 The next topic that we are going to discuss is the principle of increase of entropy. Perhaps this is the most important concern concerning second law of thermodynamics. The principle of increase of entropy has now come to be recognized as a universal principle and for that reason it is an extremely fundamental idea that comes out of second law of thermodynamics and is of fundamental importance also. So, we will discuss this in detail. So, let us start like this. Let us say that we have a thermodynamic system at temperature T and let it have a heat interaction delta Q with the surroundings at temperature T naught. So, at an instant so, we have a system like this. So, at an instant the system is at a temperature T and it has a heat interaction with the surroundings of delta Q. It could either receive heat from the surroundings or I could reject heat to the surroundings anything is possible, but it has a heat interaction with the surroundings which are at a temperature of T naught. So, it could be like this, it could be like this. Now, the change in entropy of the system as a result of this process is d s equal to delta Q over T plus delta sigma int which is the entropy generation due to internal irreversibility in the system. So, the change in entropy of the surroundings denoted d s surroundings is given simply as d s surroundings equal to delta Q surrounding over T naught. Remember there is simply a transfer of entropy from the system to the surroundings or vice versa. The surroundings are not executing a process either the surroundings are supplying an amount of heat delta Q or the system is rejecting an amount of heat delta Q to the surroundings. So, the change in entropy of the surrounding is only due to the entropy transfer. So, it is delta Q surrounding over T naught and based on our sign convention delta Q surrounding is equal to minus delta Q sis. This point is very important there is only transfer of entropy between the system and the surroundings and the system is the one that is executing a process. So, if you combine the entropy change of the system and the surroundings we get the entropy change of the universe. So, entropy change of the universe d s universe is equal to d s s plus d s surroundings and if you rewrite this expression you get this to be equal to this. So, once again there are two terms in this expression for change in entropy of the universe the first is due to external irreversibility notice that the system is at a different temperature from the surroundings when it rejects heat or when it receives heat when it is rejecting heat of course, T is greater than T naught when it is receiving heat T is less than T naught. But whatever it is there is an external irreversibility here and there is also entropy generation due to internal irreversibility. So, the entropy of the universe increases as a result of the external irreversibility plus internal irreversibility. So, any change in the entropy of the universe of the system is due to irreversibilities. If there are no irreversibilities then obviously there is no change in the entropy of the universe otherwise there is going to be a change in the entropy of the universe. Now, let us look at the various possibilities that we may encounter in this situation. The first possibility is the simplest no internal or external irreversibility which means that in this case we set this to be 0 and we set T equal to T naught no external irreversibility. So, this is also equal to 0 both the right hand side terms are 0. So, entropy of the universe remains the same. Now, let us examine the possibility when we have internal irreversibility, but no external irreversibility. So, which means that we have internal irreversibility, but no external irreversibility since delta sigma int is always greater than 0 the entropy of the universe increases in this case. So, you must remember that delta sigma int is equal to 0 or greater than 0 since we are now saying there is an internal irreversibility this is positive. So, the entropy of the universe increases. Now, the third possibility external irreversibility no internal irreversibility. So, in this case we have external irreversibility. So, that means this is present, but no internal irreversibility. So, this is 0. However, before we can ascertain the sign of this term we need to again look at two possibilities one when heat is supplied to the system another when heat is rejected by the system. Let us look at both these situations. So, system temperature is greater than the surrounding temperature this means that the system is rejecting heat. So, that means delta q is negative and the quantity within the parenthesis is also negative. So, if I look at 1 over t minus 1 over t naught since the t is greater than t naught this is also negative. So, delta q is negative and 1 over t minus 1 over t naught is also negative hence the product term on the right hand side is positive. So, d s universe is positive. Now, the other possibility is t is less than t naught this means heat is supplied to the system. So, delta q is positive and 1 over t minus 1 over t naught in this case is actually positive. So, the product term on the right hand side is positive which means that entropy of the universe increases in this case also. So, we have the last