**Leon van Dommelen**

**1/31/97**

This document describes the Reynolds transport theorem, which converts
the laws you saw previously in your physics, thermodynamics, and
chemistry classes to laws in fluid mechanics. The laws of physics
*look* somewhat different in fluid mechanics, even though they are
of course still exactly the same.

The difficulty is that your earlier classes always talked about a fixed quantity of fluid. For example, they told you that the mass of the fluid never changed (ignoring relativity effects). Also, Newton's second law said that the rate of change of the linear momentum of the fluid equals the total external force exerted on the fluid. And the first law of thermodynamics said that the total internal and kinetic energy of the fluid increases according to the work being done on the fluid and the heat being added to it. All these statements are only true of we consider a fixed quantity of fluid.

Fluid mechanics does not usually consider the same fluid at all times. For example, fluid mechanics may consider the flow through a pipe, such as maybe the funnel-shaped pipe below. The region within the shown pipe is fixed in space and is called a control volume. The fluid in a control volume changes when fluid flows in or out. In the pipe below, fluid enters one end of the pipe and leaves at the other end. The laws of physics do not directly apply to the pipe because the fluid in the pipe at one time is not the same fluid as at another time.

For most flows studied in fluid mechanics, in or outflow is common. A jet engine on a plane is another example. So is the flow around a vehicle such as a car or an airplane. Seen moving along with the vehicle; new fluid continuously enters the vicinity of the vehicle from upstream while fluid departs downstream. Gas flows out of a rocket.

For regions which contain different fluid at different times, the laws
of physics, thermodynamics, and chemistry do not directly apply; they
must be corrected for the entering and departing fluid. In fact, you
already know from calculus and thermodynamics that *derivatives may
depend on what you keep constant*.

As an example, consider the pipe flow shown in figure 1 at an
arbitrary time *t _{0}*. One of the things physics tells you is that the
net force, call it , on the fluid inside the pipe gives you
the rate of change of linear momentum

But what happens to the linear momentum *in the pipe*? In other
words, for the horizontally hatched region in figure 3? It has not
necessarily changed by . After all, the pipe now
contains some fluid which was not there before, shown as the hatched
strip in figure 4. The pipe has also lost some of its previous fluid,
shown as the shaded strip in figure 4. Mathematically, if we call the
linear momentum in the pipe , the change of linear momentum
in the pipe,

The trick is to add the momentum in the shaded strip (call it S) to the momentum in the pipe and to substract the momentum in the hatched strip (call it H). This gives us back the momentum in the fluid region, which we know. In short

A bit later we will show that the correction terms [momentum in S] -
[momentum in H] can be written as a single integral over the entire
outside surface *A* of the region within the pipe:

The bottom line is that conservation of linear momentum for a pipe, or
any other *fixed volume*, takes the form:

(1) |

If you look closer at the integral, you will see that it makes sense.
The outflow through an area element *dA* of the surface of the region
inside the pipe is obviously proportional to *dA*. It is clearly also
proportional to the component of the fluid velocity which is *
normal* to the area; motion in the direction of the surface does not
lead to in or outflow. And where the normal component of velocity
changes sign, we switch from inflow to outflow; the contribution to
the momentum equation then also changes sign as it should. The net
momentum outflow is also proportional to , which is the
linear momentum of the fluid per unit volume.

Now we will show that indeed the linear momentum in the two strips in
figure 4 can be written as the single integral over all of the surface
area of the control volume. In figure 5 we show a typical segment of
the outflow strip corresponding to an area element *dA* of the outside
surface of the control volume. We also show the local unit vector
which is normal to area element *dA*. Using this vector, we
can find the component of the fluid velocity normal to *dA* as .

The thickness of the segment is equal to the
distance the fluid has travelled in the direction normal to *dA*, so
the thickness equals . The volume
*dV*_{S} of this segment of the strip, given by the area *dA* times the
height, equals

(2) |

(3) |

To get the total contribution, we simply integrate over all outside surface
areas of our control volume. The contribution of the inflow strip in figure
4 should be negative, but
since we always take the unit vector in the direction pointing
*out* of the control volume, this is automatically taken care off.
There is no in or outflow through the solid surface of the pipe itself, but
since is here zero, that too is automatic.
So we get the single integral over all the outside surface which we
wrote down earlier.

How about conservation of mass? It is almost exactly the same story.
If we take the change in the
mass *M* inside the control volume and add to it the mass in the
shaded strip in figure 4 and substract the mass in the hatched strip,
we get the change in mass in the fluid region of figure 2;

(4) |

For the equation for the energy we take the change in
the energy *E* inside the control volume and correct for the energy in the
strips in figure 4. This gives the change in energy for the fluid which
equals the work done on the fluid and
the heat added to it:

(5) |

It is clear that we can apply this same procedure to any other quantity in the fluid for which we know a conservation law. It should also not be very difficult to modify the above formulas in case the boundary of control volume itself also moves. For example, suppose the pipe of figure 1 is not rigidly suspended but vibrates horizontally?

Some additional notes. First, note that the work term in the energy
equation is still the work done on the fluid. For example, do not
think that the pressure force on the exit surface of the pipe in
figure 1 does not perform work since the exit is fixed. The two
left-hand-side terms in equation (5) *together* are
simply the time derivative of the energy of the fluid. And we
therefor need the work done on the phsyical *fluid*, not on the
imaginary control volume. So we need, among others, the work done by
the pressure forces on the right hand fluid boundary between figures 1
and 2.

Next, note that the mass *M*, momentum , and energy *E* in the
control volume can be found by integrating the mass, momentum,
and energy per unit volume over the volume:

The approach to fluid mechanics which uses prescribed spatial regions
or positions is called a Eulerian description after the mathematician
Euler. Our pipe can be considered a Eulerian region. On the other
hand, a description using given regions or points of fluid is called a
Lagrangian description after Euler's contemporary Jean-Louis Lagrange.
The fluid region shown as shaded in figure 2 is the
Lagrangian region *L* that coincided with the Eulerian control volume
*V* at time *t _{0}*. The combined left hand sides in equations
(1), (4), and (5) are simply the
time derivatives of this Lagrangian fluid region

One other thing. So far we have only shown how derivatives of *
regions* fixed in space can be converted to derivatives of regions
of fluids by adding surface integrals. Now we want to examine how we
can convert partial time derivatives for *points* fixed in space
to time derivatives for points of fluid. For example, the
acceleration of the fluid, , is the Lagrangian time derivative of
the velocity keeping the fluid point constant. However, if
we have computed or measured a velocity field in Eulerian form as a
function , it will be the partial
derivative keeping the spatial position
constant which is immediately available. How do we find the time
derivative following the fluid? The answer is simply
the chain rule of differentiation. For any function *f*,

(6) |

(7) |