 One of the oldest and most accurate ways to measure humidity is with two thermometers. One of the thermometers you use to just measure the regular temperature. On the other thermometer, you wrap the thermometer in a wicking that holds water, and you allow as much water as can evaporate to evaporate. And if that happens steadily, the evaporation of the water will pull some energy with it because it takes some energy, the latent energy, to evaporate. Pulling that energy out of the thermometer will cause the temperature reported by the thermometer to drop, and as a result you will end up with a temperature difference between the regular thermometer and the thermometer wrapped in the wet wicking. We call the regular thermometer the dry bulb temperature and the thermometer covered in the wicking the wet bulb temperature. You can use the temperature change between the dry bulb and wet bulb to calculate how much water had to have evaporated, which you can use to figure out how much water had to go into the air, and if you assume that the air leaving the wicking is as saturated as it can be, you know it's at 100% relative humidity, so you can back calculate what the humidity had to be for it to pull that much water and drop the temperature by that much. That's the logic behind calculating a relative humidity accurately using a dry bulb and wet bulb thermometer. To explore the math, let's try a model. Instead of a wicking, let's consider a chamber and let's assume that all of the heat that is lost goes into evaporating the water. So we have an adiabatic saturation chamber where we are pushing an airstream across a body of water and allowing as much water as can evaporate to evaporate. As a result of that evaporation process, we are saying the relative humidity at state 2 is 100%. If we can figure out how much the temperature drops from state 1 to state 2, we can figure out how much water had to have been evaporated, how much water we end with at 100% relative humidity, and use that to determine what the water content was to begin with, which is our humidity at state 1. For that, let's try a mass and energy balance. When performing our mass balance, we could set up a mass balance on just the dry air, we could set up a mass balance on the water vapor, we could set up a mass balance on the atmospheric air itself. Generally speaking, doing mass balance on the individual species is a little bit more convenient. So when I look at the mass balance on the dry air, there are two opportunities for the air to cross the boundary. I have dry air entering at state 1 and dry air exiting at state 2. Therefore, because we have steady state operation of an open system with one inlet and one outlet, m.a1 is going to be equal to m.a2, which I'm just going to abbreviate m.a. On the water vapor side, I have three opportunities for the water to cross the boundary. I have water vapor entering at state 1, I have liquid water entering at state w, and I have water vapor exiting at state 2. I can rewrite those equations in terms of other quantities, like for example, I could divide everything by m.a here, at which point I would have m.v1 over m.a1 because m.a is equal to m.a1 which is equal to m.a2. Therefore, I have humidity ratio 1 plus m.w over m.a plus, excuse me, is equal to humidity ratio 2. So if it's convenient for me to write the mass flow rate of water entering as m.a times omega2 minus omega1, I can do that. But now for the energy balance. I have an open system operating steadily, which means that I have e.in is equal to e.out. e.in could be work in, could be heat transfer in, could be the sum in of m.theta. e.out could be the work out, the heat transfer out, and the sum out of m.theta. I'm assuming that my chamber is adiabatic because in reality this happens very quickly. There's no real opportunities for heat transfer. And the temperature of the air passing over the wicking is probably pretty close to the temperature of the wicking and the water anyway. Next, I have no opportunities for work. And inside of the theta, I have h plus ke plus pe. And I'm going to say that the kinetic energy change of the air is negligible. And any potential energy change is negligible. At which point I have the sum in of m.h is equal to the sum out of m.h. I have two opportunities for energy to enter, state 1 and state w. So first I will write state 1. I have m.a1 times h1 plus state w, which is m.hw. That will equal the energy of the exiting atmospheric air, which is going to be m.a2 times h2. So note that I wrote m.a1 h1 instead of m.1h1. Why did I do that? I did that because h here is the enthalpy of atmospheric air on a specific basis, which is the total enthalpy of atmospheric air per unit mass of dry air. Since what I want to write in this form of the energy balance is a rate of total energy, I need to multiply this quantity by m.a per unit time so that I get to total energy per unit time. So I wrote m.a times h because I want total energy and because the specific enthalpy of atmospheric air is defined per unit mass of dryer. Now my next step is going to be to divide everything by m.a. When I divide everything by m.a, this just becomes h1 plus m.w over m.a and then h2. And instead of writing m.w over m.a, I can write this quantity. So I can say omega2 minus omega1. Then this equation becomes h1 plus omega2 minus omega1 plus, excuse me, is equal to h1. And that doesn't work out. Oh right, because I have h.w here. So this equation becomes h1 plus omega2 minus omega1 times h.w is equal to h2. That's much more better. Now let's think through what we are likely to know in this scenario. If we know t1 and we know t2, because in this hypothetical I have measured both temperatures and I'm trying to determine what the humidity at the inlet was. Well, let's say omega1 