FCV conversion

Most of the time in fluids, we want the time changes for fixed control volumes (FCV):

The mass of a fixed control volume is not necessarily constant. For example, the fixed control volume might be a glass you are filling. Or it may be a rocket, a balloon, a pressurized aircraft cabin, or many others. We will use the glass being filled as the typical example here.

Trick: find out what happens to M_FCV from what happens to the mass of the material region that coincides with the control volume at whatever time t you are interested in. You know what happens to the mass of that material region (it remains constant).

Now use mathematics to rewrite the time derivative of the material region in terms of the time derivative of the fixed control volume. They are not the same. In the picture above, the material region keeps the same mass, while the control volume mass grows. If you do the math right, (see below for details), what you will find is that

rewrites to the following:

In words, the boxed formula says: The rate of increase in mass of the control volume plus the rate of mass flowing out of the control volume add to zero. The mass flowing out is the surface integral. It is the outflow correction to the physical law for a fixed region.

Look a bit closer at the integral. Outflow corrections are always an integral over the outside surface area S of the volume (which would include the top plane through the rim of the glass, if that is our control volume.) In the integrand, is of course the flow velocity, while is the unit vector normal to the boundary. (For the the top plane through the rim of the glass, that would be a vector of unit length sticking upwards.) So is the component of the velocity normal to the boundary (and positive outward.) It gives the speed of outflow through the boundary. Speed, area and density all increase the mass flow through the boundary.

In the picture above, the surface integral is over the entire red outside area of the fixed control volume. But almost all of it is zero. Either the density is zero (no water), or the velocity normal to the boundary is zero. The only part of the integral that is nonzero is where the water stream cuts through the red control volume surface. If the cross-sectional area of the stream is and the velocity has magnitude V_s downward, then . So in this example, mass conservation for the glass says

Bottom line: if you start applying physical laws to fixed spatial regions instead of to material regions, you better be adding outflow corrections. These outflow corrections are always integrals over the outside surface of the control volume.

Note that the mass itself is an integral over the interior volume; . Conserved properties like mass are found from integrals over the interior, while outflow corrections are integrals over the boundary.

We did not yet actually derive the above formula. To do so, we must recognize that the time derivative that we know is the time derivative of the mass of the material region. That means we must differentiate an integral over a region with moving boundaries. Such derivatives can be found using the Leibnitz rule.

• The Leibnitz rule in one dimension:

  

• The Leibnitz rule in three dimensions: The interval [a,b] becomes a volume ; the endpoints a and b become the surface S of that volume, and the sum over the two end points becomes an integral over the surface:

  

If we apply this to the integral of the mass of the material region , we get:

Since the material region was chosen to coincide with the fixed control volume at the given time, the integrals are also integral over the FCV. And since the first integral has fixed boundaries, we can take the time derivative out of it:

Exercise:

Evaluate if , , and V is the unit cube .

Instead of using the Leibnitz rule, we can derive the mass equation for a fixed control volume from more physical arguments. This is done on the class web page under Reynolds Transport Theorem.

You should now be able to do questions 5.5 and 5.1 (minus some physical interpretations.)