 Welcome back. So, we take a different kind of open system this time. So, we look at a heat exchanger. So, this does not really presuppose any knowledge about heat exchange. Such things are normally taught in a heat transfer course and we will not get into the intricacies of heat exchangers. We would just treat it like an open thermodynamic system. So, let us just read the problem here. In a heat exchanger, air is heated from 30 degree centigrade to 80 degree centigrade by means of a second air stream which enters the heat exchanger at 150 degree centigrade. Both streams have a flow rate of 2 kilograms per second and flow without any loss in pressure. Determine heat transferred between the streams and entropy production rate and we have been told to use Cp for air as 1 kilo joule per kilogram kelvin. So, let us see how do we approach this problem. So, one way to go ahead and look at this problem is to just draw some kind of a box here, a black box. This is our heat exchanger and we realize that there are two streams. So, we say that there are two inlets here. Let me call it I1 and I2. Similarly, there are two exits E1 and E2. We do not get into details such as whether this is a parallel flow heat exchanger or a cross flow or something of this sort. We just look at it as an open system. We just assume that these streams do not mix. So, we know that the first stream comes in at 30 degree centigrade and leaves at 80 degree centigrade. Whereas, the second stream comes in at 150 degree centigrade, the exit temperature is not given. It is given that both streams are at 2 kilograms per second and what else is given is that there is no pressure drop for either stream as it flows through this black box here. So, how do we go about solving this? So, a heat exchanger normally is used to exchange heat between two streams and overall as far as the outside world is concerned, it is an adiabatic system that is we do not want the heat to go elsewhere. It should only get transferred from one stream to another. So, we just show that this is an adiabatic system here and it does no work. So, we can just write the first law q dot minus w dot s and we realize that there are two inlet streams and in steady state, we call this m dot 1 and this is m dot 2. This is m dot 1 and m dot 2. m dot 1 at inlet is the same as m dot 1 at the exit. It is a steady state problem. So, we write separately for m dot 1 h e minus h i plus v e squared by 2 minus v i squared by 2 plus g z e minus g z i and all of this for stream 1. So, this is m dot 1 and then similarly for m dot 2, the second stream, if you had checked, you would have gotten a similar expression. So, you notice that if there are two streams steady, if they are not mixing, we can just write this assuming the same first law for open system and what do we have here? Overall, the system is adiabatic. So, we assume that this is equal to 0. There is no work transfer. This is 0. We assume that for both streams, there is no real change in the potential energy as the streams move from inlet to exit. It is a pretty good assumption. Put this as 0 and if the velocities do not change from inlet to exit for both streams, then there is no change in the kinetic energy. So, what we get finally is that m dot 1 h e minus h i for stream 1 is minus m dot 2 h e minus h i for stream 2. Now, you notice that up till now, we have not talked whether the streams are made up of air or steam or some other gas or water. So, if it is made up of air, h e minus h i could be written as C p delta t. This would be applicable for air or water or any other liquid. We just have to find out the C p for that fluid. So, if this is air, then this is air. If this is water, then this would be water. If this is some oil, this would be for that oil and we assume that it does not change much for this increase in delta t. Now, if it was either a two phase system that is water or steam system or purely steam, then we would have to look up h e and h i from the steam tables and not go ahead and do this calculation. So, if any one of the stream is made up of steam, then use steam table for liquids, get appropriate C p C p and use delta h is C p delta t. So, now we have in the present case m dot 1 h e 1 minus h i 1 is equal to minus m dot 2 h e 2 minus h i 2 or we could write this as m dot 2 h i 2 minus h e 2. Now, of course, if the enthalpy of stream 1 increases that is h e 1 is greater than h i 1. This automatically means that the enthalpy of stream 2 decreases and that is because it is passing on energy to stream 1 and hence the inlet enthalpy