results in isentropic works. The Carnot cycle is composed of two pairs of isothermal heat transfer steps and two isentropic work contributions. This is a rectangle on a T-s diagram with the upper horizontal line having magnitude T H and the lower horizontal line having magnitude T C (Figure 7-8). Figure 7-8. T-s diagram for a Carnot cycle. The area inside the box (T H T C ) * Ds is the net heat transfer and also the net work for the cycle. The area under the line of constant T H is the heat addition for the Carnot power cycle. Since we have to buy the energy supplied at T H and can sell the work we can define the Carnot power cycle efficiency as Note that h Carnot depends only on T C and T H so in principle it s applicable to any hardware, any working fluid, and any

18 7-18 phase(s). It also ranges from 0 to <1. Also note that the source of high temperature energy and sink of low temperature energy are usually assumed to be heat reservoirs. Finally, absolute temperatures must be used in any power cycle efficiency calculation. Returning to the above heat-transfer-through-a-wall discussion, we show that any non-ideal power cycle must have an efficiency inferior to that of a Carnot cycle when both operate at the same pair of temperatures. To accomplish this we fix the high temperature heat transfer, Q H, as well as T C and T H. For the Carnot case: For the non-ideal case there will be some non-reversibility and corresponding entropy generation. That s generation increases Q C via The non-eversible W is correspondingly smaller, so h nonreversible < h reversible. Any power cycle can, at least theoretically, be run in reverse. When so done all the energy flows reverse direction. The reversal of a Carnot power cycle becomes either a Carnot refrigerator or Carnot heat pump, depending on the energy goal of the cycle. In the case of the refrigerator the energy goal is the amount removed from the space to be cooled, Q C. The system energy that is bought is W. In the case of the heat pump the

19 7-19 purchased system energy is still W, but the energy goal is now Q H. The refrigerator and heat pump have their own versions of h Carnot, termed their coefficients of performance, COP Cycle COPs differ from power cycle efficiencies in that they can (and should) exceed unity. They don t differ from power cycle energy efficiencies in that absolute temperatures must be used during their calculation. They also don t differ from power cycle energy efficiencies in that the Carnot cycle values are always greater than the actual cycle values.

20 Examples Vapor compression refrigerator A vapor compression refrigerator uses R-134a. The evaporator temperature is -4 C, the condenser temperature 30 C, saturated vapor enters the compressor and exits at 38 C. Saturated liquid exits the condenser and a throttling valve drops the R-134a pressure from the condenser value to the evaporator value. a) Sketch the four processes on a p-h diagram. b) Calculate the enthalpy change for each component. Report your answers in kj/kg. c) Calculate the entropy change for each component. Report your answer in kj/kg-k. d) Determine the entropy generation for each component. Report you answer in kj/kg-k. e) Based on your answers to c, which component(s) would you focus engineering redesign/performance improvement efforts?

21 Gas power cycle A gas power cycle has air entering at 0.7 bar and 260 K. The compressor pressure ratio is 25:1 and its outlet temperature is 720 K. The combustor heats the air to 1700 K and the turbine exhaust temperature is 740 K. a) Sketch the three processes on a p-h diagram. b) Calculate the enthalpy change for each component. Report your answers in kj/kg. c) Calculate the entropy change for each component. Report your answer in kj/kg-k. d) Determine the entropy generation for each component. Report you answer in kj/kg-k. e) Based on your answers to c, which component(s) would you focus engineering redesign/performance improvement efforts?

22 Compressor Air flows steadily into a compressor at 295 K and 0.95 bar. The exit pressure is 9.5 bar, and operation is at steady state. a) Which of the following processes requires the minimum work? i) isentropic, ii) isothermal, iii) polytropic with 1<n< 1.4. Provide quantitative support for your choice. b) What is the air entropy generation for each case? Report your answer in kj/kg-k.

Chapter 7. Entropy. by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn

Chapter 7. Entropy. by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn Chapter 7 Entropy by Asst.Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn Reference: Cengel, Yunus A. and Michael A. Boles, Thermodynamics: An Engineering Approach, 5th ed.,