possibility is when we have both internal irreversibility and external irreversibility in this case both the terms on the right hand side are non zero and positive hence the entropy of the universe increases. So, we have considered all possible cases and we have concluded that the entropy of the universe either remains the same or increases this is known as the principle of increase of entropy. It remains the same at best remember no internal or external irreversibility is the best possible situation which means the entropy of the universe remains the same at best or increases. So, this is the principle of increase of entropy entropy of the universe always increases or at best remains the same. Now, this statement is actually considered to be the second law of thermodynamics by many many practitioners of different branches of engineering. For mechanical engineers of course, you know the pedagogy is along the lines that we have followed. We started with the Kelvin Planck statement and then we progressed from there and we have finally, arrived at the principle of increase of entropy actually these are all equivalent. In fact, we will show that they are equivalent in a minute we showed K p and Clausius to be equivalent. So, our development followed K p statement Clausius statement absolute temperature scale efficiency of a chrono cycle and then we went to entropy Clausius inequality then entropy and then from there now we are coming to principle of increase of entropy. So, this is a sequence that mechanical engineers tend to follow because we always ask the same question. What is the maximum efficiency of the engine? What is the maximum possible efficiency of the engine? If this is not possible then what is the maximum possible? So, that is the that is the path that we have gone down and then finally, arrived at this, but people in other branches of engineering and even sciences physicists for example, considered this to be the starting statement of second law of thermodynamics. So, there is a very interesting footnote here involving something called the Maxwell's demon which was for long for about a century it was actually thought to violate the principle of increase of entropy. For example, it was enunciated in 1871 and it was not until 1987 that it was satisfactorily and convincingly resolved. So, it remained like that for about or more than a century and then it was shown eventually that the Maxwell's demon does not violate the principle of increase of entropy. There are certain other examples also where it was thought that this principle is violated, but very careful analysis showed that not to be the case this some these are mentioned in my textbook if you are interested you can look it up. Now, as I said we have mentioned Kelvin Planck statement, Clausius statement, definition of absolute temperature Q H over T H equal to Q C over T C and now we are giving principle of increase of entropy or all these the same or they all equivalent statements or is there a possibility that you know there could be differences among this. For example, some engine say does not violate Kelvin Planck statement, but violates the principle of increase of entropy and so on. Let us take a look at that Kelvin Planck statement earlier we considered a direct engine that violated that violated the Kelvin Planck statement. So, basically it took an amount of heat Q H from reservoir at T H converted all of it to work without rejecting any of it. Now, if you so let us just draw the block diagram for that for that engine. So, this was so this was the engine Q H W equal to Q H and it was also said that the engine executes a cyclic process otherwise Kelvin Planck statement would not be applicable. So, the engine operates in a cycle converts all of the heat to work that is why it violates the Kelvin Planck statement. Since, the engine operates in a cycle that we take that to be the system. So, since the system executes a cyclic process delta S is 0 because S is a property we start and end at the same state. So, delta S for the system is 0. Now, the surrounding supply a certain amount of heat Q H to the engine which means the entropy of the surroundings decreases. So, it is equal to minus Q H over T H and we may then write delta S for the universe as delta S system plus delta S surroundings which is negative and clearly this engine violates the entropy increase principle also because entropy increase principle says it can be 0 or positive. So, it cannot be negative. So, clearly this violates the principle of increase of entropy. So, an engine that violates the K P statement violates the principle of increase of entropy also. The next example is a is an engine reverse engine that violates the Clausius statement. So, basically we have high temperature reservoir and engine E which operates in a cycle I am sorry. So, we have a low temperature reservoir T C. So, this moves a certain amount of heat Q C from the low temperature reservoir and without taking any work input transfers it to the high temperature reservoir and while operating in a cycle. So, again for the system the engine delta S is 0 because it is operating in a cycle and the surroundings are supplying an amount of heat Q C from a reservoir at T C. So, that means the entropy of the surroundings decrease during