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Basic state

In the basic steady state there is no fluid motion i.e. $ \boldsymbol {U}_0=0$. Now the basic presssure $ P_0(z)$ and temperature $ T_0(z)$ is given by

$\displaystyle \frac{d^2T_0}{dz^2}=0$   and$\displaystyle \quad
\frac{d}{dz}\left(\frac{P_0}{\rho_b}+gz\right) - \alpha g (T_0-T_b) = 0
$

which have solutions given by

$\displaystyle \boldsymbol {U}=0,\;T_0(z)=T_b-\beta z,\;P_0(z)=p_b-g\rho_b(z+\alpha \beta z^2/2)
$

where $ \beta=(T_b-T_u)/d$ is the basic temperature gradient. We anticipate for stability we must have $ \beta > 0$.



A/C for Homepage of Dr. S Ghorai 2003-01-16