 Example problem number four, humid South Dakota air at one atmosphere, 35 degrees Celsius and 80% humidity, enters an HVAC system at a rate of 10 cubic meters per hour, where it is to be isobarically conditioned to 22 degrees Celsius, 50% humidity. I want to know the cooling rate, if any, the heating rate, if any, and the rate of condensation required to accomplish this process. So I like to visualize things. Let's start by visualizing this on the chart. We are starting with 35 degrees Celsius and 80% humidity. So 35 degrees Celsius and 85% humidity is going to be, here I'll highlight the number, there you can see it. Right here, that's day one, and we are trying to get down to 22 degrees Celsius and 50% relative humidity. So our end condition is right here. I guess I shouldn't call that state two. I should call it state three. So in order to get from state one to state two, I am limited by my cooling mechanism. The cooling mechanism is going to move me to the left and then ride the 100% relative humidity line down. And we can't just dehumidify like we can humidification. We can't just move vertically down. What we have to do instead is follow the 100% relative humidity line all the way down until we get to the correct humidity ratio. And then we can move to the right until we get to state three. So we have a cooling process with dehumidification followed by a simple heating process that actually gets us to our desired set point. So state two is going to be about here. Let me draw that line a little bit better. So I will model this as a cooling process with dehumidification and then a simple heating process. Cooling coil and then heating coil. Water is gathering on the cooling coils as a result of the condensation process. A lot of water in fact. And that's gathering down here where it is leaving. I'm going to call that state w again just for consistency sake. Then I'm going to establish a state point between the cooling and the heating processes. That's my state point two. Egress is state point three. Ingress is state point one. So we have cooling with dehumidification from one to two and then heating from two to three. I'm going to establish two separate control volumes and two mass balances for control volume one and control volume two and two energy balances. So on control volume one I have the option of setting up a mass balance on just the dry air, on just the water vapor or on the atmospheric air itself. The atmospheric air generally speaking is not a useful mass balance. So I will recognize that for the dry air I have steady state operation with one inlet and one outlet. Therefore m.a1 is equal to m.a2 and the water enters as state one. Here let me write this a little bit more conveniently. And then because I have steady state operation of an open system, the entering water has to leave. I have water entering as water vapor at state one. I have water leaving my first control volume as state w and water vapor at state two. So m.v1 is equal to m.v2 plus m.w. What do you guys suppose I want to do next? If you've been following the previous examples, you probably recognize that I want to divide by m.a. And I'm calling m.a1 and m.a2 the same m.a quantity for convenience here. The reason I'm dividing by m.a is because it allows me to write m.v1 over m.a, which is equal to m.v1 over m.a1, which is equal to mv1 over m.a1, which is just the humidity ratio at state one. Likewise with state two, I can write that as the humidity ratio at state two. I have m.w over m.a, and m.w can be written as m.a times omega one minus omega two. For my energy balance, I have steady operation of an open system, which means my energy balance is going to simplify down to e.in is equal to e.out. I'm neglecting work and any entering heat in the first control volume. Likewise, I'm neglecting any changes in kinetic and potential energy, which means that I'm only allowing energy to enter my system as the enthalpy of the atmospheric air at state one. So I have m.a1 times h1. Play out that conversation with m.a times h1 in your head while I'm writing m.a2 plus q.out plus m.a w hw. Energy is entering as atmospheric air and leaving as enthalpy of atmospheric air q.out and the energy that leaves with the water at state point w. I can write q.out as m.a times h1 minus h2 minus m.w times hw. And then I can make the substitution from up here and write m.w as m.a times omega one minus omega two. And then I could factor out m.a if I wanted to. So with my two independent intensive properties