 Good morning. So, let us continue with the second law of thermodynamics, problem number 9. A heat pump heats a room in winter and acts as a refrigerator to cool it in summer. The room needs to be maintained at a temperature of 20 degrees centigrade in winter and 25 degrees centigrade in summer. Ok. In the required heat transfers from the room during summer and to the room during winter are the same. It is equal to 48000 kilojoule per hour. That is in the summer from the room 48000 kilojoule per hour has to be extracted and it is what is transferred to the ambient. In the winter from the ambient to the room 48000 kilojoule per hour has to be transferred. So, this is the given information. Calculate the minimum power required to drive the heat pump. If the ambient temperature is 0 degrees during winter, ok, then B, the actual power required to run the heat pump if the COP is 10. The for the same power as input in part A, that is when a reversible or see when minimum power required by heat pump that means the heat pump has to be reversible. So, if the heat pump is reversible, it has some power input, correct. So, with the same power input what is the maximum power reversible ambient temperature during the summer. So, this is these are the quantities. So, let us take winter. The room has to be maintained at 20 degrees centigrade. That is 293 Kelvin. This is heat pump which will act as a refrigerator now, refrigerator. So, the winter maintained at 20 degrees. So, this we have to transfer heat. Now, the ambient temperature in winter is given as 0 degrees, 273 Kelvin. The heat pump, ok, this is heat pump. So, now, this has to get some work from the ambient, ok. And we know that in the winter, in order to maintain the room at 20 degrees, 48000 kilo joule per hour, ok. So, in winter it will act as a heat pump and in the summer it will act as a refrigerator. So, let us first talk about winter. So, heat pump. Heat has to reject 48000 kilo joule per hour in order to maintain the temperature of the room at 20 degrees centigrade. And this heat, it has to pump using the heat which is received by the from the ambient, ok. So, this is the scenario for the winter, ok. Now, for minimum, first question is the minimum power required, minimum power required for this. So, I will say W minimum, ok. For this, heat pump should be reversible, ok. That is done. Now, in that case, COP of reversible heat pump can be calculated as T H divided by T H minus T C is in Kelvin. So, what is T H? 293 divided by 293 minus 273. So, that will be the COP that is 14.65, ok. So, that means, if HP, the heat pump is reversible, then it will work as a car not heat pump. So, the COP is calculated by the temperatures of the source and sink itself. So, from that I can calculate the COP. Now, what is Q H? Q H is given as 48000 kilo joule per hour, which is equal to 48000 divided by 3600 kilo joule per second or kilowatt, ok. Now, what is W then? W, I will say W minimum will be equal to, what is the definition of this? COP is actually Q H the heat transfer which is required divided by the, the heat transfer which is going to cost, correct. So, this, so that means, what will be W minimum? W minimum will be equal to Q H divided by COP reversible refrigerator or heat pump, ok. So, this is the value. So, what is W minimum here? From this we can calculate W minimum, W minimum equal to 0.9101 kilowatts. So, that is the first part, ok. So, first part we can see that the temperatures are fixed for the ambient as 0 degrees and at the room in the winter it should be 20 degrees. And we know how much heat has to be transferred to the room in order to maintain the room at 20 degrees, because ambient is at 0 degrees. So, we have to heat the room. So, the rate at which the room has to be heated is also given 48000 kilo joule per hour. So, now, for this the minimum power required can be calculated if the heat pump is considered as reversible or Carnot heat pump. So, the COP of the Carnot heat pump is T H by T H minus T C. So, which is 14.65. Q H is known as per the definition COP is Q H by W min. So, W min can be calculated as Q H by COP which is 0.9101 kilowatts that is part A. Part B, if COP of the heat pump is actual is 10 which is given, 10 same load Q H is same correct. So, we can see that COP HP equal to Q H by W now, W HP I will say that is the actual work required. So, what will be that? COP is 10 now. So, we can say W HP will be equal to 48000 divided by 3600 that is kilowatts divided by COP of the actual heat pump that is 10. So, which implies the work requirement for the actual heat pump with the COP of 10 will be equal to 1.334 kilowatts that is higher than this ok. This is part B. So, part C now is ok. Now, in the first part the power given to the heat pump is 0.9101. In the summer this heat pump will work as differentiator. So, we will draw that. So, here in summer ambient temperature of the summer. So, let THS is the ambient temperature during summer say 35 degrees, 40 degrees centigrade and so on. Now, room has to be