Lecture 28

Welcome to Entropic's lecture. In this lecture, we will discuss about the essence of quality which is about second law of thermomix and is related to the process. When we talk about second law, generally equality is Whether we talk about heat pump or heat pump, we said that if the heat pump is reversible, then its point of performance will be higher than the irreversible one. Wherever irreversibility comes, the performance gets reduced. If we talk about heat engine, then the efficiency of heat engine which is reversible will be more than its irreversible, so the equality is usually related to the second law of Mixture and equality happens only when the thing is reversible in nature. If it happens then we will try to understand it and another special equality which is related to the second length is the equality of xs, this is the xs equality which has been introduced, which we write like this, this is given by JE Closis in 185, so your four devices. When we use it, then when a complete cycle happens, then this term is that if your heat is being transferred on every particular process and if the temperature along the boundary is T, then we will equate it with every device. Go, you will get the complete integral through this cycle and saying that if you can calculate then this amount will be less or equal to zero, so naturally this is what we can also write like sum of which. That will be the differential term, so this is the entire cycle cycle, which is the part of every cycle completely, the T in it is the temperature at the boundary and the delta K is the amount of heat that is being transferred. Is through your boundary. Now if we understand how to prove it, then we call it validity, how is it valid? This expression which we have written, then for this we will use a reversible device and understand it like this. That this is your reversible in which you have this boiler this turbine this condenser and here is your pump and this boiler will get the energy heat which is from the reservoir so this is your thermal reservoir which is TR form and here You are losing heat from the condenser, but we said that this is the mount which will go to the system and your system is a piston cylinder which will do some work. After adding this double system heat, your system is done and here Here is your name of turbine, we have said that it is cyclic, it is reversible, this is heat engine, so it is called reversible, okay, now we can apply whatever we want to it, energy balance but we will apply this whole thing. We will take a combined system, so we have taken a combined system in which this is your combined system system plus which is a cyclic device, so what I have drawn and in it, I have kept this thermal reservoir which is here, it is outside, okay. So this is the black line, it is your combined system plus cycle device. Now you can apply energy balance on it, so this is the energy balance. Double system plus the reversible. This is yours. Plus, if there are changes in your combined system. And this will be the total change, this will be your del kl aa and it can also be written like this system plus del kr r minus s this is done your combined total work is ok off combined system and d s is done your change which is total energy. Which is the total energy of the system, which is the combined system, which is the combined system plus the cyclic device, so we can call it the combined system, if it is okay, then the change in total energy of the combined system, if it is okay, then this is yours. Finally, it came out dead. S = del k a my d So this is your total energy balance in a way Correct now if you consider this device as a reversible then if you consider it as reversible which is what we are assuming here then from the Kernot principle you will get the last In the lecture, we had seen that if we consider a reversible device, a cyclic device, in such a case, the ratio of heat to temperature is T, then the temperature of the system is T P. Okay, so its boundary, which is the system T P, so in such a case What is del k r by del k will be t bty and what is del k r bity will be del k bity okay now you can use this equation here del k r you will be in del k firm If you can write then we will try the same thing in the next, so your answer is t del k byty minus because what is del k aa, t del k baity, so this is what I have used, okay, so now this is what this expression is, which is derived from total energy. And we are considering reversible devices. Now we consider that the system which is at T, one T, you have a temperature at a boundary, this is a reversible cycle, this particular system, we can put P dot la here. Our system is in the intersex, you can imagine it, accept that it will go on a cyclic process. Cyclic process means whether it is expanded or later it can be compressed and come back to the same place, so we use this symbol. What we will do is that we can write it in such a way that it is a cyclic process, whatever it is, it will happen. Okay, so now what we have done in it, we will consider it as a system. You will accept that this system is a cycle. Whatever the cycle is, it is processed by this system and when this system does the work, then the cyclic device can also do it many times, so we will assume that the cyclic device is also a number of cycles, it does the work during this time, okay, so when This cycle has been done, that is, your system is also doing cyclic process, the device is also doing it, so in any cycle, what will be the change in total energy, that total combined energy will also come back to the same state. Now because of this, its value has become zero. Okay, and this is the total. We can also say that this is the total work done during this time also and since T is a constant, that thermal If it is about the reservoir, then this is what you are talking about. Okay, so this is the final conclusion of this process and of doing this exercise. Notice this WC, which is the total effective work in it. What is it doing? Pay attention, double what we said is that this is the combined total work. Now when we did this, this particular thing that we wrote about your combined system, what it is doing is it takes an energy from the surface and effectively does this work. If you pay attention then what is this process? This process happened that brother, I took this from you and gave a simple S.D.C. here. Okay, so this is the system, it violates yours which we People take Kelvin Planck's statement because what it is doing is exchanging only heat from the reservoir and completely it is doing work without throwing anything away, so what will happen through it that this WC is what it is. Can't be positive, okay ca n't be positive so one, it can't be positive because it is visible through this example that it is doing total net amount of work just by exchanging one ridge which is not possible. is and is positive, it means it cannot be positive then this lesson will be equal to zero and this is your XS equality. This is the relation. This equality is valid for all the cycles whether it is reversible or irreversible and whether it is refrigeration cycle. Why not? Okay, notice this. If we assume that the system that we have made inside it is also reversible, then the entire combined system will be reversible, so