at state one, which are t1 and phi1, I can determine any other psychrometric property at state one. I note that I will need the specific volume at state one in order to calculate the mass flow rate because I have a volumetric flow rate. And then I'm going to need h1, h2, omega one, and omega two. And hw will come from assuming that the water that is freshly condensed is not given the opportunity to become super cooled or compressed. And then I'm going to approximate those properties as hf at t2 because it's probably going to be closer to t2 than it is to t1. So I will need to have looked up omega one, h1, and specific volume one. I will need t3 and phi3 to look up omega three, which will give me omega two. The other thing I know about state two is that it is at 100% relative humidity. Probably need an h3 in order to answer the amount of heat transfer in. OK, one set of mass and energy balance is down. For the other set, I'm looking at control volume two. And on control volume two, I have a pretty simple mass balance because there's no opportunity for dry air to enter or exit except at states two and three. And there's no opportunity for water vapor to enter or exit except at states two and state three. Therefore, the mass balance is going to simplify down to m.a2 is equal to m.a3, which is that same m.a quantity from earlier. And on the water, I'm going to be able to say m.v2 is equal to m.v3. And this is what allowed me to write omega two is equal to omega three from earlier. And on the energy balance, I have steady operation of an open system. So e.in is going to equal e.out. Energy is able to enter as atmospheric air and heat transfer. Energy is only able to exit as atmospheric air because I'm neglecting works, changes in kinetic and potential energy. And I'm assuming that heat transfer is only in the inward direction. So it's m.a2 times h2 plus q.in is equal to m.a3 times h3. And like I've said in every example problem so far, the reason that we write m.a2 times h2 and m.a3 times h3 and m.a1 times h1 instead of just m.h is because the enthalpy of the atmospheric air on a specific basis is expressed per unit mass of dry air. Therefore, to represent it as a total energy in my energy balance, I have to multiply by the mass of the dry air. I don't multiply by the mass of atmospheric air. So q.in is equal to m.a times h3 minus h2. So I have four calculations to complete. Calculate the mass flow rate of dry air by knowing the volumetric flow rate in the specific volume. Then use that to calculate q.out, q.in, and m.w. All I have to do is use my t1 and phi1 to determine omega1, h1, and v1. Use my t3 and phi3 to determine omega3 and h3. And then I believe I am. No, I have to determine h2 as well. So from a phi value of 1 and an omega value of whatever we got at omega3, I can look up h2. And then in addition to that, I need to look up hf at t2. So would you like to determine those psychrometric properties by using the psychrometric chart or by calculating them by hand? You know what? In the interests of character building, why don't we do both? So let's start with the chart lookups. And then we will calculate the quantities by hand. That'll give us the opportunity to compare them a little bit more fairly. So t1 was 35 degrees Celsius. Phi1 is 80%, from which I want omega1, h1, and v1. Omega2 is equal to omega3. And phi2 is equal to 1, from which I want to look up h2. And at state 3, I know a temperature and relative humidity. Those are my set points of 22 and 50. And I need to look up h3 and omega3. So on state 1's lookups, I have the same lookups as example problem 2. But let's do them again. I have a specific volume of about 0.913, maybe. Because this line is 0.91, this line is 0.92. And maybe a third of the way between the two. So 0.913 ish, 913. That's in cubic meters per kilogram of dry air. h1, I'm going to say a little bit less than 110. Maybe 109.5, 109.5 kilojoules per kilogram of dry air. And for my humidity ratio, I'm saying about 0.029 kilograms per kilogram, or 29 grams of water per kilogram of dry air. Three down, three more to go. At state 2, I have a specific enthalpy of, let's call that 32. It's over here. So 32-ish kilojoules per kilogram of dry air. And then at state 3, I need an enthalpy, which is going to be about 43, 44, somewhere in that neighborhood. See, we're halfway between 40 and 50, right about here. So then I would say we're closer to 43 than to 42. 