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{{Technical|date=March 2009}}

== Mud ==
Clear as mud:) Lifted adiabatically? to dew point - then squeeze '''all''' the water vapor out (how??). Then drop it back down. Does the latent heat of evap stay with it? What is the point of this bit of jargon? Would help to know the use. --[[User:Vsmith|Vsmith]] 15:03, 6 Sep 2004 (UTC)

Also [[instability]] is a bit vague.
--[[User:DavidCary|DavidCary]] 00:37, 30 November 2005 (UTC)

"R" in this instance is not the universal gas constant, but rather the *specific* gas constant for air. This is the universal gas constant divided by the molar mass of air. changing article to reflect this.

== Formula ==
Hi all, I was just looking at the formula for theta-e and noticed two things:

(1) the link referencing the formula no longer exists
(2) is the formula correct and/or have the dimensions been given correctly? Looking at the second term in the first set of brackets, Lv is in kJ/kg and cp in J/(kgK) so is this not a factor of 10^3 out?

[[User:Funkytomatoes|Funkytomatoes]] ([[User talk:Funkytomatoes|talk]]) 17:58, 31 July 2009 (UTC)

: To answer:
:*The link to UCAR is not accessible without an account. Another should be found.
:*The formula is correct. You are forgetting that <math>L_v/c_p</math> is multiplied by the [[mixing ratio]] which in is in grams/Kilograms or <math>10^{-3}</math> kg/kg, cancelling your <math>10^{3}</math>.
:[[User:Pierre cb|Pierre cb]] ([[User talk:Pierre cb|talk]]) 00:14, 1 August 2009 (UTC)

::*It ought to be made explicit that mixing ratio here is in grams/Kilograms - if you follow the mixing ratio link, that page expresses it in kg/kg. I also work with model output that gives mixing ratio in kg/kg. I think making the units clear would help.
[[User:Funkytomatoes|Funkytomatoes]] ([[User talk:Funkytomatoes|talk]]) 23:24, 1 August 2009 (UTC)

== Explanation for adiabatic temperature change ==

"Some of the internal energy of the parcel is used up in doing the work required to expand against the atmospheric pressure, so the temperature of the parcel drops, even though it has not lost any heat. Conversely, a sinking parcel is compressed and becomes warmer even though no heat is added."

I'm 95% sure this is wrong... the energy isn't used up causing it to drop in temperature. It's an adiabatic change in temperature, the temperature change is because of the difference in pressure caused by rising. Less pressure, lower temperature. I think it may just be badly worded though. <span style="font-size: smaller;" class="autosigned">— Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/188.221.226.37|188.221.226.37]] ([[User talk:188.221.226.37|talk]]) 21:07, 24 March 2012 (UTC)</span>

Talk:Equivalent potential temperature - Revision history

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