Um espaço amplo encontra-se cheio de líquido ideal. Repentinamente, estoura uma bolha esférica de raio a. Calcular o tempo para que o líquido preencha o buraco formado. Desprezar a ação da gravidade.
Devido à simetria do problema, uma vez que a gravidade será desprezada, podemos considerar que só existirá movimento na direção radial. Assim
v
→
=
v
r
u
→
r
{\displaystyle {\vec {v}}\;=\;v_{r}\;{\vec {u}}_{r}}
ρ
g
→
−
∇
p
=
ρ
D
v
→
D
t
⇒
0
−
∇
p
=
ρ
(
∂
v
→
∂
t
+
v
→
⋅
∇
v
→
)
{\displaystyle \rho {\vec {g}}\;-\;\nabla p\;=\;\rho {\frac {D{\vec {v}}}{Dt}}\;\;\;\Rightarrow 0\;-\;\nabla p\;=\;\rho \left({\frac {\partial {\vec {v}}}{\partial t}}\;+\;{\vec {v}}\cdot \nabla {\vec {v}}\right)}
−
∂
p
∂
r
=
ρ
(
∂
v
r
∂
t
+
v
r
∂
v
r
∂
r
)
{\displaystyle -\;{\frac {\partial p}{\partial r}}\;=\;\rho \left({\frac {\partial v_{r}}{\partial t}}\;+\;v_{r}\;{\frac {\partial v_{r}}{\partial r}}\right)}
r são funções de r e t. Não podemos usar a equação de continuidade na forma canônica, pois o escoamento não está em regime permanente. Mas podemos afirmar que a conservação de massa exige que a vazão Φ de fluido fora do buraco seja independente de r, quer dizer, seja função apenas de t. Para integrar a equação de Euler e eliminar r, podemos escrever
Φ
=
4
π
r
2
v
r
⇒
v
r
=
Φ
4
π
r
2
{\displaystyle \Phi \;=\;4\pi r^{2}\;v_{r}\;\;\;\Rightarrow v_{r}\;=\;{\frac {\Phi }{4\pi r^{2}}}}
∂
v
r
∂
t
=
1
4
π
r
2
∂
Φ
∂
t
{\displaystyle {\frac {\partial v_{r}}{\partial t}}\;=\;{\frac {1}{4\pi r^{2}}}\;{\frac {\partial \Phi }{\partial t}}}
−
∂
p
∂
r
=
ρ
(
1
4
π
r
2
∂
Φ
∂
t
+
v
r
∂
v
r
∂
r
)
{\displaystyle -\;{\frac {\partial p}{\partial r}}\;=\;\rho \left({\frac {1}{4\pi r^{2}}}\;{\frac {\partial \Phi }{\partial t}}\;+\;v_{r}\;{\frac {\partial v_{r}}{\partial r}}\right)}
−
∂
p
ρ
=
1
4
π
r
2
∂
Φ
∂
t
∂
r
+
v
r
∂
v
r
{\displaystyle -\;{\frac {\partial p}{\rho }}\;=\;{\frac {1}{4\pi r^{2}}}\;{\frac {\partial \Phi }{\partial t}}\;\partial r\;+\;v_{r}\;\partial v_{r}}
−
1
ρ
∫
p
(
R
,
t
)
p
(
∞
,
t
)
d
p
=
1
4
π
∂
Φ
∂
t
∫
R
∞
d
r
r
2
+
∫
v
r
(
R
,
t
)
v
(
∞
,
t
)
v
r
d
v
r
{\displaystyle -\;{\frac {1}{\rho }}\int _{p(R,t)}^{p(\infty ,t)}dp\;=\;{\frac {1}{4\pi }}\;{\frac {\partial \Phi }{\partial t}}\int _{R}^{\infty }{\frac {dr}{r^{2}}}\;+\;\int _{vr(R,t)}^{v(\infty ,t)}v_{r}\;dv_{r}}
