Energy equation

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The energy equation is the third MHD equation . In its most succinct form it is written:

\frac{d}{dt}\left(\frac{p}{\rho^{\gamma}}\right)=0

Consider a volume with no heat transfer, that's adiabatic. Then:

d\mathcal{E}=-PdV

We know for our adiabatic fluid:

PV = NT
\mathcal{E}=\frac{3}{2}NT

So that:

d\mathcal{E}=\frac{3}{2}NdT=-PdV

Then:

\frac{3}{2}\frac{dT}{T}=-\frac{dV}{V}

We can also get from PV = NT:

\frac{dP}{P}+\frac{dV}{V}=\frac{dT}{T}

So that:

\frac{dP}{P}+\frac{dV}{V}=-\frac{2}{3}\frac{dV}{V}

Or:

\frac{dP}{P}=-\frac{5}{3}\frac{dV}{V}

By mass conservation , ρV = const, so that \frac{dV}{V}+\frac{d\rho}{\rho}=0 and:

\frac{dP}{P}=\frac{5}{3}\frac{d\rho}{\rho}

Solving this differential equation:

P = cρ5 / 3

where c is some constant. This means that we can take the derivative in the fluid frame (the Lagrangian derivative :

\frac{d}{dt}\left(\frac{P}{\rho^{5/3}}\right)=0

We can expand this derivative:

\frac{1}{\rho^{\gamma}}\frac{dP}{dt}-\frac{\gamma P}{\rho^{\gamma+1}}\frac{d\rho}{dt}=0

Using the continuity equation :

\frac{dP}{dt}+\gamma P\nabla\cdot v=0

This page was recovered in October 2009 from the Plasmagicians page on Energy_equation dated 03:32, 7 June 2006.

Retrieved from "https://qed.princeton.edu/main/PlasmaWiki/Energy_equation"
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