Reynolds Theorem

8/4/2019 Reynolds Theorem

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he Reynolds Transport Theorem

The Reynolds Transport Theorem

Leon van Dommelen

1/31/97

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

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

uid mechanics does not usually consider the same fluid at all times. For example, fluid mechanics maynsider the flow through a pipe, such as maybe the funnel-shaped pipe below. The region within theown pipe is fixed in space and is called a control volume. The fluid in a control volume changes when

uid flows in or out. In the pipe below, fluid enters one end of the pipe and leaves at the other end. Thews of physics do not directly apply to the pipe because the fluid in the pipe at one time is not the sameuid as at another time.

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

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

r regions which contain different fluid at different times, the laws of physics, thermodynamics, andemistry do not directly apply; they must be corrected for the entering and departing fluid. In fact, you

eady know from calculus and thermodynamics that derivatives may depend on what you keepnstant .

s an example, consider the pipe flow shown in figure 1 at an arbitrary time t 0. One of the things

ysics tells you is that the net force, call it , on the fluid inside the pipe gives you the rate of changelinear momentum of the fluid . That means that if the momentum of the fluid in the pipe at time t 0 i

, then the momentum of this fluid at time will be . In figure 2, the fluid at time

is shown shaded; this fluid now has additional linear momentum.

Figure 2: The fluid that was in the control volume at time t 0seen

at time .

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

e pipe , the change of linear momentum in the pipe,



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usually not equal to the change of linear momentum of the fluid, . So how do we find ?

Figure 3: The control volume at time .

Figure 4: The differences between the fluid and the control

volume at time .

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



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

here is the density of the fluid and is the component of the fluid velocity normal to the local

rface element dA.

he bottom line is that conservation of linear momentum for a pipe, or any other fixed volume , takes trm:

(1

ompared to your physics class, the only thing new is the additional second term, the surface integral. Itpresses the rate of momentum loss due to outflow and of momentum increase due to inflow. For areasoutflow, the normal component of the velocity is positive, for inflow areas it is negative. Youe that not all of the applied force goes towards increasing the momentum in the control volume;me goes towards replenishing the net momentum flowing out of the control volume.

you look closer at the integral, you will see that it makes sense. The outflow through an area elementA of the surface of the region inside the pipe is obviously proportional to dA. It is clearly alsooportional to the component of the fluid velocity which is normal to the area; motion in the directionthe surface does not lead to in or outflow. And where the normal component of velocity changes sign,

e switch from inflow to outflow; the contribution to the momentum equation then also changes sign asshould. The net momentum outflow is also proportional to , which is the linear momentum of the

uid per unit volume.

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

Figure 5: The contribution to the outflow strip due to a small area



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of the surface of the control volume dA.

he thickness of the segment is equal to the distance the fluid has travelled in the direction normal tothe thickness equals . The volume dV

Sof this segment of the strip, given by the area dA ti

e height, equals

(2)

get the momentum, we simply multiply by the local momentum per unit volume, which is :

(3)

get the total contribution, we simply integrate over all outside surface areas of our control volume.he contribution of the inflow strip in figure 4 should be negative, but since we always take the unitctor in the direction pointing out of the control volume, this is automatically taken care off. There isin or outflow through the solid surface of the pipe itself, but since is here zero, that too is

tomatic. So we get the single integral over all the outside surface which we wrote down earlier.

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

ccording to physics, this change in fluid mass must be zero. To find the mass of the strips is exactly theme as finding their momentum, except that we must multiply the volume of the segment in figure 5 bye mass per unit volume instead of the momentum per unit volume . So the equation for the mass

within the control volume becomes



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(4

r the equation for the energy we take the change in the energy E inside the control volume and correctr the energy in the strips in figure 4. This gives the change in energy for the fluid which equals the

ork done on the fluid and the heat added to it:

this case we use the energy per unit volume instead of the momentum per unit

lume , where e is the internal energy per unit mass and is the kinetic energy per unit mass.

the equation for the energy E within the control volume becomes

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

me additional notes. First, note that the work term in the energy equation is still the work done on theuid. For example, do not think that the pressure force on the exit surface of the pipe in figure 1 does notrform work since the exit is fixed. The two left-hand-side terms in equation ( 5) together are simply

me derivative of the energy of the fluid. So we need the work done by the pressure forces on the rightnd fluid boundary between figures 1 and 2. In other words, work is done on material points, not onathematical ones.

ext, note that the mass M , momentum , and energy E in the control volume can be found by

tegrating the mass, momentum, and energy per unit volume over the volume:

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the boundaries of the control volume are fixed in space, we can bring the d / dt derivatives inside the

tegrals as partial derivatives that keep the spatial position constant. However, if the boundaries

the control volume move, we cannot do so without introducing additional surface integrals.

he approach to fluid mechanics which uses prescribed spatial regions or positions is called a Eulerianscription after the mathematician Euler. Our pipe can be considered a Eulerian region. On the othernd, a description using given regions or points of fluid is called a Lagrangian description after Euler'sntemporary Jean-Louis Lagrange. The fluid region shown as shaded in figure 2 is the Lagrangiangion L that coincided with the Eulerian control volume V at time t 0. The combined left hand sides in

uations ( 1), ( 4), and ( 5) are simply the time derivatives of this Lagrangian fluid region L:

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

ne other thing. So far we have only shown how derivatives of regions fixed in space can be convertedderivatives of regions of fluids by adding surface integrals. Now we want to examine how we cannvert partial time derivatives for points fixed in space to time derivatives for points of fluid. For

ample, the acceleration of the fluid, , is the Lagrangian time derivative of the velocity

eping the fluid point constant. However, if we have computed or measured a velocity field in Eulerian

rm as a function , it will be the partial derivative keeping the spatial

sition constant which is immediately available. How do we find the time derivative

llowing the fluid? The answer is simply the chain rule of differentiation. For any function f ,

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owever, since the velocity of the fluid is defined as the time derivative of the position of the fluid, wet for the Lagrangian time derivative of any function f

(

he last three, additional, terms are called the convective terms. They describe changes that are due toe fact that the fluid changes position. Using vector notation the Lagrangian derivative can also beitten:

(7)

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Reynolds Theorem