 Humid South Dakota air at one atmosphere 35 degrees Celsius in 80% humidity enters an HVAC system at a rate of 10 cubic meters per hour Where it is to be isobarically cooled to 22 degrees Celsius I want to know the cooling rate required to accomplish this process and how much condensation occurs. I will point out before we start that I had actually intended for this number to be in cubic meters per minute But I wrote cubic meters per hour, so we just kind of have to roll with it Well, 10 cubic meters per hour is going to be a very low volumetric flow rate Which means we're going to have a very low mass flow rate, which means that everything is going to be pretty small So I guess for the purposes of this analysis just assume it's a dollhouse instead of an actual house We're entering the HVAC system for a really cool dollhouse 10 cubic meters of air every hour so I have air entering at 35 degrees Celsius and 80% relative humidity and being cooled to 22 degrees Celsius The first question I have to answer is is this simple cooling or not and to answer that question Let's look at this on the chart We are starting with a temperature of 35 degrees Celsius and a relative humidity of 80% Well, 35 degrees Celsius is going to be this line here and 80% relative humidity is going to be this line here So I see that those intersect All the way Up here So my cooling process is going to start by movement to the left Horizontally until it reaches the 100% relative humidity line If I encounter the desired set point temperature first Then I'm going to have simple cooling. So if I cooled from 35 to like 31 I Wouldn't have any condensation occurring yet But in order to cool down to 22 degrees Celsius I Have to go all the way down to this temperature Which I cannot do with direct horizontal displacement So instead my process is going to involve going to the left Until I encounter the 100% relative humidity line At which point I will follow that Down into the left Until I hit my desired temperature So this is my end-stay point here. So since I have cooling with dehumidification I'm going to draw that as a cooling coil with The opportunity for condensation to occur and for the resulting water To be able to be Collected and to leave the system at state Let's call it W. So I have atmospheric air entering as day one I have atmospheric air exiting a state two and then I have a state point W Which is the mass flow rate of condensation that we're looking for So at state two I Know the relative humidity has to be 100% because I'm cooling while I am condensing water out of the air So if I'm okay With chart look-ups I Can use these two independent intensive psychometric properties to look up any other property that I need Let's start this analysis with a mass balance and an energy balance Just like before I'm going to set up a mass balance on the dry air Which because I have steady-state operation of an open system with one entering atmospheric air and one exiting atmospheric air And no opportunities for dry air to add be added or removed throughout the process That means I'm not a one is equal to m. They too, which I can just call m.a Again for the water because I have steady-state operation The mass flow rate of water in has to equal the mass flow rate of water out I have one opportunity for water to enter my system that is as water vapor at state one And I have two opportunities for water to leave my system that says water vapor at state two And as m.w So just like we did in the adiabatic saturation analysis, it's going to be most convenient for me to write this by dividing everything by m.a Because then I can write it in terms of intensive properties Because m.v1 divided by m.a, which is the same as m.v1 divided by m.a1 Which is the same as mv over m.a, which is just the humidity ratio at state one Is equal to m.v2 over m.a, which is the same as m.v2 over m.a2, which is the same as mv2 over m.a2 Which is the humidity ratio at state two plus m.w over m.a So I will write the mass flow rate of water Is equal to m.a Times the quantity omega one minus omega two for my energy balance on my control volume. I Have a steady-state operation I have an open system, e.in Has to equal e.out. I have control volume and no opportunities for work. I'm neglecting changes in kinetic and potential energy So this is going to simplify down to m.a1 times h1 is equal to q.out plus m.a2 h2 plus m.w hw I have one opportunity for energy to enter my system. That's as atmospheric air at state one Some of that energy leaves as q.out Some of that energy leaves At state w and the rest of it leaves as atmospheric air at state two So I can write q.out is equal to m.a times h1 minus h2 minus m.w hw and then I can plug in my representation of m.w in terms of m.a and omega one minus omega two At which point I have m.a times h1 minus h2 minus m.a times omega one minus omega two times hw m.a could be factored out further if I wanted to And m.a is going to be calculated by taking the volumetric flow rate divided by the specific volume h1 omega one h2 and omega two are going to be looked up on the chart or calculated by using my psychrometric property calculations hw I'm going to be approximating by treating it as a freshly condensed substance that hasn't been given