 Now, tenth problem, consider a well insulated horizontal rigid cylinder that is divided into two compartments by a piston that is free to move, but does not allow either gas to leak to the other side ok. So, initially one side of the piston contains 1 meter cube of nitrogen at 500 kilo Pascal and 80 degree centigrade while the other side has 1 meter cube. So, it is almost divided into half years like this ok. Here 1 meter cube of nitrogen, here 1 meter cube of helium, other side has 1 meter cube of helium. So, here pressure was pressure is here nitrogen side the pressure is 500 kilo Pascal and temperature is 80 degree centigrade and in the helium side it is 500 kilo Pascal and 25 degree centigrade ok. Now, thermal equilibrium is established in the cylinder as a result of heat transfer through the piston that means, this is conducting, conducting piston ok. Determine the final equilibrium temperature in the cylinder and entropy generation due to the process ok. So, here the piston is tight enough, but free to move, tight enough means leak proof that is the nitrogen cannot mix with helium or otherwise, but it is free to move. So, since it is free to move always mechanical equilibrium will be there and it is well insulated it is saying the entire is well insulated that means, there will be no heat transfer from either of the compartment or from the piston to the ambient. But since the piston is conducting there will be heat transfer between these two ok. So, that is the problem in hand. In this we have to find the entropy generation also, so solution let T f be the final temperature, equilibrium temperature ok. Let us find the masses of nitrogen and helium ok. So, mass of nitrogen is what? Nitrogen is 500 kilo Pascal into 1 meter cube divided by A 314 by 28 is the specific gas constant of the nitrogen for nitrogen right into the temperature is 80 degrees that is 273 plus 80 353 Kelvin. So, this is the mass 4.77 kg ok. Similarly, mass of the helium chamber will be equal to pressure is same because of the mechanical equilibrium correct the pressure the piston is free to move. If it is at a particular place then it means that pressure in the both side will be the same A 314 for helium for helium the molecular weight is 4 ok. So, molecular weight of helium equal to 4 kg per kilo mole ok. So, that into temperature is 25 degree centigrade, so 298. So, that will give the mass as 0.807 kg. Now, we can also calculate the CV, CV for nitrogen will be equal to A 314 by 28 that is the specific gas constant divided by 1.4 minus 1, 1.4 is the ratio specific heat 1.4 minus 1 which is equal to 742.3 joule per kg Kelvin. Similarly, CV for helium will be equal to A 314 by 4 divided by 1.66 minus 1. So, gamma for helium equal to 1.66 ok. So, now this will be equal to 3149.2 joule per kg Kelvin. So, these are the properties. So, now I can taking the rigid vessel as the system ok apply the first law what is that q minus w equal to delta u ok. Now, what is q entire rigid vessel is insulated. So, q is 0. Since rigid vessel volume will not change in this. So, individually the volume of nitrogen or volume of helium can change based upon the movement of the piston. But the rigid the entire vessel if it is taken as a system there is no delta v. So, this also is 0 because delta v equal to 0 rigid vessel. So, entire rigid vessel is the system now. So, that means delta u will be equal to 0 correct. So, now how will you calculate this delta u because we do not know the final condition ok. So, we will apply the equation here delta u equal to 0 equal to 4.77 into 742.3 into. So, let us assume that final temperature is T f we already done that initial temperature for the nitrogen is 353 ok. So, this is the mass of nitrogen this is C v of nitrogen and this is T nitrogen 1 initial temperature of nitrogen final temperature is this plus. So, this is actually delta u is written as delta u of nitrogen plus here 0.807 into 3149.2 into T f minus 298. So, this is basically delta u of helium. So, this is mass of helium this is C v of helium ok. So, now from this I can find T f which implies because delta u equal to 0. So, T f equal to 330 Kelvin ok. So, please see this initially 353 is the temperature of nitrogen helium is 25 degrees, but due to the heat transfer it has reached a common temperature now 330 ok. So, helium temperature has raised from 298 to 330 nitrogen temperature has decreased from 352 to 330 based upon whatever is the masses etcetera ok, final temperature is got. Now, total volume is the same that is V N2 1 plus V helium 1 will be equal to V N2 2 plus V helium 2 ok. That means, here if the piston moves to the right