 Consider a 100 cubic meter room full of atmospheric air at 100 kilopascals, 35 degrees Celsius, and 70% humidity. I want to know the humidity ratio, the dew point, the mass of dry air, the water vapor, and the combined atmospheric air. To begin, let's parse out these numbers. First question is this 100 cubic meters, the volume of the dry air, the volume of the water vapor, or the volume of the atmospheric air? If you said all of the above, you are correct. Because we're using Dalton's law to model the behavior of atmospheric air, all of the substances within the mixture are assumed to take up the entire volume. Next is 100 kilobascals, the pressure of the atmospheric air, the dry air, or the water vapor. That is the atmospheric air, which does not get a subscript. That is not the same as the partial pressure of dry air nor water vapor. It is 35 degrees Celsius, the dry bulb, the dew point, or the wet bulb temperature. That is dry bulb temperature, which doesn't get a subscript, but I'm adding dry just for fun. Now is 70% humidity a relative humidity or a humidity ratio? It is a relative humidity. We know because relative humidity is expressed as a percentage, and also humidity ratios are going to be much smaller numbers than 0.7. So I'm using that information to calculate a variety of properties. First up, I want to calculate the humidity ratio. Well, we know humidity ratio is going to be 0.622 times the partial pressure of the water vapor, divided by the partial pressure of the dry air. We know the mixture pressure, and if we knew either PA or PV, we could figure out the other. But we don't have enough information to plug both of them in right now. We're going to have to come up with a method to calculate PV, and that's going to be from our relative humidity. We know our relative humidity is 0.7, and we know that that represents the proportion of vapor pressure to the saturation pressure, and the saturation pressure can be looked up. That's going to be our saturation pressure at our temperature. So if we look up PG, we can take that number multiplied by 0.7 to get PV, plug that into the numerator here, and then take 100 kPa minus PV to calculate PA, plug that into the denominator, and we'll have our humidity ratio. Our temperature is 35 degrees Celsius. So let's go find the saturation pressure corresponding to 35 degrees Celsius. I'm going to want to use table A2, specifically the bottom of table A22. I see that PSAT corresponding to 35 degrees Celsius is 0.05628 bar. Then PV is going to be 0.7 times that number. So calculator if you would please. 0.7 times 0.05628 yields a vapor pressure of 3.9 kPa. Or 0.0394 bar. So because my mixture pressure, my atmospheric air, is going to be the combination of the partial pressure of dry air and water vapor. That means PA is going to be 100 kPa minus 0.0394 bar, or 1 bar minus 0.0394 bar. So that's 0.9606 bar. Now we have everything we need to calculate the humidity ratio. So if I take 0.622 and I multiply by 0.0394 divided by 0.9606, I get 0.02551. Now what are the units on that? You might be tempted to say it's unitless, because 0.0394 bar divided by 0.9606 bar yields a unitless proportion on the right. But this number is not unitless. That number came from our division of the molar masses. Those molar masses were in kilograms per kilomole, and the kilograms were different. So as a result, I'm left with a proportion of kilograms on the right here. I have 0.02551 kilograms of water vapor per kilogram of dry air. They're different kilograms because they're referring to different things, so they don't just cancel. I will point out that while we're here, it's also pretty common to write this as 25.51 grams of water per kilogram of dry air. So the humidity ratio is 25.51 grams of water per kilogram of dry air. Part B asks for the dew point. Well, the dew point is going to be T-Sat at PV. So we need to take 0.0394 bar and look up a saturation temperature. So I will go back to my tables, hide the calculator, and we want to mosey on over to table A3 to begin with. At the top of table A3, I can see that the lowest pressure on this table is 0.04 bar. Our pressure is 0.0394 bar. So it would probably be perfectly fine for us to answer the question with, oh, it's about 28.96 degrees Celsius. But just in the interest of being as correct as possible on this approximate answer, let's talk about how we can be more accurate. First of all, we could extrapolate from 0.04 bar and 0.06 bar. We could build an equation, extrapolate down to a pressure of 0.0394. But it's better to interpolate if we can. And we can! Remember that tables A2 and A3 have the same information. One is presented in even increments of pressure. One is presented in even increments of temperature. Table A2 actually goes down to lower properties. It goes all the way down to 0.006 bar. And because it's presented in