I'm not sure if this is the correct place to ask this. Basically I have an exercise in dynamic meteorology that is set up like this:

1 kg of dry air is falling with a constant speed of 0.5 cm / s and 0.1 w/kg of heat is extracted. Does the air volume temperature rise, fall or does it remain unchanged?

Here's how I tried to solve it:

$w = 0.5 \frac{cm}{s}$

$J= -0.1 \frac{w}{kg}$

$T? $

$J = Cp \frac{dT}{dt} - \alpha \frac{dp}{dt}$

$J = Cp \frac{dT}{dt} - \frac{1}{\rho} \frac{dp}{dt}$

$\frac{dp}{dt} = \frac{\delta p}{\delta t} + u \frac{\delta p}{\delta x} + v \frac{\delta p}{\delta y} + w \frac{\delta p}{\delta z} = w \frac{\delta p}{\delta z}$

(because $\frac{\delta p}{\delta t} + u \frac{\delta p}{\delta x} + v \frac{\delta p}{\delta y} = 0$)

$ \frac{\delta p}{\delta z} = -\rho g$

$\frac{dp}{dt} = w \frac{\delta p}{\delta z} = w -\rho g$

$J = Cp \frac{dT}{dt} - \frac{-1 w \rho g}{\rho} = Cp \frac{dT}{dt} + \rho g$

$$\frac{dT}{dt} = \frac{J-gw}{Cp} = \frac{-0.1 \frac{w}{kg} - 9.81 \frac{m}{s^2} 0.5 \frac{cm}{s}}{1004.5 \frac{J}{K * kg}} $$

Here's where I get lost. Any help?

  • 1
    $\begingroup$ Been a long time since I've done stuff like this, so not sure I'm too much help. But a few things... your heat is in W\kg, right? Also, unitwise your gw term just doesn't work close with the others (you'd need to convert w to m\s, but m^2\s^3 doesn't seem to be anywhere near W\kg)? So there seems there may be something faulty. I'm also not used to seeing J be a variable for something related to heat (since it's also used for the standard unit of heat). Your starting equation looks like some alternation of the 1st Law of Thermo. And you assume hydrostatic approx (incompressible, guess is ok?)? $\endgroup$ Commented Nov 8, 2016 at 17:52

1 Answer 1


\begin{equation} \frac{w}{kg} = \frac{J}{kgs} = \frac{Nm}{kgs} = \frac{kgm}{s^2}\frac{m}{kgs} = \frac{m^2}{s^3} \end{equation} So, $J$ and $wg$ have same unit, and also your final answer has correct unit, Kelvin per second. Your final answer says the parcel's temperature will drop, even without cooling term $J$ (you will find this strange because it should be heated by adiabatic heating). It is because you set $w$=0.5 instead of $w$ =-0.5. I hope this helps you :)

  • $\begingroup$ i found a strange grammar :D $\endgroup$ Commented Nov 10, 2016 at 5:13
  • $\begingroup$ Indeed, shows how well I do! You're right, those are good matching units after all (once cm converted to m!!!). And the direction of the answer does make sense because both the rising and the diabatic term are losses (and both are similar scale, once you do cm->m). Equates to -0.5 K/hr, which does seem a quite reasonable rate for gradual widespread ascent. $\endgroup$ Commented Nov 10, 2016 at 22:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.