 Two atmospheric air streams are mixed steadily and adiabatically. The first stream enters at 32 degrees Celsius and 40% humidity at a rate of 20 cubic meters per minute. And the second stream enters at 12 degrees Celsius and 90% relative humidity at a rate of 25 cubic meters per minute. Assuming that the mixing process occurs at a pressure of one atmosphere, determine the following for the exit. The specific humidity in grams per kilogram, the relative humidity, the dry bulb temperature, and the volumetric flow rate. I will start with a diagram. So I have two air streams that are mixing together and producing an outlet. The first stream is at 32 degrees Celsius and a relative humidity of 40% and has a volumetric flow rate of 20 cubic meters per minute. The second air stream has a temperature of 12 degrees Celsius, a relative humidity of 90%, and a volumetric flow rate of 25 cubic meters per minute. I can use that information to fully define state three from which I can determine anything else that I want. And to do that, I'm going to set up a mass balance and an energy balance on my control volume. The mass balance on the control volume can consider just the dry air, just the water vapor or the atmospheric air itself. The atmospheric air mass balance isn't generally very useful, so I will split it up. I'm saying that for the water, whatever enters has to leave because it's steady state. So m.v1 plus m.v2 has to equal m.v3, likewise for the dry air, because I have steady state operation of an open system. Nothing can accumulate, so whatever mass flow rate enters has to leave, which means m.v1 plus m.v2 must equal m.v3. So unlike the previous couple of examples, it's not going to be just a matter of dividing by m.vA, but my approach is going to be the same in that I'm going to try to write this as quantities that are intensive as much as possible. So I'm going to take this equation and I'm going to recognize that for any state point, omega represents the mass of vapor per mass of dry air and the mass flow rate of vapor per mass flow rate of dry air, which means that I can write m.v as omega times m.a. So in my mass balance of the water vapor, I can write omega1 times m.a1 plus omega2 times m.a2 is equal to omega3 times m.a3. In this analysis, my goal is going to be to fully define state point 3, so I will solve for omega3 and write m.a1 over m.a3 times omega1 plus m.a2 divided by m.a3 times omega2, where m.a3 is just the sum of m.a1 and m.a2. So what I have is a mass weighted average. I end up with something similar looking if I set up an energy balance because I have an open system operating steadily. Whatever energy enters must leave and if I'm treating this mixing process as being adiabatic, that means no opportunities for heat transfer. If I'm neglecting work because there's no opportunities for work and I neglect any changes in kinetic and potential energy, then what I'm going to be left with is the sum in of m.h is equal to the sum out of m.h. So I'll write this as m.a1 h1 plus m.a3 h3 is equal to m.a2 h2 is equal to m.a3 h3 or h3 is m.a1 over m.a3 times h1 plus m.a2 divided by m.a3 times h2. So the enthalpy and the humidity ratio are mass weighted averages of states 1 and 2. Once I have h3 and omega3, I can use those two independent intensive properties to fully define state 3 from which I can determine any other psychrometric properties that I want, including but not limited to the humidity ratio, the relative humidity, the specific volume which when combined with the mass flow rate will give me volumetric flow rate and the dry bulb temperature. So my approach is going to involve determining h1 and omega1 from T1 and phi1 as well as h2 and omega2 using T2 and phi2 and then looking up the specific volumes so that I can calculate a mass flow rate from the volumetric flow rates. Using those mass flow rates to determine a mass flow rate at state 3, using my mass flow rates to average together omega1 and omega2 to determine omega3 as well as h1 and h2 to determine h3 from which I can determine v3, v3 so that I can determine volumetric flow rate from my mass and T3. That's my approach. Since we have an atmospheric pressure of one atmosphere and since I could probably get away with approximating the property determination with the psychrometric chart, we have the option of doing the psychrometric calculations by hand or with the chart. Why don't we try it both ways just to compare and contrast the process. So I will position states 1 and 2 on the psychrometric chart and then locate state 3. So at state 1, I had a temperature of 32 degrees and a relative humidity of 40 percent. 