even increments of 1 degree Celsius, it's actually going to be much more accurate for us to interpolate between 28 and 29 using 0.03782 and 0.04008 as the two pressures we are interpolating between. So let's try that. Calculator, you are needed. We have 0.0394, 0.0394, minus 0.03782. Okay, calculator, that is not even close. Divided by 0.04008 minus 0.03782. I mean, we're setting that equal to... Here, let me scroll out slightly. There we go, much more better. We're setting that equal to the thing that we're looking for. Minus 28 divided by 29 minus 28. And we get 28.6991. So 28.7. Take away from this is... Even though the saturation table by pressure seems like it's going to be more convenient for these sorts of look-ups, the saturation table by temperature goes down to lower values. And as a result, it's often more convenient when you're talking about the actual conditions of atmospheric air. Okay, part C. I want to know the mass of dry air. So we are looking for the mass of dry air in the atmospheric air. Well, we are treating each of the species in the mixture as an ideal gas, so I can just use my ideal gas law. PA is the partial pressure of dry air. VA is the volume experienced by the dry air, which is the same as the rest of the volumes because of Dalton's law. RA is the specific gas constant for dry air, and TA is just the dry bulb temperature because the air is experiencing the same temperature as everything else. So for PA, I'm going to plug in 0.9606 bar. For VA, I'm going to plug in 100 cubic meters. For RA, I'm going to go back to table A1. Specific gas constant of dry air is going to be the universal gas constant divided by the molar mass of air. So I'm going to take 8.314 kilojoules per kilomole kelvin divided by 28.97. 8.314 kilojoules per kilomole kelvin divided by 28.97 kilograms per kilomole. And then the temperature of the dry bulb was 35. Kelvin cancels kelvin, kilomoles cancels kilomoles. A bar, one bar, is 10 to the fifth pascals, which is a Newton per square meter. A kilojoule is 1,000 joules. A joule is a Newton meter. Let's see how far that gets us. Well, it doesn't get us very far because I need my bar to be on the other side. So let's try that again. Bar cancels bar. Kilojoules cancels kilojoules. Joules cancels joules. Newtons cancels Newtons. Square meters and meters cancels cubic meters, leaving me with kilograms. So calculator, wake up. We have 0.9606 times 100 times 28.97. It's 28.97 calculator times 10 to the fifth divided by 8.314 times 35 plus 273.15 times 1,000. And we get 108.623 kilograms. Next up, I want us to calculate the mass of water vapor in the air. So take a minute. Think through how we could do that. I'll give you a hint. We have two options, or at least two convenient options. The first option is to set up an ideal gas law relation again. We're treating the water vapor in the atmospheric air as an ideal gas, which means we can repeat this process using the properties of the water vapor. MV is equal to PV times VV divided by RV times TV. Instead of 0.9606, we would use 0.0394. The volume experienced by the water vapor is the same volume as everything else. The temperature experienced is the same temperature. The specific gas constant is going to be different. I'm going to take 8.314 divided by 18.02. So modifying my calculation, this will give me a quantity in kilograms. So 0.0394 times 100 times 18.02 times 10 to the fifth divided by 8.314 times 35 plus 273.15 times 1,000. And we get 2.77127. Did you spot the other way? The other way to get to that answer is this proportion. We know that there are 25.51 grams of water for every kilogram of dry air, or there are 0.02551 kilograms of water per kilogram of dry air. We know, or we knew at that point in the calculation, that we had 108.623 kilograms of dry air, which means if I were to take 108.623 multiplied by that quantity from earlier, 0.02551, we get 2.771, the same mass. Interesting. For part E, I want to know the mass of the atmospheric air. That's just going to be the mass of the water vapor plus the mass of the dryer. So if I hadn't calculated part D, I could just use 1 plus the humidity ratio multiplied by the mass of the dryer. But since I have both numbers conveniently available, let's just use both the numbers. So we get 111.394, and that gets us through A through E. But I think that we can calculate a couple more things while we're here. Just in the interest of character building, let's say that there was a part F and a part G. Let's say for part F, I want to know, oh I don't know, the specific enthalpy of the atmospheric air, and I want to know the specific volume of the atmospheric air. And remember that because these are specific quantities, they're going to be per unit mass of dry air, not per unit mass of atmospheric air. So for part F, I'm going to use the equation we built. We are taking the