Dynamic meteorology - calculating air temperature based off falling air

Question

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?

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 :)