Chalk talk: How to calculate absolute humidity
In this video, Dr. Colin Campbell discusses how to use air temperature and relative humidity to calculate absolute humidity, a value you can use to compare different sites, calculate fluxes, or calculate how much water is actually in the air.
Watch the video to find out how to calculate absolute humidity and how to avoid a common error in the calculation.
Video transcript: Absolute humidity
Hello, I’m Dr. Colin Campbell, a senior research scientist here at METER Group, and also an adjunct faculty at Washington State University where I teach a class in environmental physics. Today we’re going to talk about absolute humidity. In a previous lecture, we discussed how relative humidity was a challenging variable to use in environmental studies. So, I’m going to show the right variable to use as we talk about humidity.
Absolute humidity can be talked about in terms of vapor pressure, which is what I’m used to, or in terms of vapor density. Whatever we use, we usually start by calculating this from a relative humidity value and a knowledge of air temperature. In my relative humidity lecture, I said that Hr (or the relative humidity) was equal to the vapor pressure divided by the saturation vapor pressure. And in most field studies, we’d typically get a report of the air temperature and the relative humidity. So how do we take those two values and turn them into something we could use to compare different sites, calculate fluxes, or calculate how much water is actually in the air? We’ll need to work through some equations to get there. I’m going to take you through it and give you an example so that you know how to do that calculation.
First, we’ll talk about absolute humidity in terms of vapor pressure or Ea. If we rearrange this equation here (very simple math), the relative humidity times the saturation vapor pressure will give us our vapor pressure. And that vapor pressure is now an absolute humidity. How would we do this? Well, let’s first talk about an example in terms of vapor pressure.
Let’s say a weather report said the air temperature was 25 degrees Celsius and the relative humidity was 28% or 0.28. First, we’d use Teten’s formula which I talked about in the previous lecture. We’d say the saturation vapor pressure at the air temperature is equal 0.611 kPa times the exponential of a constant times the air temperature divided by another constant plus the air temperature. So in our case, the air temperature is 25 degrees, which we’ll add here. Remember saturation vapor pressure is a function of 25. So 0.611 kPa times the exponential of 17.502. In the previous lecture, I showed you that b value times 25 degrees divided by the c value 240.97. And then we add to that 25 degrees (this is for liquid water, of course, it’s 25 degrees Celsius because nothing’s frozen). If you were working over ice, these constants would be different. So we put this into our calculator or into a spreadsheet, and we easily calculate the saturation vapor pressure at 25 degrees C is 3.17 kPa. But we’re not done yet.
We have to go back to this equation that says the vapor pressure is equal to the relative humidity times the saturation vapor pressure. When we plug our data in, the relative humidity 0.28 times the saturation vapor pressure that we calculated right here, we get a vapor pressure of 0.89 kPa. And if we were calculating fluxes (we’ll talk about that in another lecture), this is typically the value we would use. But there are other things we can do with the absolute humidity values that might be useful.
So let’s talk about vapor density. If we had a certain volume of air, and we wanted to know how much water was in that volume of air (for example, if we were going to try to condense it out) we’d more typically use this vapor density value. But how do we get from a vapor pressure that we can easily calculate from a weather report to a vapor density that would allow me to know how much water was actually in the air?
This is our equation that says the vapor pressure times the molecular weight of water divided by the universal gas constant times the kelvin temperature of the air will give us the vapor density. So I’ll take you through an example here, just continue on the one we’ve already done, just so you can see how to calculate it and to avoid a pretty common misstep.
How to avoid a common error
Again, molecular weight of water is 18.02 g/mol. The universal gas constant R is 8.31 J/mol K. And here’s the kelvin temperature of the air. I’ve scribbled this in a little bit. That’s how I note the difference between something like this, which would be air temperature in Celsius and this air temperature in kelvin. So let’s go ahead and plug all these into our equation. There’s our vapor pressure. We’re just dragging that over here. There’s our molecular weight of water. There’s our universal gas constant. And here is the kelvin temperature of the air. So as we look at this, you immediately say, how do I cancel these units? The kilopascals and the joules are certainly not going to cancel as they are. But there are conversions we can use. A Pascal is equal to an Nm-2, and a joule is equal to an Nm.
So if we change this joule to an Nm, we change this Pascal to Nm-2, we have to pay attention here as we’re doing it that the kilo right there, don’t forget that because that can mess you up. So I’m circling that to make sure that we’ve got this. Now we cancel our N’s, and combining together we get a m-3. That’s what we’re hoping for on the bottom. The grams come out on top, they don’t cancel, but everything else does. The mols cancel mols, the kelvin cancels the kelvin there, and the N cancels the N.
And we come out with just what we were looking for, save one thing, which is a kg/m-3. And this calculation gives us point 0065. But since we actually want to do this in grams, because that’s more typical of what you find in how much water there is in air. It’s not a kg of water, but more in terms of g/m-3 of water, we get 6.5 g/m-3.
Check your calculation
One way you can check this calculation (just as a rule of thumb), is if we had a pressure of the air of 100 kPa and a temperature of 20 degrees Celsius, the multiplier to get from your vapor pressure to your vapor density is about 7.4 or so. We’ll just say around 7.0. And we’ll do a quick mental calculation, 0.89 times 7, that should give us something around 6.0. So our answer should be around 6.0, and it is. It’s certainly no orders of magnitude off. So we’ve got at least close to the right answer, by doing a mental check, and we can say this conversion works.
If you want to learn more about instrumentation to measure all kinds of atmospheric parameters, please come to our website, www.metergroup.com or you can email me to chat more about this: colin.campbell@metergroup. com.
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