 We're now going to work a second example problem involving an air conditioning process and with this one it will be at one atmospheric pressure and consequently will be able to use a psychometric chart and what it will do is it will demonstrate how useful this chart is because it enables you to do the calculations much quicker than what we saw in the previous problem where you have to evaluate all the different property values in terms so what I'll do I'll begin by writing out the problem statement and then we'll start working the problem. So there's our problem statement what we're dealing with an evaporative cooler remember this is an air conditioning system for cooling that you would use in very dry climates desert like climates we're dealing with air in one atmosphere 36 degrees c coming in so that's quite hot however if you notice the relative humidity 20 percent that's very very low so it's quite dry we're told that it's coming in a 10 meters cube per minute that is the air and it's leaving at 90 percent relative humidity so we're adding a lot of moisture to the to the air and what we want to do is we want to find the exit temperature because that would be the then temperature of the air leaving that you're using for your air conditioning process as well as how much water we need to constantly add to this system so that is the example or the problem statement and what we'll do we'll begin by writing out drawing a schematic of the problem so we'll do our common duct and we have state one here and remember I said evaporative coolers could be different types of configurations it could be spray this could be a porous media that the air is flowing through one that ideally doesn't have too much of a pressure drop but in the process water droplets are being trained into the airflow and as they evaporate they cool so what we know is we know that we have one atmosphere pressure for the air coming in 36 degrees celsius we're told the relative humidity at one is 20 and we have the air flowing through and then when it's leaving at state two we don't know the temperature but we are told the relative humidity goes up to 90 and so we have mass flow rate of water coming in here we do not know how much and so that is let me just try to clean that up a bit mass water in so those are the things that we know what we're trying to find first of all we want to find t2 we want to know what the temperature of leaving is and then we want to know mass water in how much water we need to replenish this system with so what we're going to do we're going to be able to solve this one using the psychrometric chart and recall whenever we're talking about evaporative cooler we did say that the enthalpy of this process was constant so each one is equal to h2 so we have an evaporative cooler so those of you who live in the desert you will know what these systems are and how they work for those that are in humid climates you probably haven't seen these and you wonder why anybody would want to add humidity to the air in order to cool well it works very well in dry climate so anyways what we have there h1 equals h2 so what i'm going to do i'm going to begin by writing out we have t1 equals that's going to be our dry bulb temperature at point one and that is equal to 36 degrees c and we have the relative humidity there is 20 percent so what we're going to do we're going to take these so we're going to go to our psychrometric chart and we're going to try to read off some of the values from that so let's take a look at our psychrometric chart remember we have 36 degrees c and 20 percent so here is the psychrometric chart and what i'm going to do is i'm going to write this in red so that we can see it so we're going in at 36 degrees c so 36 and we're told 20 percent so what we do is we go up vertically until we intersect with the relative humidity line at 20 percent which is right there i'll circle that in blue so it's a little easier to see so that is state one in our process now what i'm going to do is i'm going to strip off a couple of values here i'm going to take off the first of all the specific volume of the air and specific volume or the red lines that run up into the left so we have one specific volume and i'll use let's see what color we have all the different colors on here so i'll go back to black we have 0.88 and then we have 0.90 and what we can see is that we're kind of halfway between 0.88 and 0.89 so we're probably somewhere around 0.885 0.886 so that will be the first thing i'll say v10.886 and the other value that i want to strip off here is going to be and i'll do this in red again the specific humidity so i'm going to move over to the right hand side and you can see we're coming in just right below eight so we have six here that would be seven so we're probably right around 75 so specific humidity at one is 0.0075 so we have those two and what we'll do let's return back to our problem so we have specific volume at one 0.886 and that's meters cube per kilogram and then we have specific volume at one 0.0075 and i'm going to use those in order to determine the mass flow rate of dry air coming through our system and so if i look at the volumetric flow rate coming in we're told that it was 10 meters cube per minute which translates into 0.1667 meters cube per second and from that we can determine mass flow rate of dry air and that is going to be kilograms per second so that's one thing we have now another thing we can determine the mass flow rate of water at state one so that's the mass flow rate of water entering knowing the specific humidity at point one and the mass flow rate of dry air there and we get that to be 00141 kilograms per second so that's another piece of information that we have what i want to do is go back to the psychrometric chart and pull off the value of the enthalpy so let's take a look at enthalpy here and for enthalpy recall i'll use a green i'll use red because green might be hard to see but enthalpy are the green lines moving up and kind of to the left but our state is here and we're going from there up to 90 humidity and so the enthalpy in this process is not changing but i would read the enthalpy here to be partway between 50 and 60 so it's probably it's a little bit beyond the halfway point so maybe 56 kilojoules per kilogram of dry air so what i would say here is h1 equals 56 kilojoules per kilogram dry air and the way that i was able to get that i move up here and we get something that's right in about there so that's where i'm pulling that 56 from so let's go back and what we can say is h1 is 56 kilojoules per kilogram dry air h2 for an evaporative cooler is approximately h1 and 56 we know the relative humidity at 2 we were told was 90 percent so from this i'm going to go back to the chart and we can now pull off state 2 information so let's go back to our chart and if we follow up the constant enthalpy line and i will do this in red we go up and we get to our 90 percent point we're probably right in there oops sorry about that we're right in here somewhere so that would be our state and then i'm going to draw a blue circle around that so that would be state 2 and now what i'm going to do is strip off our different temperatures and so to begin with the dry bulb temperature for that i'm going to come straight down down and i get a value right around there which i would say is about 20.5 degrees c so with that temperature dry bulb 2 equals 20.5 degrees c and the other thing we want to find out is we want it to determine the amount of water being added so i'm going to pull off specific humidity at state so for that what we do is we go directly over to the right we come over here and we get 14 so i would say omega 2 it might be a little bit below i'm looking at where it is on the graph so maybe a little tiny bit below 14 0.01375 so i'm going to take those values now back to our problem and what we have is T let me write that in black temperature dry bulb at 2 which is actually the temperature coming out is 20.5 degrees c so you can see that is the cooling that we've achieved we've gone from 36 degrees celsius down to 20.5 which is really a considerable cooling and the specific humidity 0.01375 so if you can handle the increased relative humidity and it doesn't make it uncomfortable for you the the cooling is definitely going to make you happy in this climate that's at 36 degrees c last thing we want to do we want to find out how much water we need to add to the swamp cooler to make it work or the evaporative cooler so water balance and for the water balance what we can do we have water at 1 plus mass flow rate of water being injected is equal to mass flow rate of water at 2 and we can make our substitutions using the dry air and the specific enthalpy flow rates or sorry the specific humidity okay so we get that i'm going to isolate for mass flow rate in because that's what we're looking for so we have this equation here and if you notice we have everything for this equation we know the mass flow rate of the air we know omega 2 we know omega 1 so we can calculate everything and from that we get mass flow rate of water in and expressing this in kilograms per minute so that's the amount of water that we would need to add to this system in order to drop the temperature from 36 degrees c down to what do we say 20.5 but the humidity relative humidity is going from 20 up to 90 so that gives you an example of using the psychrometric chart you can see the psychrometric chart is relatively simple and straightforward to use it's a little busy but as long as you know the different things on there it's quite easy to navigate you might want to use a ruler in order to get a more accurate measure than i had here but nonetheless a pretty quick way of doing these calculations versus the last problem that we had where the calculations were a lot more involved it took a lot longer so that concludes the the lecture as well as the section on heating ventilation and air conditioning i'd like to thank you for your time