 Consider a regular room with a pressure of blank, a dry bulb temperature of blank, and a wet bulb temperature of blank. Using those blanks, determine the humidity ratio, the relative humidity, the specific enthalpy, the specific volume, and the dew point temperature. If we were in a classroom environment, I would actually use a barometer to determine the pressure in the room, and then use a dry bulb and a wet bulb thermometer to determine the actual psychrometric properties in the room. It'd be a fun in-class exercise. But since we're on the internet, we do what everyone does on the internet and just make stuff up. So let's assume that the pressure is, I don't know, about 100 kilopascals. That would be one bar. That would be nice and convenient for math purposes. Let's go with a regular dry bulb temperature, maybe something like 21 degrees Celsius. And for our wet bulb temperature, let's go with 18 degrees Celsius. So these even numbers are going to make for some nice look-ups. I don't have to do any interpolations, but the methodology is still the same, even if I had more precise measurements. Furthermore, I want to determine these psychrometric properties in two ways. I want to use the calculations to determine the best possible answers that we can come up with. And then I want to talk through how we can shortcut those look-ups by using the psychrometric chart. So we're going to be comparing and contrasting two methods to use our dry bulb and wet bulb temperatures to determine the remaining psychrometric properties. By the way, while we're here, let me point out that in addition to pressure, you need two independent intensive psychrometric properties to determine all of the rest. Theoretically, you could use essentially any combination of these seven quantities, any two of them, to determine the remaining five. So we'll start with the calculations. To do that, I'm going to have to go from my wet bulb and dry bulb temperatures to a humidity ratio, and then from that to everything else. And for that relationship, I need to use my wet bulb equations that we built in the previous video. Remember that the wet bulb equations allow us to determine the humidity ratio of the actual air, but to do that, we are looking up a bunch of properties and plugging them into equations and determining quantities that would exist if the air were fully saturated. So state one represents the actual conditions of the actual air, state two represents the wet bulb conditions, which aren't actually in existence in the actual room. Or rather, they're only in existence around the wet bulb thermometer itself. So in this equation, we need T1 and T2. Those are known. T1 is the actual temperature, which is the dry bulb temperature. T2 is the wet bulb temperature. I need the specific heat capacity of air, and we're going to use the property at 300 Kelvin. We need to look up what the specific enthalpy of a saturated liquid is at state two, and then the specific enthalpy of a saturated vapor at both states one and state two. And while we're here, we're also going to need some saturation pressures. Let's look up both of them. There'd be PSAT at T1 and then PSAT at T2. And from these quantities, we can calculate omega two, which is what the omega would be, what the humidity ratio would be at the wet bulb conditions, which aren't actually the conditions of the air, but are useful for plugging back into this equation to come up with the actual humidity ratio of the actual air in the room. I cannot underline that enough. Okay, so starting from the top here. Dry bulb temperature was known. That's 21 degrees Celsius. Web bulb temperature, 18 degrees Celsius. CP of air, we're going to jump into, I believe that's table A20, yep, in our textbook. CP of air at 300 Kelvin is 1.005. Then we want the saturation pressure at state two, the HG is state two, and the HF is state two. So we will jump into our steam tables. By temperature, state two was 18 degrees Celsius. So at 18 degrees Celsius, I want PSAT and I want both HF and HG. So HF at 18 degrees Celsius is 75.58. HG is 2534.12534.1. PSAT is 0.02064. 0.02064 bar. And then can jump into T1, which is 21 degrees Celsius. And I can look up PSAT and HG. So HG at 21 degrees Celsius is 2539.9. And again, these are convenient lookups, because I had the opportunity to pick my own temperatures. If you measured the actual temperatures, you would just interpolate for better values. So HG and then PSAT, which was 0.02487. 