is the goal and I measure t2 and I know the relative humidity at state 2 is going to be approximately 100%. Then I'm going to say h.w is a saturated liquid that is pretty close to t2. So I'm approximating state w by saying it's pretty close to the enthalpy of a saturated liquid at state 2. So in that case I'm trying to take this equation and solve for omega1. So I'm going to write omega2 minus omega1 is equal to h2 minus h1 divided by h.w. So what I'm going to want to do with this step is to solve this equation for omega1. Because at this point in the analysis I will have omega2. And as I start to plug in quantities in place of h1 and h2 I will end up with a quantity in terms of omegas. So it's not worth me doing the algebra until I have everything substituted. And for that I will go to a new page. So we define the specific enthalpy of atmospheric air as being the specific enthalpy of the dry air plus omega times the specific enthalpy of the water vapor. We are approximating the specific enthalpy of the dry air as cp of air times the temperature of the air at that state plus omega times... We are substituting in hv with hg because we are approximating the specific enthalpy of the water vapor as the specific enthalpy of a saturated vapor at that temperature. So for h1 I can plug in cp of air times t1 plus omega1 times hg at t1. For h2 I can plug in cp of air times t2 plus omega2 times hg at t2. And in place of hw I'm plugging in hf at t2. So I will make those substitutions. And note that I'm writing hg1 instead of writing hg at t1 and hg2 instead of writing hg at t2 and hf2 instead of writing hf at t2. So if I start doing some algebra I can simplify this a little bit more conveniently in terms of the things that we have and what we don't have. So I will write this as omega2 minus omega2 times hg2 over hf2 minus omega1 plus omega1 times hg1. Over hf2 is equal to cp of air times t2 minus t1 all over hf2. And if we think about what we will know at this point, if we know phi2 we can determine the humidity ratio at state 2. We can look up hf2 hg2 and hg1 as a result of knowing t1 and t2. Knowing t1 and t2 will allow us to plug in t1 and t2 and we can look up the cp of air. So at this point in our analysis what we are actually trying to solve for in order to get to the humidity at state 1 is the humidity ratio at state 1. So I am going to want to write that as h1 times hg1 over hf2 minus 1 is equal to cp of air times t2 minus t1 divided by hf2 minus omega2 times 1 minus hg2 over hf2. So then omega1 becomes cp of air times t2 minus t1 minus omega2 times 1 minus hg2 over hf2 all over hg1 over hf2 minus 1. So that is going to become cp of air times t2 minus t1 divided by hf2 times the quantity hg1 over hf2 minus 1 minus omega2 times 1 minus hg2 over hf2 divided by hg1 over hf2 minus 1. So in the first term that simplifies in a pretty straightforward way. When I bring hf2 inside the parentheses I am ending up with hg1 minus hf2 and if my goal is going to be to write that in the denominator on both sides that quantity is going to become 1 minus hg2 over hf2 times hg1 over hf2. Minus 1 to the negative 1 which is going to be equivalent to 1 over hg1 over hf2 minus 1 which is going to be equivalent to minus hg2 over hf2 times that would be hf2 over hg1 minus hf2. Then I can expand that across the parentheses and write that as hf, hf2 over hg1 minus hf2 minus hg2 over hf2 times hf2 over hg1 minus hf2. I am left with hf2 over hg1 minus hf2 minus hg2 over hg1 minus hf2 which becomes just hf2 minus hg1 minus hg2 divided by hg1 minus hf2. Aha! I am sure that is very clear. So then when I combine these together I am going to have omega1 is equal to cp of air times t2 minus t1 divided by hg1 minus hf2 minus omega2 times hf2 minus hg2. hf2 minus hg2 divided by hg1 minus hf2. So then that simplifies further to just cp of air times the quantity t2 minus t1 plus omega... excuse me... minus omega2 times the quantity hf2 minus hg1 divided by hg1 minus hf2. This is going to be the equation that allows us to determine what the humidity is at state 1. And remember that this whole thought experiment exists to be able to calculate a humidity ratio and therefore a humidity level from a dry bulb temperature and a wet bulb temperature. So with this equation state 1 is the actual conditions. State 2 is the wet bulb conditions. So t1 is the dry bulb temperature, t2 is the wet bulb temperature. Furthermore omega2 would be calculated at 100% relative humidity. So omega2 would be 0.622 times pv2 over pA2 which is equal to 0.622 times pv2 over p minus pv2, which is going to be 0.622 times pv2 times pg2 over p minus pv2 times pg2. And in that equation pv2 is 1. So the omega at state 2 is just going to be 0.622 times the saturation pressure at t2 divided by p minus the saturation pressure at t2. And again that is only for wet bulb conditions. It's only because the relative humidity at state 2 is 1 or 100%. Just to accommodate the people that only grabbed this page. I'm going to bring in the schematic. So if one knows the wet bulb temperature and the dry bulb temperature, one can look up the specific enthalpy of a saturated liquid at the wet bulb temperature, hf2, the specific enthalpy of a saturated vapor at the wet bulb temperature and the dry bulb temperature. That's hg1 and hg2, which is this one. The specific heat capacity of air and the saturation pressure at state 2. That's the saturation pressure at the wet bulb temperature. From that information a person can calculate what the humidity ratio is for the fully saturated air, leaving the saturation chamber, which is really just the air blowing past the wet bulb. And from all of that information determine what the humidity ratio is of the actual air. So now that we have all that built, let's try an example.