UNIVERSIDAD NACIONAL DE SAN CRISTÓBAL DE HUAMANGA FACULTA FA CULTAD D DE CIENCIAS CI ENCIAS AGRARIAS ESCUELA DE FORMACIÓN PROFESIONA PROFESIONAL L DE INGENIERÍA AGRÍCOLA
RESOLUCION RESOLUCIO N DE PROBLEMAS PA PARES SOBRE FLUJO EN CANALES CANALES ABIERTOS ABIERTOS Y MEDIDAS EN FLUJO DE FLUIDOS
CURSO
: HIDRAULICA
ALUMNOS
:PALOMINO HUAMANI EDISON
PROFESO R : ING. MSc. JORGE EDMUNDO PASTOR PASTOR WAT WATANABE
SOLUCION: H =
∑ h + ∑ H f
= f i
h f i
ε
loc
Li * V i
+
V s
2
= 0.075mm.
ε
= 0.00005 D numero de reynolds
2 g
2
Di * 2 g
Convirtiendo la ecuación de darcy En función de diámetro y caudal
4*Q π * v * D
Re
=
Re
= 3.13 *10 7
del
→ v = 1.142 *10 −6 m 2 / seg (T = 15º C )
diagrama
de
f = 0.0108 EFECTIVAE NTE D
40 = H = h f
= 0.0827 *
Q
2
f * L * ( D D 4
= 1.60m.↵
+ (∑ K ) + 1)
= 45m 3 / seg Krejilla Krejilla = 0.336 sea. f = 0.02 Q
40 = H = 0.0827 *
45 2 D 4
*(
0.02 * 35 D
+ (0.336 + 0.001 + 0.008) + 1)
= 1.65m. ε = 0.075mm.
D ε
= 0.00005 D numero de reynolds 4*Q
Re
=
Re
= 3.04 *10 7
del
π * v * D
→ v = 1.142 *10 −6 m 2 / seg (T = 15º C )
diagrama
de
moody :
f = 0.0108
CALCULO DE D CON EL NUEVO f 2
40
= H = 0.0827 * 454
D
= 1.60m.
D
*(
0.0108 * 35 D
+ (0.336 + 0.001 + 0.008) + 1)
moody :
SOLUCION: H =
∑ h + ∑ H f
= f i
h f i
ε
loc
Li * V i
+
V s
2
= 0.075mm.
ε
= 0.00005 D numero de reynolds
2 g
2
Di * 2 g
Convirtiendo la ecuación de darcy En función de diámetro y caudal
4*Q π * v * D
Re
=
Re
= 3.13 *10 7
del
→ v = 1.142 *10 −6 m 2 / seg (T = 15º C )
diagrama
de
f = 0.0108 EFECTIVAE NTE D
40 = H = h f
= 0.0827 *
Q
2
f * L * ( D D 4
= 1.60m.↵
+ (∑ K ) + 1)
= 45m 3 / seg Krejilla Krejilla = 0.336 sea. f = 0.02 Q
40 = H = 0.0827 *
45 2 D 4
*(
0.02 * 35 D
+ (0.336 + 0.001 + 0.008) + 1)
= 1.65m. ε = 0.075mm.
D ε
= 0.00005 D numero de reynolds 4*Q
Re
=
Re
= 3.04 *10 7
del
π * v * D
→ v = 1.142 *10 −6 m 2 / seg (T = 15º C )
diagrama
de
moody :
f = 0.0108
CALCULO DE D CON EL NUEVO f 2
40
= H = 0.0827 * 454
D
= 1.60m.
D
*(
0.0108 * 35 D
+ (0.336 + 0.001 + 0.008) + 1)
moody :
SOLUCION: tomando como referencia la plantilla del canal canal de meno denivel! denivel! Ecuación de "ernoulli:
∆ % = ∑ h f + ∑ H loc se
omi#en
iguales
en
ecuacion h f i
= f i
las
c arg as
am$os
de
Li * V i
de
!resion
y
velocidad
!un#os.
darcy : 2
calculo
Di * 2 g
hloc I
= 4m 3 / seg Q = Ai * V i Q
V 1
=
0.7854 * D1
2
en
= 2.26m / seg
la
#u$eria
L#o#al = 104.21m ε ε
= 0.00005 D numero de reynolds V * D
del
diagrama
→ v = 1.142 * 10 − 6 m 2 / seg (T = 15º C )
v Re = 2.97 * 10 6
de
moody :
f = 0.0116 Lfriccion
= 94m.