02487 bar. And that's all the property lookups I'm going to have to do for now. So my next step is to calculate what the humidity ratio is at the wet bulb conditions, which is going to be 0.622 times the vapor pressure at the wet bulb condition, which because it's at 100% relative humidity at the wet bulb, that's just the saturation pressure at state two. And then we're dividing by the partial pressure of the dry air, which we are rewriting as the total pressure at state two, which is the same as the pressure at state one, minus the vapor pressure, which again is PG because it is at 100% relative humidity. So we'll pop up our calculator, we will type 0.622 times PG at state two, divided by the total pressure at state two, which was 100 kilopascals, which is one bar, minus PG at state two, which is 0.02064. And we get a humidity ratio at state two of 0.013109. And remember that's kilograms of water per kilogram of dry air. Now with that, I can finally start plugging numbers into omega one. We have CP of air, which we know is 1.005. So 1.005 times the temperature difference, which is going to be 18 minus 21. And then we are subtracting your subtracting calculator, omega two times hf minus hg2, which is 75.58 minus 2534.1. And then we're divided by hg1 minus hf2, which would be 2539.9 minus 75.58. And we get an actual humidity ratio, the actual air of 0.011854. 0.011854 kilograms of water vapor per kilogram of dry air. So now that we have the humidity ratio, again, that's the actual humidity ratio, not omega two, which is a stepping stone to get to omega one. Now that we have the actual humidity ratio, we can calculate other quantities a little bit more conveniently. So next up, we have the relative humidity, which you'll remember is pv over pg. Well, pv isn't known. We know pg1, but we don't know pv. So what we have to do is recognize that omega represents 0.622 times pv over pa, which can be written as pv over p minus pv. So we can use omega to solve for pv, plug that into this equation, at which point we would have enough to calculate the relative humidity. So everyone over your eyes, I'm going to do algebra. p minus pv is equal to 6.22, 0.622 times pv, omega times p minus omega times pv is equal to 0.622 times pv. So omega times p is equal to pv times 0.622 plus omega. So pv is going to be omega times p divided by 0.622 plus omega. So our actual pressure was 100 kilobascals, which is one bar. So if I recognize that this quantity out front is going to be unitless, then whatever pressure unit I plug in for p is going to be what I get out for pv. So I'm going to take 0.011854 divided by 0.622 plus 0.011854, and I get 0.018702 multiplied by one bar will yield 0.018702 bar. Then p is going to be that number divided by the pg value we had looked up at state one, which was 0.02487. So when I divide those two numbers, I get a relative humidity of 71, excuse me, 75.2%. Okay, two-fifths of the way there. Next up, I wanted the specific enthalpy of the atmospheric air. Again, remember that that is total enthalpy of atmospheric air per unit mass of dry air. So I'm going to represent that as total enthalpy of dry air plus the total enthalpy of the water vapor divided by the mass of the dry air, which I can write as ma times specific enthalpy of the dry air plus mv times specific enthalpy of the water vapor divided by ma, which becomes ha plus omega times hv. I'm approximating ha with cpt at cp of air times the actual temperature in degrees Celsius, and I'm approximating hv by using hg. Now quick question, when I plug in values for t, omega, and hg, am I using the properties at state one or the properties at state two from this lookup from earlier? If you said state one because it's the actual air, you are correct. This is going to be cp of air times t1 plus omega one times hg1. Remember, state two only exists around the wicking and is only useful as a stepping stone. So 1.005 times the dry bulb temperature, which was 21 degrees Celsius, plus the actual omega value times the hg at state one, which was 2539.9. Then just to make a little bit more sense of the units, let's write that out. So because the specific enthalpy of the water vapor is expressed per unit mass of the water vapor itself, the kilograms of water cancel in the omega, and I'm left with kilojoules per kilogram of dry air, which one added to kilojoules per kilogram of dry air yields an answer in kilojoules per kilogram of dry air. Then next up, we want the specific volume of the atmospheric air per unit mass of the dry air, which again, remember, because we are modeling the behavior of the atmospheric air using Dalton's law, all of the substances within the mixture are assumed to take up the entire volume, so that could be written as the volume