v
r
(
∞
,
t
)
=
0
p
(
R
,
t
)
=
0
e
p
(
∞
,
t
)
=
c
o
n
s
t
a
n
t
e
=
p
∞
{\displaystyle v_{r}(\infty ,t)\;=\;0\qquad p(R,t)\;=\;0\qquad e\qquad p(\infty ,t)\;=\;constante\;=\;p_{\infty }}
−
1
ρ
∫
0
p
∞
d
p
=
1
4
π
∂
Φ
∂
t
∫
R
∞
d
r
r
2
+
∫
v
r
(
R
,
t
)
0
v
r
d
v
r
{\displaystyle -\;{\frac {1}{\rho }}\int _{0}^{p_{\infty }}dp\;=\;{\frac {1}{4\pi }}\;{\frac {\partial \Phi }{\partial t}}\int _{R}^{\infty }{\frac {dr}{r^{2}}}\;+\;\int _{vr(R,t)}^{0}v_{r}\;dv_{r}}
1
ρ
p
|
p
∞
0
=
1
4
π
∂
Φ
∂
t
1
r
|
∞
R
+
1
2
v
r
2
|
v
r
(
R
,
t
)
0
{\displaystyle {\frac {1}{\rho }}\left.{\frac {}{}}p\right|_{p_{\infty }}^{0}\;=\;{\frac {1}{4\pi }}\;{\frac {\partial \Phi }{\partial t}}\;\left.{\frac {1}{r}}\right|_{\infty }^{R}\;+\;\left.{\frac {1}{2}}\;v_{r}^{2}\right|_{vr(R,t)}^{0}}
−
p
∞
ρ
=
1
4
π
R
∂
Φ
∂
t
−
v
r
2
(
R
,
t
)
2
{\displaystyle -\;{\frac {p_{\infty }}{\rho }}\;=\;{\frac {1}{4\pi R}}\;{\frac {\partial \Phi }{\partial t}}\;-\;{\frac {v_{r}^{2}(R,t)}{2}}}
∂
Φ
∂
t
=
∂
∂
t
(
4
π
⋅
v
r
(
R
,
t
)
⋅
R
2
)
{\displaystyle {\frac {\partial \Phi }{\partial t}}\;=\;{\frac {\partial }{\partial t}}\left(4\pi \cdot v_{r}(R,t)\cdot R^{2}\right)}
∂
Φ
∂
t
=
4
π
(
∂
v
r
(
R
,
t
)
∂
t
R
2
+
2
R
⋅
v
r
(
R
,
t
)
⋅
∂
R
∂
t
)
{\displaystyle {\frac {\partial \Phi }{\partial t}}\;=\;4\pi \left({\frac {\partial v_{r}(R,t)}{\partial t}}\;R^{2}\;+\;2R\cdot v_{r}(R,t)\cdot \;{\frac {\partial R}{\partial t}}\right)}
∂
Φ
∂
t
=
4
π
(
∂
v
r
(
R
,
t
)
∂
R
∂
R
∂
t
R
2
+
2
R
⋅
v
r
(
R
,
t
)
⋅
v
r
(
R
,
t
)
)
{\displaystyle {\frac {\partial \Phi }{\partial t}}\;=\;4\pi \left({\frac {\partial v_{r}(R,t)}{\partial R}}\;{\frac {\partial R}{\partial t}}\;R^{2}\;+\;2R\cdot v_{r}(R,t)\cdot v_{r}(R,t)\right)}
∂
Φ
∂
t
=
4
π
(
∂
v
r
(
R
,
t
)
∂
R
v
r
(
R
,
t
)
R
2
+
2
R
⋅
v
r
2
(
R
,
t
)
)
{\displaystyle {\frac {\partial \Phi }{\partial t}}\;=\;4\pi \left({\frac {\partial v_{r}(R,t)}{\partial R}}\;v_{r}(R,t)\;R^{2}\;+\;2R\cdot v_{r}^{2}(R,t)\right)}
−
p
∞
ρ
=
1
4
π
R
4
π
(
∂
v
r
(
R
,
t
)
∂
R
v
r
(
R
,
t
)
R
2
+
2
R
⋅
v
r
2
(
R
,
t
)
)
−
v
r
2
(
R
,
t
)
2
{\displaystyle -\;{\frac {p_{\infty }}{\rho }}\;=\;{\frac {1}{4\pi R}}\;4\pi \left({\frac {\partial v_{r}(R,t)}{\partial R}}\;v_{r}(R,t)\;R^{2}\;+\;2R\cdot v_{r}^{2}(R,t)\right)\;-\;{\frac {v_{r}^{2}(R,t)}{2}}}