maintained. Now, this is a refrigerator or ok. Now, this the room, room temperature is 25 degrees centigrade ok. So, room temperature is 25 degrees centigrade or 298 Kelvin ok. So, we have to pump heat from this. This is colder now room has to be colder. So, pump heat from this that will be 48000 same as that of the ok the objective heat transfer is same. For the heat pump it is the heat transfer to the room. For the refrigerator it is heat extracted from the room and now this rejects some heat. So, I will say this is QH ok ok. Now, this is unknown temperature T, H has a unknown temperature. But what I give as input is work given to the refrigerator is 0.9101 kilowatts 0.9101 kilowatts that is the work minimum work which we supplied to the heat pump during the winter operation that is 0.9101. So, the same is given here. Now, we have to find what is the temperature here correct. So, in this case what is the COP of the refrigerator which will be equal to now QC divided by this is QC divided by W. So, which is equal to 48000 divided by 3600 for converting kilojoule power to kilojoule per second or kilowatts divided by 0.0109101 that is the COP. So, this should be what? This will be same as 14.65 because the heat load is same ok this is same. Now, how will you calculate THS? THS basically if you see the problem for the same power what is the maximum permissible ambient temperature when we can reject heat to maximum temperature in the. So, in this case if this THS if THS has to be maximum ok then for the given for the given work input then refrigerator should be reversible ok. So, in that context I can say COP of the refrigerator can be written in terms of the temperatures that is T C divided by T H minus THS minus T C. So, this is T C ok. So, we can substitute it 298 divided by THS minus 298. So, that will imply the value of T this is known 14.65. So, that is THS will come out as 318.34 Kelvin that is it ok. So, so in this problem this is done in most of the regions where heat pump is used in the summer it will act as a refrigerator in the winter it acts as a heat pump ok. The objective is to keep the temperature of the room hotter than the ambient in the winter and colder than the ambient in the summer. So, this is what the illustration is. Next problem 10 to 1. Consider two heat engines X and Y. So, let us say X. Engine X the heat received by engine X is 4 times the engine received by the heat received by the engine Y ok. So, the heat received by engine X is 4 times the heat received by engine Y ok. Let us say this is some T H it receives I will say Q H X W X Q C X T C. Let us keep like this. Now, similarly Y T H Q H Y W Y Q C Y T C ok. Now, first what is told here is Q the heat received by the engine X Q H X is 4 times the 4 times the heat received by the engine Y Q H Y. So, this is the first one. The heat rejected by engine X that is Q C X equal to 7 times the heat rejected by engine Y. If the work done by the engine X is work done by the engine X is double the work done by the engine Y ok. These are the conditions given. Determine the efficiencies of engine X and engine Y first then B. If engine Y is a reversible engine, engine Y is reversible engine and it rejects heat to a cold reservoir at 260 Kelvin. Determine the hot temperature reservoir what is this value from which it receives heat. Then C. Determine the COP of COP when engine Y is reversed and operated as a refrigerator between the cold reservoir at 260 and hot Kelvin and hot reservoir at 37 degree centigrade. So, these are the required things. So, first let us do this. So, Q H X is equal to 4 Q H Y, Q C X is equal to 7 Q C Y, W X equal to 2 W Y ok. 4 Q H Y minus 7 Q C Y equal to 2 W Y. So, let us say this is equation 3. This is equation ok. Now, I will see that I will also write Q H X minus Q C X equal to W X. So, this is 1 then Q H Y minus Q C Y equal to W Y this will be 2 ok. Now, I have three equations I just formed this. So, this is written in this way. So, now I can say 7 into 2 minus 3 will give 3 Q H Y equal to 5 W Y or efficiency of Y will be equal to what? W Y by Q H Y which is equal to 3 by 5 equal to 0.6 ok. So, with the given data. So, what you have done is I have run done first equation is the first law. First law apply to X and apply to Y. Q H X minus Q C X equal to W X, Q H Y minus Q C Y equal to W Y. Now, since Q H X is written as 4 Q H Y I can write that for this I will write like this. Similarly, Q C X equal to 7 Q C Y, W X equal to 2 W Y. So, I write this. So, from this I have two equations 2 and 3 I can use to solve, eliminate one of the variables like in this case I have eliminated by multiplying 2 by 7 and subtracting 2 3 I have eliminated Q C Y. So, I got a relationship between Q H Y and W Y and I want the ratio of W Y by Q H Y because that is the efficiency of the engine Y that has come out as 0.6. So, that is the first part. Second part is B ok A determine the efficiency of X. So, X also I have to do now. X is efficiency of X equal to W X divided by Q H X equal to what? W X is 2 W Y divided