# 3.8: Untitled Page 34

## Chapter 3

In order to relate this volume to the surface area of the control volume, we need to make use of the projected area theorem (Stein and Barcellos, 1992, Sec. 17.1).

Figure 3‐9. Volume of fluid crossing the surface dA in a time  t .

Figure 3‐10. Volume leaving at the control surface during a time t

This theorem allows us to express the cross sectional area at an exit according to

Single component systems

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dA

λ n dA ,

projected area theorem

cs

(3‐20)

and from this we see that the differential volume takes the form V

 v  t λ n dA

(3‐21)

Since the fluid velocity vector is given by,

v  v λ

(3‐22)

we see that the volume of fluid that crosses the surface area element per unit time is given by

V

volumetric flow rate across

v n dA ,

(3‐23)

t

an arbitrary area dA

This expression is identical to that given earlier by Eq. 3‐15; however, in this case we have used the projected area theorem to demonstrate that this is a generally valid result. From Eq. 3‐23 we determine that the rate at which mass crosses the differential surface area is

V

m

mass flow rate across

  v n dA ,

(3‐24)

t

t

 an arbitrary area dA

This representation for m

/ t is identical in form to that given for the special case illustrated in Figure 3‐7 where the velocity vector and the unit normal vectors were parallel. The result given by Eq. 3‐24 is entirely general and it indicates that

v n represents the mass flux (mass per unit time per unit area) at any exit.

We are now ready to return to Eq. 3‐8 and express the rate at which mass leaves the control volume according to

rate at which mass

flows out of the

 

 

v n dA

(3‐25)

 control volume 

A exit

It should be clear that over the exits we have the condition v n  0 since n is always taken to be the outwardly directed unit normal. At the entrances, the velocity vector and the normal vector are related by v n  0 , and this requires that the rate at which mass enters the control volume be expressed as

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