reem!la"an do h f 1 h f 1 h f 1
= f 1
L1 * V 1
!erdidas !erdida s
locales
2
2 g
*ALIDA → K ¨= 0.5 →
cam$io
de
direccion
= 0.49 K 2 = 0.55 K 1
2
&'()*IDAD − &ELATIVA = 0.075mm. →
Re =
de
= K V i
en#rada
Q
velocidad
!or ser
2
D1 * 2 g
94 * 2.26 2 = 0.0116 1.5 * 2 * 9.8 = 0.1894 m.
hloc I
= (0.5 + 0.55 + 0.49 + 0.5) 2.26 = 0.5316 m.
reem!la"an do
valores
ecuacion
la
de
en
19 .6 la
energia
∆ % = 0.1894 + 0.5316 = 0.721m.↵↵ &TA
Solucion: la ecuación de ener#$a y ecuación de darcy on: H =
∑ h + ∑ H f
= f i
h f i
loc
Li * V i
+
V s
2
2 g
2
Di * 2 g
calculo
= 0.03m 3 / seg Q = Ai * V i
V 2
Q
=
0.7854 * D1
=
2
Q 0.7854 * D 2
2
hloc I
= 1.53m / seg = 0.78m / seg
reem!la"an do h f 1 h f 1 h f 1
= f 1
L1 * V 1
de
!erdidas
locales
2
Q
V 1
Determinación e !"# c"e$iciente# %. !a &e!"cia en !a $"rm'!a e !a eria# !"ca!e# e# ag'a# aa" (c'an" n" #e ini+'e !" c"ntrari")
2
15 * 1.53 2 = 0.048 0.05 * 2 * 9.8 = 0.206 m. igual manera
h f 2
= 0.643m.
en#rada hloc I
→ K ¨= 0.5
= 0.0072 m.
am!liacion
en
c
2
D = 0.16 K = 2 − 1 D1 hloc = 0.005 m. I
D1 * 2 g
de
= K V i 2 g
reem!la"ando
valores
ecuacion
la
de
en
la
energia
H = ( 0 .206 + 0.643 ) + ( 0.007 + 0.005) + 0.031 = 0 .9 m.↵↵ &TA
#enemos
%&' do depóito ! cuya diferencia de nivele permanece contante e i#ual a %(m! etán comunicado por un conducto recto y )ori*ontal ! contituido por do tramo: el primero de &(m de lon#itud y %((mde diámetro+ y el e#undo de ,(m de lon#itud y ,(mm de diámetro ' a la mitad del e#undo tramo e intercala un diafra#ma de -(mm de a"ertura ' lo conducto on de acero oldado ! nuevo' Determinar el #ato .ue paa de un recipiente a otro ! a$ como la l$nea pie*ometrica ! teniendo en cuenta toda la perdida
SOLUSION: E/('0(mm D/%(cm
1%/('(%23
14/('(%23
56C7 8A9A EL DIA19A;A A-( /('(((<(< A ,(/ ('((%23 A%((/('((<0,
6C/
=-('0-
>
6C/%<'(%4 ?ALLANDO 6C
6C/
8A9A EL ES@9EC?A;IEN@O
=('&4
>
6C/('-20 ?ALLANDO CAUDAL ;EDIAN@E LA ECUACION DE ENE9IA
?/5(',=5V%(( 7474# >
>
>
>
>
%23'4/%%(%0043'44=B 4 B/('((4m- B/&'44l ENE9IA 8E9DIDA
('%,
%3' Un depóito! cuyo nivel permanece contante! alimenta al conducto de fierro fundido! motrando en la fi#ura' En C )ay un c)iflón cónico 5Cd/('2&<7 con una alida de ,( mm'
?allemo 1 en la tu"er$a' Lon#itud 5A7 / 4lo#
F
/ 4lo#
5NI6U9ADSE7'
E/('-
F
1% / ('(4-0 Lon#itud 5C7 / 4lo#
F
14 / ('(40 ?ALLE;OS 6% para
/ 02'03G
6% / (',>('-5(',02'03G>('4 6% / (',(% 64 / ('-4 6- / ('%%, E9NOULLI EN@9E A H ( J % / &/
1%= >6IF>
1%= >6IF>
14= >64F>
14= >64F>
6->%F
6->%FK
B% / B4 /B&/
5 1%= >6I7>
5 14= >647>
56->%7F
9eempla*ando valore' &/ &/
5 ,%',(%7>
B / ('((-4&3
>
>
5 %<,'-47> F
5%'%%,7F