by 4 Q X Q H Y. So, which is equal to half into 0.6 equal to 0.3 that is the efficiency of the X because from the previous I know W Y by Q H Q H Y equal to 0.6. Now, I substitute I substitute half 1 2 by 4 is 1 by 2 into this ratio is half 0.6. So, I can get 0.3 that is the efficiency of this. So, that is the part A. Then Y is reversible ok. So, that means efficiency of Y can be written as Q C by Q H Y which can be written as T C in Kelvin divided by T H in Kelvin. So, which is written as 1 minus 298 as a 260, 260 divided by T H that is what I have to find. So, go back here T C is given as 260 and T H value is asked. If engine way is a reversible engine and it rejects heat to a cold reservoir at 260, determine the hot reservoir temperature that is T H. So, now, I know the efficiency is 0.6. So, I can say 0.6 equal to 1 minus 260 divided by T H which implies T H will be equal to 650 Kelvin ok. So, that is the second part. Then third part C what is the third part? Determine the COP when engine Y is reversed. So, for example, now it will receive heat from the T C and reject heat to T H and receive work from environment. So, it is reversed. So, reversible engine can act as a reversible reversed and act as a refrigerator. Now, the temperatures are given 260 cold reservoir, hot reservoir is 37 degrees centigrade ok. Now, we can draw this. This is 37 degrees centigrade and I will say Q H, this is refrigerator, refrigerator Y I will say. Now, this is Q C, this is 260 Kelvin and now W is taken in. Now, what is asked is what will be the COP when reversible engine is reversed when Y is reversed to operate as a refrigerator ok. Then COP will be equal to what? T C divided by T H minus T C because it is reversed. So, reversible correct. So, that will be equal to 260 T C divided by T H is 37. So, 37 means what? 273 plus 37 equal to 310 Kelvin. So, this is 310 minus 260. So, this will be equal to 5.2. 5.2 is the COP that is it. That is what is asked. Ok. So, combination of engines and from one engine, we can attach the rates of heat transfers and find the values of work efficiencies etcetera. So, this is about the problem number 10. So, in all these cases, you can see that the reservoirs are at certain temperatures and the efficiency or COP can be calculated with respect to temperatures only when the engine or heat pump or refrigerator is reversible. That is the problem number 10. Problem number 11. A rigid vessel contains 2 kg of air initially at 100 kilopascals and 300 Kelvin. Ok. Let us draw this. A rigid vessel 2 kg air P1 equal to 100 kilopascals T1 equal to 300 Kelvin. The air is stirred until the temperature becomes 500 Kelvin. Determine the stirring work required assuming there is no heat loss. Ok. So, that means I put a stirrer like this and stir. So, work transfer I am doing so that the temperature rises to 500 Kelvin. Now, we can assume that in this case, the vessel is insulated. So, there is no heat loss. So, Q equal to 0. That is the first situation. Now, what is the stirring work required? That is the first part. Second part is if the same final state, that is I have to increase the temperature from 300 Kelvin to 500 Kelvin, but this final state is to be attained in a reversible manner. Determine the work required. That means what I have to do? I have to take this. It is 2 kg air. This is A. This is B. 2 kg air. And here initial pressure is 100 kilopascals. Temperature is 300 Kelvin, but I have to increase the temperature to 500 Kelvin. What I will do? Reversible manner. That means I will put a heat pump and supply heat to this. The heat pump will receive some work from the ambient. So, this is the work we are asking here. Correct? What is the work required? Now, obviously, ambient. From ambient only, I have to take heat. This ambient is 300 Kelvin, let us say. And when you do this, the temperature increases from 300 to 500. So, what is this work? What is the steering work? So, I will say here the steering work is WS. And this is the reversible work that is W. So, how to calculate this? There are two situations. So, let us say the case 1. Ok, Q minus W equal to delta U. Ok, now what is Q? Q is 0 because it is given. Assuming there is no heat loss. So, that means W equal to minus delta U equal to U1 minus U2 equal to M into Cv into T1 minus T2. That is the work required. It is negative obviously, because of the fact that work is given from the surroundings to the system. So, what is M2 kg? What is Cv? Cv of air. So, let us say R of air is 287. Or let us assume Cv itself directly. There is no problem. Cv, let us take as 0.7207 kilojoule per kg Kelvin. So, let us say that Cv is available with me. Then I can get the value of W equal to 2 into 0.7207 into, what is this? T1 300 minus 500. So, this will be equal to minus 288.28 kilojoules. So, that much work if I do, stirring work if I do, temperature will increase from 300 to 500 Kelvin in the rigid vessel, which is insulated. Second case, what I am trying to do here is, instead of doing stirring, I employ a heat pump. This heat pump