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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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