# Dynamic meteorology - calculating air temperature based off falling air

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?

• 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?)? – JeopardyTempest Nov 8 '16 at 17:52