PROYECTO

Gasto: 5,000 gal/min; presión de descarga: 110 psi; tubería de cobre; reemplazar la bomba calculada por 2 equivalentes operando en paralelo; fluido a bombear: agua; utilizar TODAS las variables de la ecuación de Bernoulli en las 3 dimensiones; calcular la eficiencia de la bomba y el diámetro más económico para la tubería y utilizar al menos 10 accesorios.



Q = 5000 gal / min = .315453 m³ / s = 11.1402 ft³ / s

Tuberia hierro fundido
Rugosidad relativa = .00085

ρ=1.94 slug/ft³
1 slug = 32.2 lbm

1.94 slug/ft³ ( 32.2 lbm/(1 slug ) )= 62.468lbm/ft³

μ=( 2x〖10〗^5 lbfs/(ft^3 ) )(( 32.2 lbmft/s^2 )/(1 lbf) )= .000644lb/(ft.s)

Ø nom = 4 in
Ced. 40
Dint = 4.026 in = .3355 ft

A=(πD^2)/4=(π(.3355 ft)^2)/4= .088405 ft^2

Q=VA

V=Q/A=(11.1402 (ft^3)/s)/(0.088405 ft^2 )=126.012ft/s

P2=(100lb/(in^2 ))((144in^2)/(1 ft^2 ))=14400lb/(ft^2 )

No.Re=DintρV/μ

No.Re=(.3355ft)(62.468 lbm/(ft^3 ))(126.012 ft/s)/(0.000644 lb/fts)=4.10087 x 〖10〗^6
Por las tablas:

f= .0235

(P2-P1)/(g/gc)+ (H2-H1)+ (V2^2-V1^2 )+ Pf+Pw=0

P1 = 0 , V1 = 0

L/D tub=(258 ft)/(0.3355 ft)=253.235

Pf=Pf et+Pf tub+Pf acc+Pfvg+Pfst

Pf=(V^2/2g Ket)+(Ket+f(L/Dtub)+f(L/D)+ Kvgb+Kst

Ket = 1 Kst = 1
Codos 90˚ = 32 x 8 = 256
Codos 45˚ = 15 x 1 = 15
Kvg = 340 ft
340 (.007) = 2.38

Pf=(((126.02 ft/2))/(2 x 32.2 ft/s))( 1+(253.353 x .0235)+ (271 x .0235)+ 2.38+1)

Pf=(246.6)(16.3498)

Pf=4031.86 ft

Wf=((14400 lb/(ft^2 )))/(62.48 lbm/(ft^3 ) x lbf/lbm) +2.33+ (216.02 ft/s)^2/(2 x 32.2 ft/s^2 )+ 4031.86 ft

Wf=230.518+2.33+724.606+4031.86

Wf=4983.31 ft


POT=Wf/(n/100) (wm/(g/gc))=4983.31ft/(90/100) ( ( 62.468 lbm/(ft^3 ) (11.1402 (ft^3)/s))/(lbf/lbm) )

POT=3853240lbfft/s=7005.89 HP

Diámetro Económico
D=1.456 (ne/f)^0.154 Q^0.46

f=35/kg

e=0.786

n=16 hrs

D=1.456((16/24)(0.786)/35) (.315453 m^3/s)^0.46

D=0.012821m= .5047 in

Reemplazar por dos bombas en serie


Pf=4031.86/2=2015.93 ft

POT=2973.38/(90/100) (( 62.468 x 11.1402)/(lbf/lbm))

POT=22991100lbft/s

POT=4180.19 HP En cada una de sus bombas