of the dry air per mass of the dry air, which means that we're really just looking for the specific volume of dry air at these conditions. The most difficult part of this calculation is remembering that we're using Pa instead of P. So in an effort to avoid calculating one more number, I'm going to write Pa as P minus Pb. So the specific gas constant for dry air is going to be the universal gas constant divided by the molar mass of dry air, 8.314 kilojoules per kilo mole Kelvin, available from the front cover of your textbook. If you don't know that number off the top of your head, available down here. And then the molar mass of dry air comes from table A1, 28.97. Then we are multiplying by the temperature of the dry air. This time it needs to be in Kelvin. So it's 21 plus 273.15 Kelvin. And we're dividing by one bar minus the actual vapor pressure at state one, which is 0.018702, 18702 bar. And then Kelvin cancels Kelvin, kilomoles cancels kilomoles, leaving me with kilojoules per kilogram bar. I'll recognize that a kilojoule is a kilonewton times a meter. A bar is 10 to the fifth newtons per square meter. And a kilonewton is a thousand newtons. Now kilodjoules cancels kilodjoules, bars cancels bars, kilonewtons cancels kilonewtons, and newtons cancels newtons, leaving me with cubic meters per kilogram. Calculator, you're needed again. Come on calculator, you can do it. I believe in you. 8.314 times 21 plus 273.15 times 1000 divided by 28.97 times the quantity 1 minus 0.018702 times 10 to the fifth. I get 0.8603. And the last thing I want to know is the dew point temperature. The dew point temperature is going to be expressed as the saturation temperature at PV, which means that I'm looking for the saturation temperature at 0.018702 bar. So back into our steam tables. If we start on the pressure table, we will note that we don't have a pressure that's low enough. Again, we're looking for 0.018702 bar. So we could extrapolate using 0.04 and 0.06. But the better thing to do is to actually jump over to table A2 because table A2 goes down to lower saturation conditions. So I have 0.02, excuse me, 0.018. Get these out of the way. So 0.0187. So that's going to occur between 16 and 17 degrees Celsius. So I can interpolate for that. So our actual PV was 0.0187, not less than or equal to calculator. Come on, 0.0187 minus 0.01818, divided by 0.01938 minus 0.01818. That's equal to the thing that I'm looking for minus 16 divided by 17 minus 16. And we get a saturation temperature of 16.435, which means my dew point temperature is going to be 16.435 degrees Celsius. So if the air in this room were cooled at a constant pressure, once it reached 16.44 degrees Celsius, it would begin to condense water out of the air. So you can think of that like cooling the entire room or you can think of that like cooling some of the air in the room. For example, if I brought in a cool glass of lemonade and set it on a table, the air around the glass would cool off once it cooled past 16.435 degrees Celsius, it would begin to condense water on the cool surface of the glass. So if I wanted to avoid creating water rings, I would just have to serve my lemonade at about, oh, I don't know, 16 and a half degrees Celsius. So we are through one half of what I want us to do. We went through and determined all of these psychrometric properties using the calculations themselves. The calculations are accurate. They will produce good numbers even though there are a couple of minor approximations in the process. But if we are okay with even more approximation, we can use the psychrometric chart. The psychrometric chart is a plot of psychrometric properties for a given pressure. It is primarily a plot of humidity ratio versus dry bulb temperature. And again, each psychrometric chart is only good for one pressure. This particular chart is generated at one atmosphere, which is 101.325 kilopascals, which is 1.01325 bar. In addition to dry bulb temperature and humidity ratio, all of the rest of our psychrometric properties are shown on the graph. So these curvy lines are lines of constant relative humidity, the dotted lines that are at a shallow angle are wet bulb temperature, the solid lines that are at a steeper angle are specific volume, the solid lines that are at a shallower angle are enthalpy. Note that the lines of wet bulb temperature and enthalpy appear parallel at first glance, but they're not quite parallel. So for our example, we have a dry bulb temperature of 21 degrees celsius and a wet bulb temperature of 18 degrees celsius. So we are assuming that our pressure of one bar is close