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⎛
⎞
πτ
R
3
1 6
τ
4
τ
Q
=
R
+
H
−
H
⎜
⎟
⎜
⎟
2
4 9
τ
7
τ
⎝
⎠
2
R
R
τ
16
1/ 2
2
ττ
=
Q
4
H
+
τ
(17)
RH
H
3
π
R
7
Equations (16) and (17) give the volumetric flow rate
2
1/ 2
⎡
⎤
⎛
23
2
⎞
2
4
πτ
R
16
R
τ
Rp
∂
∂
B
2
π
R
τ
⎛
⎞
(18)
Q
=
⎢
H
+
.
H
−
τ
R
−
−
m
+
H
⎥
⎜
⎟
⎜
⎟
H
1
m
2
7
m
2
4
∂
z
∂
z
m
π
R
⎢
⎝
⎠
⎥
⎝
⎠
⎣
⎦
2
2
2
Therefore, from Eqs. (18) and (14), the velocity of flow is
2
⎡
1/ 2
⎤
⎛
23
2
⎞
2
1
4
πτ
R
16
R
τ
Rp
⎛
∂
∂
B
⎞
2
π
R
τ
u
=
⎢
H
+
.
H
−
τ
R
−
−
m
+
H
⎥
(19)
⎜
⎜
⎟
⎟
H
1
π
R
2
m
2
7
m
2
4
∂
z
∂
z
mR
π
⎢
⎝
⎠
⎥
⎝
⎠
⎣
⎦
2
2
2
τ
Shear stress is considered as
,
∂
∂
u
r
τ
Thus,
= -k (
)
r = R
(20)
where
k
=
μ
∂
∂
u
r
τ
Therefore,
= −
μ
(
)
r = R
(21)
Differentiating Eq. (19) with respect to
r
and substituting the value in
Eq. (22), we obtain shear stress
⎡
1/ 2
⎤
⎛
4
πτ
23
R
16
R
τ
Rp
2
∂
∂
B
⎞
4
τ
72
τ
Rp
∂
∂
B
⎛
⎞
⎛
⎞
(22)
τμ
=−
⎢
H
+
.
H
−
τ
R
−
−
m
+
H
+
H
−
τ
−
−
m
⎥
⎜
⎟
⎜
⎟
⎜
⎟
H
1
H
1
m
2
7
m
2
4
∂
z
∂
z
m
π
R
3
7
m
2
2
∂
z
∂
z
⎝
⎠
⎝
⎠
⎢
⎝
⎠
⎥
⎣
⎦
2
2
2
2
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