Biomedical Engineering Reference
In-Depth Information
Seiler (1971) obtained solutions assuming that the arterial blood enters the
peripheral region at the isothermal core temperature and that the venous
blood is completely equilibrated with the tissue at the cutaneous layer.
1.5.4 Chato Model
Chato's countercurrent heat transfer model (1980) differs from Keller and
Seiler's (1971) model in its neglect of heat transfer between the blood and
the tissue. In this way, he was able to concentrate on the two temperatures
instead of three as in Keller and Seiler's model. Chato assumed that the flow
rate decreases linearly, which corresponds with the case of constant perfusion
bleed-off rate. His one-dimensional model can easily be generated from our
general expressions (1.39) and (1.42) along with (1.37) and (1.40), dropping
the transient and conduction terms as
d
dx ε a
a
a =
a
v )
a
ρ f c p f ω a
ρ f c p f
u
T
a f h f (
T
T
T
(1.50)
d
dx ε v
v
v =
v
a )+ ρ f c p f ω a
a
ρ f c p f
u
T
a f h f (
T
T
T
(1.51)
where the interfacial heat transfer coecients are assumed to be the same.
The continuity equations (1.37) and (1.40) readily provide
a = u 0
ω a x
ε a
u
(1.52)
and
v =
u 0 + ω a x
ε v
u
(1.53)
Note that u 0 is the apparent velocity at x = 0 and that the right-hand
side terms in the two equations (1.50) and (1.51) cancels out each other, as
they should for this “perfect” heat exchange system. Chato (1980) obtained
arterial and venous temperature profiles along the length of the vessels and
demonstrated that the effect of perfusion bleed-off is to increase the heat
transfer between the vessels as compared with the case of constant mass flow
rate (i.e., ω a = 0).
1.5.5 Roetzel and Xuan Model
Roetzel and Xuan (1998) used the theory of porous media to simulate a tran-
sient response of the limb to external stimulus, in which the effect of the
countercurrent heat exchange on the temperature response is expected to be
significant. Their energy equation for the tissue in our notation runs as
(1
+ a a h a (
s
s
ε ) ρ s c s
T
∂x j
ε ) k s
T
a
s )
(1
=
T
T
∂t
∂x j
v
s )+(1
+ a v h v (
T
T
ε ) S m
(1.54)
 
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