Apostila Mec. Flu

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Aula 16 Prof. MSc. Luiz Eduardo Miranda J. Rodrigues Solução do Exercício 1 Equação da energia entre (1) e (4): H + H = H + H 1 B 4 P1,4 2 1 P1 v + + γ 2 ⋅ g 2 1 P1 v + + γ 2 ⋅ g v 4 z z V 2 4 1 1 + + H H Velocidade em (4): 2 P4 v4 B = + + z 4 + H P 1,4 γ 2 ⋅ g 2 P4 v4 B = + + z 4 + H P 1,4 γ 2 ⋅ g = 4 ⋅ Q 4 ⋅ 20 ⋅10 v π 4 = ⋅ d π ⋅ 0,08 2 v = 3,98 m/s 4 −3 2 v4 = + z H B 4 P 1,4 H + 2 ⋅ g H B = 2 3,98 20 + 30 + 9 H = 0 ,792 + 39 = 39, 792m N B B N B N B γ ⋅ Q ⋅ H = η B B 10000 ⋅ 20 ⋅10 = 0,65 =12243,69W − 3 ⋅ H B Potência da Bomba: 39,792 12243,69 N B = N B =16, 62cv 736,5 Mecânica dos Fluidos
Aula 16 Prof. MSc. Luiz Eduardo Miranda J. Rodrigues Exercícios Propostos 1) Para a instalação mostrada, determine: a) A velocidade na tubulação de sucção. b) A pressão na entrada da bomba. c) Sabendo-se que N B = 10cv, calcule a altura Z 4 . Dados: Q v = 15 litros/s, γ H2O = 10000N/m³, g = 10m/s², d 1 = d 2 = 10cm, d 4 = 8cm, H P1,2 = 5m, H P3,4 = 7m, η B = 60%. (4) B Z 4 (2) B (3) M ref 2m (1) A aberto com nível constante Mecânica dos Fluidos
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Mecânica dos Fluidos Aula 1 - Defi
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Aula 1 Prof. MSc. Luiz Eduardo Mira
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Mecânica dos Fluidos Aula 5 - Man
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Mecânica dos Fluidos Aula 6 - Man
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Mecânica dos Fluidos Aula 7 - Flut
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Mecânica dos Fluidos Aula 9 - Aval
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Aula 10 Prof. MSc. Luiz Eduardo Mir
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Mecânica dos Fluidos Aula 11 - Equ
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Page 175 and 176: Aula 12 Prof. MSc. Luiz Eduardo Mir
Page 201 and 202: Mecânica dos Fluidos Aula 14 - Ins
Page 225: Aula 16 Prof. MSc. Luiz Eduardo Mir
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Mecânica dos Fluidos Aula 20 - Rec
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Apostila Mec. Flu

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