PROBLEMA RESUELTO ECUACION DE DARCY-WEISBACH

TIPO 1

SOLUCIN

Bernoulli entre 1 y 2

hlhfg

VPzH

g

VPz B

22

2

222

2

111

Despejando:

entradacodosSTvalvulacodoRLfiltroTDTSB hlhlhlhlhlhfhfzH 22 Donde:

g

V

D

Lfhf TSTS

2

2

; g

V

D

Lfhf TDTD

2

2

g

Vhl TSfiltro

210

2

g

Vfhl TScodoRL

2)20(

2

g

Vfhl TDvlvula

2)35(

2

g

Vfhl TDcodoST

2)30(

2

g

Vhl TDsalida

21

2

Hallando trminos:

sm

smVTS /863.0

1016.04

/007.0

2

3

msm

sm

g

VTS 038.0/81.9*2

)/863.0(

2 2

22

turbulentosmx

msmRe ...87681

/101

)1016.0)(/863.0(26

De igual forma:

turbulentoRmg

VsmV e

TDTD ...116967;12.0

2/535.1

2

Hallando rugosidades relativas

67 1095.21016.0

103 xm

mxDS

67

1094.30762.0

103 xm

mxDD

Encontrando los factores de friccin con Diagrama de Moody o con la ecc. de Coolebrok-White

Para tubera de succion

61095.2

0185.087681

xD

fR

S

e

0185.0

87681

74.5

7.3

1016.103log

25.02

9.0

7

x

f

y para la tubera de descarga

0175.0f Hallando las prdidas:

hfTS = 0.0185(6m/0.1016m)(0.038m) = 0.042 m

hfTD = 0.0175(40m/0.0762m)(0.12m) = 1.102m

hlfiltro = 10(0.038m) = 0.38m

hlcodoRL = 0.0185(20)(0.038m) = 0.014m

hlvalvula = 0.0175(35)(0.12m) = 0.074m

hlcodoST = 0.0175(30)(0.12m) = 0.063m

hl salida= 1*(0.12m) = 0.12m

Sustituyendo:

entradacodosSTvalvulacodoRLfiltroTDTSB hlhlhlhlhlhfhfzH 22

)12.0063.0*2074.0014.038.0()102.1042.0(6 BH

HB= 7.86 m

HP

Hps

mN

ms

mm

NQH

POT B 724.0

1745

86.73007.09810

745

3