32 degrees is going to be this line here. 40 percent is going to be this line which means that my intersection is right here. There's state 0.1. I'll switch to a different highlighter. How about blue at state 2 with blue? I have a temperature of 12 degrees Celsius and a relative humidity of 90 percent which means that my state point 2 is right here. Interesting thing about a mixing process like this. State 3 is going to lie on the line connecting states 1 and 2. If I had equal mass flow rates of 1 and 2, it would be right in the middle. If I had twice as much mass flow rate at 1 as I do at 2, it would be positioned a third of the way over. The relative positioning is based on how much mass flow rate of each there is. So using the chart lookup at state 1, I have an h1 of let's call that 63 maybe. 62 would be right here. 64 would be right here. So 63 ish. h1 is about 63 kilojoules per kilogram of dry air. Omega 1 is going to be about 12. Let's just call it 12 exactly. 12 grams of water per kilogram of dry air. And specific volume 1 is this line is 0.88. So 85, I guess 885, 8825, maybe 881, even a little bit less than that. Let's just call it 881. 0.881 cubic meters per kilogram of dry air. And then we repeat the process at state 2. So at state 2 we have an enthalpy of probably 32, maybe just a skosh more. 32 kilojoules per kilogram. Omega 2 is going to be about 8. I'll draw a straight line so that we can see a little bit more closely. But I wrote it so it'd be convenient. About 8, let's just call it 8, maybe 7.95 if we wanted to be real accurate but let's not be too arbitrarily precise. 8 grams of water per kilogram of dry air. And that specific volume is 0.818, maybe. And this is 0.81. This is 0.82. Here'd be 815, 8175. Let's call it 8. 0.818 cubic meters per kilogram of dry air. So comparing that to the hand calculations, at state 1 I had a temperature of 32 degrees celsius and a relative humidity of 40%. H1 would be cp of air times T1 in celsius plus omega 1 times hv. And we are approximating hv with hg at T1. So cp of air is going to come from table A20. That is 1.005. So if I wake up my calculator, I've got work to do. Calculator 1.005 times the temperature at state 1, just 32 degrees celsius plus omega at state 1 we don't know yet. So let's calculate omega. Omega at state 1 is 0.622 times pv1 over p minus pv1. pv1 is going to be p1 times pg1. And pg1 is psat at T1. So if we go into our steam tables and we find a temperature of 32 degrees celsius, I can see that my saturation pressure is 0.04759. 0.04759. And then we are multiplying that quantity by 0.44759 bar to get the vapor pressure. So calculator we can optimistically delete that start over 0.622 times 0.4 times 0.04759 divided by 1.01325 because one atmosphere would be 101.325 kPa which is 1.01325 bar minus 0.4 times 0.04759. And I get a syntax error as is tradition. My humidity ratio at state 1 is 0.01191. And remember that's kilograms of water per kilogram of dry air. So if we're doing an apples to apples comparison that would be 11.9 grams per kilogram instead of the 12 value that we got from the chart. Now that we have the humidity ratio we can finish our h1 calculation 1.005 times T1 in celsius plus omega 1 times hg at 32 degrees celsius back into my steam tables. Hg at 32 degrees celsius is 2559.9. So calculator if you would please 255 no two there you go 559.9 h1 is 62.65. So compare 63 versus 62.65. That's the accuracy that we get when we do the calculations by hand. Lastly I want to calculate specific volume and that would be r times T over P. Bear in mind that the specific volume of the atmospheric air is expressed per unit mass of dry air and because I'm modeling the behavior of my atmospheric air using Dalton's law the volume of the atmospheric air is equal to the volume of the dry air which means that the specific volume of the atmospheric air per unit mass of dry air is equivalent to the specific volume of the dry air which means that I'm just calculating the specific volume of dry air by using specific gas constant of air times the temperature of the air which is just T1 divided by the partial pressure of the air which is P minus PV. Since I didn't calculate PV earlier I'm going to plug in phi times Pg and in place of the specific gas constant for the air I'm using universal gas constant divided by the molar mass of dry air. So altogether this would be 8.314 kilojoules per kilo mole kelvin divided by 28.97 kilograms per kilo mole. Those two quantities come from the inside of the front cover of the textbook 8.314 kilojoules per kilo mole kelvin and table A1 respectively 