 12th problem now. Here the schematic is also given air. So, R of the air is 287 joule per kg Kelvin, gamma is 1.4. In a piston cylinder assembly is initially at 300 Kelvin and 100 kilo Pascals. So, this is P1 equal to 100 kilo Pascals, T1 equal to 300 Kelvin. And volume V initial is 0.5 meter cube. It is heated to 600 Kelvin that is T2 equal to 600 Kelvin from a by heat transfer from a reversible heat pump. This is reversible heat pump that receives energy from the ambient at 300 ok. Yeah, but the difference between the previous problem and this is, this is a piston cylinder arrangement. There it was a vessel. Find a required work input to integrate this, del W, what is the total required work input to the heat pump and its COP. So, this is the problem. So, how to do this? So, let us say this I call A, the piston cylinder arrangement I call A. So, let A be the piston cylinder device and B is ambient ok. Now, what happens here? Same thing. First we will find the mass of the air in A that will be using equation of state. So, I use P V equal to M R T. So, M equal to R 287 into 300 sorry this is P V. So, M equal to P V by R T. This P is 100 into 1000 into volume is given as 0.5 divided by 287 into 300. So, mass is 0.58072 kg ok. Now, I will now what I require is I will first try to do the first law del Q A ok is what? What is del Q A? So, first law applied to the air first I will say first law applied to A. So, I can say del Q minus del W equal to du since D ke equal to D pe equal to 0. The changes in the kinetic percentage are 0. So, D E equal to du. So, now, del Q minus what is del W? Del W is see this is a piston cylinder arrangement. So, the piston is free to move. So, constant pressure process will occur. So, I can say P dV equal to du. So, this implies del Q will be equal to du plus P dV. So, I can write this as for example, dH ok because P dV can be written as P2 V2 minus P1 V1 du is u2 minus u1. So, this can be written as h2 minus h1. So, this is dH. So, in a constant pressure process the heat interaction will be equal to change in enthalpy ok for constant pressure process heat interaction equal to change in enthalpy. So, we can write del Q A. Now, del Q A will be equal to m into Cp into D Ta ok. So, let Ta be the temperature of the air at any instant and Tb is constant, Tb is equal to T0 equal to 300 Kelvin ok. So, now, I can write this. Why I am writing del Q A is m Cp D Ta because heat interaction in a constant pressure process is changing enthalpy. So, Cp I want. So, what is Cp? Cp is equal to gamma R divided by gamma minus 1 h is equal to 1.4 into 287 divided by 1.4 minus 1 which is equal to 1004.5 joule per kg Kelvin. So, now, I know that m is also calculated here. So, D Ta. Now, D Ta is again note that note D Ta is positive because temperature increases from 300 to 600 Kelvin. So, this is the expression for the heat interaction. So, now, it is given that reversible heat pump. Since the heat pump is reversible then what? We can say that Q del Q A divided by del Q B equal to Ta by Tb in Kelvin at any time instant ok. You know that Ta is changing. So, del Q A also will change. Del Q B also will change ok. So, this is the expression. So, now, I can write del Q A by Ta equal to del Q B divided by Tb. Tb is constant or T0 I can say Tb ok. So, now, I can write this as what m A Cp into D Ta which is what I have written here. m Cp D Ta ok. So, m Cp D Ta. So, this divided by Ta. So, Tb is known. So, I can now write del Q B equal to Tb m Cp D Ta by Ta. Tb is constant. So, I can take it outside. So, now here integrating implies Q B equal to m Cp Tb into natural logarithm of Ta2 divided by Ta1 ok. So, Ta1 equal to 300 Kelvin comma Tb sorry Ta2 equal to 600 Kelvin. So, that means Q B equal to 0.58072 0.58072 into 1004.5 into 300 into natural logarithm of 600 by 300. So, that is the value of Q B which comes out. So, Q B comes out as 121 300 300 0.7 joules or 121.3 kilo joules. Now Q A, Q A is integral del Q A which is equal to m Cp integral D Ta which is equal to m Cp Ta2 minus Ta1. So, here I am integrating from Ta1 to Ta2. So, which implies this is equal to 0.58072 into 1004.5 into 600 minus 300. So, this is equal to Q A equal to 175 000 joules or 175 kilo joules. So, what is work now? From first law work equal to Q A minus Q B equal to 175 minus 121.3 which is equal to 53.7 kilo joules that is the work work. Then C O P is asked. C O P of the heat pump equal to what? What is the objective energy transfer? The energy supplied to the air. And work is the energy energy which will cost. So, that means it is Q A divided by W which is equal to Q A is 175 divided by 53.7 which is equal to 3.26. Ok. So, now these two problems, this problem and previous problem attempted in the B part. This is called finite temperature reservoir problem. What happens in the case of ambient? The massive mass and heat capacity is there. So, that means the temperature will not change even when you extract whatever amount of heat you want to extract. But when you supply heat to a finite temperature reservoir like this, ok, then the temperature increases. So, this is not a constant temperature