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Basic Thermodynamics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture - 07 Second Law and its Corollaries II I welcome you to this session of thermodynamics.

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Basic Thermodynamics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture - 07 Second Law and its Corollaries II I welcome you to this session of thermodynamics. In the last session, we discussed that if two engines operate between the same temperature limits, one is the temperature for heat addition and another is the temperature for heat rejection. If one is a reversible engine, its efficiency is higher than that of an irreversible engine. The corollary to it is that all reversible engines operating between the same temperature limits have the same efficiency. So these things we discussed and proved logically, but I think there was little confusion in its understanding through logic, that is why I will repeat this thing again, this proof. (Refer Slide Time: 01:56) You see that, let there are two thermal reservoirs one is at temperature t 1 another is at temperature t 2; t 1 is greater than t 2 and we know that the two temperature reservoirs are essential for a heat engine to operate to develop work in a continuous cycle. Let us consider one reversible heat engine, HE R, which takes heat Q 1R, this subscript R is used for reversible engine from this high temperature reservoir, source t 1, and it has to reject according to thermodynamics second law, the amount Q 2R to the thermal reservoir t 2. While doing so, it develops a work W R. Here also it is written W R which is equal to, from the first law of thermodynamics that is energy balance Q 1R minus Q 2R. Second law tells that Q 2R cannot be 0, there has to be Q 2R that is the only restriction, but first law is always valid like this. We consider another heat engine, HE, working in a continuous cycle which is not a reversible heat engine, which draws an amount of heat Q 1 which we said deliberately for the proof equals to that of the heat taken by the reversible engine, that means Q 1R. Obviously, since HE is an irreversible engine, it will reject an amount of heat Q 2 which is different from that (Refer Slide Time: 03:11 min). If the heat addition and rejections are same, the work quantity will be same by the first law. Out of these three quantities - heat addition, heat rejection, and work - two are independent and one is dependent by the conservation of energy. So, they cannot be same for a reversible engine. When it is an irreversible engine, they have to be different. That is the very first logic. Let us consider that HE rejects Q 2 which is different from Q 2R and while doing so it develops a work W which equals to Q 1 minus Q 2. Now we have to prove that, now what is eta, first of all? Definition of eta is what? Work for reversible engine, work developed by the heat added, eta R is equal to W R by Q 1R. What is eta for this irreversible engine which is a natural engine? W by Q 1R. For both it is Q 1R since Q 1 has been set Q 1. Now, we have to prove that eta R is greater than eta. To prove this, we first assume that eta R is less than eta. We first assume that and then we prove that this is not possible. How? If eta R is less than eta, then what we get? W R is less than W, eta R is less than eta, Q 1R is same, so W R is less than W, or we can write other way W is greater than W R. Now what we can do? Just see here. That is the most important thing (Refer Slide Time: 05:01 min). If HE R is a reversible heat engine, all the processes can be reversed. Reversible heat engine means, that all the processes are reversible processes, so all processes can be reversed without any other change in the surrounding. According to the requirement of a reversible process or characteristic feature of a reversible process, that means, if we reverse all the processes of this heat engine it will act as a reversible heat pump, HP R. If this reversible heat pump HP R is allowed to operate then it will give the same amount of heat Q 1R in t 1 which we use to take while acting as an engine. It will draw the same amount of heat Q 2R from the reservoir t 2 and it will demand the same amount of work which it delivered W R to the surrounding incase of an engine. That means, this work and heat quantities interactions are equal in magnitude, but opposite in direction (Refer Slide Time: 06:07). That is the characteristic feature of the reversible heat engine. But if we reverse the heat engine, HE, it will work as a heat pump. Do not consider that as it is an irreversible heat engine, it cannot be made to operate as a heat pump. In that case, the directions of the heating directions will be opposite, definitely, the working directions will be opposite. It will draw work, it will give heat; in that case, the quantities will not be same. If it has to elevate heat or pump heat Q 1 to the temperature t 1, it may not draw the work W from the surrounding, it may not draw or it may not