Ingenieria

DATOS DEL SISTEMA HIDRAULICO

Longitud efectiva 5300 ft
Velocidad:
V=QA=0.10026π.7524=0.22 ft/seg

Relación: t/d
No. Reynolds
Relación t/d→ rugosidad relativa (tablas)
t/d= 0.00080.1723ft= 0.0046
No. Reynolds
Re= VDv= 0.22(.1723)1.o5x10-5= 3610.095
Calculo f=?
f= .25logtd3.7 + 5.74Re.9
f= .25log(.00463.7 + 5.743610.095.9
f = .0202

APLICACIÓN DE LA ECUACION DELSISTEMA HIDRAULICO
Hf (Hazen – Williams)
Hf= Q1.8510.44 LRC1.85d4.87=(45)1.8510.44 (5300)(100)1.85(9)4.87= .2846 ft
Q=(45galmin)1 cfs448.8 gal/min= 0.10026 cfs
V=QA=0.10026π.7524=0.22 ft/segHv= V22g=0.22 ft/seg2x32.2= .00075ft
Ha=Z+Hv+HfQ12Q22

Ha=25+.00075+.28461752Q22













Ha=25+ 9.53^-7 +90= | 115.0001 |
Ha=25+ 9.53^-7 +92.5= |117.5001 |
Ha=25+ 9.53^-7 +95= | 120.0001 |
Ha=25+ 9.53^-7 +97.5= | 122.5001 |
Ha=25+ 9.53^-7 +100= | 125.0001 |
Ha=25+ 9.53^-7 +102.5= | 127.5001 |
Ha=25+ 9.53^-7 +105= | 130.0001 |
Ha=25+9.53^-7 +107.5= | 132.5001 |
Ha=25+ 9.53^-7 +110= | 135.0001 |
Ha=25+ 9.53^-7 +112.5= | 137.5001 |
Ha=25+ 9.53^-7 +115= | 140.0001 |
Ha=25+ 9.53^-7 +117.5= | 142.5001 |
Ha=25+ 9.53^-7 +120=| 145.0001 |
Ha=25+ 9.53^-7 +122.5= | 147.5001 |
Ha=25+ 9.53^-7 +123.5= | 150.0001 |
Ha=25+ 9.53^-7 +125= | 152.5001 |
Ha=25+ 9.53^-7 +127.5= | 155.0001 |
Ha=25+ 9.53^-7 +130= | 157.5001 |Ha=25+ 9.53^-7 +132.5= | 160.0001 |
Ha=25+ 9.53^-7 +135= | 162.5001 |
Ha=25+ 9.53^-7 +137.5= | 165.0001 |
Ha=25+ 9.53^-7 +140= | 167.5001 |
Ha=25+ 9.53^-7 +142.5= | 170.0001 |
Ha=25+9.53^-7 +145= | 172.5001 |
Ha=25+ 9.53^-7 +147.5= | 175.0001 |
Ha=25+ 9.53^-7 +150= | 177.5001 |
Ha=25+ 9.53^-7 +152.5= | 180.0001 |
Ha=25+ 9.53^-7 +157.5= | 182.5001 |
Ha=25+ 9.53^-7 +160= |185.0001 |
Ha=25+ 9.53^-7 +162.5= | 187.5001 |
Ha=25+ 9.53^-7 +165= | 190.0001 |
Ha=25+ 9.53^-7 +167.5= | 192.5001 |
Ha=25+ 9.53^-7 +170= | 195.0001 |
Ha=25+ 9.53^-7 +172.5= | 197.5001 |...