0
$\begingroup$

Beginning with the Primitive equations governing atmospheric motion for a dry gas, primarily the ideal gas law, and the conservation of mass and energy, neglecting diffusivity.

$$P=\rho R T$$ $$\omega \equiv \frac{DP}{Dt}$$ Therefore $$\omega=R(\rho\frac{DT}{Dt}+T\frac{D\rho}{Dt})$$ Since $$\frac{D\rho}{Dt}=-\rho \nabla\cdot\vec{u}$$ and $$\frac{DT}{Dt}=\frac{\omega\rho}{c_p}+\frac{Q}{c_p}$$ $$\omega=R(\frac{\rho^2\omega}{c_p}+\frac{\rho Q}{c_p}-T\rho\nabla\cdot\vec{u})$$ Distributing $R$ and applying the ideal gas law $$\omega=\frac{R\rho^2\omega}{c_p}+\frac{PQ}{Tc_p}-P\nabla\cdot\vec{u}$$ Separating $\omega$ from the right hand side of the equation yields $$\omega=(1-\frac{R\rho^2}{c_p})^{-1}(\frac{PQ}{Tc_p}-P\nabla\cdot\vec{u})$$

I call this semi-diagnostic, for $\omega$ is still a part of $\nabla\cdot\vec{u}$

Is this a valid semi-diagnostic equation for $\omega$?

$\endgroup$
  • $\begingroup$ Been a long while since I've worked in derivations and such, and so I may not be the best source of input (it gets pretty complex for me as such towards the end!). But saying rho changes over time only due to advection seems a significant simplification? $\endgroup$ – JeopardyTempest Mar 9 '17 at 21:08
  • $\begingroup$ (you may want to give more of the assumptions\sources of some of those equations around that point) $\endgroup$ – JeopardyTempest Mar 9 '17 at 21:09
  • 1
    $\begingroup$ Sure, the continuity assumption and the ideal gas assumption, are two base assumptions. I also assume a dry gas. The capital D/Dt denotes the Lagrangian derivative, such that advection is built into the derivative. The continuity equation in nonconservative form, and my definition for conservation of energy (loosely) is from "An Introduction to Dynamic Meteorology" by Holton. My definition for omega comes from "Atmospheric Science" by Wallace and Hobbes. $\endgroup$ – BarocliniCplusplus Mar 9 '17 at 21:36
  • $\begingroup$ Being that I didn't use the Navier-Stokes equations, there is no need to assume a momentum balance. If no, what did I do wrong? I am sure there is an alternate derivation- the final equation can be interpreted as a restatement of the conservation of energy with diabatic heating and work on the right hand side. $\endgroup$ – BarocliniCplusplus Mar 31 '17 at 15:26
  • $\begingroup$ You say you are starting off with the primitive equations. The primitive equations have a distinct horizontal component.So unless you clarify what is it you are trying to do I am going to wait :) $\endgroup$ – gansub Apr 1 '17 at 10:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.