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 π = 0.020 π = 0.015 π =? π =? Tubo lleno 360Β° = 6.28 rad π΄=
1 8
(π β π πππ)π2
1 (6.28 β π ππ 360)1.82 8 π΄ = 2.54 1 π πππ π
= (1 β )π 4 π π΄=
π
=
1 π ππ360 (1 β ) 1.8 π
= 0.45 4 6.28
Para n=0.020 2 1 1 π₯ π΄ π₯ π
3 π₯ π 2 π 2 1 1 π = 0.020 π₯ 2.54 π₯ (0.45)3 π₯ 0.00082 π3 π = 2.11 π ππ
π=
π=
1 1 2 π₯π
3 π₯ π 2 π
2 1 1 π₯(0.45)3 π₯ (0.0008)2 0.020 π = 0.83π/π
π=
Para n=0.015 2 1 1 π₯ π΄ π₯ π
3 π₯ π 2 π 2 1 1 π = 0.015 π₯ 2.54 π₯ (0.45)3 π₯ 0.00082 π3 π = 2.81 π ππ
π=
π=
1 1 2 π₯π
3 π₯ π 2 π
2 1 1 π₯(0.45)3 π₯ (0.0008)2 0.015 π = 1.11π/π
π=
Medio tubo 180Β° = 3.14 rad π΄=
1 8
(π β π πππ)π2
1 (3.14 β π ππ 180)1.82 8 π΄ = 1.27 1 π πππ π
= (1 β )π 4 π π΄=
π
=
1 π ππ180 (1 β ) 1.8 π
= 0.45 4 3.14
Para n=0.020 2 1 1 π₯ π΄ π₯ π
3 π₯ π 2 π 2 1 1 π = 0.020 π₯ 1.27π₯ (0.45)3 π₯ 0.00082 π3 π = 1.05 π ππ
π=
π=
1 1 2 π₯π
3 π₯ π 2 π
2 1 1 π₯(0.45)3 π₯ (0.0008)2 0.020 π = 0.83π/π
π=
Para n=0.015 2 1 1 π₯ π΄ π₯ π
3 π₯ π 2 π 2 1 1 π = 0.015 π₯ 1.27π₯ (0.45)3 π₯ 0.00082 π3 π = 1.41 π ππ
π=
π=
1 1 2 π₯π
3 π₯ π 2 π
2 1 1 π₯(0.45)3 π₯ (0.0008)2 0.015 π = 1.11π/π
π=
PROBLEMA NΒ°3: Determinar el tirante y velocidad en un colector de desagΓΌe de 16β 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: π· = 16" β π· = 0.4064π Q = 110l/s β Q = 0.110π3 βπ π == 0.008 π =? π =? π = 0.020 π = 0.010
Para tuberΓa paralelamente llena y para π = 0.020 π΄=
1 8
(π β π πππ)π2
π΄ = 0.0508(π β π πππ)
π
= 0.2 (1 β 2
π πππ ) π
1
π΄. π
3 . π 2 π= π 2
1 1 π πππ 3 0.110 = π₯ [0.0508 π₯ (π β π πππ)] π₯ [0.2 π₯ (1 β )] π₯ (0.008)2 0.020 π
ο· De donde se obtiene π: π = 2.13 πππ <> 122.04Β° ο· Reemplazando en las formulas π΄ = 0.08(π β π πππ) π΄ = 0.08(2.13 β π ππ(122.04)) = .102 π2 π πππ ) π π ππ(122.04) π
= 0.2 (1 β ) = 0.120π 2.13 π
= 0.2 (1 β
π=
2 1 1 π₯ π
3 π₯ π 2 π
π=
2 1 1 π₯ (0.12)3 π₯ (0.005)2 = 1.72 πβπ 0.010
β=
π
β π
180 β 122.04 2 β= 28.98
Calculando tirante: π· π· π = β π π πππΌ 3 3 0.5 0.5 π= β π π ππ(28.98) 2 2 π = 0.21 π