take the heat Q 2 from this t 2. That is the difference between a reversible and irreversible engine. Therefore it is always useful for us to prove W is greater than W R to make this reversible heat engine to operate in a reverse direction as a reversible heat pump, so that the quantities reveal the same magnitude, but in the opposite direction. Now, we have assumed that eta R, the efficiency of the reversible heat engine, is less than eta, which means the work developed W by the irreversible heat engine is more than that of the work developed by the reversible engine W R. Therefore in this case what we can do? We can couple these two, HP R and HE so that the work developed from the heat engine can be utilized to dive the heat pump and in spite of that, we get a net amount of work W minus W R as the work output. Since these two heat quantities, Q 1R and Q 1, are same, they can be joined by a diathermic wall (Refer Slide Time: 07:40 min). This reservoir t 1 is redundant. The working fluid of this heat pump HP R will act as the thermal reservoir for the heat engine HE, and the working fluid of this engine will act as the thermal reservoir of this heat pump HP R. The heat rejected by this heat pump will be delivered to this heat engine as the additional thing. Therefore, the entire system corresponds to a machine which interacts with only one thermal reservoir (Refer Slide Time: 08:09 min), a thermal reservoir of a single fixed temperature and developing a network which is not possible and violation of Kelvin Planck s statement. This is violation of Kelvin Planck s statement of second law. Now, this, eta R is less than eta, is not possible. Therefore there are two possibilities, eta R is greater than eta or eta R is equal to eta. Now the logic is that eta R cannot be equal to eta, because an irreversible engine and reversible engine are different, so their heating interactions and working interactions are different. So that when we reverse an irreversible engine, it will not reveal the same magnitude. It has to be a different engine than that of the reversible one, so we cut this logic, these two, eta R and eta, cannot be equal either this will be greater or less than this. That means, these two engines will be different. So with this logic, we can prove that the only go is that eta R is greater than eta. It is clear Now the next step is that if both of them, HE R and HE are reversible then they are of equal efficiencies. The next corollary to this theorem is that if there are number of reversible engines operating between the two fixed temperatures their efficiencies will be same, they cannot be different. How to prove it? Now the same thing, I am not going to a different drawing. Consider both the engines are now reversible. I am not giving this subscript 2 R W R then it will be more congested, because then R 1, R 2. Let us consider this Q 2 (Refer Slide Time: 10:02 min) is the heat rejected by another reversible engine. W is the work developed by another reversible engine. We consider two different reversible engines and we consider similar way that eta R is greater than eta, that means efficiency of the reversible engine HE R is greater than the HE. Now in this case, I consider this machine, HE, also as a reversible, eta is the efficiency of a reversible engine that is the engine at the right, right to me (Refer Slide Time: 10:27 min). Now, in this case, as I have shown, I can reverse the HE R and prove that eta R is less than eta is not possible. Now since HE is a reversible engine, in the next step, I can reverse this one and can get these quantities exactly in the same magnitude (Refer Slide Time: 10:45 min), because which I could not do in the earlier case, but now I can do it, it is also reversible. That means Q 1 and Q 2. Now when I assume this, eta R is greater than eta, I reverse the one, HE R. When I assume that, eta R is less than eta or eta is greater than eta R, then I reversehe. If I reverse this one when eta R is greater than eta then, what happened? to be proved? eta R is less than eta, then W is greater than W R. In that case, I will reverse HE R. Now I will take eta less than eta R, that means, W R is greater than is W. In that case, I will reverse the HE. First, I considered eta R is less than eta that means W is more than W R. So, I reverse the HE R and we proved that eta R is less than eta is not possible. Now I consider eta less than eta R and I reverse, when we consider eta less than eta R, means W R greater than W, I reverse HE. Now here since, it is also a reversible engine at the beginning we have considered here the both are reversible. Then I can reverse it. In the similar manner, I can prove there is a work because W R is a more than W. Network W R minus W coming out of these two, HP R and HE. We can couple these two HP R and HE and HE R will remain as a heat engine. In that case, we can couple these two, HE R and HE and we get the work, W R is greater than W, which is violation of Kelvin Planck s statement. We proved neither eta R is less than eta