43 kilojoules per kilogram of dry air. And humidity ratio, we can follow that off to the right. I have the advantage of being able to draw horizontal lines, but I will use that. I'm going to call that 0.84, maybe. This is 9. This is 85. So maybe 8.4 grams of water per kilogram of dry air. And the last lookup that I need is HF at T2. The T2 is about 11. Let's just call it 11. HF at 11 degrees Celsius. So from the steam tables at 11 degrees Celsius, I see that I have a specific enthalpy of a saturated liquid of about 46.2 kilojoules per kilogram of water. So again, my goal was to compare and contrast what we get from the chart lookups to what we would get if we calculated the saccharometric properties by hand. So we will start with a temperature of 35 degrees Celsius and a relative humidity of 0.8. So I can say omega 1 is 0.622 times PV1 divided by P minus PV1. And instead of writing PV1, I can write P1 times Pg1. And Pg1 is the saturation pressure corresponding to 35 degrees Celsius. So that's 0.8 times 0.05628 bar, 0.05628 bar. And my pressure in this problem is 1 atmosphere, which is 1.01325 bar. So my humidity ratio at state 1 is 0.622 times 0.8 times 0.05628 divided by 1.01325 minus 0.8 times 0.05628. And I get 0.0289. And that's kilograms of water per kilogram of dry air, which would be equivalent to 2.9. And we got, excuse me, that would be equivalent to 29 grams of water per kilogram of dry air, which is exactly what we got for the chart lookup. So it's what? 0.075 grams of water per kilogram of dry air, more accurate than the chart lookup. And then for H1, I'm going to use Cp of air times T1 plus omega 1 times Hg at T1. So Cp of air comes from table A20. And we see that it's 1.005. And we are multiplying that by T1, which was 35. Question mark? 35. So 1.005 times 35 plus 0.028924 times Hg at T1. So at 35 degrees Celsius, we can look up Hg and we get 2565.3. 2565.3. So our enthalpy is 109.37 kilojoules per kilogram of dry air. The last thing that I wanted was the specific volume. So the specific volume of atmospheric air per unit mass of dry air is equivalent to the specific volume of dry air because the total volume is the same if we model this using Dalton's law. So it'd be the total volume of dry air divided by the mass of dry air. So I'm going to be left with R times T over P, which is R of air times T1 over Pa1. And if specific gas constant is represented as the universal gas constant divided by molar mass, then I'm saying R bar divided by molar mass of air times T1 divided by, instead of P1, I'm going to write that as P minus Pv1, which is P minus V1 times Pg1. So the universal gas constant comes from the inside of the front cover of the textbook, which is 8.314 kilojoules per kilomole kelvin. These quantities down here, so 8.314. And then we multiply by T1 in kelvin, which is 35 plus 273.15. And then we're dividing by the molar mass of air, which comes from table A1, 28.97 times 1.01325 minus 0.8 times 0.05628. And then we have to account for the unit conversion. Here I can write that out. 8.314 kilojoules per kilomole kelvin divided by 28.97 kilograms per kilomole times 35 plus 273.15 kelvin divided by 1.01325 minus 0.8 times 0.05628 bar. And I recognize that 1 bar is 10 to the fifth newtons per square meter. And I can write 1 kilojoule is 1,000 newtons times meters because a joule is defined as a newton times a meter. Now kilomoles cancels kilomoles, kilojoules cancels kilojoules, kelvin cancels kelvin, bar cancels bar, newtons cancels newtons, which leaves me in cubic meters per kilogram. So in my denominator, I'm going to write times 10 to the fifth. And in my numerator, I'm going to write times 1,000. I could, of course, simplify and write it as 10 squared, but potato, potato. And we get 0.9134. 0.9134 cubic meters per kilogram of dry air. So 0.9134 versus 0.913. That's an extra 0.0004 cubic meters per kilogram of dry air worth of accuracy. And all we had to do was some ideogaslaw calculation. And then that gives us everything at state 1. So we need state 3 next because we can't jump to state 2. So at state 3, I have a temperature in phi again. So I'm going to use the same equation for humidity ratios. Omega 3 is going to be 0.622 times pv at state 3 divided by p minus pv at state 3. And pv is phi times pg. So we're using pg3 times phi 3. pg3 itself is going to be the saturation pressure corresponding to our temperature at state 3. And our temperature at state 3 is 22 degrees Celsius. So we go back into our steam tables. The saturation pressure corresponding to 22 degrees Celsius is going to be 0.02645, which