total, excuse me, the total enthalpy of the atmospheric air, which is the mass of dryer times the specific enthalpy of dryer plus the mass of water vapor times the specific enthalpy of water vapor divided by the mass of dry air. That simplifies to the specific enthalpy of dry air plus the proportion of the mass of the water vapor to mass of dry air times the specific enthalpy of the water vapor. We are approximating the specific enthalpy of the dry air by assuming constant specific heats. So we're using the Cp of air multiplied by the temperature of the air in degrees Celsius, remember? Plus the humidity ratio times, and then we're approximating the specific enthalpy of the water vapor by using the saturated vapor specific enthalpy at our temperature. So we're using this equation here. So for the Cp of air, we're going to have to use a Cp value from table A22, excuse me, from table A20. And it's not really worth our effort to interpolate. We're just going to use the closest Cp value to our range. We have 35 degrees Celsius, which when added to 273.15 yields about 310 Kelvin, which is about 300 Kelvin. So we're going to approximate by using the value at 300 Kelvin, which is 1.005. And then which temperature do we plug in? If you said 35 degrees Celsius, you are correct. It is not converted to Kelvin. The reason it's not converted to Kelvin is because we want the same zero point for air as we do for water. Since all of our enthalpies in the steam tables are relative to the zero being at zero degrees Celsius, which by the way, while I'm here, I might as well point out. In the steam tables at zero degrees Celsius, we are using an enthalpy of zero. So at 0.01 degrees Celsius, it's an enthalpy of 0.01. Therefore, to add together our enthalpy of the water vapor and our enthalpy of the dry air, we need the zero points, the datum points for our enthalpy difference to be the same. Or rather, it's most convenient if they are the same. We don't really need it, but we need it if we are going to be comparing this against the psychometric chart, which we're going to be doing later. Then we add our humidity ratio, which we had as 0.02551 kilograms of water vapor per kilogram of dry air. Then I'm going to get rid of these lines so it doesn't look like we're dividing. Then we are adding in Hg at temperature. So we need the specific enthalpy of a saturated vapor at 35 degrees Celsius. So 35 degrees Celsius, Hg is 2565.3. That's kilojoules per kilogram of water. So kilograms of water cancels kilograms of water. Kelvin cancels degrees Celsius. So we're left with kilojoules per kilogram of dry air. So calculator. If we take 1.005-ish multiplied by 35 plus 0.02551 times 2565.3, we get about 100.616. And that's kilojoules per kilogram of dry air. Then for part G, I want us to determine the specific volume of the atmospheric air per unit mass of dry air. So for that, we're taking the total volume divided by the mass of dry air. Interesting thing about that, though, is that because we're using Dalton's Law, the volume of the atmospheric air is the same as the volume of the dry air, which means, I can write this as the volume of the dry air divided by the mass of the dry air, which means that I'm just looking for the specific volume of dry air at these conditions. Specifically, the partial pressure of the dry air and the temperature of the air, which is the same temperature as everywhere else. So if Pv is equal to RT, I can write V as RT over P, rA times T over PA. So again, I'm going to take the universal gas constant, 8.314 kilojoules per kilomole kelvin, available from the conversion factor sheet on the inside of the front cover of your textbook. Multiply by our temperature, this time in kelvin, divided by the partial pressure of the dry air, which was 0.9606 bar. And then we want a quantity in cubic meters per kilogram. Oh, hey, I forgot to divide my universal gas constant by the molar mass. I'll just scooch that over there. Molar mass of dry air from table A1 is, as always, 28.97. 28.97 kilograms per kilomole. Then I'm going to have to do some unit conversion. So one bar is 10 to the fifth newtons per square meter. And I need a kilojoule to cancel out the newtons in meters. So a kilojoule, thousand joules. And a joule is a newton times a meter. Joule cancels joule, kilojoule cancels kilojoule. Kilomoles cancels kilomoles, kelvin cancels kelvin. Bar cancels bar, newtons cancels newtons, square meters and meters. Don't cancel, they just hang out. So we're left with cubic meters per kilogram. So calculator, if you would be so kind. We have 8.314 times 35 plus 273.15 times 1,000, divided by 28.97 times 0.9606 times 10 to the fifth. And we get 0.920622. And that's cubic meters per kilogram of dry air. And now we're really done with the problem. I mean, there's more stuff we could add to this, but I think this is the best point to break.