Uma bomba centrífuga, cuja eficiência é de 80%, possui rotor de diâmetro 20 cm e bombeia água a 1170 rpm e a uma vazão de 1500 litros por minuto. A velocidade específica da bomba é igual a 4.5.
Determine a vazão resultante quando ela funcionar a 1750 rpm;
Determine a elevação de carga obtida a 1170 e a 1750 rpm;
Determine a potência de alimentação necessária a 1170 e a 1750 rpm; Dados do problema
editar
1
2
ω
1170 rpm
1750 rpm
Φ
1500 l/min
?
h
?
?
Pa
?
?
Dados adicionais:
Eficiência da bomba (η) = 80 %
Diâmetro do rotor (D) = 20 cm
Densidade da água (ρ) = 1000 kg/m3
Velocidade específica (ωe ) = 4.5
Pelas leis de bombas
Φ
m
ω
m
D
m
3
=
Φ
o
ω
o
D
o
3
⇒
Φ
2
=
Φ
1
ω
2
ω
1
(
D
2
D
1
)
3
=
Φ
1
ω
2
ω
1
{\displaystyle {\frac {\Phi _{m}}{\omega _{m}D_{m}^{3}}}\;=\;{\frac {\Phi _{o}}{\omega _{o}D_{o}^{3}}}\;\;\;\Rightarrow \Phi _{2}\;=\;\Phi _{1}{\frac {\omega _{2}}{\omega _{1}}}\left({\frac {D_{2}}{D_{1}}}\right)^{3}\;=\;\Phi _{1}{\frac {\omega _{2}}{\omega _{1}}}}
=
1500
l
/
m
i
n
1750
r
p
m
1170
r
p
m
=
2200
l
/
m
i
n
{\displaystyle \;=\;1500\;l/min\;{\frac {1750\;rpm}{1170\;rpm}}\;=\;2200\;l/min}
ω
e
1
=
ω
1
Φ
1
h
1
3
⇒
h
1
=
(
ω
1
Φ
1
ω
e
1
)
4
3
{\displaystyle \omega _{e1}\;=\;\omega _{1}{\sqrt {\frac {\Phi _{1}}{\sqrt {h_{1}^{3}}}}}\;\;\;\Rightarrow h_{1}\;=\;{\sqrt[{3}]{\left({\frac {\omega _{1}{\sqrt {\Phi _{1}}}}{\omega _{e1}}}\right)^{4}}}}
h
1
=
(
1170
r
p
m
1500
l
/
m
i
n
4.5
)
4
3
=
(
1170
⋅
2
⋅
3.1
60
r
d
/
s
1500
⋅
0.001
60
m
3
/
s
4.5
)
4
3
=
6.9
m
{\displaystyle h_{1}\;=\;{\sqrt[{3}]{\left({\frac {1170\;rpm\;{\sqrt {1500\;l/min}}}{4.5}}\right)^{4}}}\;=\;{\sqrt[{3}]{\left({\frac {1170\cdot {\frac {2\cdot 3.1}{60}}\;rd/s\;{\sqrt {1500\cdot {\frac {0.001}{60}}\;m^{3}/s}}}{4.5}}\right)^{4}}}\;=\;6.9\;m}
h
2
=
(
1750
r
p
m
2200
l
/
m
i
n
4.5
)
4
3
=
(
1750
⋅
2
⋅
3.1
60
r
d
/
s
2200
⋅
0.001
60
m
3
/
s
4.5
)
4
3
=
15
m
{\displaystyle h_{2}\;=\;{\sqrt[{3}]{\left({\frac {1750\;rpm\;{\sqrt {2200\;l/min}}}{4.5}}\right)^{4}}}\;=\;{\sqrt[{3}]{\left({\frac {1750\cdot {\frac {2\cdot 3.1}{60}}\;rd/s\;{\sqrt {2200\cdot {\frac {0.001}{60}}\;m^{3}/s}}}{4.5}}\right)^{4}}}\;=\;15\;m}
h
o
ω
o
2
D
o
2
=
h
m
ω
m
2
D
m
2
⇒
h
2
=
h
1
(
ω
2
ω
1
)
2
(
D
2
D
1
)
2
=
h
1
(
ω
2
ω
1
)
2
{\displaystyle {\frac {h_{o}}{\omega _{o}^{2}D_{o}^{2}}}\;=\;{\frac {h_{m}}{\omega _{m}^{2}D_{m}^{2}}}\;\;\;\Rightarrow h_{2}\;=\;h_{1}\left({\frac {\omega _{2}}{\omega _{1}}}\right)^{2}\;\left({\frac {D_{2}}{D_{1}}}\right)^{2}\;=\;h_{1}\left({\frac {\omega _{2}}{\omega _{1}}}\right)^{2}}
h
2
=
6.9
m
(
1750
r
p
m
1170
r
p
m
)
2
=
15
m
{\displaystyle h_{2}\;=\;6.9\;m\;\left({\frac {1750\;rpm}{1170\;rpm}}\right)^{2}\;=\;15\;m}
P
=
Δ
E
Δ
t
=
Δ
(
m
g
h
)
Δ
t
=
g
h
Δ
m
Δ
t
=
g
h
Φ
m
=
g
h
ρ
Φ
{\displaystyle P\;=\;{\frac {\Delta E}{\Delta t}}\;=\;{\frac {\Delta (mgh)}{\Delta t}}\;=\;gh{\frac {\Delta m}{\Delta t}}\;=\;gh\Phi _{m}\;=\;gh\rho \Phi }
m é a vazão mássica do fluido. Assim
P
a
1
=
P
1
η
1
=
ρ
g
h
1
Φ
1
η
1
=
1000
k
g
/
