Biomedical Engineering Reference
In-Depth Information
ð
v
2
2
1
d
-
-
U
Q
5
W
muscle
1
u
gz
p
ν
ρ
1
1
area
v
1
2
1
v
2
2
1
Q
5
W
muscle
1
gz
1
1
p
1
ν
1
ð
2
ρ
v
1
A
1
Þ
1
gz
2
1
p
2
ν
2
ðρ
v
2
A
2
Þ
Substituting known values into this equation,
!
2
2
ð
100 mm
=
s
Þ
20 mmHg
1050 kg
75
μ
m
Q
m
3
15 W
gz
1
1
2
ð
1050 kg
=
Þð
100 mm
=
s
Þπ
5
1
1
2
=
m
3
2
!
2
2
ð
40 mm
=
s
Þ
12 mmHg
1050 kg
50
μ
m
m
3
1
1
gz
2
1
ð
1050 kg
=
Þð
40 mm
=
s
Þπ
2
=
m
3
2
15 W
1
ð
2
1
μ
W
2
ð
4
:
64
E
2
4g
=
s
Þ
g
z
1
Þ
1
ð
0
:
126
μ
W
1
ð
8
:
25
E
2
5g
=
s
Þ
gz
2
Þ
5
14
:
126 W
2
ð
3
:
814
E
2
4g
=
s
Þ
g
ð
z
2
2
z
1
Þ
5
We are making the assumption that there is no height difference between the arterial side and
venous side (
z
2
5
z
1
).
Therefore, the rate of heat transfer is 14.126
W. For every millimeter difference in height, the
rate of heat transfer would change by approximately 4 nW. The addition or subtraction of heat
would depend on if the arterial side or the venous side was higher.
μ
Example
Calculate the time rate of change of mass flow rate (
vA
) of air entering the lungs. Assume
that the lungs have a capacity of 6 L. The temperature of the lungs is 37
C. The air pressure
inside of the lungs is 0.98 atm. At the instant that air enters the lungs, the temperature of
the lungs raises by 0.0001
C/s. The height of the trachea is 20 cm. Assume that there is no work
added to the system. Assume that air behaves as an ideal gas. Assume that the velocity is slow
within the trachea.
ρ
Solution
ð
area
-
U
ð
2
W
shaft
2
W
other
5
@
@
v
2
2
1
-
dA
Q
u
gz
ρ
dV
1
1
t
V
ρ
ð
v
2
2
1
d
-
-
U
u
gz
p
ν
1
1
1
area
From the assumptions made, the given equation can simplify to
ð
V
ð
ð
area
ð
5
@
@
d
-
-
U
0
u
gz
Þρ
dV
u
gz
p
νÞρ
1
1
1
1
t
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