 Problem number three, dry winter atmospheric air at one atmosphere five degrees Celsius and 30% humidity enters an HVAC system at a rate of 10 cubic meters per hour where it is to be isobarically conditioned to 21 degrees Celsius and 50% humidity. A water line supplies saturated water vapor if necessary. I want to know the heating or cooling rate required to accomplish this process and if water has to be added how much water has to be added. Like in the previous problem we have the dollhouse volumetric flow right here so we're going to use T1 and phi1 to fully define the inlet atmospheric air. We're going to figure out how much energy difference and mass difference it takes to get to this outlet temperature and relative humidity. To figure out if we have heating or cooling and humidification or dehumidification let's look at this process on a psychrometric chart. So I'm going to start by finding five degrees Celsius and 30% relative humidity. There's five degrees Celsius and this is 10, this is 20, this is 30. So state one is right here and then state two is 21 degrees Celsius and 50% relative humidity which is going to be right here. So we have movement to the right which implies heating and up which implies humidification and just like with projectile motion even though in reality this is going to be more of a process that looks like this I'm going to split out the x component and y component and analyze them independently. So I'm going to establish a hypothetical state point here which is going to serve as the halfway point. So I'm going to call this state one, this state two and this state three and the actual process is one to three but two is a nice stepping stone from one to three. So from one to two I'm going to model this as simple heating where I'm just increasing the temperature to the desired temperature and then from two to three I'm going to call that humidification with no change in temperature. So I will call this state point W again and state point W is a saturated vapor. So note that that means we have a separate boiling process that we aren't considering in this analysis. Somewhere water is boiled or evaporated in a humidification system and the resulting vapor is being added into our ductwork to supply the increase in humidity required to go from state one to state three. So I'm splitting them into two separate analyses one changing just temperature one changing just omega. So at state one I know a temperature and relative humidity that's five degrees Celsius and 30 percent. State two is just the increase in temperature therefore omega two is equal to omega one, t two is equal to t three, t three and phi three are 21 degrees Celsius and 50 percent relative humidity. Yep. So t one and phi one fully define state one from which I can look up or calculate any psychometric property that I need. State threes, t three and phi three again fully define state three from which I can look up or calculate any psychometric property that I need and then the fact that omega two is equal to omega one and t two is equal to t three fully define state two and allows me to look up or calculate anything that I need there. Let's start this off with a mass balance and energy balance. I have three different control volumes I can analyze if I choose to. I have the control volume from one to two which I will call CV one. I have the control volume from two to three which I will call CV two and then I could set up a control volume on the process from one all the way to three if I wanted to. So I have the option of a mass balance on three different control volumes with three different substances atmospheric air dry air or water vapor. Let's start with the dry air in control volume one. I have an open system operating steadily. There's no opportunities for dry air to enter or exit my system other than state one and state two which means that m dot a one is equal to m dot a two which I can just call m dot a. The water vapor is likewise simple because there's no opportunities for water to exit or enter control volume one other than as water vapor at state one and state two and from these two quantities I concluded that omega one is equal to omega two that's how I was able to define one of my two independent intensive psychometric properties at state two. So I can set up a control volume two mass balance. I'll start with dry air. The dry air is pretty simple. There's no opportunities for dry air to enter or exit my system except for state two and state three. So I'm going to say m dot a two is equal to m dot a three which is just m dot a again. Now water is the one that's different this time. I have two opportunities for water to enter control volume two that's as water vapor at state two and as water vapor at state w. Those two sources of entering water have to leave at state three. For convenience I can divide by m dot a at which point this would become m dot w