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3
π
R
(
)
2
Q
=
τ
1/ 2
τ
1/ 2
(12)
H
2
3
2
Again, from Eqs. (9) and (10),
1
p
B
()
r
τ
=−
m
1
rr
z
z
u
1
r
p
B
=
2
ττ
τ
m
HH
1
2
rm
2
z
z
2
1
r
2
p
B
u
=
2
ττ
.
r
τ
r
m
(13)
H
H
1
m
2
4
z
z
2
Again, from volumetric flow rate and Eq. (13),
Rz
()
2
Q
=
2
π
rudr
=
π
R u
(14)
0
2
1
r
p
B
2
QR
=
π
2
ττ
.
r
τ
r
m
(15)
H
H
1
2
m
4
z
z
2
ττ
=
When r = R , then
R
2
1
Rp
B
QR
=
π
2
2
ττ
.
R R
τ
m
(16)
H
H
1
2
m
4
z
z
2
Therefore, from Eqs. (11) and (14),
R
Rdp
dz
rdp
dz
dr
=
d
τ
τ
=−
where
τ =−
and
gives
R
τ
2
2
R
τ
2
τ
R
R
2
Qf
=
πτ
()
R
d
τ
ττ
2
RR
0
 
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