One-dimensional flow with friction: Difference between revisions
From Flowpedia
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</math> | </math> | ||
==== The ideal gas law ==== | |||
For a perfect gas the ideal gas law reads | |||
<math display="block"> | |||
p=\rho R T | p=\rho R T | ||
</math> | |||
Differentiating (\ref{eqn:gaslaw:a}) gives: | |||
<math display="block"> | |||
dp=\rho R dT+RT d\rho | dp=\rho R dT+RT d\rho | ||
</math> | |||
If <math>p\neq0.</math>, we can divide (\ref{eqn:gaslaw:b}) by <math>p</math> which gives | |||
<math display="block"> | |||
\frac{dp}{p}=\frac{dT}{T}+\frac{d\rho}{\rho} | \frac{dp}{p}=\frac{dT}{T}+\frac{d\rho}{\rho} | ||
</math> | |||
which can be rearranged according to | |||
<math display="block"> | |||
\left[\frac{dp}{p}-\frac{d\rho}{\rho}\right]=\frac{dT}{T} | \left[\frac{dp}{p}-\frac{d\rho}{\rho}\right]=\frac{dT}{T} | ||
</math> | |||
Now, inserting <math>dT/T</math> from equation (\ref{eqn:ttot}) gives | |||
<math display="block"> | |||
\left[\frac{dp}{p}-\frac{d\rho}{\rho}\right]+\frac{(\gamma-1)}{2}M^2\frac{d(u^2)}{u^2}=0. | \left[\frac{dp}{p}-\frac{d\rho}{\rho}\right]+\frac{(\gamma-1)}{2}M^2\frac{d(u^2)}{u^2}=0. | ||
</math> | |||
The <math>d\rho/\rho</math>-term can be replaced using equation (\ref{eqn:cont}) | |||
<math display="block"> | |||
\frac{dp}{p}+\frac{d(u^2)}{2u^2}+\frac{(\gamma-1)}{2}M^2\frac{d(u^2)}{u^2}=0. | \frac{dp}{p}+\frac{d(u^2)}{2u^2}+\frac{(\gamma-1)}{2}M^2\frac{d(u^2)}{u^2}=0. | ||
</math> | |||
Collect terms and rewrite gives | |||
<math display="block"> | |||
\frac{dp}{p}+\left[\frac{1+(\gamma-1)M^2}{2}\right]\frac{d(u^2)}{u^2}=0. | \frac{dp}{p}+\left[\frac{1+(\gamma-1)M^2}{2}\right]\frac{d(u^2)}{u^2}=0. | ||
</math> | |||
==== Momentum equation ==== | ==== Momentum equation ==== | ||