then the volume of nitrogen will increase, but the volume of helium will decrease, but total volume should remain the same that is what the volume conservation is in this. Any compartment problem where the piston is free to move this volume conservation has to be applied. So, now we apply this. So, how the volume can be calculated? We know P V equal to M R T. So, I write the volume in terms of M R T by P ok. Now, I know the left hand side volume 1 and volume 2 it is already given. So, this what this is given as 1 meter cube and this is also given as 1 meter cube. So, that is known. So, for the right hand side I do not know the pressure. The temperature is known because the final state temperature is 330 we have just now calculated, but I do not know where is the piston now. I do not know the pressure also. So, I cannot calculate this. So, I try to eliminate one of the variable ok. Write it in only the pressure terms. So, I will write this as 1 plus 1 ok the volume conservation 1 plus 1 meter cube equal to 2 equal to here V N 2 2 that is written as M N 2 R N 2 T final divided by P final ok plus M helium R helium T final divided by P final. So, now, I know this values I can directly get the P final implies P final will be equal to 510460.3 Newton per meter square or Pascal that is it. So, that is what is asked here. Final equipment temperature and pressure or the pressure also is asked we do this. Why we want pressure because you have to calculate the entropy change entropy generation basically. What is entropy generation? Entropy generation. So, entropy generation basically can be written as delta S system plus delta S surroundings. Entropy generation is delta S universe that is it. So, delta S system plus delta S surroundings. Now, what is delta S system delta S of N 2 plus delta S of helium that is delta S of system. What is delta S surroundings plus 0? Why there is no heat transfer here. See, please understand that the entire rigid vessel is insulated. So, there is no entropy transfer. So, we can say that the rigid vessel is isolated basically. It is not having any connection mechanically or thermally. So, there is no entropy change for the surroundings. So, if I calculate entropy change for the system which is the entropy change for nitrogen and that of helium that is enough for us. So, what I do here? I can apply first if I want to find the Cp value of nitrogen equal to 1.4 R nitrogen divided by 1.4 minus 1 which is equal to 1039.2 joule per kg Kelvin. Similarly, Cp helium can be got as 1.66 R helium divided by 0.66 which is equal to 5227.67 joule per kg Kelvin. So, now, what is delta S of nitrogen? Delta S of nitrogen equal to m N 2 into Cp ln T 2 by T 1. I will say T nitrogen 1. T 2 is final temperature Tf minus R ln R nitrogen this is Cp N 2 ln P 2 by P 1, P 2 by P 1 or P 2 is Pf, P 1 is basically nitrogen 1 that is it. Okay. Similarly, delta S of helium can be written as m He into Cp He ln Tf by T He 1 minus R helium ln Pf by P He 1. So, substituting these two I will get S generation S generated that is delta S universe equal to S gen equal to 32 joule per Kelvin. So, you have to understand that entropy generation is delta S system plus delta S surroundings. Okay. So, this is the problem which has basically why there is a entropy generation because of the irreversibility in the this isolated system basically. So, you can see that this isolated system there is no heat transfer, but due to some irreversibility see here clearly it is not given that the friction is the piston is frictionless it is free to move that is it. So, probably there is some friction in the system due to which there is a irreversibility that has caused the entropy generation. So, entropy generation is caused only due to irreversibility. This irreversibility basically can be internal irreversibility which is due to friction in this case what we have seen also it can be external irreversibility, but is if there is a heat transfer between system and surrounding if that occurs in a finite temperature difference then that will also cause this. Okay. So, in this case basically there is no heat transfer. So, the delta S surrounding is zero if there is a heat transfer then in the previous problem we have done know the delta S here delta S surrounding equal to Q surrounding by T surrounding we have written. If there is any heat transfer that can contribute to the change in the entropy of the surrounding. Since it is an isolated system in this case we cannot have any change in the entropy for the there is nothing entropy transfer due to heat transfer is not there plus there is no internal irreversibility in the surroundings also. So, this entropy change of universe will take care of both internal irreversibility and external irreversibility. Okay. This is about the problem 10. So, 11th problem carbon steel bolts density is 7833 kilogram per meter cube and specific heat is 0.465 kilo joule per kg Kelvin. 