anymore. So, when when you add for a finite mass and finite heat capacity material, if you want to add heat, this temperature increases. If you remove heat, its temperature will decrease. So, that is what you have seen here. Similarly, in this case also we have taken a question syndrome arrangement and the material is heated because of its finite heat capacity, its temperature increases, ok. But if it is reservoir, the temperature increase will be very less because you know that MCB DTA. So, if MCB is finite, then DTA will will be there. There will be a definite change in temperature. But if MCB is very very high, then DTA will tend to 0 or very low value, ok. So, this is what is basically seen. This is called finite temperature reservoirs. Problem number 13 is now here we are considering first a constant volume tank containing 20 kg air at 1000 Kelvin and a constant pressure device, Swissam cylinder containing 10 kg air at 300 Kelvin. So, initially the air temperature in the constant volume tank is 1000 and constant pressure tank is 300. Now, a heat engine is placed between the tank and the device, the constant pressure device like this I have shown in the figure. Now, the heat engine extracts heat from the high temperature tank. Initially, the temperature is high. But please understand that it is a finite reservoir again because its mass is 20 kg only. So, as you extract heat, the temperature will decrease and it will reach a final temperature Tf, ok. Now, produces work and rejects heat to the low temperature device. Now, low temperature device that is persistence in the arrangement. Now, here also the mass is finite, capacity is heat capacity is finite. So, temperature as you add the heat will increase to Tf, ok. Why I am saying Tf? Because Tf is the equivalent temperature after which the temperature difference ceases in this case, correct. So, what happens? The engine will not function, ok. Now, go back in this problem ambient is there. I can increase. Now, I can continue to operate this. I can increase the temperature to 300, 600, 900 etcetera because this is constant, ok. So, from the low temperature I can reject extract heat through a reversible heat pump and supply heat to a any device is constant volume or constant pressure device and increase this temperature. But in this case, it is not so. What happens is from 1000 Kelvin as you extract heat, the temperature drops to Tf. From 300 Kelvin, when you add heat, the temperature increases to Tf. Once the temperature in the constant volume tank and the constant pressure device becomes same value of Tf, then the engine will not operate. So, let us say del Qh is the heat extractor at any small time duration. Similarly, del Qc and del W is the work done and at any time instant. So, let us call this tank constant volume tank as A and the pistons in ambient as B. So, I will say temperature at any instant in the constant volume tank is Ta and here it is Tb, ok. So, initially Ta is 1000 and finally, Ta is Tf. Initially Tb is 300, finally it is Tf. That is what is given here. Now, determine the maximum work that can be produced by the heat engine and the final temperatures of the air in the tank and the device. So, we have after the logical argument, this temperature increases, this temperature decreases. Once they reach a value of Tf, temperature difference between these two will cease, ok. Then engine cannot operate, correct. Engine can operate only when there is a temperature difference. So, that is not that. So, it does not operate and it is all maximum work that can be produced. So, to produce maximum work, the engine should be reversible, ok. Engine should be reversible. So, here there are two sources and sink, ok, one source and one sink where the temperature will change. There is finite temperature reservoir problem again. So, now that means, I can say del QA, so I will say this, I will say A and this is also I will say B to be consistent with the notation of the last problem. So, Ta del QA is extracted, Tb is the temperature, del QB is rejected by the engine. I can say this, correct, ok. Now, what is del QA? Del QA is equal to this is a constant. So, you have to apply constant volume for A del W equal to 0 because dV equal to 0, correct. So, del QA will be equal to del UA, ok, del Ke equal to del Pe equal to 0, kinetic energy, but the changes are also d0. So, this I can write as ma into Cv into dt A, correct. So, now why I am saying Cv? Because del U is mCv dt, that is it. So, now, but what happens to temperature A? It decreases from 1000 Kelvin because you are extracting heat. So, that means, negative sign will appear here, ok. Since, dTa is less than 0 because temperature decreases here, negative sign is used to show the magnitude, ok. This is one. Now, what is del QB? Del QB equal to, so you can say that for B del W equal to PDV. So, you can say del QB equal to PDV in plus. So, here also I have to put d, ok. So, dUB, so which is equal to dHB, it is equal to mCp dtB. Now, here note that dTB is greater than 0 because temperature is increasing. So, I can just say del QB equal to mCp dtB. So, this is the equations I have. Now, del