I'm having trouble with the difference between saturated and unsaturated adiabatic lapse rates.

I know the unsaturated adiabatic lapse rate is approximately 5.4°F/1000ft (9.8 °C/1000 metre), in which a "dry" parcel of air cools at that rate. The unsaturated adiabatic lapse rate depends on what?

Why is the unsaturated adiabatic lapse rate lower than the saturated? From searching, I found that the unsaturated adiabatic lapse rate is lower because the air parcel releases its latent heat, which causes the air parcel to cool more slowly, but I don't seem to get it. When a "dry" parcel of air is cooling adiabatically, are we assuming 0% moisture?


1 Answer 1


Let's start with the simplest fact. With nothing else going on, air in our atmosphere will always be cooled at 5.4°F/1000ft (or the more widely used 9.8°C/km), due to it expanding in size (often times we'll throw a negative sign: -5.4°F/1000ft... this is just to emphasize that these lapse rates are about the air COOLING). Every bit of the rising air cools at that rate (the oxygen, the nitrogen, even any water vapor in the air). All of it will cool at 5.4°F/1000ft. All rising air always expands at this same rate.

However sometimes there is a second complicating factor: the rising air may cool to the point where it is holding its maximum quantity of water vapor (because when colder, air cannot hold as much water vapor).

When it reaches this point, it is saturated, and any more cooling/rising will force some of its water vapor to condense instead into liquid droplets. For this gaseous water vapor to become a liquid, its particles must expel quite a lot of energy. And that energy gets released into the air.

So once air hits this saturation point, though its expansion continues to have a -5.4°F/1000ft impact on the temperature, it also has a second factor, that released phase change energy, warming the air some. The result is that saturated air doesn't cool as fast.

  • Whenever air isn't saturated, it cools at the 5.4°F/1000ft (9.8°C/km). We call this the dry adiabatic lapse rate because nothing else is altering the process. We aren't assuming 0% moisture. It doesn't matter, as long as it's not 100% moisture. Water vapor cools just like every other molecule unless saturated. The first 6 pages of these notes go into intimate detail for why the value is 9.8°C/km, but basically it depends only on the values for gravity and the specific heat capacity of water, plus any error in the hydrostatic approximation... all of which are minuscule for large air masses in the troposphere. For all intents and purposes, the dry (unsaturated) adiabatic lapse rate on Earth depends on nothing, it is always 5.4°F/1000ft (9.8°C/km)
  • Then when air is saturated, because the condensation process hinders the cooldown by adding its energy, we call it the moist adiabatic lapse rate to distinguish it. Picture moist as having an image of water DROPS, which you only get when saturated, due to condensation. Moist and dry both have some water vapor. But only moist has water droplets. The moist adiabatic lapse rate has a smaller value because it is cooling LESS overall (due to the warming impact from the latent heat release). But it is still cooling, just less. The moist adiabatic lapse rate varies depending on how much water is actually released (which is based upon the amount of water held, or consequently, the temperature). If you really wish to dig into the fine details, this AMS link gives the formula. But generally the moist adiabatic lapse rate is around 5-7°C/km, or 2.5-4.0°F/1000ft, in the low-levels of the atmosphere.
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    $\begingroup$ To the strict meteorologists: I apologize, I know I breeched into some imperfect vocabulary, such as the term "holds". I did it attempting to make this more approachable for students without overstuffing the description with complex vocabulary. If this bothers you and you think you can better balance precision with ease, please take a shot! For students: if you're attempting more than a cursory understanding of the topic, further research into saturation vapor pressure would behoove you, especially when it comes time for topics like supersaturation, deposition, and cloud physics. $\endgroup$ Commented Apr 13, 2017 at 7:43
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    $\begingroup$ I like the "in bold" distinction between strict meteorologists and students. $\endgroup$
    – user1066
    Commented Apr 13, 2017 at 8:06
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    $\begingroup$ @JeopardyTempest, another point I think you could have touched on is that it takes energy to turn water from a liquid to a gas. That energy doesn't go away. Once it cools back into a liquid, it has to give up that energy, and it does to its environment. This is also called latent heat of evaporation. Otherwise, upvote for a lucid answer. $\endgroup$
    – BillDOe
    Commented Apr 14, 2017 at 20:36
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    $\begingroup$ I curious where the authors of hs.umt.edu/physics/documents/BOREALIS/… were sourcing their value of Cp for dry air. With g = –9.80665 m/s^2, you'd need Cp = 999.7 J/(kg K) to get their quoted g/Cp = –9.81 K/km, which has become the de facto standard. But 999.7 J/(kg K) is below any literature value I've seen for Cp of dry air between 250 K and 300 K. The values I typically see over this range are 1003 to 1005 J/(kg K), (ohio.edu/mechanical/thermo/property_tables/air/air_Cp_Cv.html), which would give g/Cp = –9.78 to –9.76 K/km. $\endgroup$
    – theorist
    Commented Feb 14, 2022 at 0:30
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    $\begingroup$ @theorist sounds like a good separate question to me, I certainly don't know off the top of my head, and am likewise a big proponent of proper sigfig use and not using errant values when possible $\endgroup$ Commented Feb 14, 2022 at 3:44

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