Entonce la prdida por fricción on: 8rimero )allemo v%! v4! v-' /
/
/ ('4%0
/
/
/ ('3&0,
/
/
/ %'3-
8erdida por friccion )
/
=
=
/ ('%%,,m
)
/
=
=
/ -'<%m
)allemo la perdida en acceorio y curva' )
/
=
/ ('((%%-m
)
/
=
/ ('((30(m
)
/
=
/ ('(%,,m
/
/ ('%-,&m
/ -'2&&
&m'
%0' a7 El tan.ue de a#ua motrado en la fi#ura alimenta al conducto A de %(( mm de diámetro y decar#a al am"iente por un orificio! de pared del#ada! de ,( mm de diámetro 5vane coeficiente en la fi#'3'4,7' a7 Determinar el #ato en el conducto' "7 Se deea colocar en C una to"era para la medición del #ato cuyo diámetro en la alida ea de ,( mm' Dic)a to"era eta perfilada de manera .ue la ección contracta coincide con la ección de alida' para compenar la reitencia uplementaria! de"ida a eta to"era! ea decidido o"reelevar el tan.ue de a#ua en la dirección de la tu"er$a vertical! in modificar la altura )' Calcular la o"reelevacion del tan.ue! neceario para conervar el #ato ori#inal' El factor de fricción en el conducto e f/('(4 y la perdida local en la curva e ('4 V44#'
DA@OS ?/%4m Cd/('3&0 #/2'0%m4 Solución a7 B/CdMAM B/('3&0M
M
B/('(%2, "7 ?allemo la perdida por la tu"er$a ?/1M M ?%/('(4M
M
?%/('3432m ?4/('(4M M ?4/4('%(m La perdida en la "o.uilla e ?-/
M
?-/ ?-/3'2,m
M
cv/('3&0
?&/('4 ?4/('4M ?&/('(34<2m la car#a en la alida al uperficie li"re
?,/ ?,/,'(4,m LA AL@U9A @O@AL ES ?@/ ?%> ?4 >?-> ?&> ?, ?@/-4'0m la tu"er$a .ue e aumenta e ?/-4'00 / 4&'0m
4(' La tu"er$a maetra .ue aparece en la fi#ura tiene una lon#itud de %((( m! un diámetro de 4(( mm! y un factor de fricción f/('(4,' Con eparacione de ,( m )ay alida laterale .ue derivan un #ato ./4 lte#'
a7 Determinar en denivel ) .ue de"e tenere para .ue el #ato Bt a la alida de la tu"er$a! ea de &( lte#' "7 Determinar ) i e deea .ue Bt aumenta a 0( lte#' c7 Determinar ) i e mantiene Bt/&( lte# y e aumenta . a & lte#'
1/('(4, /(,(m Bu%/4l Bu4/&l ,) V%/
> BS
V%/
>
V%/4',&3m ?1/ M M 5%> 7 ?1/ M
M
5%>
7
?1/ ?1/ ?
"7 V%/
?/&-'--m
> BS
V%/
>
V%/,'(24m ?1/ M M 5%> 7 ?1/ M
M
5%>
7
?1/,3'&& ?1/ ? c7 V%/
?/%32'-4m > BS
V%/
>
V%/4'(%m ?1/ M M 5%> 7 ?1/ M
M
5%>
7
?1/0'<2&m ?1/ ?
?/43'-0m
44'8ara la tu"eria motrada en la fi#ura! e pide: a7' cuando L/ %(6m! tu"o de fierro fundido nuevo5?/4(m! Diametro/('&(m7! cualcular el #ato' "7' 8ara L/%(6m: tu"o de acero oldaod nuevo5?/4(m! B/%((l7+ calcular D 'c7' 8ara L/,6m: tu"o de a"etocemento 5diámetro (',(m! B/4((l7! calcular ? d7' i para el tu"o de fierro fundido! L/%(((m+ ?/4(m' B/4(l y el diámetro de la "o.uilla del c)iflon'
SOLUCION a7 L = 10+m L = 10000m D = 0.40m
→
L D
=
10000 0.40
= 25000