this part of the process and then a certain amount of heat Q C is rejected to the surroundings by the engine during another part of the cycle which means entropy of the surroundings go up during this part of the cycle. So, that is plus Q C over at T H. So, delta S universe is delta S system plus delta S surroundings and you can see that this is equal to Q C times 1 over T H minus 1 over T C. Again this quantity since T H is greater than T C this quantity comes out to be negative which violates the principle of increase of entropy. So, any engine that violates the Clausius statement also violates the principle of increase of entropy. The last one thermodynamic temperature scale. So, here we are looking at a corner engine. So, let us just catch the PV diagram T S diagram anything is ok. So, let us say that this is a corner engine that operates between reservoirs T H and T C direct engine that operates like this. So, delta S for the system is 0 because the corner engine is executing a cyclic process. So, delta S system as usual is equal to 0 delta S surroundings a certain amount of heat Q H is supplied at temperature T H and a certain amount of heat Q C is rejected at temperature T C. So, the entropy of the surrounding decreases when heat is supplied to the system and the entropy of the surroundings increases when heat is rejected by the system. So, the change in entropy of the universe delta S universe is delta S system plus surroundings is equal to this and since there are no internal or external irreversibilities. Remember, we are talking about a corner engine with no internal or external irreversibilities the entropy change of the universe should be 0. So, Q H over T H equal to Q C over T C. So, the thermodynamic temperature scale definition of that is also consistent with the principle of increase of entropy. So, what we will do in the next lecture is to work out examples which involve calculation of entropy change and and as a result we will calculate or and from that we will calculate entropy generated in the universe as a result of the process that that we look at. So, we will do several examples and calculate entropy change for each process and then evaluate entropy change of the universe as a result of each one of these processes. So, in this lecture we will work out a few examples that demonstrate the principle of increase of entropy and we will calculate the entropy change of the system plus surroundings. Now, before we do that let us just take a final look at this expression. So, here we have shown that DS universe is made up of two terms. One is this term here which is due to external irreversibility and the other one is this term which is due to internal irreversibility. Now, notice that both these actually are entropy generation terms. So, this is also entropy generation this term denotes entropy generation due to external irreversibility and this is as we have already discussed entropy generation due to internal irreversibility in the system. So, in fact what we can actually say is that you know this itself may be interpreted as as an entropy generation in the universe. For instance, we may write it like this. So, we say that this is delta sigma int and this entire term may be written as delta sigma ext denoting external irreversibilities and instead of writing it as DS universe we may actually interpret this and write this as delta sigma which is the total entropy generated in the universe due to internal irreversibility in a system and external irreversibility due to its interactions with the surroundings. Notice that the external irreversibility is coming from the I mean it is coming from the interaction of the system with the surroundings. So, it is external to the system but it is a consequence of the interaction of the system with its surroundings. So, in that sense it is connected to the system it is external to the system but it is a consequence of the interactions of the system with the surroundings. So, we may actually write this term as delta sigma which essentially says that this is entropy generated in the universe as a result of the process that the system undergoes. So, the system undergoes a certain process and there is a certain amount of entropy generated due to the internal irreversibility and there is a certain amount of entropy generated due to external irreversibility. These two together contribute to the change in universe of the system or alternatively we may say that this is the amount of entropy that is generated in the universe. So, in fact, this is the notation that we will use in our examples. We will say that sigma is equal to so much which means that this is entropy change in the universe or entropy generation in the universe. That is what we are going to do and this is actually a let us say an interesting and quite powerful way of interpreting this particular expression here. Since both these are entropy generation terms the entropy change of the universe is actually nothing but entropy generated in the universe due to internal and external irreversibilities. That is how we will interpret this. The first example that we are going to look at reads like this. So, we have a steel casting specific heat is given and the mass is 20 kilogram at 200 degree Celsius. So, it is to be cool to the