28.97 and then we are multiplying by T1 which is 32 plus 273.15 kelvin and we are dividing by P minus phi times Pg. Okay calculator let's see if we can't do this actually let's write that out so that I don't lose track of the unit conversions 8.314 kilojoules per kilo mole kelvin divided by 28.97 kilograms per kilo mole multiplied by T1 which is 32 plus 273.15 kelvin divided by 1.01325 minus 0.4 times 0.4759 0.4759 let's definitely make sure that I don't interpret that as a 6 0.04759 still looks like a 6 fingers crossed and since my goal is going to be to represent the specific volume in cubic meters per kilogram here let's give myself a little bit more space I'm going to have to write one bar is 10 to the fifth newtons per square meter and a kilojoule is a thousand newtons times meters so kilojoule cancels kilojoule newtons cancels newtons bar cancels bar kelvin cancels kelvin leaving me with cubic meters per kilogram so 8.314 times 32 plus 273.15 times a thousand divided by p no 28.97 times 1.01325 minus 0.4 times 0.04759 0.04759 look at that didn't interpret a single 6 times 10 to the fifth and we get a specific volume of 0.881 so comparing contrast 0.881 against 0.881 look at the magnificence of our extra what ten thousandths place okay state one done now we are going to repeat essentially the same process for state two because it's the same given independent intensive psychrometric properties my procedure for calculating state two's properties is going to be the same so I'm just going to start with copying and pasting the framework and see if I can't properly substitute everything t2 is now 12 and 90 12 degrees Celsius 0.9 so h2 is going to be dp of air times t2 plus omega 2 times hg of t2 and there's where I would forget that I need omega first so I will calculate that omega 2 is equal to 0.622 times pb2 divided by p minus pb2 and in order to be efficient and real lazy I'm looking in phi times pg instead of dv v2 is going to be r times t2 over p2 okay I think that's everything so first up p sat at 12 degrees Celsius back to our steam tables 12 degrees Celsius has a p sat of 0.01402 0.01402 01402 so omega 2 is going to be 0.622 times 0.9 times 0.01402 divided by 1.01325 minus 0.9 times 0.01402 and I get a humidity ratio of 0.00784 0.00784 which I can use to calculate the enthalpy at state 2 which is going to be 1.005 times 12 because I need them to be added together they need to have the same arbitrary zero point so the enthalpy of water is evaluated relative to a zero point at zero degrees Celsius therefore the enthalpy of the dryer has to be evaluated relative to a zero point at zero degrees Celsius then 0.00784 times the specific enthalpy of a saturated vapor at 12 degrees Celsius back to the steam tables hg at 12 is going to be 25 23.4 25 23.4 and we get 31.851 so 31.851 let's compare those two numbers against what we got in the chart and we get 32 instead of 31.85 and we get 0.0078 instead of 8 so these were conservative we didn't even try to use a ruler or a scale to try to come up with better values and these are still pretty close all things considered lastly I'm going to be plugging in 12 because that's my temperature here and then 0.9 times 0.0142 so for my specific volume I'm going to get 8.314 times 12 plus 273.15 times a thousand divided by 28.97 times 1.01325 minus 0.9 and 0.0142 times 10 to the fifth and I get 0.81783 1783 cubic meters per kilogram of dry air versus 0.818 which is what we got when we use the chart so again I'll ask is that fourth decimal place really that important probably not but hey we are being as accurate as possible in order to understand the approximations that we make when we use the chart maybe I'll put that down here just to make that a little bit more clean okay so now we can use our specific volumes and our volumetric flow rates to calculate a mass flow rate so for that I'm going to use the calculated values just because those are better but if you were working this problem it would be completely reasonable to use the specific volume for the purposes of this example so m.a1 is going to be v1 divided by a specific volume 1 because that specific volume is expressed per unit mass of dry air and volumetric flow rate at state 1 was 20 cubic meters per kilogram and our specific volume no not cubic meters per kilogram what was the time unit minute 20 cubic meters per minute that's better okay 20 cubic meters per minute divided by our specific volume at state 1 which was 0.88084 0.88084 cubic meters per kilogram of dry air presumably I want kilograms per second later on although it doesn't really matter here because