QA by Ta equal to del QB by TB implies minus mA Cv into dTA by Ta equal to mB Cp dtB by TB, ok. This is mB I will put, mB, ok. So, now, this is the expression I have got. I know the value mA, I know, ok. What is CpC? So, R is 287 joule per kg Kelvin Cp and Cv, so gamma equal to 1.4. So, that means, Cp equal to gamma R by gamma minus 1 equal to 1004.5 joule per kg Kelvin and Cv equal to R by gamma minus 1 equal to 717.5 joule per kg Kelvin. So, this we will keep. So, we want all the values of cv, Cp, etcetera. And what is mA? So, from the mA is 20 kgs, mB is 10 kgs. So, that is known to us. So, I can write minus dTA by Ta equal to mB by mA, just I am just rearranging, mB by mA Cp by Cv into dTB by TB equal to 0.7 dtB by TB, that is it, ok. Now, integrating, integrating between the initial and final states, ok. Initial states are different, you know, for both. So, I will say minus natural volume of Tf divided by 1000. Why? Because Ta initial equal to 1000 Kelvin, Ta final equal to Tf, ok. So, that is what is this. So, this is equal to 0.7. I am substituting value, mB mA Cp by Cv is 1.4 and then 0.7 comes. So, substitute the values I get this. Now, 0.7 into natural volume of Tf by 300. So, here you can see that TBi equal to 300 Kelvin, TBf equal to Tf, that is it. So, now, in this equation, the unknown is Tf. So, I can find Tf equal to 609.11 Kelvin, ok. Now, temperature in the A decreases from 1000 to 609.11. Simultaneously, temperature in B increases from 300 to 609.11 and once the temperature in these two constant volume and constant pressure systems are going to be same, then engine cannot work. So, till that time it will work, correct. It will do the extraction from the constant volume and rejection to the constant pressure devices and also deliver work, ok. Now, work delivered by the engine is calculated by using first law applied to the engine, that is del W equal to del QA minus del QB. This is actually sigma W equal to sigma Q operating in a cyclic process. So, now, this is equal to what? Del QA. Del QA is minus m A CV DTA minus mB Cp DTB. Please understand that this del QA is positive. Since I am putting minus here, that should be negative. So, that is this again both should be positive here. So, why I am putting minus in here? Because DTA is negative. So, to give the magnitude of del QA, I am putting negative sign. Since DTB is positive, I need not put negative sign. So, this is the equation. So, now, what is W? W will be equal to integral of state 1 to state 2 del W which is equal to minus 20 into CV is 717.5 into final temperature minus initial temperature for the A minus mB is 10 into Cp into initial temperature say final temperature is this minus initial temperature is 300. So, when you calculate this, W will be equal to 250 4261.5 joules or 2504.26 kilojoules. So, that is the work. So, this is also a finite reservoir problem. In this case, both are finite reservoirs. So, in the previous cases as I told you, one was infinite reservoir. That is, the temperature will not change by extracting heat from it. The other one where the temperature was changing due to the heat transfer, that is the finite reservoir. But in this case, both are finite reservoirs. When you extract it from the finite reservoir, its temperature decreases. When you add heat to the finite reservoir, its temperature increases. So, we have to consider a small duration in which small amount of heat transfer or work transfer occurs, then you convert, integrate it to find the total heat transfer. That means, the heat transfer and work transfer will vary and we have to find the total by integrating it. So, this problem 14th one, a work developed by a reversible heat engine E operating between two identical metal blocks A and B. So, let us draw this. E is the engine, reversible heat engine. Identical metal blocks A and B. So, A and another block is B. So, it rejects heat here and rejects heat here. So, this is the configuration. Okay. So, this develops some work W and I can say. So, let us say D del W, you can say del QA, we can say del QB here and so on. This is reversible. Now, the work developed is used to drive a reversible heat pump. So, let us say reversible heat pump HP, which extracts heat from the metal block B. That is metal block B, it extracts heat at the same rate at which engine E rejects to heat. So, engine E rejects some heat. No, same rate it takes. That is, I will say del QB is taken by this also. So, the heat pump rejects the heat to another metal block C, del QC, this is C, which is identical to A and B. So, all the three are identical metal blocks. Mass, I can say mass of this is given, mass of ABC is M and specifically it is C. So, M comma C we will put for this M comma C. This is also M comma C. So, initially the temperatures of A, B and C are 300 Kelvin, 100 Kelvin and 300 Kelvin respectively. What are the final temperatures of A, B and C? Now, one thing we have to understand here. The blocks are having finite mass and finite heat capacity. So, thermal capacity is finite. So, it is when the heat is extracted, the temperature will decrease and heat is added, temperature