Deduzir o perfil de velocidades para escoamento laminar de um líquido Newtoniano em um tubo cilíndrico horizontal de diâmetro D.
Aqui, não podemos modelar as paredes do tubo como duas placas paralelas, e será preciso empregar as equações em coordenadas cilíndricas.
1
r
∂
∂
r
(
r
v
r
)
+
1
r
∂
∂
θ
(
v
θ
)
+
∂
∂
z
(
v
z
)
=
0
{\displaystyle {\frac {1}{r}}\;{\frac {\partial }{\partial r}}\;(rv_{r})\;+\;{\frac {1}{r}}\;{\frac {\partial }{\partial \theta }}\;(v_{\theta })\;+\;{\frac {\partial }{\partial z}}\;(v_{z})\;=\;0}
ρ
0
(
∂
v
r
∂
t
+
v
r
∂
v
r
∂
r
+
v
θ
r
∂
v
r
∂
θ
−
v
θ
2
r
+
v
r
∂
v
r
∂
z
)
=
{\displaystyle \rho _{0}\left({\frac {\partial v_{r}}{\partial t}}\;+\;v_{r}\;{\frac {\partial v_{r}}{\partial r}}\;+\;{\frac {v_{\theta }}{r}}\;{\frac {\partial v_{r}}{\partial \theta }}\;-\;{\frac {v_{\theta }^{2}}{r}}\;+\;v_{r}\;{\frac {\partial v_{r}}{\partial z}}\right)\;=\;}
=
ρ
0
g
r
−
∂
p
∂
r
+
μ
0
[
∂
∂
r
(
1
r
∂
∂
r
(
r
v
r
)
)
+
1
r
2
∂
2
v
r
∂
θ
2
−
2
r
2
∂
v
θ
∂
θ
+
∂
2
v
r
∂
z
2
]
{\displaystyle \;=\;\rho _{0}g_{r}\;-\;{\frac {\partial p}{\partial r}}\;+\;\mu _{0}\;\left[{\frac {\partial }{\partial r}}\left({\frac {1}{r}}\;{\frac {\partial }{\partial r}}(rv_{r})\right)\;+\;{\frac {1}{r^{2}}}\;{\frac {\partial ^{2}v_{r}}{\partial \theta ^{2}}}\;-\;{\frac {2}{r^{2}}}\;{\frac {\partial v_{\theta }}{\partial \theta }}\;+\;{\frac {\partial ^{2}v_{r}}{\partial z^{2}}}\right]}
ρ
0
(
∂
v
θ
∂
t
+
v
r
∂
v
θ
∂
r
+
v
θ
r
∂
v
θ
∂
θ
+
v
r
v
θ
r
+
v
r
∂
v
θ
∂
z
)
=
{\displaystyle \rho _{0}\left({\frac {\partial v_{\theta }}{\partial t}}\;+\;v_{r}\;{\frac {\partial v_{\theta }}{\partial r}}\;+\;{\frac {v_{\theta }}{r}}\;{\frac {\partial v_{\theta }}{\partial \theta }}\;+\;{\frac {v_{r}v_{\theta }}{r}}\;+\;v_{r}\;{\frac {\partial v_{\theta }}{\partial z}}\right)\;=\;}
=
ρ
0
g
θ
−
1
r
∂
p
∂
θ
+
μ
0
[
∂
∂
r
(
1
r
∂
∂
r
(
r
v
θ
)
)
+
1
r
2
∂
2
v
θ
∂
θ
2
+
2
r
2
∂
v
θ
∂
θ
+
∂
2
v
θ
∂
z
2
]
{\displaystyle \;=\;\rho _{0}g_{\theta }\;-\;{\frac {1}{r}}\;{\frac {\partial p}{\partial \theta }}\;+\;\mu _{0}\;\left[{\frac {\partial }{\partial r}}\left({\frac {1}{r}}\;{\frac {\partial }{\partial r}}(rv_{\theta })\right)\;+\;{\frac {1}{r^{2}}}\;{\frac {\partial ^{2}v_{\theta }}{\partial \theta ^{2}}}\;+\;{\frac {2}{r^{2}}}\;{\frac {\partial v_{\theta }}{\partial \theta }}\;+\;{\frac {\partial ^{2}v_{\theta }}{\partial z^{2}}}\right]}
ρ
0
(
∂
v
z
∂
t
+
v
r
∂
v
z
∂
r
+
v
θ
r
∂
v
z
∂
θ