8 mm in diameter or annealed. So, annealing is a heat treatment process. And annealing is the heat treatment process. What they do here is they heat the balls to 900 degrees centigrade in a furnace in hot environment and allow them to cool slowly to 100 degrees centigrade in the ambient air which is at 35 degrees centigrade. So, once you do the structure the the atomic structure of this will change and you will get the required toughness etcetera. So, this is a metal processing or metal treatment process. If 2500 balls are to be annealed per hour determine the rate of heat transfer from the balls to the air. So, basically we are not caring about how it has reached 900. The cooling process is the annealing process. So, how it cools from 900 to 100 degrees in an air environment of 35 degrees. Okay. So, first one is rate of heat transfer from the balls to the air. Second is the rate of entropy generation due to the heat loss from the balls to the air. Okay. So, heating is not the main point here. The cooling is the main point here. Okay. So, solution. So, this is the entropy generation in solid, basically solid this mass of one ball, one steel ball. Okay. That will be what density into volume which is equal to 7 833 into 4 by 3 pi into radius square, radius cube that is 8 into 10 power minus 3 by 2 cube that is volume. Okay. So, now that will be 2.1 into 10 power minus 3 kg that is the mass of one ball. So, heat transfer to the air from one ball is what equal to m c delta t which is equal to 2.1 into 10 power minus 3 into c is 465 I have written in joule per kg Kelvin into 900 minus 100 that is what is the heat loss in air. Correct. So, that will be equal to 781.6 joules. So, rate of heat transfer. So, here you can see this is the amount of heat which is transferred basically, but I want the rate of heat transfer because I wish to anneal 2500 balls per hour. So, time is given now. So, rate of heat transfer for 2500 balls in 1 hour what is that that will be 781.6 into 2500 is the number of balls right 2500 divided by 1 hour is what 3600 seconds. So, this is in seconds. So, this is in total joules. So, joule per second so, it is in watts that is 542.5 watts that is the rate of heat transfer first part. So, if you want to anneal 2500 balls per hour then this much heat should be lost. So, that means, you have to design the environment with some convection etcetera to facilitate this much heat loss the rate of heat losses in watts. Now, what is second part what is entropy generation entropy generation is nothing but what entropy change for the system which is I will say ball plus or balls plus entropy change for the surrounding which is air ok. Now, this is basically we have to write in rate at which. So, rate at which entropy is generated rate of the so, I can say this I put a overhead dot to represent that it is a rate at which it is done. So, not for the balls it is basically mass 2.1 into 10 power minus 3 into 2500 into 465 into ln of 100 Kelvin 100 degrees centigrade that is 373 divided by 900 is 1173 ok. So, this plus this is a rate at which so, 542.5 it will be ambient temperature is what 273 plus 35. So, now, here it is mass of one ball it is number of balls n this is c this is ln of t 2 by t 1 ok. Now, this is heat rejected see this is for surroundings this is heat received by the surroundings basically correct because ball is cooling the heat is rejected to the surrounding. So, the heat gained by the surrounding is q dot surrounding this is 542.5 watts divided by the temperature of the surrounding which is 35 degree centigrade so, 273 plus 35. So, this will totally give s gen as 0.985 watt per Kelvin which is the entropy gen rate of entropy generation ok due to the heat loss from the balls there. So, this is the a simple problem which demonstrate the entropy generation for the solids ok. Here we have to divide by 3,600 ok. So, here this divided by 3,600 we have rate no. So, rate at which this is the entropy change of the balls, but by dividing by 3,600 we will have the rate at which it is done ok. So, the rate at which the entropy changes in the balls plus the rate at which entropy changes in the surrounding that will be the entropy generation that is the entropy rate at which entropy changes for the universe that will be the rate at which the entropy is generated. This can be again positive or 0.