Sunday, 18 December 2016

fluid dynamics - Transport theorem derivation question


I am trying to rigorously go through some fluid mechanics proofs and theorems. I am currently going through a proof related to the transport theorem and I am having trouble with a step.



The steps in question are the following, first the variables are transformed according to:


$$ \frac{d}{dt}\int_{W_t} \rho \mathbf{u} dV = \frac{d}{dt}\int_W (\rho \mathbf{u})(\phi (\mathbf{x},t),t)J(\mathbf{x},t)dV $$


where $\phi(\mathbf{x},t)$ is the trajectory function of the particle found at $\mathbf{x}$ at time $t=0$. $W$ is the volume of the fluid under consideration at $t=0$ and $W_t$ is that volume at time $t$. $J$ is the Jacobian determinant of the transformation, followed by differenciating under the integral sign (RHS)


$$ \frac{\partial}{\partial t}(\rho \mathbf{u}) (\phi(\mathbf{x},t),t) = \left( \frac{D}{Dt} \rho \mathbf{u} \right) (\phi (\mathbf{x},t),t) $$


In the context of the first equation, why is the second equation true? When I try to prove it to myself I get:


$$ \frac{D}{Dt}(\rho \mathbf{u}) = \frac{\partial}{\partial t}(\rho \mathbf{u}) + \mathbf{u}\cdot \nabla (\rho \mathbf{u}) $$


and I don't know how to eliminate the second term. I think it has something to do with the trajectory function $\phi$ but I am not familiar enough with this idea to know how to proceed.




No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...