 So, we will do tutorial number 7 now. This is on second law of thermodynamics. The first problem, a Carnot engine operates between two reservoirs at temperatures of T H Kelvin and T C Kelvin. Okay, so let us draw this. There is a reservoir at T H Kelvin and another one is at T C Kelvin, a Carnot engine. That is reversible engine. Reversible engine operates between these two. The work output of the engine, so actually it gets some heat, so let us say it is Q H and rejects some heat, that is Q C and does some work, that is W. Now the work output of the engine is 0.6 times the heat rejected. So we can say that the work output is 0.6 times the heat rejected is Q C. So per cycle, so let us say Q C kilo joules per cycle. Now given that the difference between the temperature, between the difference in temperature between the source and sink, that is T H, source is T H, high temperature reservoir and the T C is the sink temperature. So these are reservoirs basically. So reservoirs means the temperature will be constant. If you extract heat normally from a finite temperature body, that is finite temperature material, then the temperature will decrease. When you reject heat to a say a better material block, then the temperature will increase. But when you say reservoir, then the temperatures will not change. It has huge mass or thermal capacity or we can say that MC, that will be very huge. So for example, a furnace where continuous burning of a fuel and air takes place, the temperature will be always at a very high value, say 1500 Kelvin, something like that. Similarly, the ambient is at say 300 Kelvin, that will be the reservoir at a lower temperature. So the temperature difference between these two reservoirs, T H minus T C is given as 200 degree centigrade. So calculate the source temperature, T H, sink temperature and the thermal efficiency. And this is the Carnot engine. So that is the engine is reversible. Then this means the efficiency, thermal efficiency can be written as 1 minus Q C by Q H equal to 1 minus T C by T H, where both T C and T H are in Kelvin. So this is because the engine is reversible. Now, W from the first law, first law, W sigma W equal to sigma Q for the cycle. Because the Carnot engine operates in a thermodynamic cycle. So what is the W here? W is the heat which is developed by the engine that will be equal to net Q, that is, Q supplied is positive plus minus Q C, which is the Q rejected. So Q H minus Q C. Now W is 0.6 times Q C which is given in the problem. So we can say 1.6 Q C equal to Q H or Q C by Q H equal to Q C by Q H equal to 1 by 1.6. Okay, now from this, basically I can find the thermal efficiency. To find the temperatures we can say Q H by Q C equal to T H by T C for a reversible engine. So this is, we know this, where again we have to stress thus the temperature should be in Kelvin or else it will not, it is not correct. Okay, so temperature should be in Kelvin. So in this case you can say T C by T H equal to 1 by 1.6. Do you understand? Then this is one equation. Second equation is already given that is T H minus T C equal to 200. So combining these two, I can find T H as 533.33 Kelvin and T C as 333.33 Kelvin. So using these two equations. So using equations 1 and 2, we can get T H and T C values. Now what is thermal efficiency? It will be equal to 1 minus 333.33 divided by 533.33 equal to 0.375. So that is the current efficiency. So a simple problem, there are important concepts here. First one is the first law, sigma W equal to sigma Q. So that will give you this relationship. And for the reversible engine, I can write the ratio of heat transfers to equal to the ratio of the temperatures, corresponding temperatures of the reservoirs where the heat is exchanged. But this temperature should be in Kelvin. Okay, so that is this first problem. The heat engine is used to drive a heat pump. So let us draw a heat engine and the heat pump, HP. A heat engine is used to drive a heat pump. That means the work done by the heat engine is supplied to the heat pump. That is called driving it. So like for example, turbine drives a compressor. The heat transfers from the heat engine and from the heat pump. Okay, from the heat engine. So for example, heat rejected by the heat engine, I will say Q C E. Similarly, the heat rejected by the heat pump, that will be Q, I will say P H. Pumpers, I will say P and H is the heat rejected because this, when the heat is rejected by the pump, this should be rejected to a higher temperature reservoir. So these two are, the, these two heat transfers are used to heat water circulating to radiators of the building. So it is a radiators, radiator of the building. Okay, now efficiency of the heat engine. That means I have one reservoir at say T H and one cold temperature space, from which heat engine takes heat. I will say heat pump takes the heat. So I will say Q PC and this I will say Q H engine. That is Q H is taken by the engine. Q P, P is the pump which takes the heat, Q PC. So some nomenclature we should give. Now efficiency of the engine is given as 0.29 or 29 percentage and COP of the pump, heat pump, heat pump is given as 3.5. Evaluate the ratio of heat transfer to the radiator, water to the heat transfer to the engine. That is what ratio I want, ratio of heat