• Author / Uploaded
• Yuliza Lucio Roque

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Pb (hp)=

ρ . g .Q . hb   746 e

Pb: potencia teórica de la bomba (hp) Ρ: es la densidad del fluido (kg/m3) g: la gravedad (9.81 m/s2) Q: es el caudal (m3/s)

Hb sale de la ecuación de bernoulli

Hb: Cabeza de la bomba (m) e: Eficiencia de la bomba 745.7w = 1hp 3

v 12 p1 v 2 p3 h1 + + +hᵇ =h₃+ + + hƒ ₁͢͢ ₃ 2g γ 2g γ

m

3 139 L m3 min =2,32∗10¯ s min 1000 L 60 s

3

v2 hb =h ₃+ +hƒ ₁ ͢͢ ₃ 2g

43,68 mm

h ₃=4 m+25 m=29 m

v 2=

3

h =29m 3

v=

Q Q 4Q = = A πD ² πD ² 4

( )

V 21 L hf = + F + ∑kc +kv 2g D

[

ℜ=

]

E D

]}

VDp μ

1,548

m kg ( 0,04358 m ) 998 3 s m

1,005 x 10 ¯ 3 pa∗s

ℜ=67155,2 ℜ=¿ 4000 es turbulento

VDp μ

F

1,325 c 5,74 −ln + 0 ² 3,7 D R e ’ ⁹

{ [

ℜ=

m s

P=0,04368 m

L V 21 V 22 V 22 hf =F +∑ kc + kv D2g 2g 2g

R e=

π D2 m =1,548 4 s m L=LD + L s=38

V =1,548

LD LS

F=

m =0,0 4368 m 1000 mm

V =1,,548

Rugosidad relativa

m s

D=0,04368 m

E 0,0015 mm = =3,43∗10 ¯ 5 D 43,68 mm F=

F=

hf =F

1,325 c 5,74 −ln + 0 ² 3,7 D R e ’ ⁹

{ [

hf =

]}

1,325

{ [ −ln

5

3,43∗10 ¯ 5,74 + 3,7 ( 67155,2 )0 ’ 9

2

]}

L∗V 21 V2 V2 +∑kc 2 + kv 2 D∗2 g 2g 2g

V 21 L + F + ∑ kc +kv 2g D

[

1,548 m s hƒ ₁͢͢ ₃= 2g

(

]

2

)

[

0,0196

38 m + ( 0,9∗2 )+ 10 0,04368 m

hƒ ₁ ͢͢ ₃=3,53 m

F=0,0196 3

v2 hb =h ₃+ +hƒ ₁ ͢͢ ₃ 2g

Pb (hp)=

ρ . g .Q . hb   746 e

( 2,32∗10 ¯ 3 ) ( 998 )( 9,8 )( 32,65 )

1.5482 h b=29 m+ +3,53 m 2g

Pb ( hp )=

hb=32,65

Pb ( hp )=1.32 hp

Q=2,32∗10¯ ³ p=998

kg m3

m3 s

e=0,75 hb=32,65

746 ( 0,75 )

]