nor eta is less than eta R. The only possibility is the eta is equal to eta R. Try to understand the logic, so that if two of the reversible engines, then if I assume that. One of these engines for example, eta R is less than this, then I reverse this engine, HE R and prove that this is a violation of Kelvin Planck s statement. In another case, if I consider HE is less than the engine HE R, the efficiency of the engine, HE is less than HE R then I reverse HE and I can prove that the equivalent one is a machine which interacts with a thermal reservoir at a single fixed temperature and developing a network violation of Kelvin Planck s statement. So, only opportunity is eta is equal to eta R. So, I think logic is convincing, but it is not at all essential to memorize this thing, because nobody will ask us for this how did you prove it. It is extremely important to know this theorem throughout the course of thermodynamics wherever you go even at the higher level. What is this theorem? If there are two thermal reservoirs at different temperatures t 1 and t 2 a heat engine will operate, it will take heat from the higher temperature reservoir, it will reject heat at the lower temperature reservoir and will operate in a continuous cyclic process and will develop work which according to first law of thermodynamics, will be the difference between the heat added and the heat rejected. This is number one. Number two, if there is a reversible engine operating between the same temperature limits its efficiency is the highest among all other irreversible engines. Next is, all reversible engines have the same efficiency if they operate under the same temperature limits. The efficiency of reversible engine is unique when the temperature limits are prescribed. This acts as the maximum performance criterion or the ideal value for any natural engine. That means the efficiency of any natural engine, so at the same time you can think that is true that all natural engines will be having different efficiencies, operating between the same temperature limits. Even the maximum of these efficiencies is lower than the efficiency of a reversible heat engine operating between the same temperature limits. (Refer Slide Time: 15:16) So with this, I will go to a very interesting thing that concept of absolute thermodynamic scale of temperature. Now, so far what you have read, if we have a thermal reservoir at any temperature t 1, at any conventional scale. Let us consider Celsius scale and if there is a heat engine which is a reversible one H ER. If there is a reversible heat engine and it takes Q 1 amount of heat. Now this is the block diagram of any heat engine. Now, we know that the efficiency of a reversible heat engine is fixed when the two temperatures are fixed. When the two temperatures are fixed the efficiency of a reversible engine is uniquely fixed. All reversible engines have the same efficiencies. The second part of the Carnot s theorem which can be expressed mathematically that efficiency of reversible engine, I can give this eta R is therefore a function of these two temperatures, eta R is equal to function of temperatures t 1 and t 2. Now, how do we define the efficiency? Now in all cases the efficiency is defined by the unique manner that is work done by heat added. Now I am not using the subscript R. Let us use the subscript R for your understanding, for a reversible engine. Now the efficiency is same for reversible or irreversible engine. It is the work divided by the heat added. Similarly, the work is the difference between the two heat quantities, it is also same for both the engines, because it is the law of conservation of energy. There is no restrictions for friction and other conditions that irreversibility. Therefore, it is valid; you have to remember that. In this case, Q 1R minus Q 2R by Q 1R, so we can write 1 minus Q 2R by Q 1R. We see that, eta R is a function of temperature t 1 and t 2 from which we can tell this is a very important conclusion. That the ratio of the heat interactions, I write in this fashion it looks nice, that heat added to heat rejected is some function of two temperatures. It is a very important conclusion that the ratio of heat added to heat rejected is a function of two temperatures (Refer Slide Time: 18:25 min), the ratio of the heat interactions Q 2R or Q 1R also. That means the ratio of the two heat quantities of interactions by the reversible heat engine are a function of temperatures only. This is valid only for a reversible engine. For irreversible engine, it is not so. This is because, for irreversible engines if t 1 t 2 are fixed, different irreversible engines will depict different efficiencies. It is because of the fact that all reversible engines have the same efficiency when t 1 and t 2 are fixed. The efficiency of a reversible engine is a function of the two temperatures only. Now we will exploit this definition to define thermodynamic scale of temperature. At this moment, we do not know anything about this functional relationship of t 1 and t 