I'm not doing anything with the result so minute is 60 seconds and it's canceled minutes cubic meters canceled cubic meters and I'm left with kilograms per second so 20 divided by let me just double check that I wrote down my specific volume correctly 0.880836 multiplied by 60 and I get 0.37843 0.37843 cubic meters per kilogram kilograms per second try John kilograms of dry air specifically per second and I can repeat the process for m.a2 which is going to be v2 divided by the specific volume of dry air at state 2 which is 25 cubic meters per minute so confident am I in that number then I'm going to go back and check it yep divided by 0.81783 0.81783 cubic meters per kilogram of dry air and one minute is 60 seconds cubic meters cancels cubic meters therefore I get kilograms per second so 25 divided by 0.81783 times 60 and I get 0.50948 so m.a1 is 0.37843 kilograms per second and m.a2 is 0.50948 kilograms of dry air per second and then from my mass balance of the dry air I know that m.a3 has to be the sum of those two numbers which is going to be 0.888 0.88791 kilograms of dry air per second so now I have everything I need to determine h3 and omega 3 using the mass weighted averages that I determined back on this page so I'm going to take 0.378 divided by 0.888 times omega 1 which is 0.011909 plus 0.5095 divided by 0.888 times omega 2 which is 0.00784 let me just scroll through and make sure that that looks correct and it does look correct to me so I get an omega 3 of 0.00784 0.00784 that'd be kilograms of dry air no kilograms of water per kilogram of dry air and I repeat the process for h3 so 0.3 and then some numbers divided by 0.5 and then some numbers multiplied by h1 which was 62.65 let's go 647 too many decimal places and we are adding to that 0.50948 divided by 0.888 times h2 which was which was 31.851 and I get 64.8088 that's kilojoules per kilogram of dry air from these two quantities I can fix my position on the chart and determine anything else that I want or I can determine those with hand calculations so 0.0078 is going to be that can be right so with a value of omega 3 that is actually the correct value and not this incorrect value that was 0.009574 I wonder why I wrote down the wrong one no one will ever know let's see 0.9574 would be about nine and a half go a smidge and more and then because I know that has to be connecting the two that's actually enough to position it but I will point out while we're here that that's also about 45 which is nowhere near what I got for my enthalpy why is that nowhere near oh that's why okay we're divided by m dot three which was 0.888 that gives us about 44.97 I could go back I could edit out that mistake and pretend like it never happened but I think that that's a good way to remember that even if you're not using the psychometric chart as a direct lookup it can be a convenient way of sanity checking your calculations I mean if we're doing a mass weighted average we know that we should end up somewhere between this number and this number and the fact that I ended up with an enthalpy that was higher and also a specific volume that was the same as the specific volume at day one those were strong indications to me that those numbers were not correct keep in mind you shouldn't assume that you will make no mistakes you should build in mechanisms that catch the mistakes when they happen anyway now that I have a specific enthalpy and a humidity ratio I can determine the rest of them so t3 from the chart it's going to be about 20.3 the specific volume from the chart is going to be about 0.843 and the relative humidity is going to be about 62 63 percent but in the interests of calculating these quantities by hand for practice let's recognize that omega three is going to be 0.622 times pv3 over p minus pv3 so again it's algebra time omega three times p minus omega three times pv3 equal to 0.622 times pv3 so pv3 times the quantity 0.622 plus omega three and then I take omega three times p divided by that quantity and I get 0.622 plus omega three in the denominator so omega three times the pressure I'm saying that the mixing process is happening isobarically so p3 is equal to p1 which is equal to p2 therefore if I take 0.009574 multiplied by the pressure in bar which is 1.01325 divided by 0.622 plus 0.009574 that will give me a vapor pressure in bar h3 is equal to cp of air times t3 plus omega three times hg at t3 this is a little bit more complicated because I have to work my way to an answer as a result of a guess and check process I mean this is essentially three equations and three unknowns I'm saying 0.0154 bar is equal