the opportunity to be supercooled or compressed So hw is about hf And I'm going to say the temperature at state w is closer to t2 than it is to t1 So let's use hf at t2 So dollhouse volumetric flow rate of 10 cubic meters per hour Tiny tiny number And again, that's volumetric flow rate of atmospheric air and dry air and water vapor And then I'm going to divide by the specific volume at state one So I need to determine h1 omega one And the specific volume of the atmospheric air at state one And I need to look up h2 and omega two So there's my setup. I use t1 and phi1 to determine h1 omega one and v1 I use phi2 and t2 to determine h2 and omega two and then I'm assuming that the Substance that leaves at state w is approximately a saturated liquid at t2 So either I use my psychrometric calculations to determine those five properties and then look up hf Or I can assume that the chart is close enough Look up those five properties on the chart and then look up hf on the steam tables in the interests of Time Let's use the chart And let's look up hf My temperature is day two is 22 degrees celsius. So I'm looking up the hf value at 22 which is 92.33 and then for my chart lookups At state one I have a specific volume of Oh 0.912 0.913 Let's call it 0.9125 just for arbitrary precision And apply that we're extra confident in our number that we're not that confident about Omega one is going to be about 29 Actually, if I erase this I can probably read it a little bit better 29.1 One maybe and again, that's grams of water per kilogram of dry air on this particular chart H1 is going to be so this is the 100 line. This is the 110 line So it's going to be really close to 110 Yeah, let's just call it 110 110 kilojoules per kilogram of dry air And then at state two we have a relative humidity of 100 The temperature 22 degrees celsius So I'm going to call my enthalpy 65.6 65.5 And my humidity ratio at state two going to be 17 essentially on the nose Let's call that 17 Grams of water per kilogram of dry air Now we have everything we need to be able to finish the question. I will start with my mass flow rate of dry air That's going to be 10 cubic meters per hour divided by 0.9125 cubic meters cancel I'm left with kilograms of dry air per hour One hour is 3600 seconds Therefore my mass flow rate of dry air is going to be 0.003 Kilograms per second Then my total rate of heat rejection is going to be That mass flow rate multiplied by h1 minus h2, which is going to be 110 minus 65.5 That's going to be kilojoules per kilogram of dry air multiplied by kilograms of dry air per second, which is going to yield kilojoules per second Which is a kilowatt, and then I'm going to subtract I'm going to A again kilograms of dry air per second multiplied by the enthalpy difference, excuse me multiplied by the humidity ratio difference, which is going to be 20 29.1 Come on calculator 29.1 minus 17 And that's in grams of water per kilogram of dry air So kilograms of dry air are going to cancel and I'm going to be left with grams of water per second Multiplied by a specific enthalpy of the water in kilojoules per kilogram of water So I need to Divide that quantity by 1000 In order to get that into kilograms of water per second Then I can multiply by 92.33 kilojoules per kilogram of water to yield kilojoules per second as this tradition We forget at least one of the parentheses Let's double check that that's actually written correctly You know that is good enough calculator and we get a rate of heat rejection of 0.132 kilowatts That's a pretty small rate of heat rejection for a house, but probably pretty normal for a souped-up dollhouse Then for part B, we're calculating the mass flow rate of water That's just this equation right here So I'm going to take my mass flow rate of dry air I'm going to multiply by my difference in humidity ratios again And this wants grams per second So I'm taking kilograms of air per second multiplied by grams of water per kilogram of dry air Which means I'm going to yield an answer in grams of water per second And I get 0.0368 So before we leave this question, let me point out that a 100 relative humidity at the outlet of your air conditioning process is awfully uncomfortable To make humans comfortable, we would typically over cool this And then reheat it to get it to a more reasonable humidity I mean if we were considering the same exact example problem But instead of just cooling until it reached 22 degrees Celsius and then stopping If we wanted to hit 22 degrees Celsius and 50 percent relative humidity What we'd actually have to do is cool until we got to the correct humidity ratio And then heat back up Until we get to the correct temperature So we'd have to cool all the way down to What a little under 11 degrees Celsius And then heat it back to 22 degrees Celsius That's why actual air conditioning processes are so inefficient Relatively speaking because you are spending so much energy to cool so much And then spending more energy to heat it back up again You can get a little bit of return on investment by recycling some of the heat But generally speaking it is a very expensive energetically speaking process to do this cooling To a relative humidity less than 100 percent When we have dehumidification occurring