of stream 2 would be greater than the exit enthalpy of stream 2 and vice versa. You could have as well assumed that stream 1 is the one which is losing energy and stream 2 is the one which is having its enthalpy raised. Now, in the present case m dot 1 is equal to m dot 2 is 2 kilograms per second and hence we remove m dot 1 and m dot 2 we get h e 1 minus h i 1 is h i 2 minus h e 2. Now, this we have C p air delta t air stream 1 is C p air delta t air stream 2 both streams are made up of air here we could as well remove this. So, we get t e 1 minus t i 1 is equal to t i 2 minus t e 2. Now, t because this is a difference I could have used degrees centigrade or Kelvin sometimes it is safer to operate in Kelvin, but I will leave it to you. We know let us say stream 1 is the one which is getting heated then it is getting heated from 30 degrees centigrade to 80 degrees centigrade. So, this is 80 and 30 and this here is actually going from 150 to a lower temperature t e 2, but we can now find out because everything else is known and t e 2 can be found out. So, this is how you go ahead and find out what is the exit temperature and if you want q dot then q dot is just m dot 1 C p delta t 1 and this is same as m dot 2 C p delta t 2. Of course, there would be a negative sign here because one is losing heat the other one is gaining heat and we can find q dot. What about entropy production rate? So, this is part b this is entropy production rate. You see that we have assumed that q dot for the entire this q dot is between stream 1 and 2, but q dot for the entire open system we assumed is equal to 0. So, for this entire black box that we drew here q dot is equal to 0 because there is only an exchange of q between 1 and 2 in between the 2 streams which are flowing here. I do not know how they are flowing, but there is an exchange this is entirely internal here, but now if q dot system is equal to 0 you know that the entropy production rate s dot p is equal to m dot s e minus s i if there is only one stream. In this case it would be m dot 1 s e 1 minus s i 1 plus m dot 2 s e 2 minus s i 2 and you notice that here there is no change in pressure I could have as well used a property relationship t d s is equal to d h minus v d p and since d p is 0 in this process we get d s is d h by t is equal to c p d t by t. If we integrate d s from inlet to exit we get c p ln t exit by t inlet this is of course integrating c p d t by t from inlet to exit we get this expression. And now you realize that in this case if we have net s dot p this would be m dot 1 c p ln t e 1 upon t i 1 plus m dot 2 c p ln t e 2 upon t i 2. Of course, in this case both the c p's are same in this case both m dots were also same. If one of the streams was made of steam then you would have to find out the actual entropy from the steam tables you would not do it as an integral of c p d t by t but you could have just found it out from the steam tables the s e 2 and s e 1 depending on the state. But in this case this is the equation you get and now notice that you can no longer use degrees centigrade. You have to use Kelvin when you are taking a log this is a ratio. So, you change 30 degrees centigrade and 80 degrees centigrade to Kelvin you notice that the exit temperature here is greater than the inlet temperature here in stream 1. So, this is going to be a positive quantity this is going to be a negative quantity. But overall if you check this will be a bigger number than this and s dot p will turn out to be greater than 0 as it should be. So, if you just go ahead and do it you will get 2 kilograms per second for both of them multiplied by 1.0 kilo joule per kg Kelvin for both of them this is m dot multiplied and then you have ln for the log of 353.15, 303.15 is the inlet stream and log 473.15 and 523.15. This is stream 2 and this is stream 1 and we can just put in the values this of course here is kilo joule per second kilowatt it would be 2 kilowatt per Kelvin and you have the 2 terms here 1 is plus 0.306 other is minus 0.201 and you would end up with plus 0.105 multiplied by 2 kilowatt per Kelvin. So, this is the entropy production rate you should get a positive number and that will be a good check. So, see what we have assumed here we had an air stream. So, we know what the change in enthalpies are. So, we know what the heat transfer between the two streams are and we can also find out the change in entropy and as I told you if one of the streams was steam or if both were steam then you would have to look up the steam tables and not go with cp delta t. Thank you.