-n 'n t'" !arg" #e e#recia !a# eria# !"ca!e# 〉 5000 '" ierr" $'ni" n'e&" 35
@omando ernoulli A y :
⇒ H =
V ,2 2 g
+ ∑ hf A − ,
f =
2 g
( 8.85 !"g D + N ) 2
f =
2 * 9.81
( 8.85 !"g 0.40 + 35 ) 2
= 0.0198
⇒ H =
⇒ H =
⇒ H =
V ,2 2 g
+ f * L
V ,2
D 2 g
V ,2
1 + f * L 2 g D 0.0198 * 10000 1 + 2 g 0.40
V ,2
= V * A
Q
El #ato e :
⇒ V , = 0.8893m / seg
= V * Π d = 0.7854 * V * d 2 = 0.7854 * 0.8893 * 0.40 2 2
Q
4
Q
= 0.1118 m 3 / seg = 111.8l# / seg
b) ?/4(m
-
L/%((((m B/ ('%m e#' @u"o de acero oldado nuevo N / -& 6o*eny @omando ernoulli A y :
⇒ H =
V ,2
2
V + f * L , 2 g D 2 g
L V ,2 2 g * * ⇒ H = + 2 g ( 8.86 !"g D + N ) 2 D 2 g V ,2
@omando ernoulli A y :
⇒ H = ⇒ H =
⇒ H =
V ,2 2 g
+ ∑ hf A − ,
V ,2
2
V + f * L , 2 g D 2 g
V ,2
1 + f * L 2 g D
@omando ernoulli A y :
⇒ H = ⇒ H =
⇒ H =
⇒ H =
V ,2 2 g V ,2 2 g
+ ∑ hf A − ,
+ f * L
V ,2
D 2 g
V ,2
1 + f * L 2 g D 0.0827Q 2 D
4
+ 0.0827
LQ 2 D
5
*
1 D ( 8.86 !"g D + N )
2
RTA
D/
0.0827Q 2
⇒ H =
D 4
⇒ 20 =
0.0827( 0.1) D
2
4
⇒ 24183.7969 = D
1 1 + 2 + ( ) D 8 . 86 !"g D N
1 1 + 2 ( ) + D 8 . 86 !"g D 34
1 + 1 D 4 D ( 8.86 !"g D + 34 ) 2 1
= 0.528m RTA.
C.- @u"o a"eto cemento E= 0.025mm
L=5000m
Q=0.2m 3/ seg.
D=0.50m
H=?
⇒ E = 0.00005 D
V =
4Q
π D
4
⇒ V =
0.2 * 4
π * 0.5
= 1.0185m / seg
2
,, 3 D- -
⇒ H =
⇒ H =
V ,2
2
V + f * L , 2 g D 2 g
1 + 0.0198 * 5000 2 g 0.50
V ,2
H = 10.53m+
) tu"o de )ierro fundido'
L/%(((m
?/4(m
RTA. 4
B/('(4m e#'
D/ %"en
Depreciando la perdida locale:
⇒ H =
V ,2
2
L V , + f * 2 g D 2 g
f =
2 g
( 8.85 !"g D + N )
Q 2 L ⇒ H = 0.0827 4 1 + f * 5 * Q 2 D D , 256 2 g * L ⇒ 604594 .9214 = 4 + 2 D ( ) + 8 . 86 !"g D 35 , 9eempla*ando y evaluando D = 0.1434m D = 143.4mm
D/ %,( mm e eli#e y e re#ula con una llave'
2
D$o-uilla
=
1 4
D
4-' En la o"ra de toma! motrada en la fi#ura! el tu"o e de acero in cotura nuevo+ u diámetro %'&(m y la lon#itude: LA/4(((m+ LC / 2m' Determinar el #ato .ue tranporta y la preión en ' Si dic)a preión no e tolera"le! indicar .ue medida de"en tomare para ae#urar el #ato calculado! in coniderar la prdida menore'
Calculando la velocidad por la teorema de torrecelli' V /
/
V / 3'%44
e#'
Aplicamo la ecuación de ernoulli a$:
)f / <3'&,22
/ -',-
V / %'-4&,me#' A /
/
/ %',-03
B / A=V / %'-4&,=%',-03 B / 2'&4
e#'
e#'
1inalmente calculando la preión en el punto D e: 8 /
/
8 / %'<-%P#c 4,' Determinar el #ato .ue tranporta cada una de la tu"er$a! del itema motrado en la fi#ura! a$ como la prdida total de A a ' La lon#itude y diámetro on:
L% / L, / <,( m+ L4 / L& / ,(( m+ L- / -(( m+ D% / D, / (',( m+ D4 / D& / ('&( m+ D- / ('3( m'
Solución: B / %!,((lite#' / %',
e#'