transfer to the radiator. That is Q CE plus Q PH divided by, this is required ratio. Required ratio will be equal to this, divided by the heat transfer to the engine, heat engine. That is Q HE, that is what I want. Do you understand? So this problem, let us do this. So apply the first law to the engine sigma, Q equal to sigma W. So we can say Q H engine minus Q C engine equal to W. Now W for the pump, for the pump, I can say minus W equal to, because pump receives the work. So heat added is Q PC minus Q PH. So R can say Q PH minus Q PC equal to W. So these are equal, these two are equal. So now efficiency of the engine equal to 1 minus Q CE divided by Q HE equal to 0.29, given. So that implies Q CE divided by Q HE equal to 1 minus 0.29 equal to 0.71. So this is equation 3. Do you understand? Similarly, please understand that the engine or the heat pump need not be reversible, because nowhere in the problem it is given that they are reversible, correct? It is say it is given that an heat engine, a heat engine and the heat pump, that is it. So that means we should not use the temperature ratio here. Obviously, the temperature ratio also is not, cannot be used because it is not given. Then COP of the heat pump, by definition will be what? For the heat pump, the objective heat transfer is the heat which is rejected by the heat pump. So Q PH divided by, for the heat or energy transfer which costs is W. So that will be the, so we can write this as Q PH divided by W, I can write Q here, HE minus Q CE, but this is given as 3.5. So that means I can say, this is say equation 4. Now there are 4 equations. What we require is this ratio, okay? So I will try to solve this. Now we can say that from COP of heat pump which is equal to, we can write this as Q H Q PH divided by Q H, again I will say or in the terms of this itself we will write first, because we can eliminate pH now. So divided by Q C, Q C is sorry, Q PC, okay, that also we can write. So from this I can write as 1 minus, sorry, 1 divided by, so divide this, so 1 minus Q PC by Q PH. So this is equal to 3.5. That means 1 minus Q PC divided by Q PH will be equal to 1 by 3.5. Or I can say Q PC divided by Q PH will be equal to 0.7143. I can now eliminate this. I want Q PC. So I can write Q PH in terms of Q. So this implies Q PH can be written as Q PC divided by 0.7143, okay. So this is the, this is the relationship between these two heat transfers related to the pump. Now the required ratio, first of all I have to find the Q ratio between the Q PH and Q HE. So we can combine this basically. So 0.29 Q H, that is Q HE equal to, that is the work, correct. See, this is thermal efficiency into work. So the work, now I connect this to work equal to 0.29 thermal efficiency into the heat received by the engine. That will be equal to Q PH minus Q PC, okay. Now this means, this implies Q 0.29 Q HE divided by Q PH, sorry, yeah, PH equal to 1 minus Q PC divided by Q PH, correct. So that, so this is, I am just taking this to the left hand side and dividing both side by Q PH. So now this, from this I get the relationship between Q PH divided by Q HE equal to 1.015. So that means the required ratio will be what? Required ratio is required ratio equal to Q PH plus Q HC divided by Q HE. So that means it will be equal to 1 point because this, the first ratio 1.015 plus second ratio is Q HC by Q HE, correct. That will be what? That is 0.71. That is 1 minus thermal efficiency. So the required ratio will be 1.725. So this problem we can see that there are two engines which are not actually reversible. We want to calculate, but the rejected heat from both heat engine and heat pump are given to radiator. So that total heat divided by the heat received by the engine, that is basically is asked. So this is the procedure. So here first law is applied, then definitions for efficiency and COP are applied since these values are given and we have got the connecting the heat transfer between the engine and the pump. We have got the answer. Third one, a reversible engine operates between two systems at constant temperatures of 800 degrees centigrade and 40 degrees centigrade. So constant temperatures. A reversible heat engine is also given. So what is the first one? 800 degrees centigrade. So I will say 800 plus 273. So 800 plus 273 because writing in Kelvin is important. So this is in Kelvin and this is 40 plus 273. That is 313. A reversible heat engine acts, acts like runs between these two. So receiving some heat and rejecting some heat. Now it also does some work. The engine drives a reversible refrigerator which operates between systems at constant temperature of 40 degrees. So 40 degrees is 313 Kelvin. So there is a reversible refrigerator. So let us say R and so this takes, refrigerator actually takes some, say I will say Q, C, R. Here it is Q, C, E. This is Q, H, E and W. So this is, this is running this. Now in this case, unlike the previous case, both are given as reversible, reversible heat engine and reversible. And now it rejects to 40 degrees. Sorry, this is actually 40 degrees. So it should reject here. It should reject here, Q, R, H. And this is 5 degrees. That is 278 Kelvin. Now it receives heat here, Q, R, C. And this is the W which is given here. But not all the work is given. So that is what