over m dot a is equal to omega three minus omega two. So m dot w is going to equal m dot a multiplied by omega three minus omega two. And if I were to set up a control volume three which was the entire analysis from one all the way to three the combination of control volume one plus control volume two. I would have entering dry air at state one and exiting dry air at state three which means m dot a one is equal to m dot a three and my water vapor would look essentially the same as it does in control volume two. I would have m dot v one plus m dot w is equal to m dot v three which we know anyway because m dot v one is equal to m dot v two. I have the option of setting up an energy balance on control volume one or control volume two or the combination of control volume one and control volume two but let's think about what we actually want. All I asked for in terms of energy is the heating or cooling rate so all I'm looking for is this q in here which means if I'm doing as little work as possible to try to answer the question all I need to do is an energy balance on one to two. I have an open system operating steadily so e dot in is equal to e dot out. I have no opportunities for work I'm neglecting changes in kinetic and potential energy. I have no heat transfer out therefore m dot a one h one plus q dot in is equal to m dot a two h two so q dot in is equal to m dot a times h two minus h one and imagine in your mind that I had asked you hey you guys why is it m dot a one times h one and not just m dot one h one and then for you to apply with because we define the atmospheric air per unit mass of dry air therefore we need to multiply by the mass of dry air to get it into total energy to combine with the heat transfer in and our energy balance okay glad we're getting good at that so m dot a times h two minus h one we can calculate m dot a if we know the specific volume at state one because we have dollhouse volumetric flow rate we're going to want h one and we are going to want omega one and then at state two we're going to want h two at state three we are going to want omega three so I believe I only need those five quantities to answer everything I need for this question so again I have two independent intensive psychrometric properties from which I can determine anything that I need either with the psychrometric calculations or the psychrometric chart so would you like to do the lookups on the chart or calculate them by hand well we've done the chart lookups for the previous two example problems and just to try to underline the point that we shouldn't get too comfortable with those let's try it with the hand calculations so I'm going to use t one and f one to determine the specific volume of dry air at state one the specific enthalpy of atmospheric air per unit mass of dry air at state one and the humidity ratio at state one I'm going to use that humidity ratio with the temperature at state three to determine the specific enthalpy at state two and then I'm going to use the the temperature at state 3 and the relative humidity at state 3 to determine the humidity ratio at state 3. And for that let's step to another sheet of paper just to try to prevent as much of a mass as possible. So let's start with VA1. For VA1 I'm going to take the total volume divided by mass and that total volume is the volume of the atmospheric air which is the same as the dry air because of Dalton's law and the mass is per unit mass of dry air which means that this is the specific volume of dry air which is just going to be r air times t divided by the partial pressure of the dry air. And in an effort to reduce how much arithmetic I'm doing I'm going to write that as p minus pv at state 1. r air is the universal gas constant divided by the molar mass of dry air, t1 is known, p is known, pv at 1 can be determined by recognizing that phi1, the relative humidity at state 1 is the vapor pressure divided by the saturation pressure at state 1. So I need to look up the saturation pressure corresponding to t1, multiply that by my relative humidity at state 1, that'll give me pv1 from which I can calculate the specific volume at state 1. So p sat at t1 is going to come from my saturation tables, I have a temperature of 5 degrees Celsius. So at 5 degrees Celsius I have a saturation pressure of 0.00872 when I multiply that by my relative humidity of 0.3 I will get pv which I can plug into my calculation. So I will involve my calculator as little as possible because you guys know my calculator. I'll write that as pv1 times pg1. And then instead of writing the specific gas constant of dry air I will write the universal gas constant 8.314 kilojoules per kilo mole kelvin times t1 which is 5 degrees Celsius converted to kelvin. And then our pressure was one atmosphere which is equal to 101.325 kilopascals which is equal to 1.01325 bar minus v1 which is 0.3 times the saturation pressure which