Aplicando la Ecuación de Continuidad' B / B% > B4 > B- > B& > B, / %',
e#' QQQQQQQQQQQQ 5%7
8ero a"emo .ue B% / B,! B4 / B& + 9eempla*ando valore en Ec' 5%7 Se tiene' B / B% > B4 > B- > B& > B, / %', e# B / 4B% > 4B4 > B- / %',
e#QQQQQQQQQQQQQQQ'' 547
8or etar en paralelo la prdida de car#a on i#uale' )f% / )f4 / )f-/ )f& / )f, / ?t'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''' 5-7 Aumimo:
/ ('(4,! f4 / ('(44! f- / ('(-( eto dato e o"tienen de la
@a"la por tanteo' )f% /
/
/ <3'&,22
/ )f,''''''''''''''''''''''''''''' 5&7
)f4 /
/
/ 3-'<%(
/ )f&''''''''''''''''''''''''''''' 5,7
)f- /
/
/ 4,'&0&
/ )f&''''''''''''''''''''''''''''' 537
B% /
QQQQQQQQQQQQQQQQQQQ' 5<7
B4 /
QQQQQQQQQQQQQQQQQQQ' 507
B- /
QQQQQQQQQQQQQQQQQQQ' 527
9eempla*ando lo Ec' < ! 0 y 2 en la Ec' 4' B / 4B% > 4B4 > B- / %', e#
B/4
>4
>
/ %',
e#
)f / -'-,-m' Calculando la velocidade:
)f / <3'&,22
/ -',-
V% / %'-4&,me#' A% /
/
/ ('%23-,
B% / A%=V% / %'-4&,=('%23-, B% / B, / ('43(
)f / 3-'<%(
e#'
e#' / -',-
V4 / %',&3
/
/ ('%4,33
B% / A%=V% / %',&3<=('%4,33 / ('%2& B4 / B& / ('%2&
e#'
B / 4B% > 4B4 > B- / %',
e#
45('43(7 > 45('%2&7 > B- / %', B- / (',24
e#'
e#
e#
4<' El ifón motrado en la fi#ura tiene la i#uiente #eometr$a: L% / ,( m! L4 / %(( m L- / %,(m! D% / <,mm! D4 / ,(mm! D- / <,mm' A demá f% / ('(4,+ f4 / ('(40 y f- / ('(4,' a7' Determinar la car#a ?! necearia para .ue B- / - LitSe#' "7' Si ) / 4 m y la lon#itud del tramo C D de 4( m! Determinar en .u punto 5C o D7 e preente la m$nima preión+ Calcular la ma#nitud de ta'
Solucion: )f4 /
/
/
)f4 / 3'33(03&m )f% / B% / ('(((%,
/ e#
/
/ 3'33(03&
B% > B4 / B- / %', ('(((%,
e#
e# > ('((-
e# / B- / ('((-,
e#
a7' Calculando La prdida de car#a )f- /
/
/
/ )f- / %'420m
? / )f% > )f- > )c)orro > )válvula ? / 3'3(03& > %'42<<%(- > -'(&%&4,0 ? / %%m "7' 1inalmente calculando la preión en el punto D e: 8D /
/
8D / ('4<-P#c
SOLUCION: H = h f i
20 Q 20
∑ h + ∑ H
= f i
f
Li * V i
loc
+
V s
2
2 g
2
LLEVANDO EN 1UNCION DE DIA;E@9O H CAUDAL
Di * 2 g
= H = h f = 0.0827 *
Q2 D
4
*(
f * L D
+ (∑ K ) + 1)
= 0.03m 3 / seg = H = 0.0827 *
0.03 2 D
4
*(
f * 820 D
+ 4.3)
sea. f = 0.02 D
= 0.22m.
a!"re# e ;. -ntraa: 0.5 <"": 0.9 a!ia:1 e: 0.9 =!&'!a aierta: e#recia!e
ε
= 0.2mm.
ε
= 0.0009 D numero de 4 *Q
Re
=
Re
= 1.74 *10 5
del
π * v * D
→ v = 0.01cm 2 / seg
diagrama
f = 0.02 en#onces D
reynolds
= 0.22m.
de
moody :
!". E# $% &'(# *+,&/, $# %/ ('01/) &$ $&$/ c,#,c$: /) $% 0/&, ,/% 21$ (%13$ $ A / B4 &' L 5 677+8 D5 677++8 Ѵ57.76c+9 &$08 ; 5 7.9++8 H 5< + . b)C1/#, $b$ &$ =4 $ +/#$/ 21$ %/ >$&'# $# C #, &$/ '#($', / - 7." ?0c+9 .
F'01/ $% >,b%$+/ !". S,%1c'#. /) B$#,1%%' $#$ A 3 B.
