Environmental Engineering Reference
In-Depth Information
4
R
2
0
(4.12)
v
pR v
Q
1
0
0
P
R
If
(
)
4
R
<< (no nucleus) and
v
0
D
pR
2
/8
l
h (no slip), then Eq. (4.12) coincides
with Poiseuille formula (the Eq. (3.5)). When
/
0
(no of a viscous liquid in-
R
≈
0
R
Q
≈
π
R
2
v
terlayer in the nanotube), the flow rate
Q
is equal to volumetric flow
0
of fluid for a uniform field of velocity (full slip).
Accordingly, flow rate of the viscoplastic fluid flowing with a velocity (4.7),
is equal to:
3
4
πτ
�
R
3
R
R
�
�
�
Q
=−
0
1
−
00
+
Q
1
−
00
-
-
€
‚
€
‚
(4.13)
P
ƒ
„
ƒ
„
3
h
R
R
-
-
Comparing the Eqs. (4.7), (4.8), (4.9), (4.10) and (4.12), (4.13), we can see
that the structure of the flow of the liquid through the nanotubes considering the
slippage, is similar to that of the flow of viscoplastic fluid in a pipe of the same
radius
R
.
Given that the size of the central core flow of viscoplastic fluid (radius
R
00
) is
defined by:
τ
=
D
2
l
R
0
00
p
(4.14)
for viscoplastic fluid flow we obtain Buckingham formula:
4
1
�
2
l
τ
�
4
�
2
l
τ
�
QQ
=
1
+
0
−
0
-
€
‚
€
‚
(4.15)
P
3
ƒ
Rp
D
„
3
ƒ
Rp
D
„
-
We'll establish conformity of the pipe that implements the flow of a visco-
plastic fluid with a fluid-filled nanotube, the same size and with the same pressure
drop. We say that an effective internal critical shear stress
0
e
t
of viscoplastic
fluid flow, which ensures the coincidence rate with the flow of fluid in the nano-
tube. Then from Eq. (4.15) we obtain equation of fourth order to determine
0
e
t
:
4
2
l
τ
2
l
τ
�
�
�
�
(
)
(4.16)
0
ef
0
ef
−
4
=
AA
,
= e− e=
3
1,
Q Q
/
€
‚
€
‚
P
ƒ
Rp
D
„
ƒ
Rp
D
„
The solution of Eq. (4.16) can be found, for example, the iteration method of
Newton:
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