Alcantarillado.docx

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PROBLEMA NΒ°2: Una galerΓ­a circular de ladrillo tiene un diΓ‘metro de 1.8m con una pendiente de 0.0008, calcular la capacidad y velocidad cuando trabaja a tubo lleno y medio tubo. Considerar n = 0.020 y n=0.015 Datos: D = 1.8m S = 0.0008 𝑛 = 0.020 𝑛 = 0.015 𝑉 =? 𝑄 =? Tubo lleno 360Β° = 6.28 rad 𝐴=

1 8

(πœƒ βˆ’ π‘ π‘’π‘›πœƒ)𝑑2

1 (6.28 βˆ’ 𝑠𝑒𝑛 360)1.82 8 𝐴 = 2.54 1 π‘ π‘’π‘›πœƒ 𝑅 = (1 βˆ’ )𝑑 4 πœƒ 𝐴=

𝑅=

1 𝑠𝑒𝑛360 (1 βˆ’ ) 1.8 𝑅 = 0.45 4 6.28

Para n=0.020 2 1 1 π‘₯ 𝐴 π‘₯ 𝑅3 π‘₯ 𝑆 2 𝑛 2 1 1 𝑄 = 0.020 π‘₯ 2.54 π‘₯ (0.45)3 π‘₯ 0.00082 π‘š3 𝑄 = 2.11 𝑠𝑒𝑔

𝑄=

𝑉=

1 1 2 π‘₯𝑅 3 π‘₯ 𝑆 2 𝑛

2 1 1 π‘₯(0.45)3 π‘₯ (0.0008)2 0.020 𝑉 = 0.83π‘š/𝑠

𝑉=

Para n=0.015 2 1 1 π‘₯ 𝐴 π‘₯ 𝑅3 π‘₯ 𝑆 2 𝑛 2 1 1 𝑄 = 0.015 π‘₯ 2.54 π‘₯ (0.45)3 π‘₯ 0.00082 π‘š3 𝑄 = 2.81 𝑠𝑒𝑔

𝑄=

𝑉=

1 1 2 π‘₯𝑅 3 π‘₯ 𝑆 2 𝑛

2 1 1 π‘₯(0.45)3 π‘₯ (0.0008)2 0.015 𝑉 = 1.11π‘š/𝑠

𝑉=

Medio tubo 180° = 3.14 rad 𝐴=

1 8

(πœƒ βˆ’ π‘ π‘’π‘›πœƒ)𝑑2

1 (3.14 βˆ’ 𝑠𝑒𝑛 180)1.82 8 𝐴 = 1.27 1 π‘ π‘’π‘›πœƒ 𝑅 = (1 βˆ’ )𝑑 4 πœƒ 𝐴=

𝑅=

1 𝑠𝑒𝑛180 (1 βˆ’ ) 1.8 𝑅 = 0.45 4 3.14

Para n=0.020 2 1 1 π‘₯ 𝐴 π‘₯ 𝑅3 π‘₯ 𝑆 2 𝑛 2 1 1 𝑄 = 0.020 π‘₯ 1.27π‘₯ (0.45)3 π‘₯ 0.00082 π‘š3 𝑄 = 1.05 𝑠𝑒𝑔

𝑄=

𝑉=

1 1 2 π‘₯𝑅 3 π‘₯ 𝑆 2 𝑛

2 1 1 π‘₯(0.45)3 π‘₯ (0.0008)2 0.020 𝑉 = 0.83π‘š/𝑠

𝑉=

Para n=0.015 2 1 1 π‘₯ 𝐴 π‘₯ 𝑅3 π‘₯ 𝑆 2 𝑛 2 1 1 𝑄 = 0.015 π‘₯ 1.27π‘₯ (0.45)3 π‘₯ 0.00082 π‘š3 𝑄 = 1.41 𝑠𝑒𝑔

𝑄=

𝑉=

1 1 2 π‘₯𝑅 3 π‘₯ 𝑆 2 𝑛

2 1 1 π‘₯(0.45)3 π‘₯ (0.0008)2 0.015 𝑉 = 1.11π‘š/𝑠

𝑉=

PROBLEMA NΒ°3: Determinar el tirante y velocidad en un colector de desagΓΌe de 16” de diΓ‘metro que estΓ‘ funcionando paralelamente lleno con una carga de 110 l/s . Y una pendiente de 0.008. Considerar: n= 0.020 y n=0.010 Datos: 𝐷 = 16" β†’ 𝐷 = 0.4064π‘š Q = 110l/s β†’ Q = 0.110π‘š3 ⁄𝑠 𝑆 == 0.008 𝑉 =? 𝑇 =? 𝑛 = 0.020 𝑛 = 0.010

Para tuberΓ­a paralelamente llena y para 𝑛 = 0.020 𝐴=

1 8

(πœƒ βˆ’ π‘ π‘’π‘›πœƒ)𝑑2

𝐴 = 0.0508(πœƒ βˆ’ π‘ π‘’π‘›πœƒ)

𝑅 = 0.2 (1 βˆ’ 2

π‘ π‘’π‘›πœƒ ) πœƒ

1

𝐴. 𝑅 3 . 𝑆 2 𝑄= 𝑛 2

1 1 π‘ π‘’π‘›πœƒ 3 0.110 = π‘₯ [0.0508 π‘₯ (πœƒ βˆ’ π‘ π‘’π‘›πœƒ)] π‘₯ [0.2 π‘₯ (1 βˆ’ )] π‘₯ (0.008)2 0.020 πœƒ

ο‚· De donde se obtiene πœƒ: πœƒ = 2.13 π‘Ÿπ‘Žπ‘‘ <> 122.04Β° ο‚· Reemplazando en las formulas 𝐴 = 0.08(πœƒ βˆ’ π‘ π‘’π‘›πœƒ) 𝐴 = 0.08(2.13 βˆ’ 𝑠𝑒𝑛(122.04)) = .102 π‘š2 π‘ π‘’π‘›πœƒ ) πœƒ 𝑠𝑒𝑛(122.04) 𝑅 = 0.2 (1 βˆ’ ) = 0.120π‘š 2.13 𝑅 = 0.2 (1 βˆ’

𝑉=

2 1 1 π‘₯ 𝑅3 π‘₯ 𝑆 2 𝑛

𝑉=

2 1 1 π‘₯ (0.12)3 π‘₯ (0.005)2 = 1.72 π‘šβ„π‘  0.010

∝=

πœƒ

∝ 𝑇

180 βˆ’ 122.04 2 ∝= 28.98

Calculando tirante: 𝐷 𝐷 𝑇 = βˆ’ 𝑋 𝑠𝑒𝑛𝛼 3 3 0.5 0.5 𝑇= βˆ’ 𝑋 𝑠𝑒𝑛(28.98) 2 2 𝑇 = 0.21 π‘š

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