Deber Ejer Cengel

7/21/2019 Deber Ejer Cengel

http://slidepdf.com/reader/full/deber-ejer-cengel 1/10

ESCUELA POLITECNICA NACIONAL FACULTAD DE INGENIERÍA MECÁNICA

TURBOMAQUINAS

Nombre: José Luis ErazoCurso: Gr-2Fech: 14/11/2015

• Resolver los siguientes ejercicios:

Datos:gua 20!"

ρ=998 kg

m3

a# ηbomba=?

$# CaudalQ=?

Desarrollo:

! ηbomba= ρ∗g∗V ∗ H

bhp



6< ¿min

∗46,2

m∗min

60 s ∗m

3

1000<¿

142W ∗kg∗m

2

W ∗s3

998 kg

m3∗9.8

m

s2∗¿

ηbomba=¿

ηbomba=31.82

12< ¿min

∗42,5

m∗min60 s ∗m3

1000<¿

153 W ∗kg∗m

2

W ∗s3

∗100

998 kg

m3∗9.8

m

s2∗¿

ηbomba=¿

ηbomba=54.33

ηbomba= ρ∗g∗V ∗ H

bhp

18< ¿min

∗36.2

m∗min

60 s ∗m

3

1000<¿

164 W ∗kg∗m

2

W ∗s3

∗100

998 kg

m3∗9.8

m

s2∗¿

ηbomba=¿

ηbomba=64.76


bhp



24< ¿min

∗26.2

m∗min

60 s ∗m

3

1000<¿

172W ∗kg∗m

2

W ∗s3

∗100

998 kg

m3∗9.8

m

s2∗¿

ηbomba=¿

ηbomba=59.59


bhp

30< ¿min

∗15

m∗min

60 s ∗m

3

1000<¿

174 W ∗kg∗m

2

W ∗s3

∗100

998 kg

m3∗9.8

m

s2∗¿

ηbomba=¿

ηbomba=42.16

b!

Desarrollo:

V ( Lpm) %&'# (%)&*# +

0 4,5 1.. 0 42 142 .1212 425 15. 54..1 .2 14 4, $e 24 22 1,2 55.0 15 1,4 421. 0 1,4 0



Patm=14.696 psi=2116.2

lf

ft 2 ,

μ=6.002 x10−4[

lbm

ft ∗s]¿

ρ=62,24 lbm

ft 3

Pv=66.19 lf

ft 2

P1

ρg+

V 12

2 g + z1=

P2

ρg+

V 22

2 g + z2+ht

P1= Patm

V 1=0

Patm

ρg

+ z1= P2

ρg

+V 2

2

2g

+ z2+ht

P!H =( P

ρg+

V 2

2g )− P v

ρg

P!H = Patm− Pv

ρg +( z1− z2 )−h¿

hl=( f L

"+∑ #L) V

2

2 g





P!H = (2116.2−66.19 )lf / ft

2

(62,24 lbm/ ft 3∗32.174 ft /s

2)∗(32.174 lbmft

s2

lf )+ (20 ft )−17.79

P!H =35.1 ft

P!H ()*=1 ft +(0.0054 ft

gpm2) V

2

P!H ()*=1 ft +(0.0054 ft

gpm2)(402)

P!H ()*=9,64 ft po( lo tantola bomba no+avita

Desarrollo

Patm=1o1.3 [ #Pa ]=¿

- =25.C *u) μ=8.81 x10−4[

#g

m∗s]¿

ρ=997 #g

m3

Pv=3.169 #Pa

P1

ρg+

V 12

2 g + z1=

P2

ρg+

V 22

2 g + z2+h¿

P1= Patm

V 1'0



Patm

ρg + z1=

P2

ρg+

V 22

2g + z2+h¿

P!H =

( P

ρg

+V

2

2g

)❑

− Pv

ρg

P!H = Patm− Pv

ρg +( z1− z2 )−h¿

h¿=( f L

"+∑ #L) V

2

2g

V = V %= 4 V

$ "2= 4 (40 lpm)

$ ((0.024m)2)∗(

1m3

1000 l )(1min60 s )

V =1.474 [m /s]

∑ #L=0.85+0.3=1.15

ℜ=V"

u =

1.474m /s∗0.024 m

(8,81 x10−4 /997)

=40033.9069, &

"=0

1

f 1 /2 '−1.8log

2.51

ℜ f 1 /2+(

&

"

3.7 )1.11

f '0.022

h¿=( f L

"+∑ #L) V

2

2g=(0,022 2.8

0.024+1.15) 1.147

2

2(9.8)=0.41m

h¿=0.41m

P!H = Patm− Pv

ρg +( z1− z2 )−h¿

P!H =(101.3−3.169 ) /m2

(997 #g

m3∗9.8m /s

2)∗( #gm

s2 )−(2,2m )−0.41m

P!H =7.42m



P!H ()*u)=2.2m+(0.0013 mlpm

2 ) V 2

P!H ()*u)=2.2+(0.0013 ft

gpm2)(402)

P!H ()*u)=4.28m po(lo tanto nu)st(a bombano +avita

Desarrollo

V+)((ado=4Vlobulo

V+)((ado=4∗0.145 gal

V+)((ado=0.58gal

V+)((ado=4Vlobulo

V =n∗V+)((ado

n

V =300(pm∗0.58 gal

1()v

V =174 gal

min



/omba d)l p(obl)ma14.91

V a=18 lpm

Ha=16m)n)l/0P

na=1200(pm

La mitad d) las (pm:nb=600(pm

V b=V a 1b

1a ( "b

"a )3

V b=18 lpm 600

1200 (1)3

V b=9[ lpm]

Hb= Ha( 1b

1a)2

( "b

"a )2

Hb=1.6m( 6001200 )2

(1 )2

Hb=0.4 [m]

sp= 2 V

1/2

( gH )3/4=

(600 (ad

min )( 2 $

60∗(ad /s

(pm )(9 l

min

1m3

1000

1min

60 s )1/2

(( 9,8m2

s2 )(0.4 m))3/4

sp=0,276



Documents

Deber Ejer Cengel