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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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