+
v
z
∂
v
z
∂
z
)
=
{\displaystyle \rho _{0}\left({\frac {\partial v_{z}}{\partial t}}\;+\;v_{r}\;{\frac {\partial v_{z}}{\partial r}}\;+\;{\frac {v_{\theta }}{r}}\;{\frac {\partial v_{z}}{\partial \theta }}\;+\;v_{z}\;{\frac {\partial v_{z}}{\partial z}}\right)\;=\;}
=
ρ
0
g
z
−
∂
p
∂
z
+
μ
0
[
1
r
∂
∂
r
(
r
∂
v
z
∂
r
)
+
1
r
2
∂
2
v
z
∂
θ
2
+
∂
2
v
z
∂
z
2
]
{\displaystyle \;=\;\rho _{0}g_{z}\;-\;{\frac {\partial p}{\partial z}}\;+\;\mu _{0}\;\left[{\frac {1}{r}}\;{\frac {\partial }{\partial r}}\left(r\;{\frac {\partial v_{z}}{\partial r}}\right)\;+\;{\frac {1}{r^{2}}}\;{\frac {\partial ^{2}v_{z}}{\partial \theta ^{2}}}\;+\;{\frac {\partial ^{2}v_{z}}{\partial z^{2}}}\right]}
∂
η
∂
θ
=
0
{\displaystyle {\frac {\partial \eta }{\partial \theta }}\;=\;0}
v
r
=
v
θ
=
0
{\displaystyle v_{r}\;=\;v_{\theta }\;=\;0}
∂
η
∂
t
=
0
{\displaystyle {\frac {\partial \eta }{\partial t}}\;=\;0}
0
+
0
+
∂
∂
z
(
v
z
)
=
0
⇒
∂
v
z
∂
z
=
0
{\displaystyle 0\;+\;0\;+\;{\frac {\partial }{\partial z}}\;(v_{z})\;=\;0\;\;\;\Rightarrow {\frac {\partial v_{z}}{\partial z}}\;=\;0}
ρ
0
(
0
+
0
−
0
+
0
)
=
ρ
0
g
r
−
∂
p
∂
r
+
μ
0
[
0
+
0
−
0
+
0
]
⇒
∂
p
∂
r
=
ρ
0
g
r
{\displaystyle \rho _{0}\left(0\;+\;0\;-\;0\;+\;0\right)\;=\;\rho _{0}g_{r}\;-\;{\frac {\partial p}{\partial r}}\;+\;\mu _{0}\;\left[0\;+\;0\;-\;0\;+\;0\right]\;\;\;\Rightarrow {\frac {\partial p}{\partial r}}\;=\;\rho _{0}g_{r}}
ρ
0
(
0
+
0
+
0
+
0
+
0
)
=
ρ
0
g
θ
−
1
r
∂
p
∂
θ
+
μ
0
[
0
+
0
+
0
+
0
]
⇒
1
r
∂
p
∂
θ
=
ρ
0
g
θ
{\displaystyle \rho _{0}\left(0\;+\;0\;+\;0\;+\;0\;+\;0\right)\;=\;\rho _{0}g_{\theta }\;-\;{\frac {1}{r}}\;{\frac {\partial p}{\partial \theta }}\;+\;\mu _{0}\;\left[0\;+\;0\;+\;0\;+\;0\right]\;\;\;\Rightarrow {\frac {1}{r}}\;{\frac {\partial p}{\partial \theta }}\;=\;\rho _{0}g_{\theta }}
ρ
0
(
0
+
0
+
0
+
v
z
∂
v
z
∂
z
)
=
0
−
∂
p
∂
z
+
μ
0
[
1
r
∂
∂
r
(
r
∂
v
z
∂
r
)
+
0
+
∂
2
v
z
∂
z
2
]
{\displaystyle \rho _{0}\left(0\;+\;0\;+\;0\;+\;v_{z}\;{\frac {\partial v_{z}}{\partial z}}\right)\;=\;0\;-\;{\frac {\partial p}{\partial z}}\;+\;\mu _{0}\;\left[{\frac {1}{r}}\;{\frac {\partial }{\partial r}}\left(r\;{\frac {\partial v_{z}}{\partial r}}\right)\;+\;0\;+\;{\frac {\partial ^{2}v_{z}}{\partial z^{2}}}\right]}
∂
v
z
∂
z
=
0
⇒
μ
0
r
∂
∂
r
(
r
∂
v
z
∂
r
)
=
∂
p
∂
z
{\displaystyle {\frac {\partial v_{z}}{\partial z}}\;=\;0\;\;\;\Rightarrow {\frac {\mu _{0}}{r}}\;{\frac {\partial }{\partial r}}\left(r\;{\frac {\partial v_{z}}{\partial r}}\right)\;=\;{\frac {\partial p}{\partial z}}}
