fluid_mechanics

170 Chapter 4 ■ Fluid Kinematics
If we combine Eqs. 4.10, 4.11, 4.12, and 4.13 we see that the relationship between the time
rate of change of B for the system and that for the control volume is given by
or
DB sys
Dt
0B cv
dt
B # out B # in
(4.14)
The time derivative
associated with a
system may be different
from that for
a control volume.
DB sys
Dt
0B cv
0t
r 2 A 2 V 2 b 2 r 1 A 1 V 1 b 1
(4.15)
This is a version of the Reynolds transport theorem valid under the restrictive assumptions
associated with the flow shown in Fig. 4.11—fixed control volume with one inlet and one outlet
having uniform properties 1density, velocity, and the parameter b2 across the inlet and outlet with
the velocity normal to sections 112 and 122. Note that the time rate of change of B for the system
1the left-hand side of Eq. 4.15 or the quantity in Eq. 4.82 is not necessarily the same as the rate
of change of B within the control volume 1the first term on the right-hand side of Eq. 4.15 or the
quantity in Eq. 4.92. This is true because the inflow rate 1b 1 r 1 V 1 A 1 2 and the outflow rate 1b 2 r 2 V 2 A 2 2
of the property B for the control volume need not be the same.
E XAMPLE 4.8
Use of the Reynolds Transport Theorem
GIVEN Consider again the flow from the fire extinguisher
shown in Fig. E4.7. Let the extensive property of interest be the
system mass 1B m, the system mass, or b 12.
FIND Write the appropriate form of the Reynolds transport
theorem for this flow.
SOLUTION
Again we take the control volume to be the fire extinguisher, and
the system to be the fluid within it at time t 0. For this case
there is no inlet, section 112, across which the fluid flows into the
control volume 1A 1 02. There is, however, an outlet, section 122.
Thus, the Reynolds transport theorem, Eq. 4.15, along with Eq.
4.9 with b 1 can be written as
0 a
Dm sys
r dVb
cv
r (1) (Ans)
Dt 0t
2 A 2 V 2
COMMENT If we proceed one step further and use the basic
law of conservation of mass, we may set the left-hand side of this
equation equal to zero 1the amount of mass in a system is constant2
and rewrite Eq. 1 in the form
0 a cv
r dVb
r
0t
2 A 2 V 2
(2)
The physical interpretation of this result is that the rate at which
the mass in the tank decreases in time is equal in magnitude but
opposite to the rate of flow of mass from the exit, r 2 A 2 V 2 . Note
the units for the two terms of Eq. 2 1kgs or slugss2. Note that
if there were both an inlet and an outlet to the control volume
shown in Fig. E4.7, Eq. 2 would become
0 a r dVb
cv
r (3)
0t
1 A 1 V 1 r 2 A 2 V 2
In addition, if the flow were steady, the left-hand side of Eq. 3
would be zero 1the amount of mass in the control would be constant
in time2 and Eq. 3 would become
r 1 A 1 V 1 r 2 A 2 V 2
This is one form of the conservation of mass principle discussed in
Sect. 3.6.2—the mass flowrates into and out of the control volume
are equal. Other more general forms are discussed in Chapter 5.
Right
Atrium
Left
Atrium
Right
Ventricle
Left
Ventricle
Equation 4.15 is a simplified version of the Reynolds transport theorem. We will now derive
it for much more general conditions. A general, fixed control volume with fluid flowing through
it is shown in Fig. 4.12. The flow field may be quite simple 1as in the above one-dimensional flow
considerations2, or it may involve a quite complex, unsteady, three-dimensional situation such as
the flow through a human heart as illustrated by the figure in the margin. In any case we again
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 1region II2 has exited from the control volume and additional fluid
1region I, not part of the original system2 has entered the control volume.