will increase. So, first we have to make an assumption. So, let us assume that A, B and C interact only with E and HP that is engine and the heat pump. Now, this work which is extracted that is given to the heat pump that is also written drive. So, entire work developed by the engine is given to the heat pump also. So, this is the scenario here. So, let us assume that the blocks interact only with E and HP and not with surroundings, not with anything else. So, now what happens? Please see that for B, it is taking heat from the engine, but rejecting the same amount to the heat pump. So, that means its temperature will not change, correct. Since B interacting with E and HP only, ok, takes del QB from E and rejects the same amount of del QB to HP, comma, its temperature will not change. So, TB final will be equal to 100 Kelvin. So, one waveform, ok, one value waveform. So, now what about the initial temperature is 300, what is the final temperature of A and C? Obviously, we can note easily that temperature of A from 300 reduces to a final value. Similarly, this, so I will say final value 1. We do not know whether it is the same, but since C receives heat from the heat pump, its temperature will increase to a final value say 2, ok. So, now we will see how to calculate this. What is del QA? From this, del QA will be equal to minus MC DTA since DTA is less than 0, because temperature decreases. So, I have to put a minus sign here. The minus sign appears because of the fact that DTA is 0 and I want absolute value of this, ok. Similarly, I can say del W equal to del QA minus del QB which is also equal to 1 minus 100 divided by TA. So, let us say that the TA be the temperature at a time instant, any time instant in A. Similarly, TC is the temperature at any time instant in C and you know that for B it is only 100. So, I can say the efficiency, these are reversible engine reversible heat pump correct, reversible heat pump, reversible heat engine. So, efficiency can be written in terms of temperature. So, 1 minus 100 by TA into del QA. I can write this correct, del W equal to del QA minus del QB or I can write 1 minus 100 by TA into del QA, ok. So, now we can say del QB equal to 100 into del QA by TA from that expression. Similarly, for heat pump I can say del QC equal to MC DTC because DTC is positive correct, because the temperature increases in C. Now, I can say the work first law minus W because it is given to the heat pump equal to del QB minus del QC, ok. Or I can say del W will be equal to del QC minus del QB which is equal to del QC divided by COP which is equal to del QC divided by what is COP? TC divided by TC minus 100, ok. So, from then I can calculate del QB equal to 100 into del QC divided by TC. Similarly, I have calculated del QB in this. So, please understand the work is same, del QB also should be same, ok. So, I can say that this is one expression. I can say 1 minus the work, ok 1 minus 100 by TA into del QA equal to this expression del QC divided by TC divided by TC minus 100. So, which is one expression? Then second one is work is same. So, del. So, this is what is this. And second expression is these two. So, these are the three equations I have. These are the three equations. From these three equations, we can conclude that del QA will be equal to del QC or minus MC DTA will be equal to MC DTC or minus DTA equal to DTC, ok. So, this means engine, no. So, the rate at which the temperature decreases in A will be the same rate at which temperature increases in C, ok. Because of the heat transfer from A to engine, temperature in the A increases. That will, that decrease will be same as the rate at which temperature increases due to the rejection of heat from the heat pump to the block C. So, we can say that finally, finally when TA is equal to 100 Kelvin, engine cannot operate. So, the heat pump also because no work is delivered. So, heat pump cannot operate, ok. So, in this case, final temperature of A will be equal to 100 Kelvin, ok. So, that means I can say delta TA equal to 100 minus 300 equal to minus 200 Kelvin, which is equal to minus delta TC or delta TC should be equal to 200 Kelvin. So, that means, TC final will be equal to what? 300 initial temperature of C plus delta TC equal to 300 plus 200 equal to 500 Kelvin. So, that is the answer. So, again a finite reservoir problem, but we have to see that the block B, all the blocks only interact with the engine and heat pump, not with surroundings or any other system. So, in that case, it is very simple to understand that the temperature of the B will remain constant because it receives heat from the engine and rejects the same value to the heat pump. So, from that applying the first law etcetera, we can arrive at this answer, ok. So, initial temperature of A is 300, it will reduce to 100 Kelvin because of extraction. Since B is at 100, it cannot operate, engine cannot operate. So, the delta T for A is minus 200, that should be negative of delta TC. So, delta TC would be 200 Kelvin. So, initial temperature of C is 300 plus delta TC will give the 500 Kelvin. So, this is the answer. So, this covers the tutorial.