Um cilindro de 12 cm de raio gira no interior de outro, que está fixo, e cujo raio mede 12.6 cm. Os eixos dos cilindros são concêntricos e ambos têm 30 cm de comprimento. É necessário aplicar um torque de 9.0 kg.cm para manter a velocidade de rotação em 60 rpm. Determinar a viscosidade do fluido que preenche o espaço entre os cilindros.
Dados do problema
editar
r1
12 cm
r2
12.6 cm
l
30 cm
ω0
60 rpm
Ω
9.0 kg.cm
μ0
a calcular
Esse problema já foi resolvido anteriormente , considerando-se valores médios para todas as variáveis, tendo-se obtido o valor de 0.025 kg.s/m2 . Com a equação obtida para o fluxo entre dois cilindros para a tensão cisalhante τrΘ e a velocidade vΘ :
τ
r
θ
=
2
μ
0
ω
0
R
2
1
−
(
R
R
+
h
0
)
2
1
r
2
{\displaystyle \tau _{r\theta }\;=\;{\frac {2\mu _{0}\omega _{0}R^{2}}{1\;-\;\left({\frac {R}{R\;+\;h_{0}}}\right)^{2}}}\;{\frac {1}{r^{2}}}}
v
θ
=
ω
0
1
−
(
R
R
+
h
0
)
2
(
r
−
R
2
r
)
{\displaystyle v_{\theta }\;=\;{\frac {\omega _{0}}{1\;-\;\left({\frac {R}{R\;+\;h_{0}}}\right)^{2}}}\left(r\;-\;{\frac {R^{2}}{r}}\right)}
P
=
∫
d
P
=
∫
v
θ
d
F
r
θ
=
∫
v
θ
τ
r
θ
d
A
r
θ
{\displaystyle P\;=\;\int dP\;=\;\int v_{\theta }\;dF_{r\theta }\;=\;\int v_{\theta }\;\tau _{r\theta }\;dA_{r\theta }}
P
=
∫
[
ω
0
1
−
(
R
R
+
h
0
)
2
(
r
−
R
2
r
)
]
[
2
μ
0
ω
0
R
2
1
−
(
R
R
+
h
0
)
2
1
r
2
]
[
l
d
r
]
{\displaystyle P\;=\;\int \left[{\frac {\omega _{0}}{1\;-\;\left({\frac {R}{R\;+\;h_{0}}}\right)^{2}}}\left(r\;-\;{\frac {R^{2}}{r}}\right)\right]\;\left[{\frac {2\mu _{0}\omega _{0}R^{2}}{1\;-\;\left({\frac {R}{R\;+\;h_{0}}}\right)^{2}}}\;{\frac {1}{r^{2}}}\right]\;\left[ldr\right]}
P
=
2
l
μ
0
ω
0
2
r
1
2
(
1
−
(
r
1
r
2
)
2
)
2
∫
r
1
r
2
(
1
r
−
r
1
2
r
3
)
d
r
{\displaystyle P\;=\;{\frac {2l\mu _{0}\omega _{0}^{2}r_{1}^{2}}{\left(1\;-\;\left({\frac {r_{1}}{r_{2}}}\right)^{2}\right)^{2}}}\;\int _{r1}^{r2}\left({\frac {1}{r}}\;-\;{\frac {r_{1}^{2}}{r^{3}}}\right)\;dr}
P
=
2
l
μ
0
ω
0
2
r
1
2
(
1
−
(
r
1
r
2
)
2
)
2
[
ln
r
+
r
1
2
2
r
2
]
|
r
1
r
2
{\displaystyle P\;=\;{\frac {2l\mu _{0}\omega _{0}^{2}r_{1}^{2}}{\left(1\;-\;\left({\frac {r_{1}}{r_{2}}}\right)^{2}\right)^{2}}}\left.\left[\ln r\;+\;{\frac {r_{1}^{2}}{2r^{2}}}\right]\right|_{r1}^{r2}}
P
=
2
l
μ
0
ω
0
2
r
1
2
(
1
−
(
r
1
r
2
)
2
)
2
[
ln
(
r
2
r
1
)
+
1
2
(
(
r
1
r
2
)
2
−
1
)
]
{\displaystyle P\;=\;{\frac {2l\mu _{0}\omega _{0}^{2}r_{1}^{2}}{\left(1\;-\;\left({\frac {r_{1}}{r_{2}}}\right)^{2}\right)^{2}}}\left[\ln \left({\frac {r_{2}}{r_{1}}}\right)\;+\;{\frac {1}{2}}\;\left(\left({\frac {r_{1}}{r_{2}}}\right)^{2}\;-\;1\right)\right]}
0 ,
μ
0
=
Ω
(
1
−
(
r
1
r
2
)
2
)
2
2
ω
0
l
r
1
2
[
ln
(
r
2
r
1
)
+
1
2
(
(
r
1
r
2
)
2
−
1
)
]
{\displaystyle \mu _{0}\;=\;{\frac {\Omega \left(1\;-\;\left({\frac {r_{1}}{r_{2}}}\right)^{2}\right)^{2}}{2\omega _{0}lr_{1}^{2}\left[\ln \left({\frac {r_{2}}{r_{1}}}\right)\;+\;{\frac {1}{2}}\;\left(\left({\frac {r_{1}}{r_{2}}}\right)^{2}\;-\;1\right)\right]}}}
=
9.0
k
g
⋅
c
m
⋅
(
1
−
(
12.0
c
m
12.6
c
m
)
2
)
2
2
⋅
60
r
p
m
⋅
30
c
m
⋅
(
12.0
c
m
)
2
⋅
[
ln
(
12.6
c
m
12.0
c
m
)
+
1
2
(
(
12.0
c
m
12.6
c
m
)
2
−
1
)
]
