Calcular a vazão através do tubo do exercício E11 , para uma profundidade do reservatório igual a 5 m.
Do exercício anterior , sabemos que
g
Δ
h
=
1
2
(
N
f
i
+
N
f
l
L
D
+
1
)
v
¯
2
⇒
v
¯
2
=
2
g
Δ
h
N
f
i
+
N
f
l
L
D
+
1
{\displaystyle g\;\Delta h\;=\;{\frac {1}{2}}\;\left(N_{fi}\;+\;N_{fl}\;{\frac {L}{D}}\;+\;1\right){\bar {v}}^{2}\;\;\;\Rightarrow {\bar {v}}^{2}\;=\;{\frac {2g\;\Delta h}{N_{fi}\;+\;N_{fl}\;{\frac {L}{D}}\;+\;1}}}
fl é função da velocidade. Para resolver a equação, será preciso empregar um processo iterativo.
No primeiro passo, consideremos fluxo plenamente turbulento; para a rugosidade relativa de 0.0015, o fator de fricção será considerado como igual a 0.022 e
v
¯
=
2
⋅
9.8
m
/
s
2
⋅
5
m
0
,
5
+
0.022
⋅
50
m
100
m
m
+
1
=
2
⋅
9.8
m
/
s
2
⋅
5
m
0
,
5
+
0.022
⋅
50
m
0.10
m
+
1
{\displaystyle {\bar {v}}\;=\;{\sqrt {\frac {2\cdot 9.8\;m/s^{2}\cdot 5\;m}{0,5\;+\;0.022\cdot {\frac {50\;m}{100\;mm}}\;+\;1}}}\;=\;{\sqrt {\frac {2\cdot 9.8\;m/s^{2}\cdot 5\;m}{0,5\;+\;0.022\cdot {\frac {50\;m}{0.10\;m}}\;+\;1}}}}
v
¯
=
2.8
m
/
s
{\displaystyle {\bar {v}}\;=\;2.8\;m/s}
N
R
e
=
ρ
0
v
¯
D
μ
0
=
1000
k
g
/
m
3
⋅
2.8
m
/
s
⋅
100
m
m
0
,
0010
k
g
⋅
m
−
1
⋅
s
−
1
{\displaystyle N_{Re}\;=\;{\frac {\rho _{0}\;{\bar {v}}\;D}{\mu _{0}}}\;=\;{\frac {1000\;kg/m^{3}\cdot 2.8\;m/s\cdot 100\;mm}{0,0010\;kg\cdot m^{-1}\cdot s^{-1}}}}
N
R
e
=
1000
k
g
/
m
3
⋅
2.8
m
/
s
⋅
0.10
m
0
,
0010
k
g
⋅
m
−
1
⋅
s
−
1
=
280000
{\displaystyle N_{Re}\;=\;{\frac {1000\;kg/m^{3}\cdot 2.8\;m/s\cdot 0.10\;m}{0,0010\;kg\cdot m^{-1}\cdot s^{-1}}}\;=\;280000}
Re não corresponde a turbulência plena. Será preciso calcular um novo Nfl pela fórmula de Miller:
N
f
=
0.25
[
l
o
g
(
0.15
m
m
3.7
⋅
100
m
m
+
5.74
⋅
280000
−
0.9
)
]
−
2
=
0.023
{\displaystyle N_{f}\;=\;0.25\left[log\left({\frac {0.15\;mm}{3.7\cdot 100\;mm}}\;+\;5.74\cdot 280000^{-0.9}\right)\right]^{-2}\;=\;0.023}
Com o fator de fricção considerado como igual a 0.023
v
¯
=
2
⋅
9.8
m
/
s
2
⋅
5
m
0
,
5
+
0.023
⋅
50
m
0.10
m
+
1
=
2.7
{\displaystyle {\bar {v}}\;=\;{\sqrt {\frac {2\cdot 9.8\;m/s^{2}\cdot 5\;m}{0,5\;+\;0.023\cdot {\frac {50\;m}{0.10\;m}}\;+\;1}}}\;=\;2.7}
N
R
e
=
1000
k
g
/
m
3
⋅
2.7
m
/
s
⋅
0.10
m
0
,
0010
k
g
⋅
m
−
1
⋅
s
−
1
=
270000
{\displaystyle N_{Re}\;=\;{\frac {1000\;kg/m^{3}\cdot 2.7\;m/s\cdot 0.10\;m}{0,0010\;kg\cdot m^{-1}\cdot s^{-1}}}\;=\;270000}
fl pela fórmula de Miller será então:
N
f
=
0.25
[
l
o
g
(
0.15
m
m
3.7
⋅
100
m
m
+
5.74
⋅
270000
−
0.9
)
]
−
2
=
0.023
{\displaystyle N_{f}\;=\;0.25\left[log\left({\frac {0.15\;mm}{3.7\cdot 100\;mm}}\;+\;5.74\cdot 270000^{-0.9}\right)\right]^{-2}\;=\;0.023}
Φ
=
π
D
2
v
¯
4
=
3.14
(
100
m
m
)
2
⋅
2.7
m
/
s
4
=
0.022
m
3
/
s
=
22
l
/
s
{\displaystyle \Phi \;=\;{\frac {\pi \;D^{2}\;{\bar {v}}}{4}}\;=\;{\frac {3.14\;(100\;mm)^{2}\cdot 2.7\;m/s}{4}}\;=\;0.022\;m^{3}/s\;=\;22\;l/s}
![{\displaystyle N_{f}\;=\;0.25\left[log\left({\frac {0.15\;mm}{3.7\cdot 100\;mm}}\;+\;5.74\cdot 280000^{-0.9}\right)\right]^{-2}\;=\;0.023}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b1f7ba9ed61c674c6557cd7cd525b880e5e6737)
![{\displaystyle N_{f}\;=\;0.25\left[log\left({\frac {0.15\;mm}{3.7\cdot 100\;mm}}\;+\;5.74\cdot 270000^{-0.9}\right)\right]^{-2}\;=\;0.023}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf060252e0707f21c41479496505910f20ff5338)