is 0.02645 bar. And phi 3 was 0.5. So my omega 3 is going to be 0.622 times 0.5 times 0.02645 divided by 1.01325 minus 0.5 times 0.02645. And we get a syntax error. Thank you calculator. One too many parentheses, I think. So our humidity ratio at state 3 is 0.0082300823 kilograms of water per kilogram of dry air, compare and contrast to 8.4. So that would be 8.2 grams of water per kilogram of dry air versus 8.4. And we want the specific enthalpy while we're here. So Cp of air times T3 plus the humidity ratio we just calculated and we're approximating Hv by using Hg at T3. So Cp of air that same 1.005 quantity times 22. Again, we're using Celsius instead of Kelvin because we want to be able to add together quantities relative to the same 0 point. And we are adding 0.00823 times the enthalpy of a saturated vapor at 22 degrees Celsius, which is 2541.7. So calculator 2541, nope, 2541.7. And we get 43 almost on the nose. So compare 43.02 against 43. That's an extra 0.02 kilojoules per kilogram. Then state 2. The reason state 2 is complicated is because we are going to need to determine a temperature. So we have a humidity ratio at state 2 that is the same as it is at state 3. And we know V2 is 1. And we're trying to get to specific enthalpy and temperature. Because remember, if we're not using the chart, we need the temperature in order to look up Hw. So we need temperature and our specific enthalpy at state 2. So let's start by exploring what we know. Omega 2, which we know, is 0.622 times pV at state 2 divided by p minus pV at state 2. And pV is phi times pG. And since phi is 1, that means pV2 is equal to pG. So in this specific case, this is pG divided by p minus pG, which means that I can calculate what pG is at state 2. For that, I'm going to have to do some algebra. Everyone, avert your eyes, a minus pG2 is 0.622 times pG2. So in the ratio 2 times p minus in the ratio 2 times pG2 is equal to 0.622 times pG2. In the ratio 2 times p is equal to pG2 times 0.622 plus in the ratio 2, therefore, pG2 is equal to omega 2 times p divided by 0.622 plus omega 2. So our omega at state 2 is omega at state 3, which is 0.00823. Regulator, come on, that's not a 0. That's less than or equal to. And whatever pressure unit I put in for p is going to be when I get out for pG, because the quantity omega 2 divided by 0.622 plus omega 2 is unitless. So if I'm plugging in 1.01325 bar, I will get out an answer in bar 0.622 plus 0.00823. And I get a pG2 of 0.013232. And from that saturation pressure corresponding to my temperature, I can determine a temperature. Because I'm looking for the temperature that has a saturated pressure of 0.013232, which means I can look up the saturation temperature corresponding to that pressure in table A2. So 0.013232, 0132 is going to be between 11 and 12 degrees Celsius. So my interpolation is going to go, calculator. My pressure, which is 0.013232 minus 0.01312, calculator. That's not a 0. Come on. Divided by 0.01402 minus 0.01312. And I'm saying that's equal to x minus 11 divided by 12 minus 11. And that yields a temperature of 11.1244 degrees Celsius. And while we're here, we can use that setup to interpolate for our value for hw because we're using hf. So I can set up this same interpolation except using hf values instead of temperature values to come up with hw. So I will replace 11 with hf at 11, which is 46.2. And I will replace 12 with hf at 12, which is 50.41. And that yields a specific enthalpy of state w of 46.724. And is it worth it to be that accurate when we are approximating the temperature and the phase of the freshly condensed water leaving? Probably not. But now we have everything we need except for h2. So I can use my temperature to calculate an h. h2 is going to be Cp of air times t2 plus omega 2 times hg at t2. So I'm going to have to interpolate for an hg at t2 in addition to an hf at t2. Luckily for me, I have that same interpolation already set up. So instead of 46.2, I'm going to be plugging in hg at 11 degrees Celsius, which is 2521.6. And then 46.2 is also 2521.6. And in place of 50.41, I'm plugging in the hg value at 12 degrees Celsius, which is 2523.4. That gives me an h value of a saturated vapor at t2. It's 2521.82 kilojoules per kilogram. And now, with much ado, I can take 1.005 times my temperature at state 2, which is 11.1244. Again, that needs to be in Celsius. Plus omega 2, which was equal to omega 3, which is 0.00823 times my shiny new hg value 2521.82. And that gives me a specific enthalpy at state 2 of 31.935. So instead of 31.935, we would have been using 43. No, we would have been using 32. That's much more reasonable. 