is assumed to be reversible, ok, because I want to find the minimum work required for this heat pump. So, let us say assume this heat pump to be reversible. And now ambient temperature is 300. Initially, the air is also at 300 Kelvin. Now, when I take heat from the ambient to the heat pump and supply it to the rigid vessel, then its temperature increases slowly. Ambient is a reservoir, so its temperature remains constant, correct? So, let us say at a small time interval, I will supply a heat to the vessel, which is delta Q H. And let T H be the temperature of the vessel at any time instant. Similarly, del Q C is the heat I remove from the ambient. Ambient temperature is always at 300 Kelvin. And during this period, del W is supplied to the engine, ok. So, I can say a reversible heat pump is used. So, B used to heat or supply heat to the vessel containing air at 300 Kelvin, 100 kPa, ok, comma by extracting heat from ambient at 300 Kelvin. Now, it requires minimum work W to accomplish this, because it is reversible, correct? So, that work only we will compare now, we will try to get, ok. So, since the heat pump is reversible, I can say Q H by Q C equal to T H by T C or I can say here del Q C, ok. Del Q H by T H equal to del Q C by T C. I can write this, correct? So, ratio of heat transfers is equal to ratio of the temperatures in Kelvin. So, that is what we are trying to write here. So, from this I can write this equation. Del Q H by T H equal to del Q H by T C. Now, T C is say I will say T naught ambient temperature. So, I will say T naught T C equal to T naught constant ambient temperature. Now, you have to understand that T H is not a constant, ok. T H, this is constant. T H is not a constant, because it varies from 300 to 500 Kelvin, ok. Now, first law for air as a system, air in the vessel as a system, we can say d U equal to del Q minus del W. Since I can assume kinetic energy changes and potential energy changes are 0 and d E equal to d U now, ok. But rigid vessel constant volume, ok. So, that implies del W, which is equal to P d V equal to 0. We have now that form of work here, no stirring, etcetera, correct? Just we have to heat this. So, there is no other form of work. This place of work is also 0, because there is no change in the volume. This means del Q will be equal to d U which is equal to m C V d T, ok. So, now I can specifically write as what? Del Q H equal to m mass of air into C V into d T H, ok. Note that d T H is positive. Why? Because it is increasing, temperature increases from 300 to 500 Kelvin. Now, we can write del Q C by T naught equal to del Q H by T H which is equal to. Now, instead of del Q H, I can write this, correct? So, m C V d T H by T H. So, that means del Q C will be written as T naught into m into C V into d T H by T H. Now, this I can integrate, integrate between initial and final temperatures of air, ok. So, Q C equal to integral T 1 to T 2 del Q C equal to T naught m C V integral T 1 to T 2, ok. This is the temperature, initial temperature and final temperature d T H by T H. So, which is equal to m C V T naught natural logarithm of T 2 by T 1, ok. Now, I can calculate, how will I calculate this now? Q C is calculated now because m is known, it is given in the problem, ok, 2 kg. So, m is known, C V from this we have already taken C V as 0.7207, T 1 is 300 Kelvin and T 2 is 500 Kelvin. So, that I know, so I can calculate this, ok. What is W? W equal to Q H minus Q C. So, how can I get Q H? Q H del Q H I can write as m C V d T H. That means, Q H can be written as m C V T 2 minus T 1, ok. T 1 equal to T naught. Note that T 1 equal to T naught equal to 300 Kelvin. So, that we can use. So, now I can write W as m C V T naught into T 2 divided by T naught minus 1 minus natural logarithm of T 2 by T 1. So, this is the expression. So, you can substitute and see. So, T naught T 2, T naught, T 2 by T naught into T naught will be T 2 and minus T naught. So, that will be this. So, this is the Q H and the Q C is m C V T naught into natural logarithm of T 2 by T 1. When I substitute this, I will say this will be 0.7207 into T naught is 300 into 500. T 2 is 500 Kelvin divided by 300 minus 1 minus logarithm of 500 by 300 which is equal to 67.4 kilojoules, ok. Now, I get 67.4 kilojoules, but previously when I want to heat this from temperature as well as from 300 to 500 by stirring, about 288.28 kilojoules has to be supplied. Now, if I do this using a reversible heat pump, I only supply 64.67.4 kilojoules, 67.4 kilojoules. So, see that tremendous reduction in the work from 288 to 67 kilojoules has been accomplished. So, that is a better way to do. Even, see this is a reversible heat pump. Even if the reversible heat pump is replaced by actual heat pump, then also we can see that the COP if the COP is not so low, then we will supply only a smaller amount of work to accomplish this heat. So, that is when you want to heat the room, better to use a heat pump than a direct, for example, electrical heater etcetera. So, this is the problem, love and problem.