The Reynolds Transport Theorem

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.

 

Figure 1: A typical control volume for flow in an funnel-shaped pipe is bounded by the pipe wall and the broken lines.

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 t₀. One of the things physics tells you is that the net force, call it ${\bf F}$, on the fluid inside the pipe gives you the rate of change of linear momentum of the fluid. That means that if the momentum of the fluid in the pipe at time t₀ is ${\bf I_0}$, then the momentum of this fluid at time $t_0+\delta t$will be ${\bf I}_0 + {\bf F} \delta t$. In figure 2, the fluid at time $t_0+\delta t$ is shown shaded; this fluid now has ${\bf F} \delta t$ additional linear momentum.

 

Figure 2: The fluid that was in the control volume at time t₀ seen at time $t=t_0+\delta t$.

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 ${\bf F} \delta t$. 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 ${\bf I}(t)$, the change of linear momentum in the pipe,

$$\frac{d{\bf I}}{dt}\delta t$$

is usually not equal to the change of linear momentum of the fluid, ${\bf F} \delta t$. So how do we find $d{\bf I}/dt$?

 

Figure 3: The control volume at time $t_0+\delta t$.

 

Figure 4: The differences between the fluid and the control volume at time $t_0+\delta t$.

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

$$\frac{d{\bf I}}{dt} \delta t + \mbox{ [momentum in S] } - \mbox{ [momentum in H] } = {\bf F} \delta t$$

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:

$$\mbox{ [momentum in S] } - \mbox{ [momentum in O] } = \delta t \int_A \rho {\bf u}\ {\bf u} \cdot {\bf n}\ dA$$

where $\rho$ is the density of the fluid and ${\bf u} \cdot {\bf n}$is the component of the fluid velocity ${\bf u}$ normal to the local surface element dA.

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

$$\frac{d{\bf I}}{dt} + \int_A \rho {\bf u}\ {\bf u} \cdot {\bf n}\/ dA = {\bf F}$$ (1)

Compared to your physics class, the only thing new is the additional second term, the surface integral. It expresses the rate of momentum loss due to outflow and of momentum increase due to inflow. For areas of outflow, the normal component of the velocity ${\bf u} \cdot {\bf n}$ is positive, for inflow areas it is negative. You see that not all of the applied force ${\bf F}$ goes towards increasing the momentum ${\bf I}$ in the control volume; some goes towards replenishing the net momentum flowing out of the control volume.

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 $\rho{\bf u}$, 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 ${\bf n}$ which is normal to area element dA. Using this vector, we can find the component of the fluid velocity normal to dA as ${\bf u} \cdot {\bf n}$.

 

Figure 5: The contribution to the outflow strip due to a small area of the surface of the control volume dA.

The thickness of the segment is equal to the distance the fluid has travelled in the direction normal to dA, so the thickness equals $\delta t {\bf u}\cdot{\bf n}$. The volume dV_S of this segment of the strip, given by the area dA times the height, equals  

$$dV_S = \delta t \ {\bf u}\cdot {\bf n}\ dA$$ (2)

To get the momentum, we simply multiply by the local momentum per unit volume, which is $\rho{\bf u}$: 

$$d{\bf I}_S = \delta t \ \rho{\bf u}\ {\bf u}\cdot {\bf n}\ dA$$ (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 ${\bf u} \cdot {\bf n}$ 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;

$$\frac{dM}{dt} \delta t + \mbox{ [mass in S] } - \mbox{ [mass in H] }.$$

According to physics, this change in fluid mass must be zero. To find the mass of the strips is exactly the same as finding their momentum, except that we must multiply the volume of the segment in figure 5 by the mass per unit volume $\rho$ instead of the momentum per unit volume $\rho{\bf u}$. So the equation for the mass M within the control volume becomes  

$$\frac{dM}{dt} + \int_A \rho\ {\bf u} \cdot {\bf n}\ dA = 0$$ (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 $\dot W\delta t$ done on the fluid and the heat $\dot Q\delta t$ added to it:

$$\frac{dE}{dt} \delta t + \mbox{ [energy in S] } - \mbox{ [energy in H] } = \dot W\delta t + \dot Q\delta t$$

In this case we use the energy per unit volume $\rho(e+\frac12{\bf u}\cdot{\bf u})$ instead of the momentum per unit volume $\rho{\bf u}$, where e is the internal energy per unit mass and $\frac12{\bf u}\cdot{\bf u}$ is the kinetic energy per unit mass. So the equation for the energy E within the control volume becomes  

$$\frac{dE}{dt} + \int_A \rho(e+{\textstyle\frac12}{\bf u}\cdot{\bf u}) \ {\bf u} \cdot {\bf n}\ dA = \dot W + \dot Q$$ (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 ${\bf I}$, and energy E in the control volume can be found by integrating the mass, momentum, and energy per unit volume over the volume:

$$\frac{dM}{dt} = \frac{d}{dt} \int_V \rho \ dV$$

$$\frac{d{\bf I}}{dt} = \frac{d}{dt} \int_V \rho{\bf u} \ dV$$

$$\frac{dE}{dt} = \frac{d}{dt} \int_V \rho(e+{\textstyle \frac12}{\bf u}\cdot{\bf u})\ dV$$

If the boundaries of the control volume are fixed in space, we can bring the d/dt derivatives inside the integrals as $\partial/\partial t$ partial derivatives that keep the spatial position constant. However, if the boundaries of the control volume move, we cannot do so without introducing additional surface integrals.

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 t₀. The combined left hand sides in equations (1), (4), and (5) are simply the time derivatives of this Lagrangian fluid region L:

$$\frac{dM}{dt} + \int_A \rho\ {\bf u} \cdot {\bf n}\ dA = \frac{DM_L}{Dt}$$

$$\frac{d{\bf I}}{dt} + \int_A \rho{\bf u}\ {\bf u} \cdot {\bf n}\ dA = \frac{D{\bf I}_L}{Dt}$$

$$\frac{dE}{dt} + \int_A \rho(e+{\textstyle\frac12}{\bf u}\cdot{\bf u}) \ {\bf u} \cdot {\bf n}\ dA = \frac{DE_L}{Dt}$$

It is conventional in fluid mechanics to use D instead of d in time derivatives if it is the derivative of a Lagrangian quantity.

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, $D{\bf u}/Dt$, 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 ${\bf u} = {\bf u}(x,y,z,t)$, it will be the partial derivative $\partial{\bf u}/\partial t$ keeping the spatial position constant which is immediately available. How do we find the time derivative $D{\bf u}/Dt$ following the fluid? The answer is simply the chain rule of differentiation. For any function f,

$$\frac{df}{dt} = \frac{\partial f}{\partial t} + \frac{\parti... ...y} \frac{dy}{dt} + \frac{\partial f}{\partial z} \frac{dz}{dt}.$$

However, since the velocity of the fluid is defined as the time derivative of the position of the fluid, we get for the Lagrangian time derivative of any function f  

$$\frac{Df}{Dt} = \frac{\partial f}{\partial t} + u \frac{\... ...rac{\partial f}{\partial y} + w \frac{\partial f}{\partial z}.$$ (6)

The last three, additional, terms are called the convective terms. They describe changes that are due to the fact that the fluid changes position. Using vector notation the Lagrangian derivative can also be written:  

$$\frac{Df}{Dt} = \frac{\partial f}{\partial t} + ({\bf u} \cdot \nabla) f.$$ (7)