32 versus 31.93. Was that extra 10 minutes of calculation worth it? Probably not. But it's important to know how to calculate the quantities so that when we need to be accurate, we can be. So with these properties, I can go back and finish the problem. Let's start with our Q dot in. Q dot in is going to require that I calculate the mass flow rate of dry air. So m dot a is equal to m dot a1, which is v1 divided by a specific volume 1. So I'm going to take 10 cubic meters per hour because it's a dollhouse. 10 cubic meters per hour. And then we divide by our specific volume at state 1, which I will use the calculated value because we happen to have it, 0.9134, 0.9134. And that's cubic meters per kilogram dry air. And then one hour is 3,600 seconds. Hour cancels hour. cubic meters cancels cubic meters, leaving me with kilograms per second. So m dot a is going to be 10 divided by 0.9134 times 3,600. 0.00304. And then Q dot in is going to be that mass flow rate times h3 minus h2. And again, we have the chart lookups and the hand calculated values. I will use the hand calculated values just because we happen to have them already. But in this circumstance, using the chart lookups would be perfectly fine. I would go as far as to say, for the purposes of an exam in my class, you can use the chart if you have a pressure of one atmosphere and I don't tell you explicitly not to use the chart. Spoiler alert, I'm going to tell you not to use the chart at least once. And then h2 was 2521.82. Nope, that's not right. 31.935. So Q dot in is 0.337 kilowatts. No, excuse me, 0.0337 kilowatts or 33.7 watts. And then Q dot out, I'm going to start with my mass flow rate of dry air. And then we're going to take h1 minus h2, which is 109.37 minus h2, which is 31.935. And then we are subtracting omega 1 minus omega 2. Omega 1 is 0.0289 minus omega 2, which is equal to omega 3. 0.00823. And I'm multiplying by hw, which is 46.724. And let's be careful with our units there because it happens to work out, but it works out for a specific reason. I have kilograms per second for my mass flow rate of dry air, which means everything inside of these square brackets has to be in kilojoules per kilogram of dry air. The specific enthalpies of the atmospheric air are already in kilograms or kilojoules per kilogram of dry air. But with the humidity ratio difference, but with the difference in humidity ratios, I need that to be in kilograms of water per kilogram of dry air because I'm multiplying by kilojoules per kilogram of water. If I had used the chart lookups, I would have had to convert those to kilograms per kilogram. Otherwise, I would end up with an incorrect answer. So I have 0.23255 kilowatts. We can call that 233 watts. So if you're the person operating the HVAC system for this dollhouse, it has to be consistent. Should just square the top value. So if you're the person operating the HVAC system for this dollhouse, you would have to pay for the electricity to run the heat pump or whatever device you're using to supply the heat for the makeup in addition to the electricity required to operate the refrigeration system that is yielding a Q-dot out of 233 kilowatts. Or if you're just burning natural gas, you have to pay for electricity and natural gas. In a lot of these reheating processes, it's pretty common to just use a electric resistance heater because it happens to be a convenient way to get a very precise amount of heat added when you're talking about a relatively small amount of heat transfer. The last quantity I wanted was the rate of condensation for which we're going to use this equation here. So the mass for rate of condensation is what I'm calling m.w, which is m.a times the difference in omega 1 minus omega 2. And I want that answer in grams per second, which means I'm going to be taking grams per kilogram as my humidity ratio. So m.a was 0.003041. And then I'm multiplying by omega 1 minus omega 2, 0.0289 minus omega 2, which is equal to omega 3, 0.00823. Then I have to multiply by 1,000 to get it into grams per kilogram. And I get 0.06286. So the condensation from that cooling system is dripping out of some sort of collection tube. And the drips that it gathers are about a gram. That means that a drip is going to occur, maybe every 17, 18 seconds or so. That's still not much condensation. But keep in mind, this is a teeny, tiny volumetric for rain.