to the saturation pressure times the relative humidity which I don't know and I'm saying that for a given t I can look up hg and I'm using h3 to determine t3 and hg at t3 so I essentially have four equations and four unknowns it's just that two of the equations are property lookups which means I can't actually do the algebra to solve for those equations so instead I have to guess and check my way through so the way I'm going to approach this is guessing a t3 looking up an hg and then calculating what that would yield for an h3 and then repeating until my h3 is what I know the actual h3 is which is 44.976 I mean again we know from the chart lookup t3 is just 20 and change 20.2 maybe but in a situation where we didn't have the chart this is how we would have to solve for that temperature so the way that we would start this process is by assuming temporarily that t3 was a mass weighted average of t1 and t2 it's not because the relationship isn't just linear but let's assume that it was in order to get a starting point so my mass flow rate is day one you guys know how good I am at this calculation actually you know what since we have one that works let's just start there and plug in t2 which is 12 and t1 which is 35 question mark 32 so that gives me a starting temperature of 30 no that's still wrong I didn't correct this okay the denominator should have been 0.88791 and 0.5 okay so the starting temperature we're going to use is 20.524 20.524 and then we're going to interpolate a value for hg so 20.524 is going to occur between 20 and 21 so 2538.1 and 2539.9 so calculator if you would please it's going to be 20.524 minus 20 divided by 21 minus 20 is equal to x minus 2538.1 divided by 2539.9 minus 2538.1 we get a value of 25 39.04 and then using those quantities to calculate h3 I get 1.005 times 20.524 plus my omega 3 which I know 0.009574 times 2539.04 and we get 44.935 so that's not exactly the same it's probably close enough to be fine but in the interest of demonstrating this process let's think through what we would do next our h3 didn't sense our h3 isn't perfect we are repeating the gas process and think about whether or not we should increase or decrease t3 bear in mind that hg is a function of temperature and cp times temperature is a function of temperature so in order to increase h3 we're going to have to increase the temperature so let's increase by oh let's go with 0.1 shall we I mean we could be a little bit more scientific about why we're increasing the number to the quantity that we are but just in the interest of getting somewhere to start let's see use 20.624 so I will interpolate again using 20.624 to get a new hg which is 25 39.22 and in fact we could rewrite this on my ti 89 so that the temperature was a thing that we plugged in at the end and that would make that process go a little bit faster but we are in this for the guess and check part we might as well do the guess and check part manually I mean if we go that far we might as well just open up matlab so 45 0.0376 oh look we went up a little bit too much so let's back off a little bit maybe 20.55 that by the way is where the power in starting with a mass weighted average really shows itself because we ended up real close 2539.09 2539.09 and you know we could also think through if we keep this process up a couple of times we could generate a line and then use a line of best fit that would be another way to try to be as accurate as possible 39.09 44.96 44.962 so do you guys think that's close enough I mean probably but hey I got the space maybe five six what do you think it's five six going to be the number I think it's going to be the number but let's see what happens five six 2539.1 2539.11 and ladies and gentlemen boys and girls I just deleted the equation great 1.005 times 25 excuse me 20.56 plus the humidity ratio which was 0.009574 times 2539.11 and we get 44.9722 so you guys think that's accurate enough I mean probably let's see we increased by 0.01 and that increased by almost a tenth so if we want to increase by about a third of a tenth about half of a tenth which increased by half so 20.565 let's go with 566 I mean we're really getting into the weeds here but this is the process 2539.12 9.12 okay now 566 and 2539.12 44.9784 oh man a little bit too much 44.9784 should have just done 20.565 okay whatever this is I'm gonna call it because you guys get the idea right hey look it's that number that we wanted close enough 44.9774 so we're calling t3 20.565 that's what we've decided and we wanted relative humidity as well so using t3 pg3 it's going to be p sad add t3 so we can interpolate