!A γ
+
VA2 2 g
+ %A =
!,
+
γ
V, 2 2 g
+ %, + h f −
A ,
l V12 V 22 + ÷ 5 = h f → h f = f D 2 g 2 g l V12 V 2 2 + ÷ ......................I 5 = f D 2 g 2 g P$, /+, 9 3 ! $&/# $# >//%$%, $#,#c$&: &,# '01/%$& $# %,& $& /+,&4 >, %, /#,:
→ f
l V2
2
f
= h f 8 /$+/& L4 D. = f1 = f 2 = f 3 h f2
3
2
=
f
l V2
→ V2 = V3........II
D 2 g D 2 g Q1 = Q2 + Q3 → V1 A = V 2 A + V3 A A$+@& Q2 >, %, /#, V2 = V 3
V1 = V2 + V3 → V1 R$$+>%//#,
= Q3
= 2V2........................ ( α )
( α ) $#
*II)
l V12 V2 2 5 = f + ÷ → 5 = D 2 g 2 g
l V12 V12 + ÷ f D 2 g 8g
l 5V 12 2 5 = f → 5 = f * 63.71( V1 ) ..............IV ÷ D 8 g e 0.2 C,+,: = ÷ = 0.002 D
100
♣ S1>,#'$#, 21$: V 1
& E =
E#,#c$&8
V *D
= 1.5 m seg 150 *10
= = 1.5*105 0.01 f = 0.0243
ν D$% DIAGRAMA A - 6 : R$$+>%//#, $# %/ $c1/c'# *IV)4 &$ '$#$8
♣ S1>,#'$#, 21$: V 1
& E =
= 2.0 m seg
200*10
= 2.0*10 5 0.01 A 6: f = 0.024
E#,#c$&8
< !.
D$% DIAGRAMA R$$+>%//#, $# %/ $c1/c',# *IV)4 &$ '$#$8
5 > 6.116
In#er!olando &$ '$#$8 V 1 = 1.79 m seg P, %, /#, $$+>%//#,
f1 = 0.0242 = f 2
T/+b'#:
Q1 = V1 * A = Q2
V 1 $# ( α ) &$ '$#$: V2
= Q3 =
Q1 2
=
( 1.79 ) * π ( 0.1)
f 3 2 3 m = 0.014
4
=7
seg .
l . seg
b). c/%c1%, $ =. B$#,1%%' $#$ A 3 C
!A γ
2
+
VA
2 g
+ %A =
!C γ
2
+
VC
2 g
+ %C + h f −
l V 12 0 = −6 + + h + f ÷ 2 g D 2 g V12
= V 3 = 0.895
A C
= 14
l . seg
l V 12 6 = h + ( f + 1) ÷ D 2 g l 100 P$,8 f = 0.0242 = 24.2 D
01
V 12 E#,#c$&8 6 = h + (24.2 + 1) ÷ 2 g V 12 6 = h + (24.2 + 1) ÷ 2 g 1.72 6 = h + (25.2) ÷ 2*9.81 h = 3.40m.
!. E# $% &'&$+/ $ 1b,&4 +,&/, $# %/ ('01/4 c/%c1%/ H4 $ +/#$/ 21$ 6 5 69 %&$0. P// %,& &'01'$#$& /,&: L6 5 L! 5 <7 + . L9 5 977 + 8D 5677 ++ 8 ; 57.9 ++ 8 Ѵ 5 7.76c+9 &$0.
S,%1c'#. B$#,1%%' $#$ A 3 C.
!A γ
+
VA2 2 g
+ %A =
!C γ
+
VC 2 2 g
+ %C + h fA−C
A$+@& +
l1 > D1 = l3 > D3 3 >, %, /#,: f1 = f 3 8E#,#c$&8 K1 = K 3
V 21 V 23 → h f = K1 * + K 3 * H = h f + 2 g 2g 2 g V 21 V 23 H = K1 * + ( K 3 + 1) * 2 g 2 g V32
R$$+>%//#,8 ?6 5 ?!4
V 23 V 21 E#,#c$&8 H = K1 * + ( K 1 + 1) * 2 g 2 g
V 21 V 23 + ÷ H = ( 2 K1 + 1) 2 g 2 g
!, γ
P, %, /#,8
.................I
B$#,1%%' $#$ B 3 C.