room temperature which is 27 degree Celsius. We do this in different stages. So, initially the casting at 200 degree Celsius is kept in a furnace that is maintained at 140 degree Celsius. So, once it is the temperature of the casting reaches 140 degree Celsius we take it out and put it in a water bath. This is not an infinite water bath. This is a water bath with a finite mass 80 kilograms of water which is initially at the room temperature. So, we put the casting at 140 degree Celsius in the water bath and then allow it to cool down. Since the water bath is of a finite mass the final temperature there will be an increase in temperature of the water bath as the casting cools down. So, it cools down to a certain temperature. Once it is at that temperature we then take it out and allow it to cool in the ambient air which is at 27 degree Celsius. We are asked to calculate the entropy generated during this process. So, that would be sigma entropy generated due to internal plus external irreversibilities and the system here is the casting. So, we take the casting as the system and calculate its entropy change and and treat the rest of the the universe as the surrounding. So, initially the part of the universe that interacts with the casting is the furnace. Then the part of the universe that interacts with the system is the water bath. Then it is the ambient. So, we are doing this in 3 stages. So, step 1 delta S system for step 1 may be calculated quite easily the casting cools from 200 degree Celsius to 140 degree Celsius and because it is losing heat the entropy of the system decreases. Now, from first law we can easily calculate the amount of heat that is lost by the system that is nothing but mass of the system times its specific heat capacity times change in temperature. So, 600 kilojoules of heat is transferred to the surrounding. So, the entropy of the surrounding increases. So, the surroundings are at a temperature of 140 degree Celsius. In fact, the furnace is at a temperature of 140 degree Celsius and 600 kilojoules of heat is transferred to the furnace. So, that means entropy changes the surroundings is positive in this step. So, in the next step we take the the casting and put it in the water bath. The final temperature of the bath plus the system may be determined by using first law. So, if you apply first law to the bath plus the casting delta E equal to delta U no change in K E or P E is equal to Q minus W. W is 0 there is no displacement or any other form of work. So, we can easily calculate and delta U is nothing but delta U casting plus delta U of the bath. Since delta U is M times C times delta T because this is a casting is a solid the bath which is water is actually in liquid. So, we can easily calculate the final temperature of the bath to be 30.266 degree Celsius. So, the bath initially was at 27 degree Celsius as a result of putting the casting which is at a temperature of 140 degree Celsius into the bath its temperature increases which means that the casting now is cooled from 140 to to 30.266 and the bath temperature of the bath increases from 27 to 30.266. So, delta S for the casting may be evaluated based on its initial temperature which is 140 degree Celsius and the final temperature which is 30.266. And the surroundings in this case is the bath. So, the change in entropy of the surroundings may be evaluated based on its initial temperature and final temperature. Since heat is transferred to the surroundings the entropy change in the system is negative and since the surroundings receive heat the entropy change of the surroundings is positive. Step 3 the casting is now taken out of the water bath and allowed to cool in the ambient. So, its temperature changes from 30.266 to 266 to 27 degree Celsius. Again heat is lost to the ambient. So, the entropy of the casting decreases. And the amount of heat that is lost to the surroundings may be easily calculated by applying first law. So, the Q comes out to be equal to delta U which is nothing but M times C times C of the casting times delta T. So, this much heat is transferred to the surroundings 32.66 kilo joules. So, the entropy change of the surroundings may be evaluated like this. Since the temperature of the surroundings remains a constant we may evaluate this simply as Q over T naught. Now the total entropy generated during the process may be evaluated quite easily as total entropy change for the system plus total entropy change for the surroundings. So, we add them up algebraically and we end up with a positive number 0.6469 as we showed because entropy of the universe increases or at best remains the same. So, sigma should always be positive and we get this to be 0.6469. So, this is actually the entropy change of the universe or alternatively the entropy that is generated in the universe as a result of this quenching process. The next example is something that we have looked at earlier. You may recall that 2 kg of air that was contained in the piston cylinder assembly was compressed from 100 kPa to 800 kPa is initially at a temperature of 300 Kelvin and it underwent a polytropic process exponent was also given. So, we are now asked to calculate the amount of entropy generated in the universe as a result of this polytropic process. So, we considered the air to