Proyecto Problema Personal
Datos
SAE 10
µ = 4.1 cst
Densidad= .91
Φ= 2 in cedula 40
A= πd2 /4= π(.172 Ft)2 /4 A= .023 ft.
Di= 2.067 in/12 in = 1 ft= Di= .172 ft.
Q= 8.88 ft cubicos/seg
V= Q/A V= 8.88/.023 ft V= 386.08 ft/seg.
F= .25/ [Log 7.8567 x10 -5 + .0007] 2
F= .25/16.84 F= .0148
Accesorios
5 codos de 90 grados
2 codos de 45 grados
2 niples
1 T
Tuberias distancias= 15.82 m = 51.9 ft.
V2/2g = 386.08/2(32.2 ft/ 2seg ) = 2314.56 ft.
Perdidas
P codos 90 = (2314.56) (.0148)(32)(5)= 5480.8 ft.
P codos 45= (2314.56) ( .0148)(15)(2) = 1027.66 ft.
P nicles = 0 no existen perdidas
P T= (2314.56)(.0148)(20) = 685.11 ft.
P ft = 2314.56(.0148) 51.9 ft / .172 ft = 10336.4 ft
N de Re= VDint/Viscosidad
N de Re= (386.08 ft/seg)(.172 ft)/.0036 ft2 = 20772.8
P2-P1/Densidad g/gc + (h2-h1) + V22 -V12 /2g + Pf+wf= 0
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