Considerando a tubulação do exercício anterior , calcular o maior comprimento admissível para o tubo sem que a vazão de água diminua, se a profundidade do reservatório for mantida no mínimo em 5 m.
A perda total será dada por
Δ
H
t
=
Δ
H
i
+
Δ
H
t
+
Δ
H
o
=
−
1
2
(
N
f
i
+
N
f
l
L
D
+
1
)
v
¯
2
{\displaystyle \Delta H_{t}\;=\;\Delta H_{i}\;+\;\Delta H_{t}\;+\;\Delta H_{o}\;=\;-\;{\frac {1}{2}}\;\left(N_{fi}\;+\;N_{fl}\;{\frac {L}{D}}\;+\;1\right){\bar {v}}^{2}}
Δ
h
=
−
Δ
H
t
g
{\displaystyle \Delta h\;=\;-\;{\frac {\Delta H_{t}}{g}}}
g
Δ
h
=
1
2
(
N
f
i
+
N
f
l
L
D
+
1
)
v
¯
2
⇒
L
=
D
N
f
l
(
2
g
Δ
h
v
¯
2
−
N
f
i
−
1
)
{\displaystyle g\;\Delta h\;=\;{\frac {1}{2}}\;\left(N_{fi}\;+\;N_{fl}\;{\frac {L}{D}}\;+\;1\right){\bar {v}}^{2}\;\;\;\Rightarrow L\;=\;{\frac {D}{N_{fl}}}\;\left({\frac {2g\;\Delta h}{{\bar {v}}^{2}}}\;-\;N_{fi}\;-\;1\right)}
L
=
D
N
f
l
(
g
Δ
h
8
(
π
D
2
Φ
)
2
−
N
f
i
−
1
)
=
100
m
m
0.023
(
9.8
m
/
s
2
⋅
5
m
8
(
3.14
⋅
(
100
m
m
)
2
10
l
/
s
)
2
−
0.5
−
1
)
{\displaystyle L\;=\;{\frac {D}{N_{fl}}}\;\left({\frac {g\;\Delta h}{8}}\;\left({\frac {\pi D^{2}}{\Phi }}\right)^{2}\;-\;N_{fi}\;-\;1\right)\;=\;{\frac {100\;mm}{0.023}}\;\left({\frac {9.8\;m/s^{2}\cdot 5\;m}{8}}\;\left({\frac {3.14\cdot (100\;mm)^{2}}{10\;l/s}}\right)^{2}\;-\;0.5\;-\;1\right)}
L
=
0.10
m
0.023
(
9.8
m
/
s
2
⋅
5
m
8
(
3.14
⋅
(
0.10
m
)
2
0.010
m
3
/
s
)
2
−
0.5
−
1
)
=
220
m
{\displaystyle L\;=\;{\frac {0.10\;m}{0.023}}\;\left({\frac {9.8\;m/s^{2}\cdot 5\;m}{8}}\;\left({\frac {3.14\cdot (0.10\;m)^{2}}{0.010\;m^{3}/s}}\right)^{2}\;-\;0.5\;-\;1\right)\;=\;220\;m}