enough to the pressure of 1.01325 bar that the error introduced by using this chart is okay. Furthermore, the values that we look up from the chart aren't going to be quite as precise as the values we get from our calculation. So in order to use the chart, we have to be okay with a little bit of approximation in our analysis. So our dry bulb temperature, which I will show in red, is going to be here and the line corresponding to it is here. And then our wet bulb temperature was 18 degrees celsius, so wet bulb temperature are these dashed lines. They are right off of here, here, here and here. So I have 15, 16, 17, 18. So the line of wet bulb temperature of 18 degrees celsius is this line here, which means that my intersection occurs right here. So that's my state point. So from that intersection, I can read off the rest of the sacrometric properties. First of all, I can follow that all over to the right and read off my humidity ratio. Let's color code this process to hopefully make it easier to follow. Dry bulb temperatures are down here and are in red. Dry bulb thermometer temperatures are here and shown in light blue. Let's go with green for humidity ratio. So if we follow our line over to the right using a green line, I can see that my humidity ratio is going to be a little bit more than 11 and a half. Because this would be 10. This would be 11. This would be 12. This line would be 11 and a half. So if we were looking this up, we might say, oh, about 11.6. So let's write out our lookups chart. Lookups humidity ratios. Oh, about 11.6 grams of water per kilogram of dry air fee our relative humidity. Let's go with a purple color for that. These curved lines are my lines of constant relative humidity. And I can see that I'm Oh, I would say a little bit under halfway between the 70% line and the 80% line. So we might draw that like this. So I could say 74 or 75. Let's just call it 75. As a general rule of thumb, we shouldn't try to interpolate visually unless we have a measuring device for that specific enthalpy. Let's go with a bright yellow, maybe. Enthalpy is right off of these solid lines in the upper left. So I see that my enthalpy is going to be about 51, maybe. This would be 50. This would be 52. 54, 56, 58. So let's call it 51. And again, that's kilograms of dry air. For the specific volume, I'm looking for my steep solid lines. And let's indicate those in orange, maybe. So I can see that my state point is a little bit less than 0.85. I would say it's maybe okay, orange. If this is 0.84, this is 0.85. This would be 45, 475, maybe 48 cubic meters per kilogram of dry air. Go with 848, John. That's 748, 848. And lastly, I want dew point temperature. So for dew point temperature, I don't have a direct look up on the chart. What I have to do instead is follow what would happen if I cooled until condensation began to occur. So on the saccharometric chart, heating represents movement to the right, cooling represents movement to the left. So in this situation, I'm cooling at a constant pressure. I'm moving straight to the left because no water is added nor removed. And I'm doing that until I reach the 100% relative humidity line. And then I'm reading the temperature corresponding to that position. So I can say that's 15, 16, maybe 16.2. Let's call it that 16.2. So comparing and contrasting, I got 11.85 grams of water per kilogram of dry air using the calculation and 11.6 ish using the chart. I had a relative humidity of 75.2% instead of 75 ish percent at a specific volume of 0.86 versus 0.848. A dew point temperature of 16.44 versus 16.2. And a specific enthalpy of 51.2 versus 51 on the nose. So now the question becomes, are those chart look ups accurate enough? And the answer to that question, unsurprisingly, is it depends on what you're doing. If all you're trying to do is eyeball a process, just to get you started on something in the field, then the chart look ups are probably fine. Generally speaking, if you're trying to be exact, you would use the calculations. And in reality, you would probably use a computer to do those calculations for you. Alternatively, there are about 1000 apps on the market that will perform these psychometric calculations for you. However, the chart is still useful. The chart allows us to see trends, we can see kind of what's happening visually, and we can keep track of our processes in a visual manner. Furthermore, if we just need approximate numbers to get us started on something, we can use the chart look ups for that. For the purposes of my class, I expect you to be able to handle the calculations and the chart look ups both.