We will now analyze the case where our function $\phi$ is equal to the density of our fluid. Again, this is a property defined over all the region $\Omega$ and it might change in time, and be different at different places inside the fluid.
From our last post, we had an equation, which after replacing $\phi\rightarrow\rho$ looks like this:
\[\frac{d}{dt}\int_{V_f(t)}\rho\,d\Omega=\frac{d}{dt}\int_{V_c(t)}\rho\,d\Omega+\oint_{\partial V_c(t)}\rho \left[(\vec{u}-\vec{u}_c)\cdot\vec{n}\right]\,d\sigma\]
We can inmediatly identify the LHS of the equation with the time derivative of the total mass inside our fluid volume.
\[M=\int_{V_f(t)}\rho\,d\Omega\]
However, we must remember that one of the basic properties of the fluid volume, is that it's mass remains constant. Therefore the time derivative of such quantity is zero. We are left with:
\[0=\frac{d}{dt}\int_{V_c(t)}\rho\,d\Omega+\oint_{\partial V_c(t)}\rho \left[(\vec{u}-\vec{u}_c)\cdot\vec{n}\right]d\sigma\]
This is a very important equation, and can help us understand many interesting properties of the system we are studying. For example, as a general consequense, we can say that the change in mass inside a given volume is equal to the mass-flow through the surface of such volume. Something rather trivial, but that can be mathematically quantified in this particular equation.
The last equation is also called the Integral Formulation of the Law of Conservation of Mass. As any other integral formulation or representation, it encodes a macroscopic condition on the physics of the fluid. It provides a set of rules that restrict the behavior from a macroscopic point of view. What I mean by that is, The conditions imposed on the fluid take the form of conditions or regulations over fluid as a whole (integrals over Vc and it's surfaces). The integral formulation offers no local conditions for the fluid. The fluid, as far as the integral formulation is concerned, can have "holes" in it. As far as the final number, resulting from the integration over the whole volume is in agreement with the equation, everything might seem OK.
My point here is the following. Given a control volume, the integral formulation of the conservation of mass doesn't really impose, by itself, conditions on the internal, and local characteristics of the fluid. The fluid might have points where the density is incredibly high, like a black hole, then mysteriously disappear and appear instantaneously in some other point (of course, also inside Vc), and still, the integral formulation will not have any idea of whats going on inside.
The last argument is perfectly valid, no flaws. However, if you add a small detail (a very important, and must never be forgotten) you can actually convert the integral formulation into a set of restrictions for the fluid at the local level, at the microscopic level. The trick goes somehow like this: The integral formulation is correct, and it is correct for ANY CONTROL VOLUME you can think off.
Now think about this, if it is valid for any control volume, you can shrink your control volume (The shrinking is not as if Vc depends on time, and shrinks while you integrate, no. Its more like everytime you are about to integrate, you pick a smaller and smaller Vc), almost to the point where is seems like a point. At this level, you can see the integration formulation of such a tinny volume as a local formulation, a local restriction on the fluid.
Is kinda hard to imagine at first, but think about it. Given that the integral formulation offers you info about the macroscopic state of the fluid, and since it is valid for any Vc, and it's respective surface, you could shrink Vc to the point where is so small that you could identify it with a microscopic condition on the fluid. The math behind this is consistent with what we have talked here. The resulting microscopic formulation is usually called the differential formulation of the conservation of mass, or simply (and perhaps more familiarly) the continuity equation.
To get there we will use the well known Stoke's theorem, to transform the surface mass fluxes into an integral over a volume. The most general formulation of this theorem is something like this:
\[\int_{\Omega}d\omega=\int_{\partial\Omega}\omega\]
Where $\omega$ is n-form in a n-dimensional space, and $d$ is the exterior derivative. $\Omega$ is an n-dimensional volume, and $\partial\Omega$ is the (n-1) dimensional boundary of $\Omega$. Green's Theorem, The gradient Theorem, Gauss-Ostrogradsky theorem, or the 3D-Stokes theorem are all specific cases of this more general version. Each one of them is the result of the correct choice of the n-form $\omega$. Check out this guy, he's got everything you need to know about this topic in one pdf:
For our purposes, we will pick n=3, and
\[\omega=A_1dy\wedge dz+A_2dz\wedge dx+A_3dx\wedge dy\]
Which corresponds to: $\vec{A}\cdot\vec{n}\,d\sigma$.
Applying the exterior derivative to this form, we have:
\[d\omega=\left({\frac{dA_1}{dx}+\frac{dA_2}{dy}+\frac{dA_3}{dz}}\right)dx\wedge dy\wedge dz\]
Which corresponds to $\nabla\cdot\vec{A}\,dV$
For this specific form, Stoke's theorem reads:
\[\oint_{\partial V} \vec{A}\cdot \vec{n}\,d\sigma = \int_V \nabla\cdot\vec{A}\,dV\]
This is called, the divergence theorem.
Now, back to our mass equation with 0 on the LHS. If we look at the RHS, more specifically, the integral over $\partial\Omega$ we can apply the divergence theorem simply by setting:
\[\vec{A}=\rho\vec{u}\]
More over, if Vc is such that it is stationary, ie $V_c(t)=V_c$ and $\vec{u}_c=0$. Then we can put the time derivative in our first term inside the intergral at the cost of making it a partial derivative. This is:
\[\frac{d}{dt}\int_{Vc}\rho\,d\Omega = \int_{Vc}\frac{\partial\rho}{\partial t}\,d\Omega\]
Finally, putting all together, we have:
\[\int_{Vc}\left(\frac{\partial\rho}{\partial t}+\nabla\cdot (\rho\vec{u})\right)\,d\Omega=0\]
Now, back to our naive argument "valid for any Vc you can think off". In our case, this argument is extremely powerfull. It means, that no matter which Vc we pick, we will always have this integral equal to zero. No matter how small, or large Vc is, we will have this integral being zero. Now, if you think about this for about $ to \times 10^{15}$ minutes, you will convince yourself that the only possible explanation for this, rather curious, fact is that the integrand itself must be equal to zero.
\[\frac{\partial\rho}{\partial t}+\nabla\cdot (\rho\vec{u})=0\]
This can also be written in a nicer way, using the identity: $\nabla(\psi\vec{A})=\psi\nabla\cdot\vec{A}+\vec{A}(\nabla\psi)$, and our previously derived Lagrangian/Convective/Total/Material derivative:
\[\frac{D\rho}{Dt}+\rho\nabla\cdot\vec{u}=0\]
This is what we call, the continuity equation, or the differential formulation of the conservation of mass.With this I finish the treatment of the special case $\phi=\rho$. I hope you enjoyed as much as I did, and let's start counting how many times you will go back to this blog in your life..hehe.
No comments:
Post a Comment