deber ejer cengel
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Deber Ejer CengelTRANSCRIPT
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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
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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
ηbomba= ρ∗g∗V ∗ H
bhp
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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
ηbomba= ρ∗g∗V ∗ H
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
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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
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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
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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
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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
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/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
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