Civil Engineering Reference
In-Depth Information
If
υ
3
depends also on
x
2
,X
3
is equal to
η
∂
2
υ
3
∂
2
υ
3
∂x
2
X
3
=
∂x
1
+
(4.8)
This simple description will be used to evaluate the effect of viscosity in cylindrical
pores, the radial velocity components being neglected, and the pressure depending only
on the
x
3
direction of the pores.
Cylindrical tubes having a circular cross-section
A cylindrical tube having a circular cross-section is represented in Figure 4.2. The axis
of the cylinder is 0
x
3
. By using Equation (4.8), Newton's law reduces to
η
∂
2
υ
3
∂
2
υ
3
∂x
2
∂p
∂x
3
+
jωρ
o
υ
3
=−
∂x
1
+
(4.9)
ρ
o
being the density of air, and
p
the pressure.
The geometry of the problem is axisymmetrical around 0
x
3
, and Equation (4.9) can
be rewritten
r
∂υ
3
∂r
∂p
∂x
3
+
η
r
∂
∂r
jωρ
o
υ
3
=−
(4.10)
The velocity
υ
must vanish at the surface of the cylinder, where the air is in contact
with the motionless frame. The solution of Equation (4.10), where the velocity vanishes
at the surface
r
=
R
of the cylinder is
1
−
1
jωρ
o
∂p
∂x
3
J
o
(lr)
J
o
(lR)
υ
3
=−
(4.11)
In this equation,
l
is equal to
jωρ
o
/η)
1
/
2
l
=
(
−
(4.12)
and
J
o
is the Bessel function of zero order.
Both determinations of the square root in Equation (4.12) give identical results,
because the Bessel function
J
0
is even. The mean velocity
υ
3
over the cross-section
is equal to
o
υ
3
2
πr
d
r
πR
2
υ
3
=
(4.13)
X
3
0
0
v
3
v(r)
r
X
1
x
3
M(r,x
3
)
R
Figure 4.2
A cylindrical tube having a circular cross-section of radius
R
.
