 Moist air enters a duct at 10 degrees Celsius, 80% relative humidity, and a volumetric flow rate of 150 cubic meters per minute. The mixture is heated as it flows through the duct and exits at 30 degrees Celsius. No moisture is added nor removed, and the mixture pressure remains approximately constant at one atmosphere. Determine first the mass flow rate of dry air through the duct, the total rate of heat transfer into the duct, and the relative humidity at the exit. So I will begin to model this, and I will recognize that I have what's called a simple heating process. When I say simple heating, I'm referring to the fact that we are adding heat and not adding or removing any water. All we are doing is adding heat. I'm going to call the inlet, stay one, the outlet stay two. I know the temperature and relative humidity at the inlet. T1 is 10 degrees Celsius. T1 is 80%, and I know the volumetric flow rate. Quick question for you is that the volumetric flow rate of the dry air, the atmospheric air, or the water vapor. Secret to question, the answer is yes, it is all of them. Because when we're modeling psychrometric behavior, we are using Dalton's law to approximate the behavior of our air and water vapor mix together. Dalton's law assumes that all of the species in the mixture occupy the same volume. Therefore, the volumetric flow rate of the atmospheric air is 150 cubic meters per minute. The volumetric flow rate of the dry air is also 150 cubic meters per minute. And the volumetric flow rate of water vapor is 150 cubic meters per minute. At state 2, I know that the temperature is 30 degrees Celsius. And I need two independent intensive psychrometric properties to fully define my state point from which I can look up anything else. And the other state point comes from our mass balance. Just like in the adiabatic saturation chamber, I can set up a mass balance on just the dry air, just the water vapor, or the atmospheric air as a whole. The atmospheric air, generally speaking, isn't very useful. So if I set up a mass balance on the dry air, I recognize that I have steady state operation of an open system with one inlet and one outlet. Therefore, m.a1 is going to equal m.a2, which I'm just going to call m.a. If I set up a mass balance on the water, I can say that the entering water has to leave again. The only opportunity for water to enter is as water vapor at state 1. The only opportunity for water vapor to exit is as water vapor at state 2. So the mass flow rate of dry air doesn't change, and the mass flow rate of water vapor doesn't change. That means the humidity ratio doesn't change. So the other psychrometric property is the fact that m.a2 is equal to m.a1. I like to visualize these processes on the psychrometric chart. I think that that is the easiest way to kind of keep track of what's happening in your head. And on the psychrometric chart, heating represents movement to the right. Cooling represents movement to the left. Those are horizontal processes. Humidification is vertical movement up, or vertical displacement up. Dehumidification is vertical displacement down. So if I have a process where heat is added and nothing else, no water is added, no moisture is removed, I have a horizontal line going to the right. If I have a cooling process, I'm going to go left until I reach 100% relative humidity, at which point I would follow the 100% relative humidity line down until I reach my desired temperature. The fact that there's some vertical displacement in the downward direction implies that dehumidification would occur. We would be removing water from the air. If I have a heating process with humidification, that's going to be movement to the right and movement up. And just like when we analyze projectile motion, it is usually more useful for us to model the horizontal component and the vertical component separately, even though they actually don't behave separately in the real process. So now let's set up an energy balance. We have studied the operation of an open system with one inlet and one outlet. There are no opportunities for work and heat transfer is only in the inward direction. Furthermore, I'm neglecting changes in kinetic and potential energy. So this is going to simplify all the way down to just m dot h in plus q dot in is equal to m dot h out. So if I account for the enthalpy of the atmospheric air as it enters, I'm going to write that as m dot a1 h1. Why did I write m dot a1? Because the specific enthalpy of the atmospheric air is defined per unit mass of dry air. So to write it as a total energy term, I have to multiply the specific enthalpy of the atmospheric air by the mass of the dry air. Same goes for state 2. I can write the total energy leaving at state 2 as m dot a2 times h2. Since the