is given here. See the heat transferred to the engine, this value is given as 2500 kilowatts. And 30 percent of the work output of the engine is supplied to the refrigerator. So that means I can say here 0.3 W. And 0.7 W comes out of the engine for some other purpose. Only 30 percent is supplied to the refrigerator here. So that you have to understand. If the heat transfer from the refrigerator is used to heat water in the radiator, heat transfer from the refrigerator. That is this. Heat transfer from the refrigerator is Q, R, H. It is used to heat water in the radiator, evaluate the flow rate of radiator water. So that means here there is a radiator connector. What is the flow rate of radiator water? If the temperature of the radiator water increases only by 20 degrees and specific heat of the water is 4.2 kJ per kg Kelvin. Like that the kinetic percentage changes. So this is this scenario which is deputed in this figure here. Now both are reversible. Engine and refrigerator are reversible. And it is given, one data is given, the heat transfer to the engine. Heat transfer to the engine is given. Okay. So Q, H, E equal to 2500 kilowatts or you can say kilojoule per cycle also. Now this, now efficiency of the engine will be equal to, because of the engine is reversible engine, efficiency can be written as 1 minus Tc by Th, which is equal to 1 minus 313 divided by 1073. Note that the temperature should be, even though temperatures are given in degrees integrated, we have to first convert it into Kelvin. So what is thermal efficiency? Thermal efficiency will be equal to 0.7083. Okay, now from this I can find the work developed by the engine. That is the total work developed by the engine is what that will be efficiency of the engine into the heat received by the engine. So this will be equal to 0.7083 into 2500, which is equal to 1770.736 kilowatts. Now 30 percent of this is given to the refrigerators. So what does the work received by the refrigerator? That will be 0.3 times We, which is equal to 0.3 into 1770.736, which is equal to 531.22 kilowatts. So now we have found the heat received by the refrigerator, that is work received by the refrigerator, because that is 30 percent of the work developed by the engine. And work developed by the engine can be found by the efficiency of the engine, which is calculated by using temperature ratios, because the engine is given to be reversible engine. So reversible engine, efficiency is 0.7083, the heat received is given as 2500. So we can find the work developed by the engine. 30 percent of the work is given to the refrigerator, that is this value. Now I can find the COP of the refrigerator. What is that? That will be refrigerator. So that is Q, go to the diagram. The heat objective energy transfer for the refrigerator is the heat, which it receives from the cold space, that is QRC. So QRC divided by WR. Now WR can be written as what? QRC divided by WR can be written as Q, QRH minus QRC. QRH minus QRC. Now this can be written as 1 divided by QRH divided by QRC minus 1. Now refrigerator is reversible. So which implies QRH divided by QRC, the ratio of the heat transfers can be written as the ratio of the temperatures, that is QRC receives, is received by the refrigerator from a space which is at 278 Kelvin. Similarly, QRH is rejected by the refrigerator to a reservoir at 313 Kelvin. So 278, so this is 313 divided by 278. So that means I can substitute and get the value of COP of the refrigerator as 1 divided by 313 divided by 278 minus 1. So what? That will be equal to 7.943. Okay. So now I know WR, correct? So from this equation, I can say COP of the refrigerator equal to 7.943, which is equal to QRC divided by WR. WR is 533.22. Okay, that is what we have calculated here, 531.22. 1.22, which implies QRC will be now equal to 7.943 into 531.22, which is equal to 4219.487 kilowatts. So that is done. And now apply the first law, QH QRH, QRH minus QRC equal to WR. That is the first law. So from this, I can find QRH equal to WR plus QRC, which is equal to 531.22 plus 4219.487, which is equal to 4750.707 kilowatts. So this is the heat, which is rejected by this and by this refrigerator. And this heat transfer from the refrigerator is used to heat water in the radiator. So that means 4750.707, that is QRH will be equal to m dot of water in the radiator Cp of water into delta T. So now delta T is given as 20 degrees centigrade. Cp is given as 4.2 kJ per kg Kelvin. So I want to find this. So this implies the water flow rate to maintain the temperature rise within 20 degrees or at 20 degrees will be equal to 4750.707 divided by Cp is 4.2. So this is in kilowatts. So 4.2 kJ per kg you can use into 20, you can find, write this. So now that will be the mass flow rate of water. When you substitute it, you will get 56.556 kg per second. That is the flow rate of water in the radiator in order to maintain the delta T at 20 degrees. So this is the problem where the engine reversible, both are reversible, engine refrigerated are reversible. Engine drives the reversible refrigerator by supplying 30 percent of the work it has developed. And all the reservoir temperatures are given from which it was easy to find the efficiency and COP. So from that, by reference we have found the other energy sensors.