fluid_mechanics

5.1 Conservation of Mass—The Continuity Equation 199
The velocity of the
surface of a deforming
control
volume is not the
same at all points
on the surface.
must be determined with the relative velocity, W, the velocity referenced to the control surface.
Since the control volume is deforming, the control surface velocity is not necessarily uniform and
identical to the control volume velocity, V cv , as was true for moving, nondeforming control volumes.
For the deforming control volume,
V W V cs
(5.18)
where V cs is the velocity of the control surface as seen by a fixed observer. The relative velocity, W,
must be ascertained with care wherever fluid crosses the control surface. Two example problems that
illustrate the use of the continuity equation for a deforming control volume, Eq. 5.17, follow.
E XAMPLE 5.8
Conservation of Mass—Deforming Control Volume
GIVEN A syringe 1Fig. E5.82 is used to inoculate a cow. The
plunger has a face area of 500 mm 2 . The liquid in the syringe is
to be injected steadily at a rate of 300 cm 3 min. The leakage rate
past the plunger is 0.10 times the volume flowrate out of the
needle.
FIND
With what speed should the plunger be advanced?
Plunger
motion
Q leak =
0.1 Q 2 Q 2 =
A 300 cm 3 /min
p =
V
500 mm 2
p
Section (1)
Control volume
F I G U R E E5.8
Section (2)
SOLUTION
The control volume selected for solving this problem is the deforming
one illustrated in Fig. E5.8. Section 112 of the control surface
moves with the plunger. The surface area of section 112, A 1 , is
considered equal to the circular area of the face of the plunger, A p ,
although this is not strictly true, since leakage occurs. The difference
is small, however. Thus,
A 1 A p
(1)
Liquid also leaves the needle through section 122, which involves
fixed area A 2 . The application of Eq. 5.17 to the contents of this
control volume gives
0
0t
cv
r dVm # 2 rQ leak 0
Q leak
Even though and the flow through section area are
steady, the time rate of change of the mass of liquid in the
shrinking control volume is not zero because the control volume
is getting smaller. To evaluate the first term of Eq. 2, we note
that
A 2
(2)
Note that
0/
(5)
0t V p
where V p is the speed of the plunger sought in the problem statement.
Combining Eqs. 2, 4, and 5 we obtain
However, from Eq. 5.6, we see that
and Eq. 6 becomes
Solving Eq. 8 for
rA 1 V p m # 2 rQ leak 0
rA 1 V p rQ 2 rQ leak 0
V p
yields
m # 2 rQ 2
V p Q 2 Q leak
A 1
Since Q leak 0.1Q 2 , Eq. 9 becomes
(6)
(7)
(8)
(9)
cv
r dVr1/A 1 V needle 2
where / is the changing length of the control volume 1see Fig.
E5.82 and V needle is the volume of the needle. From Eq. 3, we
obtain
0
(4)
0t r dVrA 0/
1
cv 0t
(3)
and
V p Q 2 0.1Q 2
A 1
1.1Q 2
A 1
V p 11.121300 cm3 min2 1000 mm3
a b
1500 mm 2 2 cm 3
660 mmmin
(Ans)
E XAMPLE 5.9
Conservation of Mass—Deforming Control Volume
GIVEN Consider Example 5.5. FIND Solve the problem of Example 5.5 using a deforming control
volume that includes only the water accumulating in the bathtub.