Biomedical Engineering Reference
In-Depth Information
To account for the bottom half of the flow,
s 2
198 g
m
=
t
v ball 5
2
25 g
132 g
=
s
t
1
At t
0.5 sec,
5
s 2
198 g
m
=
0
:
5s
v ball 5
2
218 cm
=
s
5
=
:
25 g
1
132 g
s
0
5s
This is consistent with a rapid opening of the valve, but the total length that the ball would
traverse would only be approximately 4 cm (at most). Over time, the velocity of the ball would
follow a logarithmic relationship ( Figure 3.17 ), if there was no mechanism to stop the ball from
moving (i.e., the cage).
FIGURE 3.17
300
Velocity of the ball
with respect to time.
250
200
150
100
50
0
0
0.25
0.5
0.75
1
1.25
1.5
1.75
2
Time (sec)
We made the assumption in the derivation of Equation 3.40 , that the flow was irrota-
tional, and therefore it only experienced pure translation. We will not show the derivation
of the formula here, but the most general formula for the conservation of momentum must
include all possible velocity components. This formula takes the form of
ð
V ð
- b 1
- s 2
- r 1
- 3
- xyz 1 - 3 ð - 3
-
Þ 1 - 3
-
2
Þρ
dV
ð
ð
ð
3
:
42
Þ
5 @
@
d -
- xyz
- xyz
- xyz
ρ
dV
ρ
1
U
t
area
V
-
- xyz is the Coriolis acceleration,
-
-
-
where
ω
is the angular velocity (note:
ω
ω
3 ðω
Þ
is
3
3
- is the tangential acceleration due to angular veloc-
ity). This formula would be used if the fluid elements rotate and translate about each other
or some reference coordinate axis ( XYZ ).
-
the centripetal acceleration, and
ω
3
Search WWH ::




Custom Search