and I will start with this equation and replace 2538.1 with the saturation pressure at 20 degrees Celsius which is zero two three three nine zero two three three nine calculator that's not a zero what you're doing and then 0.02487 that gives me a saturation pressure of 0.024226 0.024226 then phi three is going to be pv3 0.01536 divided by pg3 which is 0.024226 63.4 percent 63.4 percent let's add in a degree Celsius and this oh you can't see what I'm doing and then t3 and phi 3 what else do we want we have the humidity ratio we have everything we need except for the volumetric flow rate so that's going to require that we calculate v3 which is v a3 which is the same exact process as the last two times let's do that one more time just for good measure 0.0154 since we actually have pv3 might as well use it so our specific volume at state 3 is going to be 8.314 times 20.565 plus 233.15 times 1000 divided by 28.97 times 1.01325 times excuse me minus 0.0154 because we happen to have pv3 times 10 to the fifth unnecessary parentheses just to make me feel better I get 0.78565 0.78565 let's compare that against the chart shall we 0.5 0.78565 versus 0.844 what's up with you specific volume you shouldn't be less than any of those numbers ah okay missing a closing parentheses here which means that this parentheses at the ends actually was unnecessary okay 0.844 I think if we've taken anything away from this example problem so far it's that despite the fact that the calculated results should be closer as a result of human error sometimes they're not 0.84484 is everything we need to calculate a volumetric flow rate so we take our mass flow rate at state 3 which is which is 0.88791 0.88791 and we multiply by our new specific volume and that's in kilograms of dry air per second but multiplied by cubic meters per kilogram of dry air the kilograms of dry air cancel and we're left with cubic meters per second and we get 0.75 we probably want an answer in cubic meters per minute because that's what everything else was in one minute is 60 seconds we multiply by 60 we get 45.003 cubic meters per minute so we have the humidity ratio at state 3 in kilograms per kilogram which is what we wanted we have a relative humidity in percentage we have a temperature it's a dribble temperature in degrees Celsius and a volumetric flow rate in cubic meters per minute so we have everything we need to consider this question answered I will point out we had a volumetric flow rate at one and two why didn't we just take volumetric flow rate at state three is equal to volumetric flow rate at state one plus the volumetric flow rate at state two so the reason that we don't do that is because there is no such thing as the law of conservation of volumetric flow rate it's the law of conservation of mass flow rate it's not that the volumetric flow rates are added together to yield approximately 45 which is essentially what we got by the way it's that if the densities are pretty equivalent then the mass flow rate that I get at the end yields a volumetric flow rate at the end that is pretty close to what it would have been with the volumetric flow rates at one and two added together so it is a result of how the mass flow rates and specific volumes combined and not 20 plus 25 the other thing I'll point out while we're looking at this problem is why do we care why are we mixing air streams together how is that relevant to HVAC analysis the reason this is relevant is because it costs a lot of money to heat and cool air it's in our best interest to avoid heating and cooling air as much as possible if you're operating an office building in the middle of winter and it's five degrees celsius outside and you're trying to keep it at 21 it's a lot easier and more efficient energy wise to mix some of the air that's already in inside with some outside air and then you don't have to heat it as much to hit your target set point we mixed together some inside air and some outside air to run through our HVAC system so that we don't have to provide as much of the energy that would be required to bring the outside air to its set point conditions every single time the real question then becomes how many air changes per hour do you need to accommodate the ashway standard that's specific for the type of environment you're analyzing if it's a classroom or a hospital you need a lot of intake air from outside if it's a private residence you need almost no outside air i have lived in several college apartment buildings that didn't have any fresh air intake from outside the HVAC system just processed inside air to keep it warmer or cooler