∂
∂
r
(
r
∂
v
z
∂
r
)
=
1
μ
0
Δ
p
L
r
⇒
r
∂
v
z
∂
r
=
k
1
r
2
+
k
2
(
k
1
=
1
2
μ
0
Δ
p
L
)
{\displaystyle {\frac {\partial }{\partial r}}\left(r\;{\frac {\partial v_{z}}{\partial r}}\right)\;=\;{\frac {1}{\mu _{0}}}\;{\frac {\Delta p}{L}}\;r\;\;\;\Rightarrow r\;{\frac {\partial v_{z}}{\partial r}}\;=\;k_{1}r^{2}\;+\;k_{2}\qquad \left(k_{1}\;=\;{\frac {1}{2\mu _{0}}}\;{\frac {\Delta p}{L}}\right)}
∂
v
z
∂
r
=
k
1
r
+
k
2
1
r
{\displaystyle {\frac {\partial v_{z}}{\partial r}}\;=\;k_{1}r\;+\;k_{2}\;{\frac {1}{r}}}
2 deve ser nulo, caso contrário a velocidade seria infinita em r = 0 (centro do tubo). Assim,
∂
v
z
∂
r
=
k
1
r
⇒
v
z
=
k
1
2
r
2
+
k
3
{\displaystyle {\frac {\partial v_{z}}{\partial r}}\;=\;k_{1}r\;\;\;\Rightarrow v_{z}\;=\;{\frac {k_{1}}{2}}r^{2}\;+\;k_{3}}
z (R) = 0, temos
v
z
(
R
)
=
0
⇒
k
1
2
R
2
+
k
3
=
0
⇒
k
3
=
−
k
1
2
R
2
{\displaystyle v_{z}(R)\;=\;0\;\;\;\Rightarrow {\frac {k_{1}}{2}}R^{2}\;+\;k_{3}\;=\;0\;\;\;\Rightarrow k_{3}\;=\;-\;{\frac {k_{1}}{2}}R^{2}}
v
z
=
k
1
2
(
r
2
−
R
2
)
=
1
4
μ
0
Δ
p
L
(
r
2
−
R
2
)
{\displaystyle v_{z}\;=\;{\frac {k_{1}}{2}}\left(r^{2}\;-\;R^{2}\right)\;=\;{\frac {1}{4\mu _{0}}}\;{\frac {\Delta p}{L}}\;(r^{2}\;-\;R^{2})}
N
R
e
=
ρ
v
¯
D
μ
0
{\displaystyle N_{Re}\;=\;{\frac {\rho \;{\bar {v}}\;D}{\mu _{0}}}}
Φ
=
∫
A
v
z
d
A
=
∫
0
R
1
4
μ
0
Δ
p
L
(
r
2
−
R
2
)
(
2
π
r
d
r
)
=
π
2
μ
0
Δ
p
L
∫
0
R
(
r
3
−
R
2
r
)
d
r
{\displaystyle \Phi \;=\;\int _{A}v_{z}dA\;=\;\int _{0}^{R}{\frac {1}{4\mu _{0}}}\;{\frac {\Delta p}{L}}\;(r^{2}\;-\;R^{2})(2\pi r\;dr)\;=\;{\frac {\pi }{2\mu _{0}}}\;{\frac {\Delta p}{L}}\int _{0}^{R}(r^{3}\;-\;R^{2}r)\;dr}
Φ
=
π
2
μ
0
Δ
p
L
(
r
4
4
−
R
2
r
2
2
)
)
|
0
R
{\displaystyle \Phi \;=\;{\frac {\pi }{2\mu _{0}}}\;{\frac {\Delta p}{L}}\left.\left({\frac {r^{4}}{4}}\;-\;{\frac {R^{2}r^{2}}{2}})\right)\right|_{0}^{R}}
Φ
=
−
π
R
4
8
μ
0
Δ
p
L
{\displaystyle \Phi \;=\;-\;{\frac {\pi R^{4}}{8\mu _{0}}}\;{\frac {\Delta p}{L}}}
v
¯
=
P
h
i
A
=
−
π
R
4
8
μ
0
Δ
p
L
1
π
R
2
=
−
R
2
8
μ
0
Δ
p
L
{\displaystyle {\bar {v}}\;=\;{\frac {Phi}{A}}\;=\;-\;{\frac {\pi R^{4}}{8\mu _{0}}}\;{\frac {\Delta p}{L}}\;{\frac {1}{\pi R^{2}}}\;=\;-\;{\frac {R^{2}}{8\mu _{0}}}\;{\frac {\Delta p}{L}}}
