Biomedical Engineering Reference
In-Depth Information
and so
δρ
δ
t
+
∼
T
(
ρ
∼
)
=
0.
(11.27)
For the balance of momentum this yields
,
v
T
+
δ
v
δ
t
∇·
σ
+
ρ
∇
q
=
ρ
·
v
(11.28)
and in column/matrix notation:
.
T
T
T
∼
+
δ
∼
T
∼
σ
+
ρ
q
∼
=
ρ
∼ ∼
(11.29)
δ
t
For the special case of a stationary flow of a material, the following balance
equation results for the mass balance:
∇·
(
ρ
v
)
=
0,
(11.30)
and also
T
(
ρ
∼
)
=
0.
∼
(11.31)
The momentum equation reduces to:
∇
v
T
∇·
σ
+
ρ
q
=
ρ
·
v
,
(11.32)
and also
T
T
T
∼
.
T
∼
σ
+
ρ
q
∼
=
ρ
∼ ∼
(11.33)
Exercises
11.1 Compressible air is flowing through the bronchi. A bronchus is modelled
as a straight cylindrical tube. We consider a stationary flow. For each cross
section of the tube the velocity of the air
V
and the density
ρ
is constant
over the cross section. In the direction of the flow, the velocity of the air and
the density vary, because of temperature differences. Two cross sections A
and B at a distance
L
are considered. See the figure below.