+
V,
2
2 g
+ %, =
!C γ
2
+
VC
2 g
+ %C + h f,−C
V 23 V 2 2 → h f = K 2 * + K 3 * P, %, /#,8 H = h f + 2 g 2g 2 g V 23 V 2 2 H = K 2 * + ( K 3 + 1) * ................II 2 g 2 g A$+@&: Q3 = Q1 + Q2 → Q3 − Q2 = Q1 V1 A = V3 A − V2 A → V1 = V3 − V 2 V32
C,+,4
Q1 = 12
l
3 m . = 0.012
seg Q1 = E#,#c$&: V3 − V 2 A V3 − V 2
=
seg . 0.012*4 = 1.5279m / seg . 2 π ( 0.1 )
= 1.5279m / seg . ..*III)
I01/%/#, %/& $c1/c',#$& I 3 II4 &$ '$#$:
V 21 V 23 V 23 V 2 2 2 ( K 1 + 1) 2 g + 2 g ÷ = K 2 * 2 g + ( K 3 + 1) * 2 g V32 V 23 V12 V 2 2 + ( K 3 + 1) * ( K 3 + 1) * + K1 * = K 2 * 2 g 2g 2g 2 g
2
K1 *
V1
2 g
K1 = f1 R E = D
%$2
V
= K2 * l 1
2 g
= K 3
D1 V1 * D1
=
2
→ K1V12 = K 2V 2 2 ..*IV)
P, %, /#,4
152.7 *10
ν 0.01 DIAGRAMA A - 6 : f1
K1 = K 3
= 0.0242
100 0.1
=
f1 = f 3
= 1.53*105 f 3 = 0.0242
= 24.2
K 1 4 $# %/ $c1/c'# *IV) (24.2)*(1.5279) = K 2V 2 → 56.4943 = K 2V 2 l 56.4943 = f 2 2 V 2 2 → 0.028 = f 2V 2 2.......................(V ) D
R$$+>%//#,
♣ S1>,#'$#, 21$: V 1
R E =
V2 * D2
ν D$% DIAGRAMA
=
= 1.5 m seg
150 *10
= 1.5*105 0.01 A 6: f 2 = 0.0243
R$$+>%//#, $# %/ $c1/c',# *V)4 &$ '$#$8
♣ S1>,#'$#, 21$: V2
R E =
V2 * D2
=
100 *10
ν 0.01 D$% DIAGRAMA A 6: f 2
= 1.5m / seg.
0.0282 < 0.054675
= 1.0 *105
= 0.0247
0.0282 > 0.247 In#er!olando. &$ '$#$ 21$: V2 = 1.055m / seg. 3 f 2 = 0.0246 C/%c1%, $ %/ V 3 R$$+>%//#, $# %/ $c1/c',# *V)4 &$ '$#$8
= Q1 + Q2 = 0.012 + 0.008286 Q3 = 0.020m3 / seg. Q3
V3
=
Q3 A
= 2.58m / seg.
REEMPLAANDO4
K1 >V1 >V 3 EN LA ECUACIÓN *I).
1.52792 2.582 + H = ( 2 * 24.2 + 1) ÷ 2*9.81 2*9.81 H = 22.64m.
6. E# $% >,b%$+/ .67 $$+'#/ %/ '&'b1c'# $ 0/&, $# %,& 1b,& 4 c1/#, $% c,$('c'$#$ $ >$'/ $# %/ K@%K1%/ &$/ V57. .67. E# $% &'&$+/ +,&/, $# %/ ('01/ '$#$ %/ &'01'$#$ 0$,+$/ 8H59+ 8L65L95L!5L5677+ 8/$+@& 4 ( 65( 95( 57.79< 3 ( !57.798 $% c,$('c'$#$ $ >$'/ $# %/ K/%K1%/ V5!7. C/%c1%/ %,& 0/&,& $# c// 1b,4 $&>$c'/#, %/& >$'/& %,c/%$&. S,%1c'#:
L/ >$'/ $ $#$0/ $#$ B 3 C $& : f L 0.025 × 100 K 2 = K 1 = 2 2 = = 25 D2
K 4
0 .1
= K 1 = 25
En.el .#u$o.#res .>.#enemos K 3
=
f 3 L3 D3
+ K V =
0.02 × 100 0 .2
+ 0 = 10
!or .la.ecuacion.de. !erdida . se.#iene
∆ H =
n
∑ ( Di i =0
8× Q2 2 n π ∑ ( Di / i =0 2
/ Ki )
=
Ki )
0.01 25
..............(α )
+ 0.04 = 0.0126 10
2 n π ∑ ( Di i = 0
= (3.14 × 2.0126) 2 = 39.937....
/ Ki )
Re em!la"ando.en.(α ) 8 × Q4
∆ H =
2
= 0.0204Q 4 2
39.937 × 9.81 Adem/s. D1 = D 4 !or .la.ecuacion.de.con#inuida d . se.o$#iene V 1
2
2
D V = ( 4 )2 4 2 g D1 2 g
=
V 4
2
2 g
>→
V 4
2
2 g
=
Q4
2 2
2 g (π D 4 / 4)
2
=
Q4
2
19.6(0.7854 × 0.01)
= 828Q 4 2
La.ecuacion.de.energia.en#re. A. y. D.nos.da : 24 = + 1
V 4
2
+ 0.0204Q 4 + + 4 2
2 g
V 4
2
2 g
+
V 4
2
2 g
> reem!la"an do.los.valores
24 = ( 2 × 25 × 828 + 0.0204 + 828)Q 4 >. → Q 4 2
= 0.0238m 3 / seg .