2. We can only express in an implicit form like this (Refer Slide Time:19:11 min) that some function of t 1 and t 2, unknown function, but let us utilize the concept of heat engine to go one step further to give some particular form of the function. What is that? We remember this definition and then we construct this. (Refer Slide Time: 19:29) Let us consider a reversible engine, there are three temperatures. One, now here all reversible engines, I am not using subscript R, so one reversible engine is heat engine, HE A which takes heat Q 1 from a thermal reservoir t 1 and rejects heat Q 2 to a thermal reservoir t 1 small t 1 you give because, now it is a conventional scale and let us consider a Celsius scale. Therefore, t 1 has to be greater than t 2 and HE A develops a work W A which must be equal to Q 1 minus Q 2. Now we consider another heat engine, HE B, which takes this heat Q 2 from this thermal reservoir, t 2. This thermal reservoir, t 2, is in contact with both the engines. So, this heat engine B takes this heat Q 2 from the thermal reservoir t 2 and delivers or rejects heat Q 3 at another thermal reservoir t 3, where t 3 is less than t 2 obviously, and it develops a work W B which is equal to Q 2 minus Q 3. Now, according to this relation that the ratios of heat interactions are the functions of the temperatures, we can write for the heat engine HE A, Q 1 by Q 2, all are reversible engines. So I am not writing the subscript R any more, is equal to what? Same, function of t 1 t 2. Q 1 by Q 2 is equal to the functions of t 1 and t 2. For the reversible engine HE B, what it will be? Q 2 by Q 3 and it will be equal to the same function with the argument change, because here the temperatures are t 2, t 3. Same function F. If we consider these two heats, the heat rejected by the heat engine A and the heat taken by the heat engine B, now we can remove this reservoir, t 2, because, we can make these two heat engines, A and B, connected. So that heat rejected by this heat engine A will be taken by this heat engine B. That means the working fluid of heat engine A will be the source or thermal reservoir for the working fluid of heat engine B and the vice versa. So that we can get rid of this t 2 and we can connect this heat engines A and B. I am not drawing this separate figure. In that case, the combination of A and B (Refer Slide Time: 22:14 min), these are known as heat engines in series, will be an equivalent heat engine developing a work of W A plus W B and interacting with thermal reservoirs t 1 and t 3. They are taking heat that means equivalent engine is taking heat Q 1 from t 1 and rejecting heat Q 3 to t 3. For that equivalent engine also one can write Q 1 by Q 3 is the same function t 1, t 3. Now this gives a further clue (Refer Slide Time: 22:44 min) to the type of this function. We may not know this function explicitly, but at least a type of the function is known the particular type how we can write this? If we see the left hand side, we can write this function of t 2, t 3, Q 2 by Q 3. is what? How to write it? I think it will be better if we write this way Function of t 1, t 2 that will be better becomes is equal to Q 1 by Q 2, means what? Q 1 by Q 2 means Q 1 by Q 3 divided byq 2 by Q 3 that means function of, anyway we can write, there are various ways we can write this, is now very simple t 2, t 3 (Refer Slide Time: 23:36 min). That means, Q 1 by Q 2 is equal to Q 1 by Q 3 divided by Q 3 by Q 2. Anyway we can write, Q 2 by Q 3 also the similar way this into this (Refer Slide Time: 23:47 min). So, if we write this way, then we see that the function should be such that the function of t 1, t 2 must come out as the quotient of the two functions one is t 1, t 3 and other is the function t 2, t 3. So, what is the alternative? So, the alternative is this, it can only happen if and only if this function can be expressed as a function of t 1 and t 2, like this, in terms of quotients (Refer Slide Time: 24:22 min). So that it cancels out, so this is very important. Where it comes from? That if this, function of t 1 t 2 is equal to function of t 1 t 3 by function of t 2 t 3 has to be true, this is only possible if this function and relationship, function of t 1 t 2, is of this form, function of t 1 by function of t 2. If it can be expressed by this, only we can get this relationship. Any of these combinations, Q 2 by Q 3 is equal to function of t 2 t 3 or Q 1 by Q 3 is equal to function of t 1 t 3, will give like this provided, this becomes a function of t 1 by function of t 2. Clear? Okay, very good. Now this function of t 1 divided by t 2 is the base, first we started with that the heat interactions by a reversible engine between the two thermal reservoirs is a function of two temperatures. Then we came to a conclusion using two reversible heat engines in series that this function will have a shape like this, function of t 1 by function of t 2, in terms of quotients. The type of function will be like these that fun

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