be the system and we will continue to do that. The final temperature of the air was determined to be 467 Kelvin and the heat that was transferred to the ambient was minus 116.06 Kelvin I am sorry 116.06 kilojoules. So, taking the air as the system we may evaluate delta S for the system using this formula since the final temperatures and final temperature and final pressure are both known we may evaluate the change in entropy of the system. Since heat is lost by the system its entropy decreases as seen from this negative sign. The surroundings comprise of the ambient at 300 Kelvin. So, the entropy change of the surroundings may be evaluated as Q surrounding divided by T naught and this comes out to be plus 386.87 joule per Kelvin because heat is transferred to the surroundings its entropy increases. So, the entropy change of the universe delta S universe is nothing but entropy that is generated which is equal to delta S cis plus delta S surroundings and that comes out to be plus 80.56 joule per Kelvin. The last example that we will look at in this on this topic is this 5 kg of saturated R134A vapor minus 15 degree Celsius is contained in a rigid vessel. The R134A is stirred by transferring an amount of work equal to 500 kilojoules. So, basically we have R134A in a rigid vessel and we have a stirrer and it is initially at minus 15 degree Celsius saturated vapor and we transfer an amount of work 500 kilojoules by stirring and simultaneously heat transferred to the ambient also takes place. So, the ambient is at 30 degree Celsius it is also losing heat to the ambient as stirring process takes place and then initially the R134A vapor is at minus 15 degree Celsius. So, it actually is at a temperature less than the ambient temperature but as we stir it its temperature increases stirring is highly irreversible process its temperature increases. So, there will be heat loss from the R134A to the surrounding and we take this to be our system. So, the initial state it is given that it is a saturated vapor at minus 15 degree Celsius. So, we may easily retrieve property values from the temperature table specific volume specific insulin energy and specific entropy. At the final state again two property values are known temperature is given to be 30 degree Celsius since the volume of the vessel is constant specific volume also remains constant. So, V2 equal to V1 equal to 0.12066 and temperature is also known. So, and we can easily show that this is a superheated state by noticing that or by noting that V2 is greater than Vg at 30 degree Celsius. So, we can retrieve specific insulin energy and specific entropy from the superheated tables. Now, we have already shown that we have taken the R134A in the vessel as our system. If you apply first law we get delta E equal to delta U equal to Q minus W no change in K or P E. So, Q may be evaluated like this notice that W here is stellar work PDV work is absent because the volume is constant and 500 kilojoules is supplied which means W is minus 500. So, we get the heat loss from the system to the surroundings as minus 332.45 kilojoules. Entropy change in the system may be evaluated as M times the change in the specific entropy. So, that comes out to be plus 0.599 kilojoule per Kelvin. Notice that although heat is being lost by the system to the surroundings its entropy still increases because of the internal irreversibility which is stirring. So, the internal irreversibility is so high that the entropy increases in spite of the fact that heat is lost to the ambient. So, the entropy generation due to internal irreversibility is very high. So, delta S system comes out to be positive in this case and delta S surroundings may be easily evaluated as the heat that is transferred to the surroundings. Since heat is transferred to the surroundings the entropy of the surroundings increases. So, it is Q divided by T naught in Kelvin. So, 273 plus 30. So, the total change in entropy of the universe delta S universe is same as entropy generated in the universe and that comes out to be a positive number 1.6962. So, heat is lost to the surroundings as a result of which the temperature of the surroundings increases and its entropy I am sorry heat is lost to the surroundings as a result of which the entropy of the surroundings increases. Although heat is lost to the surroundings from the system its entropy still goes up because of the internal irreversibility associated with the stirring work. So, the entropy of the system increases, entropy of the surroundings also increases. So, we have a very special process here where the system temperature the entropy of the system increases and the entropy of the surroundings also increases and so we have total entropy of the universe increasing as a result of both this. This is somewhat unusual because when a system is losing heat to the surroundings usually the entropy of the system will go down as we would expect. You recall dS is delta Q dS for a system is delta Q over T plus delta sigma int. So, when delta Q is negative we usually expect dS to go down unless delta sigma int is so high and positive that the overall entropy increases which is what is happening in this case. So, this is very high and positive this is less than 0 but the sum eventually comes out to be greater than 0.