−
p
∞
ρ
=
R
v
r
(
R
,
t
)
∂
v
r
(
R
,
t
)
∂
R
+
3
v
r
2
(
R
,
t
)
2
{\displaystyle -\;{\frac {p_{\infty }}{\rho }}\;=\;R\;v_{r}(R,t){\frac {\partial v_{r}(R,t)}{\partial R}}\;+\;{\frac {3v_{r}^{2}(R,t)}{2}}}
2 (R,t) e integrando, temos
−
p
∞
ρ
=
R
2
d
u
d
R
+
3
u
2
{\displaystyle -\;{\frac {p_{\infty }}{\rho }}\;=\;{\frac {R}{2}}\;{\frac {du}{dR}}\;+\;{\frac {3u}{2}}}
(
2
p
∞
ρ
+
3
u
)
d
R
=
−
R
d
u
{\displaystyle \left(2\;{\frac {p_{\infty }}{\rho }}\;+\;3u\right)\;dR\;=\;-\;R\;du}
d
u
2
p
∞
ρ
+
3
u
=
−
d
R
R
{\displaystyle {\frac {du}{2\;{\frac {p_{\infty }}{\rho }}\;+\;3u}}\;=\;-\;{\frac {dR}{R}}}
∫
0
u
d
u
2
p
∞
ρ
+
3
u
=
−
∫
a
R
d
R
R
{\displaystyle \int _{0}^{u}{\frac {du}{2\;{\frac {p_{\infty }}{\rho }}\;+\;3u}}\;=\;-\;\int _{a}^{R}{\frac {dR}{R}}}
1
3
l
n
(
2
p
∞
ρ
+
3
u
)
|
0
u
=
−
l
n
(
R
)
|
a
R
{\displaystyle {\frac {1}{3}}\;\left.ln\left({\frac {2p_{\infty }}{\rho }}\;+\;3u\right)\right|_{0}^{u}\;=\;-\;\left.{\frac {}{}}ln(R)\right|_{a}^{R}}
1
3
l
n
(
2
p
∞
ρ
+
3
u
2
p
∞
ρ
)
=
l
n
(
a
R
)
{\displaystyle {\frac {1}{3}}\;ln\left({\frac {{\frac {2p_{\infty }}{\rho }}\;+\;3u}{\frac {2p_{\infty }}{\rho }}}\right)\;=\;ln\left({\frac {a}{R}}\right)}
(
1
+
3
ρ
u
2
p
∞
)
1
3
=
a
R
{\displaystyle \left(1\;+\;{\frac {3\rho u}{2p_{\infty }}}\right)^{\frac {1}{3}}\;=\;{\frac {a}{R}}}
u
=
2
p
∞
3
ρ
(
(
a
R
)
3
−
1
)
{\displaystyle u\;=\;{\frac {2p_{\infty }}{3\rho }}\;\left(\left({\frac {a}{R}}\right)^{3}\;-\;1\right)}
Δ
t
=
∫
d
t
=
∫
a
0
d
R
v
=
∫
a
0
d
R
−
u
{\displaystyle \Delta t\;=\;\int dt\;=\;\int _{a}^{0}{\frac {dR}{v}}\;=\;\int _{a}^{0}{\frac {dR}{-\;{\sqrt {u}}}}}
Δ
t
=
∫
0
a
d
R
2
p
∞
3
ρ
(
(
a
R
)
3
−
1
)
=
3
ρ
2
p
∞
⋅
∫
0
a
R
3
2
⋅
(
a
3
−
R
3
)
−
1
2
⋅
d
R
{\displaystyle \Delta t\;=\;\int _{0}^{a}{\frac {dR}{\sqrt {{\frac {2p_{\infty }}{3\rho }}\;\left(\left({\frac {a}{R}}\right)^{3}\;-\;1\right)}}}\;=\;{\sqrt {\frac {3\rho }{2p_{\infty }}}}\;\cdot \;\int _{0}^{a}R^{\frac {3}{2}}\cdot (a^{3}\;-\;R^{3})^{-\;{\frac {1}{2}}}\;\cdot \;dR}
Δ
t
=
3
ρ
2
p
∞
⋅
∫
0
a
(
R
a
)
3
2
⋅
(
1
−
(
R
a
)
3
)
−
1
2
⋅
d
R
{\displaystyle \Delta t\;=\;{\sqrt {\frac {3\rho }{2p_{\infty }}}}\cdot \int _{0}^{a}\left({\frac {R}{a}}\right)^{\frac {3}{2}}\;\cdot \;\left(1\;-\;\left({\frac {R}{a}}\right)^{3}\right)^{-\;{\frac {1}{2}}}\;\cdot \;dR}
