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Hi guys i have some problems starting from $(1)$ deriving the $(2)$:

$$\frac{d}{dt}\int_V \rho e dV=\int_V \rho \pmb{f}\cdot \textbf{u}dV+\int_{V}\rho~s~dV+\int_{\partial V}(\pmb{\sigma} \cdot \pmb{u}) \cdot \pmb{n} dA+\int_{\partial V}k \nabla T \cdot \pmb{n} dA~~~(1)$$

using the Reynold and Divergence Theorem:

$$\frac{\partial \rho e}{\partial t}+div(\rho\pmb{u}-\pmb{\sigma}\cdot \pmb{u}-k\nabla T)=\rho\pmb{f}\cdot \pmb{u}+\rho s ~~~(2)$$

Now doing calculations i feel comfortable with the "div" terms, what I don't get is the presence of $\frac{\partial \rho e}{\partial t}$ because I have a total time derivative in the $(1)$ and a partial time derivative of the right-hand side of the $(2)$ that completely disappears (partial time derivative of the RHS, together with $dV$)

I can quite accept thought is non-rigorous that because we are considering the whole volume we can simplify the $dV$ together from the LHS and RHS

but for the time derivative, what I see is something like $$\frac{d}{dt}\frac{\partial t}{\partial}=\frac{\partial}{\partial t} (?)$$ and to me this looks seriously non sense.

Could some one make some clarifications? Thanks in advance

MatteoX
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1 Answers1

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The PDE (2) is a continuity equation which we can write more shortly as $$\tag{2*} \partial_t q+\operatorname{div}\boldsymbol{v}=g\,. $$ In $\int_V q\,dV$ it is presumably the case that the volume $V$ over which you are integrating does not depend on time. Then it is trivial that the total time derivative of that integral is the partial time derivative and the rest follows from Gauss' divergence theorem: \begin{align} \partial_t\int_V q\,dV&=\int_V \partial_tq\,dV\stackrel{(2*)}{=}\int_Vg\,dV-\int_V\operatorname{div}\boldsymbol{v}\,dV\\ &\stackrel{Gauss}{=}\int_Vg\,dV-\int_{\partial V}\boldsymbol{v}\cdot\boldsymbol{n}\,dA\,.\tag{1*} \end{align} If I am not mistaken, in your version of (1) there could be a term $$ -\int_{\partial V}\rho\boldsymbol{u}\cdot\boldsymbol{n}\,dA $$ missing.

Kurt G.
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