m
3
⋅
9.8
m
/
s
2
⋅
6.9
m
⋅
1500
l
/
m
i
n
0.8
{\displaystyle P_{a1}\;=\;{\frac {P_{1}}{\eta _{1}}}\;=\;{\frac {\rho gh_{1}\Phi _{1}}{\eta _{1}}}\;=\;{\frac {1000\;kg/m^{3}\cdot 9.8\;m/s^{2}\cdot 6.9\;m\cdot 1500\;l/min}{0.8}}}
P
a
1
=
1000
k
g
/
m
3
⋅
9.8
m
/
s
2
⋅
6.9
m
⋅
1500
⋅
0.001
60
m
3
/
s
0.8
=
2100
W
{\displaystyle P_{a1}\;=\;{\frac {1000\;kg/m^{3}\cdot 9.8\;m/s^{2}\cdot 6.9\;m\cdot 1500\cdot {\frac {0.001}{60}}\;m^{3}/s}{0.8}}\;=\;2100\;W}
P
m
ρ
m
ω
m
3
D
m
5
=
P
o
ρ
o
ω
o
3
D
o
5
⇒
P
2
=
P
1
ρ
2
ω
2
3
D
2
5
ρ
1
ω
1
3
D
1
5
{\displaystyle {\frac {P_{m}}{\rho _{m}\omega _{m}^{3}D_{m}^{5}}}\;=\;{\frac {P_{o}}{\rho _{o}\omega _{o}^{3}D_{o}^{5}}}\;\;\;\Rightarrow P_{2}\;=\;P_{1}\;{\frac {\rho _{2}\omega _{2}^{3}D_{2}^{5}}{\rho _{1}\omega _{1}^{3}D_{1}^{5}}}}
P
a
2
=
P
2
η
=
ρ
1
g
h
1
Φ
1
η
(
ω
2
ω
1
)
3
=
1000
k
g
/
m
3
⋅
9.8
m
/
s
2
⋅
6.9
m
⋅
1500
l
/
m
i
n
0.8
(
1750
r
p
m
1170
r
p
m
)
3
{\displaystyle P_{a2}\;=\;{\frac {P_{2}}{\eta }}\;=\;{\frac {\rho _{1}gh_{1}\Phi _{1}}{\eta }}\;\left({\frac {\omega _{2}}{\omega _{1}}}\right)^{3}\;=\;{\frac {1000\;kg/m^{3}\cdot 9.8\;m/s^{2}\cdot 6.9\;m\cdot 1500\;l/min}{0.8}}\;\left({\frac {1750\;rpm}{1170\;rpm}}\right)^{3}}
=
1000
k
g
/
m
3
⋅
9.8
m
/
s
2
⋅
6.9
m
⋅
1500
⋅
0.001
60
m
3
/
s
0.8
(
1750
r
p
m
1170
r
p
m
)
3
=
7000
W
{\displaystyle \;=\;{\frac {1000\;kg/m^{3}\cdot 9.8\;m/s^{2}\cdot 6.9\;m\cdot 1500\cdot {\frac {0.001}{60}}\;m^{3}/s}{0.8}}\;\left({\frac {1750\;rpm}{1170\;rpm}}\right)^{3}\;=\;7000\;W}
P
2
=
ρ
2
g
h
2
Φ
2
⇒
P
a
2
=
1000
k
g
/
m
3
⋅
9.8
m
/
s
2
⋅
15
m
⋅
2200
l
/
m
i
n
0.8
{\displaystyle P_{2}\;=\;\rho _{2}gh_{2}\Phi _{2}\;\;\;\Rightarrow P_{a2}\;=\;{\frac {1000\;kg/m^{3}\cdot 9.8\;m/s^{2}\cdot 15\;m\cdot 2200\;l/min}{0.8}}}
=
1000
k
g
/
m
3
⋅
9.8
m
/
s
2
⋅
15
m
⋅
2200
⋅
0.001
60
m
3
/
s
0.8
=
7000
W
{\displaystyle \;=\;{\frac {1000\;kg/m^{3}\cdot 9.8\;m/s^{2}\cdot 15\;m\cdot 2200\cdot {\frac {0.001}{60}}\;m^{3}/s}{0.8}}\;=\;7000\;W}
![{\displaystyle \omega _{e1}\;=\;\omega _{1}{\sqrt {\frac {\Phi _{1}}{\sqrt {h_{1}^{3}}}}}\;\;\;\Rightarrow h_{1}\;=\;{\sqrt[{3}]{\left({\frac {\omega _{1}{\sqrt {\Phi _{1}}}}{\omega _{e1}}}\right)^{4}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ce3d694d33c5471c0f898da4ef9ec8591b16a820)
![{\displaystyle h_{1}\;=\;{\sqrt[{3}]{\left({\frac {1170\;rpm\;{\sqrt {1500\;l/min}}}{4.5}}\right)^{4}}}\;=\;{\sqrt[{3}]{\left({\frac {1170\cdot {\frac {2\cdot 3.1}{60}}\;rd/s\;{\sqrt {1500\cdot {\frac {0.001}{60}}\;m^{3}/s}}}{4.5}}\right)^{4}}}\;=\;6.9\;m}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7da52254b385010b9c637ed9e514294d30627fb2)
![{\displaystyle h_{2}\;=\;{\sqrt[{3}]{\left({\frac {1750\;rpm\;{\sqrt {2200\;l/min}}}{4.5}}\right)^{4}}}\;=\;{\sqrt[{3}]{\left({\frac {1750\cdot {\frac {2\cdot 3.1}{60}}\;rd/s\;{\sqrt {2200\cdot {\frac {0.001}{60}}\;m^{3}/s}}}{4.5}}\right)^{4}}}\;=\;15\;m}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c3d5177ba3c285002ebe7347e668ed075457903)