mass flow rate of dry air is the same, I can factor that out and write that as m dot a times h2 minus h1. So quick question. We have a heating process of air. Can we just plug in cp of air times t2 minus t1? I mean, that's what we did in thermal 1, right? We can't because we have to account for how much energy it takes to increase the temperature of the water as well. And even though there's not much water, the water has a higher heat capacity, which means that we have to account for the energy increase and not just neglect it. That's the whole reason that we are modeling this as atmospheric air instead of just dry air. So we need to determine m dot a and we need to look up h1 and h2. So using my two independent intensive psychrometric properties t1 and phi1, I can determine omega1 and h1. I can either use the calculations for that or the psychrometric chart. And then using t2 and omega1, we can determine h2. Again, I can use the equations for that or the psychrometric chart. I'm going to assume that it's okay for me to use the psychrometric chart as a lookup because I have a pressure of one atmosphere and because that'll get us close enough for the purposes of this analysis. But don't get too comfortable with the psychrometric chart. It's useful. It'll help us work through more example problems. But you still need to know how to calculate these psychrometric properties by hand. So from my mass flow rate of dry air, I'm going to take the volumetric flow rate of dry air and divide it by the specific volume of dry air. The total volume of dry air is going to be the same as the total volume of the atmospheric air, which is just 150 cubic meters per minute. And the specific volume of the dry air is going to be what I get when I look up the specific volume of the atmospheric air because that's expressed per unit mass of dry air. So I can add to my list of lookups specific volume at state 1. That will give me enough information to determine m.a1, which gives me m.a, which allows me to determine the total rate of heat transfer in. So T1 of 10 degrees Celsius and a phi 1 of 80% relative humidity. That's going to occur right about here. Highlight those numbers. 80% 10 degrees Celsius. And we are heating to 30 degrees Celsius by a direct horizontal process until we encounter the 30 degrees Celsius line. So at state 1, I can look up my specific volume first just because that happens to be a direct line. So I know that's 0.81 cubic meters per kilogram of dry air. H1 is going to be about 26-ish. And the humidity ratio at state 1 is going to be... I'll follow that all the way to the right. It's going to be 6.1, 6.2. I mean, this is 6. This is 7. This is 6.5. This would be 6.25. So maybe 6.125. So 6.125. That's going to be grams of water per kilogram of dry air. And at state 2, we want the specific enthalpy. So I'm going to follow that back right about here-ish. I'll say that's 42, 44, 46, maybe 47. Let's call that about 47 kilojoules per kilogram. So plugging in my specific volume of 0.81. Again, that's cubic meters of atmospheric air per kilogram of dry air which is equivalent to cubic meters of dry air per kilogram of dry air. These cubic meters are going to cancel leaving me with kilograms per minute. And let's convert that to kilograms per second. Calculator, if you would please. It's 150 divided by 0.81 times 60. And we get 3.086 kilograms of dry air per second. And then armed with that mass flow rate, we can multiply by the difference in specific enthalpy. 0.8642 kilograms of dry air per second. Multiply by 47-ish minus 26-ish kilojoules per kilogram of dry air. Kilograms of dry air cancels leaving me with kilojoules per second, which is a kilowatt. So if I take 3.08642, that very accurate number, and multiply by about 47 minus about 26, I get 64.8. So this heating process would draw about 64.8 kilowatts. So that gives me the answer to part A and B. The last thing I wanted to know was the relative humidity at the exit. Let me pose that to you as a question. Is the relative humidity at the exit the same? Is it still 80%? It is not. The humidity ratio doesn't change because the mass of water and the mass of air don't change. But remember that the relative humidity is the proportion of water that's in the air to how much water the air can hold. Even though the water that's in the air doesn't change, the amount of water that the air can hold does change. As the air is heated up, it can hold more water, which means that the relative humidity is going to drop. I can see that on the psychometric chart. As I go to the right horizontally, I'm dropping my relative humidity. From the chart, I can see my relative humidity at state 2 is going to be about... What do you think, 23%? Let's call it that. Arbitrary precision B2 is about 23%.