G.MATH
FLUID MECHANICS
Reynolds Transport Theorem complex case

Previously we derived a simplified version of the Reynolds transport theorem. How easy our analysis would be if all fluids move in one dimension, which we all know is not true. We will now derive it for much more general conditions.

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LESSON
A general fixed control volume with fluid flowing through it is shown below.

The flow field may be quite simple, or it may involve a complex, unsteady, three-dimensional situation. Whatever the case, we consider the system to be the fluid within the control volume at the initial time t. A short time later, a portion of the fluid, marked as region II, has exited from the control volume and additional fluid, marked as region I and not part of the original system, has enter the control volume.

As before, we consider an extensive fluid property B and seek to determine how the rate of change of B associated with the system is related to the rate of change of B within the control volume at any instant.

Although this ‘complex’ case may seem difficult handle, it turns out that we can adopt our previous equation

and give correct interpretations to the terms  and  such that it suits this complex case. In general, the control volume may contain more (or less) than one inlet and one outlet. Taking a pipe system with several inlets and outlets for example, we argue that rate of inflow of property B comes from all the inlets grouped together  and likewise for the outflow, all the outlets are grouped together .

The term  represent the net flowrate of the property B from the control volume. Just think of its value as arising from the addition, or integration, of the contributions through each infinitesimal area of element of size  on the portion of the control surface dividing region II and the control volume. We term this surface as .

 

In time  the volume of fluid that passes across each area element is given by , where  is the height (normal to the base, ) of the small volume element, and  is the angle between the velocity vector and the outward point normal to the surface, . Thus, since , the amount of property B carried across the area element  in the time interval  is given by

The rate at which B is carried out of the control volume across the small area , denoted , is

By integrating over the entire outflow portion of the control surface, , we obtain

The quantity  is the component of the velocity normal to the area element . From the definition of the dot product, this can be written as . Hence, an alternate form of the outflow rate is

In a similar fashion, by considering the inflow portion of the control surface, , as shown below, we find that the inflow of B into the control volume is

You might be wondering about the negative sign in the above equation. We are using the standard notation that the unit normal vector to the control surface, , points out from the control volume.  for outflow regions, the normal component of  is positive; . For inflow regions , the normal component of  is negative; . The value of  is, therefore, positive on the  portions of the control surface and negative on the  portions.

 

Finally, the net flux or flowrate or parameter B across the entire control surface is

where the integration is over the entire control surface.
By combining the equations, we have

We write it in a slightly different form by using  so that

This is the general form of the Reynolds transport theorem for a fixed, nondeforming control volume. It took us a while to get to this result but it is sure an important one in our analysis of fluids.

All information presentated, less questions and exercises, is original content of Donny, with slight references to various books.
Video courtesy of YouTube.com service.
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