if the system is reversible, then it means that Combined system is also reversible, so if we reverse the whole system, we propose to reverse the whole combined system, that is, if we change the direction of the cycle itself, then it becomes a reverse cycle, then in the case of reverse cycle, whatever quantity we use. The magnetia of is gets reversed, negative will become positive, positive will become negative and when the reverse cycle happens, then in such a case, like in the regular case, we said that db cannot be greater than zero, similarly in the reverse case. DBC lesson cannot be zero for the reverse case. Okay, this means that if your system plus device is reversible, the system is also reversible, then the entire internal system has become reversible. The combined system inside the dotted line has become reversible. In such a case, DBC because both these conditions will be valid only when DBC is internally reversible. This is the case we are discussing. It means that it will become zero because this is your regular case and reverse it. If you give then this is the condition, it means that if your system plus device combined system is reversible then this is your condition, it means this is your condition which you had used OK del ka baati which is your closure inequality, finally this expression is your When it becomes zero, then when it becomes internally reversible, then your it becomes zero, then it means that the equality which is equality means the expression which is of classes OK, the equality which has equal sign in it should either be totally reversible. Or it is okay if it is internally reversible and whatever quality you have will be irreversible. In the cases, there will be equality among the cases. Equality will be totally reversible and intermediate. Let us understand through the example of equality of classes and how we can use it to check. To find out whether the process we have shown violates the Second Law of India or not, we have shown a process diagram of yours. This is your proper process in which there is a turbine, a boiler, a pump and so on. Condenser means it is a heat engine. Okay, now this condition has been given in it that brother, what came out of your turbine is at 90 kilo pascal in quality. After getting condensed in the condenser, at 10 it has come in quality at 15 kilo pascal. After going to the pump, mega pascal was made from this saturated liquid and from here it got boiled, it got boiled and became vapor, the vapor which went into the turbine, now this cycle which we made of the heat engine, what does it slow down, what does it mean to say second leg thermodynamics? This is whether it follows equality and inequality, does it follow this expression or not? To check this, you will have to note two things, first, in which device your heat is being transferred, then in the boiler, your heat is If you say that there is no proper, there is no loss, then it will be equal. The amount will be your delta H, that is to say, this will be your delta H. Similarly, you can say this too, this will also be delta H, between this and this and the rest is like admatic, meaning there is no heat interaction here. So, out of these four devices, only two are your devices with heat interaction, so it can be properly written that this is your Dell K, this is your boiler, this is your condenser. Okay, now I can write this Dell K like this, Boiler. The one which is the boiler is here, so this is your part integration which is because the temperature will be fixed, so we can write it like this is your / t del k plus / t is the temperature of the boiler which is the value, since This is one and one and two are going to be the same, so this is because it will be the constant temperature which will boil at this condition and t1 is that t will be set, so t1 del kps t3 integral del k is okay, so we can also write it like this. We can say that if its integral is the total amount of boiler, then this is your k which is the total which is effectively this amount which is because we can also assume because in this you can also see that this is the unit or amount which is Instead of del k, we can also write small k because the unit can also be mass, so KG is your fluid, which is working fluid which is working all around in the cycle, so we can also use KG, okay. So this can also be small K, so I will write this back in the way that it will be divided into two and D K is plus D K is 3 integral which says and this is your D K is your total amount heat, this means. a2 mine a2 a2 mine av bav plus loss hua it means h3 minus a4 but note one thing in this that is it correct to write it in this way or not? Pay attention to this, this is del k which is this sign which is not there by default in this. Okay, we are not saying how much is the loss or how much is more, it requires a sign convention, so if I have taken in this that h2 min h1, then here I have used that this is your direction, final minus. If the initials are seen in this way, then h4 and h3 h4 means h3, if your del Kadel comes out, then of course there is excess heat in your condenser, so here you can either write it as a minus or you can write it like this. h2 min a1 p t1 p h4 my a1 p t3 h4 So your negative will be like this, so you will sign it in this way, your automatic check will be done that this part is heat loss and this part is heat gain, so this expression of yours Now this is the delta H jove, which is T, your T set is coming at mega pascal and T which is T is your T set is at 15 kilo pascal. Now the main thing is that you have to If you have to find out what these values ​​are then you will have to look at the table, you will get a5 in a5, if you look at the table then you will get t set, you will get t1 = t set, you will get 160 4 95 degree Celsius, you should have this. 53.9 and h4 h1, you will also get it from table a5, h2 and h1 because it is super heated, so you will get it from 7 mega pascal table, you will get it from your steam table, so if you enter its value, then h2 - h1 will come out. It is 206 6.3 kilojoules on kg and h4 minus a1 comes out as nickel minus 98.4 kg on water kg is fine and plug it in, you get nickel on it, plug it in, enter all the values, then it will come out on If your mine is 1.87 kg of water per kg Kelvin, then it comes negative. If it comes negative, it means that this is your process, it is your right, which has been shown to you, so I have not shown you this table, if you have seen the table before then you should know. You should know how to calculate the table. First of all, in the case of boiler, pure liquid has changed from pure vapor, so anyway you can actually see it from Table A5. Look at the table of 700 Kilo Pascal and from that you will get Delta H. You will get a simple HFG and you have to use it directly. In this case, since the quality is 90, you will get this h3 which is and you will get the HFG plus x HFG and also h4 which is you will get the hps x edge. If the quality is seen in this way then h4 my aa will be your minus p8 hfg because This is fine, so in this way you can easily calculate it. I hope you have understood this, so in this lecture we will do only this much. Now from this particular equality and the class equality, we will calculate entropy in the next lecture. Till then bye bye avi