was 0.00872 bar. And then I want that quantity to be in cubic meters per kilogram. So I will write one kilojoule is a thousand joules which is a thousand Newton times meters. And then a bar is 10 to the fifth Newtons per square meter. Newtons cancel Newtons, kilojoules cancel kilojoules, kilo moles just hang out because I didn't involve the molar mass yet. So to get the specific gas constant I have to divide by the molar mass of dry air which comes from table A1 which is 28.97 kilograms per kilo mole. And now kilo moles cancel and then bars cancel, kelvin cancels kelvin leaving me with meters and square meters per kilogram which is going to be cubic meters per kilogram. And if you're thinking to yourself hey John why aren't you just combining this with the volumetric flow rate to calculate a mass flow rate directly that would eliminate steps of arithmetic which I know you're all about. Well that's true but my goal here is going to be to determine the specific volume and the specific enthalpy and the specific humidity i.e. the humidity ratio so that I can compare them to the chart look-ups. So we get, that can't be right. Yes, I forgot to tend to the fifth. This is why I can't talk and type at the same time. I can barely type in this calculator. So we get 0.7898, 0.78, 985 cubic meters per kilogram of dry air. And then let's compare that to the chart look-up. So at state one we would have gotten oh about 0.79. Honestly essentially exactly 0.79. So look guys instead of using 0.79 we're using 0.7899. Much better, right? Much more accurate. Next up I want H1. H1 is going to be Cp of air times T1 in degrees Celsius plus omega 1 times Hg1 which means that I need humidity ratio first. Humidity ratio is 0.622 times pv divided by p minus pv. Well instead of pv I'm going to write phi times pg because I'm all about avoiding arithmetic when I can. So pg1 divided by p minus phi1 times pg1. And I just need the proportion on the right to cancel internally. So as long as I plug in bar for everything. 0.622 times 0.3 times 0.00872 divided by 1.01325 minus 0.3 times 0.00872 gives me a humidity ratio of 0.00161. 0.00161 kilograms of water per kilogram of dry air. That would be 1.6 grams of water per kilogram of dry air. Compare that to the chart. So let's say 1. That's a little bit above the third line from the bottom. So a little bit above the third line from the bottom. It might be about here. So we probably would have said about 1.6 ish grams per kilogram. And if you compare our calculated value of 1.61 to what we would have eyeballed maybe 1.6 or 1.7 that accuracy is what we are expending all of this extra work for. That's why the chart lookups are so okay for these sorts of calculations. CPU of air 1.005 kilojoules per kilogram Kelvin. Again that comes from table A20. CPU of air at 300 Kelvin is 1.005 kilojoules per kilogram Kelvin. And then we are multiplying by our temperature which is 5. And then because we want to add together these two quantities I need the zero points to be the same. So I am using 5 degrees Celsius not 5 converted to Kelvin. And then we are adding in our shiny new omega which is 0.00161 kilograms of water per kilogram of dry air. And then we are multiplying by hv which we are approximating by using hg at 5 degrees Celsius. So back to the steam tables. We get an hg value of 2510.6. 2510.6 kilojoules per kilogram of H2O. That leaves me with kilojoules per kilogram of dry air because this is the heat capacity of the dry air itself. Which means that I will have, come on and calculate it. You can do this. I believe in you. 1.005 times 5 plus 0.00161 times 2510.6. We get an h1 value of 9.067 kilojoules per kilogram of dry air. Let's compare that to what we would have gotten on the chart. We would have gotten a number that's off the chart. Well, this line is 10. You can read it down here by the way as a pro tip. 9, 8 because it's going to be the same at zero. Because at zero humidity ratio we have pure dry air for which the h value is going to be 1.005 times the temperature therefore h1 is almost exactly the same as T1. That minor difference is why they aren't exactly on top of each other, especially as we get further to the right. See over here it's almost half a degree off. But it's a good way to keep track of where your enthalpies are. Actually, while I'm rambling like that, h is also written over here on the right side. So if you're trying to follow a line above the chart, this is 120 as opposed to trying to go, oh, I don't know, about 120. Anyway, so we're at, oh, about 9? I mean, it's easy to say that having just calculated the actual quantity, but we probably would have said, well, we're not quite 2. So we're not quite 8. They're about halfway between 8 and 10, which is about 9 kilojoules per kilogram dryer. Whatever the case, we now have everything we need for day one. We have three fifths of our quantities calculated. So without much ado, let's