1 / 9

Ejercicio 2

Ejercicio 2. 40. 39. 38. Se pide: tiempo de riego considerando una eficiencia de riego de 85%. Terciarias: PE DN40, DI 35,2mm. Q terciaria=0,00693 m3/s. 50/0,80= 63 laterales. 50m. 70m, goteros a 0,30m. 70/0,30= 233 goteros. 233 x 1,7l/h= 396 l/h= Q lateral = 0,00011 m3/s.

kosey
Download Presentation

Ejercicio 2

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Ejercicio 2

  2. 40 39 38

  3. Se pide: tiempo de riego considerando una eficiencia de riego de 85%

  4. Terciarias: PE DN40, DI 35,2mm Q terciaria=0,00693 m3/s 50/0,80= 63 laterales 50m 70m, goteros a 0,30m 70/0,30= 233 goteros 233 x 1,7l/h= 396 l/h= Q lateral = 0,00011 m3/s Laterales: DN 16, DI 13,6

  5. Análisis laterales Hf= 10,679 x 70 x 0,000111,852= 4,0 m 1501,852 x 0,01364,87 Hf x CSM = 4 x 0,352= 1,41m Presiones en el lateral (hg= 0,7m) P inicial = Pa + ¾ hf - hg/2 = 10 + 0.75(1,41) - 0.7/2=10,7m P min = P max - t’hf = 10,7 – t’ (1,41) = 10,7 – 0.622(1,41)=9,82m DPlateral= 10,7 – 9,82= 0.88m

  6. Análisis Terciarias Hf= 10,679 x 50 x 0,006931,852= 59 m 1501,852 x 0,03524,87 Hf x CSM = 59 x 0,354= 21,0m Presiones en la terciaria, Sin pendiente P MAX (t) = P inical (l) + ¾ hf (t) (terciaria a nivel) P MAX (t) = 10,7 + ¾ 21,0 = 26.5m P final(t) = P max(t) - hf(t) P final(t) = 26.5 – 21,0= 5,45 PMIN del SECTOR= Pfinal (t) – t’ hf(l) PMIN del SECTOR= 5,45 – 0,622 (1,41) = 4,6m

  7. P maxlat: 10,7 P maxterc: 26.5m P min terc: 5,45m Pa: 10m P min: 9,82m P min sector: 4,6m

  8. Pmin del sector Q min Qmin = 0,59 Hmin0.46 Qmin = 0,59 x (4,6)0.46 Q min = 1,2 l/h

  9. Necesidades netas = 6,5 / (0,85x0,67) = 11,4 mm/d Cálculos por superficie Supuestos: Ancho de cantero a mojar: 0.6m Área en 1m de cantero= 0,6 x 1,0m= 0,6 m2 dosis por metro lineal = 0,6 m2 x 11,4mm= 6,8 litros Caudal por metro lineal= 1m/0,30 x 1,7 =5,67 l/h/m Tiempo de riego = 6,8 / 5,67= 1,2 horas = 1 h 12 min

More Related