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From force balance in a circular tube as shown in Figure 10.3,
dP
1 ∂
=
(rτ)
dx
r ∂r
dP
d
r
=
(rτ)
dx
dr
dP
d
r
dr =
(rτ) dr
dx
dr
Therefore,
1 dP
τ = r
∼ r
2 dx
From BCs,
r = 0, τ = 0,
r = R, τ = τ w
One obtains
r
τ = τ w
(10.18)
R
(
)
r
R − y
y
∂u
τ = τ w =
τ w = 1 −
τ w =
ρ(ν + ν t )
R
R
R
∂y
(
) (
)
∂u
ν t
y τ w
ν
1 +
= 1 −
∂y
ν
R ρ
y +
(
)
(
) du +
ν t ∂u
ν
ν t
1 −
= 1 +
√
√
= 1 +
R +
ν ∂y (τ w /ρ) τ w /ρ
ν dy +
where
∗
+
u
+
yu
R
+
Ru ∗
∗
τ w
u =
y =
,
=
, u =
u ∗ ,
v
v
ρ
ΔP
y
r
P 1
P 2
R
τ w
FIGURE 10.3
Force balance in a circular tube.
202
Analytical Heat Transfer