Δ
t
=
3
ρ
2
p
∞
⋅
∫
0
a
(
R
a
)
−
1
2
⋅
(
1
−
(
R
a
)
3
)
−
1
2
⋅
a
3
⋅
3
a
(
R
a
)
2
d
R
{\displaystyle \Delta t\;=\;{\sqrt {\frac {3\rho }{2p_{\infty }}}}\cdot \int _{0}^{a}\left({\frac {R}{a}}\right)^{-\;{\frac {1}{2}}}\;\cdot \;\left(1\;-\;\left({\frac {R}{a}}\right)^{3}\right)^{-\;{\frac {1}{2}}}\;\cdot \;{\frac {a}{3}}\;\cdot \;{\frac {3}{a}}\;\left({\frac {R}{a}}\right)^{2}\;dR}
Δ
t
=
3
ρ
2
p
∞
⋅
∫
0
a
[
(
R
a
)
3
]
−
1
6
⋅
(
1
−
(
R
a
)
3
)
−
1
2
⋅
a
3
⋅
d
[
(
R
a
)
3
]
{\displaystyle \Delta t\;=\;{\sqrt {\frac {3\rho }{2p_{\infty }}}}\cdot \int _{0}^{a}\left[\left({\frac {R}{a}}\right)^{3}\right]^{-\;{\frac {1}{6}}}\;\cdot \;\left(1\;-\;\left({\frac {R}{a}}\right)^{3}\right)^{-\;{\frac {1}{2}}}\;\cdot \;{\frac {a}{3}}\;\cdot \;d\left[\left({\frac {R}{a}}\right)^{3}\right]}
Δ
t
=
a
3
⋅
3
ρ
2
p
∞
⋅
∫
0
1
w
−
1
6
⋅
(
1
−
w
)
−
1
2
⋅
d
w
{\displaystyle \Delta t\;=\;\;{\frac {a}{3}}\cdot {\sqrt {\frac {3\rho }{2p_{\infty }}}}\cdot \int _{0}^{1}w^{-\;{\frac {1}{6}}}\cdot \left(1\;-\;w\right)^{-\;{\frac {1}{2}}}\cdot \;dw}
Wikipedia
Wikipedia
B
(
α
,
β
)
=
∫
0
1
t
α
−
1
(
1
−
t
)
β
−
1
d
t
.
{\displaystyle \mathrm {B} (\alpha ,\beta )=\int _{0}^{1}t^{\alpha -1}\,(1-t)^{\beta -1}\,dt.\!}
5
6
{\displaystyle {\frac {5}{6}}}
1
2
{\displaystyle {\frac {1}{2}}}
Δ
t
=
a
3
⋅
3
ρ
2
p
∞
⋅
B
(
5
6
,
1
2
)
=
a
ρ
p
∞
⋅
1
6
⋅
B
(
5
6
,
1
2
)
{\displaystyle \Delta t\;=\;\;{\frac {a}{3}}\cdot {\sqrt {\frac {3\rho }{2p_{\infty }}}}\cdot \mathrm {B} \left({\frac {5}{6}},{\frac {1}{2}}\right)\;=\;a\;{\sqrt {\frac {\rho }{p_{\infty }}}}\cdot {\frac {1}{\sqrt {6}}}\cdot \mathrm {B} \left({\frac {5}{6}},{\frac {1}{2}}\right)}
Neste site pode-se obter o valor de 2,24 para a função Beta com esses argumentos. Assim, pode-se escrever
Δ
t
=
0.914
a
ρ
p
∞
{\displaystyle \Delta t\;=\;0.914\;a\;{\sqrt {\frac {\rho }{p_{\infty }}}}}
![{\displaystyle \Delta t\;=\;{\sqrt {\frac {3\rho }{2p_{\infty }}}}\cdot \int _{0}^{a}\left[\left({\frac {R}{a}}\right)^{3}\right]^{-\;{\frac {1}{6}}}\;\cdot \;\left(1\;-\;\left({\frac {R}{a}}\right)^{3}\right)^{-\;{\frac {1}{2}}}\;\cdot \;{\frac {a}{3}}\;\cdot \;d\left[\left({\frac {R}{a}}\right)^{3}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/79b653bd1fe120b154333aa83d8844160a9e7031)