![{\displaystyle \;=\;\rho _{0}g_{r}\;-\;{\frac {\partial p}{\partial r}}\;+\;\mu _{0}\;\left[{\frac {\partial }{\partial r}}\left({\frac {1}{r}}\;{\frac {\partial }{\partial r}}(rv_{r})\right)\;+\;{\frac {1}{r^{2}}}\;{\frac {\partial ^{2}v_{r}}{\partial \theta ^{2}}}\;-\;{\frac {2}{r^{2}}}\;{\frac {\partial v_{\theta }}{\partial \theta }}\;+\;{\frac {\partial ^{2}v_{r}}{\partial z^{2}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf0babac13b1af4287b826a2087130ff581e40b2)
![{\displaystyle \;=\;\rho _{0}g_{\theta }\;-\;{\frac {1}{r}}\;{\frac {\partial p}{\partial \theta }}\;+\;\mu _{0}\;\left[{\frac {\partial }{\partial r}}\left({\frac {1}{r}}\;{\frac {\partial }{\partial r}}(rv_{\theta })\right)\;+\;{\frac {1}{r^{2}}}\;{\frac {\partial ^{2}v_{\theta }}{\partial \theta ^{2}}}\;+\;{\frac {2}{r^{2}}}\;{\frac {\partial v_{\theta }}{\partial \theta }}\;+\;{\frac {\partial ^{2}v_{\theta }}{\partial z^{2}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3d17419830bab90839ece906c94ec3650441c43f)
![{\displaystyle \;=\;\rho _{0}g_{z}\;-\;{\frac {\partial p}{\partial z}}\;+\;\mu _{0}\;\left[{\frac {1}{r}}\;{\frac {\partial }{\partial r}}\left(r\;{\frac {\partial v_{z}}{\partial r}}\right)\;+\;{\frac {1}{r^{2}}}\;{\frac {\partial ^{2}v_{z}}{\partial \theta ^{2}}}\;+\;{\frac {\partial ^{2}v_{z}}{\partial z^{2}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/03e54d7ff3c519d30a831944d9392be0fa2e3fe8)
![{\displaystyle \rho _{0}\left(0\;+\;0\;-\;0\;+\;0\right)\;=\;\rho _{0}g_{r}\;-\;{\frac {\partial p}{\partial r}}\;+\;\mu _{0}\;\left[0\;+\;0\;-\;0\;+\;0\right]\;\;\;\Rightarrow {\frac {\partial p}{\partial r}}\;=\;\rho _{0}g_{r}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/650c87622deedbf9d229bcb79622fe1f25453e2b)
![{\displaystyle \rho _{0}\left(0\;+\;0\;+\;0\;+\;0\;+\;0\right)\;=\;\rho _{0}g_{\theta }\;-\;{\frac {1}{r}}\;{\frac {\partial p}{\partial \theta }}\;+\;\mu _{0}\;\left[0\;+\;0\;+\;0\;+\;0\right]\;\;\;\Rightarrow {\frac {1}{r}}\;{\frac {\partial p}{\partial \theta }}\;=\;\rho _{0}g_{\theta }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4674ecdde0f86624341fb981ada8b25ab86e3037)
![{\displaystyle \rho _{0}\left(0\;+\;0\;+\;0\;+\;v_{z}\;{\frac {\partial v_{z}}{\partial z}}\right)\;=\;0\;-\;{\frac {\partial p}{\partial z}}\;+\;\mu _{0}\;\left[{\frac {1}{r}}\;{\frac {\partial }{\partial r}}\left(r\;{\frac {\partial v_{z}}{\partial r}}\right)\;+\;0\;+\;{\frac {\partial ^{2}v_{z}}{\partial z^{2}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bf873ca68b6a5d9ea4700a5cc421901cc6a11e25)