determine h2 at Cp of air times t2 plus the humidity ratio at state two times hg at t2. Because t2 is equal to t3, that means we're using 21 degrees Celsius for everything here. And because omega two is the same as omega one, we're using 0.00161 again. So our h2 calculation goes 1.005 times 21 plus, oops, 0.00161 times hg at 21 degrees Celsius, which is 2539.9. Calculator, come on. 2539.9. So we get an h2 value of 25.1943 kilojoules per kilogram of dry air. So comparing 25.2 to what we would have gotten on the chart, we would have gotten, well, that's about year-ish. So we probably would have said maybe 26, maybe, depending on our ability to draw a straight line. I mean, you guys probably have something, some straight edge that you can use when you're physically reading the chart. I'm using a PDF here. So best I can do is draw a straight line and rotate it and try to get it close to parallel. So like something like that, maybe. Where are you trying to pop to an angle? That's not great. I mean, probably would have said 26 or so. We would have missed out on those 0.8 kilojoules per kilogram. Then at state three, we are trying to determine the humidity ratio at state three. So for the humidity ratio, we have a calculator. We have 0.622 times the pv value at state three divided by p minus the pv value at state three, which I'm going to substitute in, phi times pg in an effort to try to reduce the steps of calculation, v times pg at state three. It's a lot of rhyming donations, p minus phi three times pg three. We need pg at state three, which is going to be p sat at 21 degrees Celsius. Back to our steam tables, we get 0.02487 bar 0.02487. And then P is 1.01325. And as long as the quantity on the right is all in the same units of pressure that will cancel, leaving me with a unit less proportion multiplied by the units of 0.622, which are kilograms per kilogram. So calculator, you're needed again, zero points. Nope, not less than or equal to 0.622 times v three, which I believe was 0.5. Yes, 0.5 times 0.02487 divided by 1.01325 minus 0.5 times 0.02487. And I get 0.00773. And that's kilograms of water per kilogram of dry air. And if we compare that to what we would have gotten on the chart, this time I can draw an actual straight line. Move that to about there. So had I been reading this, I probably would have said, well, here's seven, here's seven and a half, or 7.75. So we're probably about 7.8. And 7.8 versus 7.73. Is that enough to matter? Depends on the circumstance. But now that we have those five quantities, we can actually calculate what we're trying to do way back over here, which you'll remember is what we're actually trying to calculate. So instead of mass flow rate of dry air, I'm going to calculate volumetric flow rate divided by specific volume. I could combine these together. But this will give me another vector to compare. Actually, I guess it won't because it's a different endless state. Let's just combine symbolically. So I'm done a is the total volume divided by the specific volume of dry air at state one. So in our Q in calculation, I guess we're going to need that for the water. Anyway, let's just calculate the mass flow rate guys. We have a volumetric flow rate of 10 cubic meters per hour. And we are dividing by the specific volume that we just calculated, which was over here somewhere 0.78985 0.78985 cubic meters per kilogram of dry air cubic meters are going to cancel cubic meters one hour is 3600 seconds, leaving me with kilograms of dry air per second. We should get a really small number because it's a dollhouse 78985 times 3600. And we get 0.003517 kilograms per second. And then we take that number multiplied by our difference in enthalpy. So that would be 0.003517 kilojoules, excuse me kilograms per second multiplied by H2, which is 25.1943 minus H1, which is 9.067. That's in kilojoules per kilogram of dry air. So the kilograms of dry air will cancel leaving me with kilojoules per second and we get 0.0567 kilowatts. And I believe I asked for kilowatts. Yep, I did. So the best distancer is probably 56.7 watts, but got a circle both. And then the mass flow rate of water required here is going to be 0.03517 multiplied by the difference in omegas. Omega three we got as 0.00773 minus Omega one, which is 0.00161. Note that we are taking kilograms of water per kilogram of dry air multiplied by kilograms of dry air per second, which leaves us with kilograms of water per second. I wanted an answer in grams per second. So we actually have to divide that by 1000. Excuse me, multiply that by 1000. And we get 0.022 grams of water per second. And those are our two answers. It's a shame we didn't look up the psychometric properties first on the chart and then calculate the quantities so that we could do like a percent error introduced by using the chart. I guess maybe take that as an opportunity to try a percent error calculation on your own.