 We're now going to work an example problem involving an adiabatic heat exchanger and what we're going to do is we're going to calculate the energy exchange between the two fluid streams and then we're going to use the entropy balance equation in order to determine the amounts of entropy generation in the heat exchange process. So I'll begin by writing out the problem statement. So that is our problem statement. Whenever you're given the problem statement what you should be doing is taking a look at the wording and pull out the pertinent information. To begin with we are dealing with an adiabatic heat exchanger and what that is implying is that there is no heat exchange from the outside of the heat exchanger to the surrounding world. So outside of our control volume there is no heat loss. There is heat exchange between the two fluid streams however there is no heat transfer to the outside world. So that is an important piece of information that you should note. We're dealing with two fluids. One is ethylene glycol which is being cooled. We're given some property information and we're given the mass flow rate of the ethylene glycol and it is being cooled by another fluid and that is water. We're given property information of the water. It's going from 80 to 40, sorry, that the water is increasing. It's the ethylene glycol that is going through that. So the water enters at 20 degrees C and it exits at 55 and the mass flow rate of water has not been given for this particular problem and so we'll have to think about that and determine whether or not that is something that we need to solve for. But what it is telling it wants us to calculate, first of all it wants us to calculate the heat transfer and so that means what is the heat transfer from the ethylene glycol to the water and then the second thing it wants us to do is calculate the entropy generation due to the heat transfer. So just reading this problem and calculating rate of heat transfer that sounds to be the first law that we would want to invoke. The second part of doing the entropy generation that would be using the entropy balance equation for an open system and if you recall the entropy generation or the entropy balance equation did have the mass flow rate of both of the fluid streams in it. So that tells us that we probably are going to want to calculate the mass flow rate of the water as part of the solution as well even though it is not explicitly asked for. So let's go ahead and write out what we know and then we'll start working the problem. So with the problem information written out one thing to note is I have written out the temperatures in Kelvin because I know when we use the entropy balance equation we are going to have to evaluate the entropy generation due to heat transfer and that is divided by the temperature in Kelvin and so I just listed that out there. Another thing to note I'm using kilojoules per kilogram Kelvin. There is, oops I apologize about that, no difference between the term kilojoules per kilogram Kelvin or kilojoules per kilogram degrees C which they have in the problem statement. So don't get those two confused they're the same thing because a degree cell is Celsius and Kelvin are the same. So let's go ahead and try to work out the solution here. What are we after? We're looking for one thing is rate of heat transfer. The second thing we're looking for entropy generation in the heat exchanger. So the first part for calculating the rate of heat transfer what we'll do is we will perform an energy balance on the heat exchanger itself and this is basically just applying the first law but I'm going to write it in a form of energy flowing in minus energy flowing out is going to be equal to the change of energy in the system. Now we are given no information about transients and we're actually told how the system is operating at steady state. Consequently the system itself is not changing in the amount of energy with time and so this term here drops out and what we're left with then is just energy in equals energy flowing out. So let's evaluate those and what we're going to be doing is we're using the representation for energy flowing as being m dot ccp delta t and so what we'll do is we'll begin with the glycol flow and that will be equal and so that is the glycol flowing in minus the glycol flowing out and that will be equal to the water flowing in minus the water flowing out. So that is a balance that we have. Now looking at this equation we know the glycol flow rate. We know the specific heat of the glycol. We know its temperature and again we know the mass flow rate of glycol, the specific heat, the exit temperature. We do not know the mass flow rate of water so that's an unknown. We know the specific heat of water and we know the inlet and the exit temperature for the water. So we know everything with the exception of the mass flow rate of water so we can plug numbers in and this equation will then enable us to determine the mass flow rate of the water which is not specifically asked for in the question. However you'll see when we do the entropy balance we will need it. So now the heat exchange from one fluid stream to the other is basically just the mass flow rate times the specific heat of a fluid times the change in the temperature. So you can do this for either the water stream or for the glycol stream but what we will do we will take a look at the glycol stream and given that the glycol is cooling we'll start with the inlet temperature minus the exit and when we plug values into that we find that the heat transfer out of the glycol going into the water is 204.8 kilowatts. Now you could have done this the other way by looking at the energy gain of the water and you would get the same number by going through that process. So that gives us the answer to the first part of the the question which is the heat transfer. Let's now take a look at the entropy balance and we're going to do this over the entire heat exchanger. So writing out the entropy balance equation for an open system in rate form. So what we can look at here first of all remember we said that it was an adiabatic heat exchanger. Well with that that term goes to zero. The other thing that we can say is that our heat exchanger is operating at steady state so the entropy of our control volume is not changing with time and so that term disappears as well. So the terms that we're left with are the ones that we will now work with and what that can be rearranged into is entropy generation is that. So the fluid exiting what we can do is look at the change of the mass flow rate and what we'll do is we'll take this on the glycol stream and the water stream each of them by themselves and these are little s's that I'm writing here and the rate form equation. It's a big s on the left hand side of the equation and little s's on the right hand side. So what we can now do we need to evaluate the change in entropy of both the glycol stream as well as the water stream and so for that what we can do is use the equation that we derive for change of entropy of either a liquid or a solid and this is where you would want to ensure that your temperatures were in degree Kelvin. The other place would have been had we had heat exchange and we had to use the first term in the entropy balance equation this one up here we don't need that though so but we do need it when we're evaluating the change in entropy of either of the fluid streams so with that what we can now do we can go and plug our numbers in and we calculate that the entropy generation for the heat transfer within the heat exchanger is 0.0446 kilowatts per Kelvin and so that would be the answer to the second part of the problem.