{\displaystyle \;=\;{\frac {9.0\;kg\cdot cm\cdot \left(1\;-\;\left({\frac {12.0\;cm}{12.6\;cm}}\right)^{2}\right)^{2}}{2\cdot 60\;rpm\cdot 30\;cm\cdot (12.0\;cm)^{2}\cdot \left[\ln \left({\frac {12.6\;cm}{12.0\;cm}}\right)\;+\;{\frac {1}{2}}\left(\left({\frac {12.0\;cm}{12.6\;cm}}\right)^{2}\;-\;1\right)\right]}}}
=
0.090
k
g
⋅
m
⋅
(
1
−
(
12.0
c
m
12.6
c
m
)
2
)
2
2
⋅
60
⋅
2
⋅
3.1
60
r
d
/
s
⋅
0.30
m
⋅
(
0.12
m
)
2
⋅
[
ln
(
12.6
c
m
12.0
c
m
)
+
1
2
(
(
12.0
c
m
12.6
c
m
)
2
−
1
)
]
{\displaystyle \;=\;{\frac {0.090\;kg\cdot m\cdot \left(1\;-\;\left({\frac {12.0\;cm}{12.6\;cm}}\right)^{2}\right)^{2}}{2\cdot 60\cdot {\frac {2\cdot 3.1}{60}}\;rd/s\cdot 0.30\;m\cdot (0.12\;m)^{2}\cdot \left[\ln \left({\frac {12.6\;cm}{12.0\;cm}}\right)\;+\;{\frac {1}{2}}\left(\left({\frac {12.0\;cm}{12.6\;cm}}\right)^{2}\;-\;1\right)\right]}}}
=
6.3
k
g
⋅
s
/
m
2
{\displaystyle \;=\;6.3\;kg\cdot s/m^{2}}
![{\displaystyle P\;=\;\int \left[{\frac {\omega _{0}}{1\;-\;\left({\frac {R}{R\;+\;h_{0}}}\right)^{2}}}\left(r\;-\;{\frac {R^{2}}{r}}\right)\right]\;\left[{\frac {2\mu _{0}\omega _{0}R^{2}}{1\;-\;\left({\frac {R}{R\;+\;h_{0}}}\right)^{2}}}\;{\frac {1}{r^{2}}}\right]\;\left[ldr\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d801047726563fc20adf9481a62a6e2d7b72ec78)
![{\displaystyle P\;=\;{\frac {2l\mu _{0}\omega _{0}^{2}r_{1}^{2}}{\left(1\;-\;\left({\frac {r_{1}}{r_{2}}}\right)^{2}\right)^{2}}}\left.\left[\ln r\;+\;{\frac {r_{1}^{2}}{2r^{2}}}\right]\right|_{r1}^{r2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b2d57eef8cd0b1635a88fb887e5b118d6d163de2)
![{\displaystyle P\;=\;{\frac {2l\mu _{0}\omega _{0}^{2}r_{1}^{2}}{\left(1\;-\;\left({\frac {r_{1}}{r_{2}}}\right)^{2}\right)^{2}}}\left[\ln \left({\frac {r_{2}}{r_{1}}}\right)\;+\;{\frac {1}{2}}\;\left(\left({\frac {r_{1}}{r_{2}}}\right)^{2}\;-\;1\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d1af2fd804e355d15ed318bacab0a1cdd17b18ba)
![{\displaystyle \mu _{0}\;=\;{\frac {\Omega \left(1\;-\;\left({\frac {r_{1}}{r_{2}}}\right)^{2}\right)^{2}}{2\omega _{0}lr_{1}^{2}\left[\ln \left({\frac {r_{2}}{r_{1}}}\right)\;+\;{\frac {1}{2}}\;\left(\left({\frac {r_{1}}{r_{2}}}\right)^{2}\;-\;1\right)\right]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/84b4c693b059072e08142da92c87b49314b1c6e2)
![{\displaystyle \;=\;{\frac {9.0\;kg\cdot cm\cdot \left(1\;-\;\left({\frac {12.0\;cm}{12.6\;cm}}\right)^{2}\right)^{2}}{2\cdot 60\;rpm\cdot 30\;cm\cdot (12.0\;cm)^{2}\cdot \left[\ln \left({\frac {12.6\;cm}{12.0\;cm}}\right)\;+\;{\frac {1}{2}}\left(\left({\frac {12.0\;cm}{12.6\;cm}}\right)^{2}\;-\;1\right)\right]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/24ac464b5197fc34e79affe60719381d82d7d2f7)
![{\displaystyle \;=\;{\frac {0.090\;kg\cdot m\cdot \left(1\;-\;\left({\frac {12.0\;cm}{12.6\;cm}}\right)^{2}\right)^{2}}{2\cdot 60\cdot {\frac {2\cdot 3.1}{60}}\;rd/s\cdot 0.30\;m\cdot (0.12\;m)^{2}\cdot \left[\ln \left({\frac {12.6\;cm}{12.0\;cm}}\right)\;+\;{\frac {1}{2}}\left(\left({\frac {12.0\;cm}{12.6\;cm}}\right)^{2}\;-\;1\right)\right]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a01c6dacddb376001e0f1f82c839d4184d94edeb)
