I am studying Aerodynamics, to be more precise, the fundamentals of Aerodynamics. The first law is the continuity equation, for which it is explained in the book that I am using. However, I wished to derive this on my own... with the consequence that I cant. Therefore I hope that somebody might spot the mistake I did or corrects me in my attempt.
Basically, the continuity equation states:
\begin{equation} m`_{out} - m`_{in} =\frac{dm}{dt} \end{equation}
Now what I did is I assumed an imaginary small cube with lengths of a. Three of those surfaces will be the entry of the flow, the other three the outlet of the flow.
Therefore we can write the equation as: \begin{equation} ((\rho+d\rho)(v+dv)(3a^2))-(\rho v (3a^2)) = \frac{\delta}{\delta t} (p dV) \end{equation}
Rewriting this, we get to: \begin{equation} (3a^2)((\rho+d\rho)(v+dv) - (\rho v)) = \frac{\delta}{\delta t} (p dV) \end{equation} \begin{equation} (3a^2)(\rho v + \rho dv + vd\rho + d\rho dv - \rho v) = \frac{\delta}{\delta t} (p dV) \end{equation} \begin{equation} (3a^2)(\rho dv + vd\rho) = \frac{\delta}{\delta t} (p dV) \end{equation}
Since we are talking about a tiny volume, we can express $3a^2$ as dA. This would result in: \begin{equation} (\rho dv + vd\rho)dA = \frac{\delta}{\delta t} (p dV) \end{equation}
Now I would notice that $(\rho dv + vd\rho) = (\rho v)`$, but this is exactly where the equation suddenly turns out to be wrong. I should obtain: \begin{equation} (\rho v)dA = \frac{\delta}{\delta t} (p dV) \end{equation} from which I can then take the corresponding integrals for the Area and the Volume. But with my method, I arrive to the wrong result of: \begin{equation} (\rho v)` dA = \frac{\delta}{\delta t} (p dV) \end{equation}
What am I doing wrong???
Thanks for your input!