Biomedical Engineering Reference
In-Depth Information
We can directly apply
Equation 3.26
because we have made the assumption that the density
does not change with time, that the volume is not deformable, and that the flow is steady:
ð
d
-
-
U
5
0
ð
3
:
26
Þ
area
Equation 3.26
will further simplify to the volume flow rate at each location because of the
same assumptions. Therefore,
v
1
A
1
1
v
2
A
2
1
v
3
A
3
1
vA
4
1
v
5
A
5
5
0
2
The area at each location is
Location Area
2
3cm
2
069 cm
2
1
π
7
:
5
2
1
:
5cm
2
767 cm
2
2
π
1
:
5
2
0
:
8cm
2
503 cm
2
3
π
0
:
5
2
1
:
1cm
2
950 cm
2
4
π
0
:
5
2
2
:
7cm
2
726 cm
2
5
π
5
:
5
Substituting the known values into the previous equation,
v
1
A
1
2
v
2
A
2
2
v
3
A
3
2
vA
5
v
4
5
A
4
069 cm
2
767 cm
2
503 cm
2
726 cm
2
v
4
5
ð
120cm
=
s
Þð
7
:
Þ
2
ð
85cm
=
s
Þð
1
:
Þ
2
ð
65cm
=
s
Þð
0
:
Þ
2
ð
105cm
=
s
Þð
5
:
Þ
0
:
950 cm
2
54
cm
s
67
:
5
This velocity would flow at a 75
angle off the positive
x
-axis. The inflow at location 1
was negative because the velocity vector acts in an opposite direction to the area normal
vector (at all other locations they act in the same direction). To simplify this procedure, if
it is known that the flow is inflow, you can assume that it has a negative sign associated
with this term, whereas outflow can be assumed to have a positive sign. Density was not
used in any of the formulations because it would cancel out in each term. This problem
also illustrates that it does not matter which direction (
x
or
y
) the velocity is acting,
because all of the mass needs to be conserved.
Example
Calculate the time rate of change in air density during expiration. Assume that the lung (see
Figure 3.11
) has a total volume of 6000 mL, the diameter of the trachea is 18 mm, the air flow
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