<p><span class="math-container">\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}</span>
So, <span class="math-container">$J$</span> and <span class="math-container">$wg$</span> 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 <span class="math-container">$J$</span> (you will find this strange because it should be heated by adiabatic heating). It is because you set <span class="math-container">$w$</span>=0.5 instead of <span class="math-container">$w$</span> =-0.5. I hope this helps you :)</p>

CC BY-SA 3.0
2016-11-10T05:13:38.487Z
i found a strange grammar :D
Jaemin Hong
CC BY-SA 3.0
2016-11-10T22:20:51.553Z
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.
JeopardyTempest
CC BY-SA 4.0
2016-11-10T04:58:19.547Z
2021-09-05T13:39:57.703Z
2021-09-05T13:39:57.703Z
Jaemin Hong
<p>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:</p>

<blockquote>
  <p>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?</p>
</blockquote>

<p>Here's how I tried to solve it:</p>

<p>$w = 0.5 \frac{cm}{s}$</p>

<p>$J= -0.1 \frac{w}{kg}$</p>

<p>$T? $</p>

<p>$J = Cp \frac{dT}{dt} - \alpha  \frac{dp}{dt}$</p>

<p>$J = Cp \frac{dT}{dt} - \frac{1}{\rho}  \frac{dp}{dt}$</p>

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

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

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

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

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

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

<p>Here's where I get lost. Any help?</p>

CC BY-SA 3.0
2016-11-08T17:52:33.823Z
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
<meteorology><temperature><homework>
Dynamic meteorology - calculating air temperature based off falling air
PadaKatel
# Dynamic meteorology - calculating air temperature based off falling air

(Asked by: username_0 on 2016-11-08)

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?

**Tags:** meteorology, temperature, homework

> **Comments:**
> **username_1**: 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?)?  

---

## 1 answers

### Answer by username_2 on 2016-11-10. Score: 2 | Accepted answer

\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 :)

> **Comments:**
> **username_2**: i found a strange grammar :D  
> **username_1**: 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.  

---