La. !erdida.de.c arg a.en#re. ,. y.C
∆ H = 0.0204(0.0238) 2 = 0.0000115 m...(es#a?. !.rdida. sera.igual . !ara.las.dos.ramas> !or .lo. tan #o. se.#iene.
∆ H = K 2 .Q 2
= π ×
V 2
2
2 g
> de.donde.la.Q 2 .es :
(0.1) 2
2 × g × ∆ H
4
K 2
= π ×
(0.1) 2
2 × 9.81× 0.0000115
4
25
= 0.00324m 3 / seg .
De.la.misma.manera. !ara.el .caudal ..Q 3 Q2
= π ×
(0.2) 2
2 × 9.81× 0.0000115
4
10
= 0.0205m 3 / seg .
9. C/%c1%/ %/ >,$#c'/ %/ b,+b/ 21$ '$#$ 1#/ $('c'$#c'/ #5 <4 >// 21$ $% 1b, 9 %%$K$ 1# 0/&, $ < %&$0. L/ 0$,+$'/ $&: L 65< +8 D65 <++ 8 f1 f 2 f 3 7.7!8 L9 5 L! 5 677 +8 D9 5 D! 5 <7 +8 H 5 67 + 3 V 56<.
=
SOLUCIÓN.
=
=
K
=
l f D 100
75
= K 3 = 0.03 = 60 3 K 1 = 0.03 = 30 0.05 0.075 Q1 = Q2 + Q3 → Q1 = 0.05 + Q3....................I
K 2
B$#,1%%' $#$ A 3 B
!A γ
H , H , H ,
2
VA
+
+ %A + H , =
2 g
+5+
V12 2 g
= 10 + K1
= 10 − 5 − = 5 + 29
P$,8
Q2
V12 2 g
V12 2 g
+ 30
+ 60
!, γ
V12 2g V12 2g
V 2 2 2 g
2
2 g
+ K 2 + 60
+ %, + h fA− ,
V 2 2
2 g V 2 2 2 g
.....................II
= V2 * A2 → V2 =
R$$+>%//#,
+
V,
0.005*4 π *(0.05
2
)
= 2.5465m / seg.
V 2 $# I
2
H ,
= 29
V 1
2 g
+ 24.83.......................III
B$#,1%%' $#$ A 3 C
!A
+
VA2
+ %A + H , =
!C
+
VC 2
+ %C + h fA−C
2 g γ 2 g 2 2 2 2 V1 V 1 V3 V 3 = K1 + K3 + K V H , + 5 + 2 g 2g 2g 2 g V12 V 12 V32 V 32 + 30 + 60 + 15 H , = −5 − 2 g 2g 2g 2 g 2 2 V 1 V 3 + 75 ................IV H , = −5 + 29 2 g 2 g γ
I01/%/#, %/ $c1/c'# III 3 IV
29
V12 2 g
+ 24.83 = −5 + 29
V 12 2g
+ 75
V 32 2 g
29.83 = 75
V 3
2
2 g C/%c1%, $ Q3
Q3
→ V3 = 2.79m / seg.
= V3 A3 = 2.79 *
π (0.05
P, %, /#,.
2
)
4
→ Q3 = 0.005478m3 / seg.
Q1 = Q2 + Q3 = 0.005 + 0.005478 → Q1 = 0.010478 m 3 / seg. Q1 = V 1 A1 → V1 = R$$+P%//#,
H ,
= 29
(2.3717) 2
=
π *(0.075
2
)
= 2.3717 m / seg.
V 1 $# %/ $c1/c'# III.
2 g H , = 33.144m. 0o#.CV
0.010478*4
γ QH ,
76*0.85
+ 24.83 → 0o#.CV = 5.38C V
!. C/%c1%/ %/ >$&'# 21$ $b$ %$$&$ $# $% +/#+$, M4 $ +,, 21$ $% #'K$% $ %/ &1>$('c'$ %'b$ $% $c'>'$#$ A &$/ $% +'&+, 21$ $% $% $c'>'$#$ B8 /&'+'&+,4 95<%&$0. U'%'/ %,& &'01'$#$& /,&: L65<+8 D65<++8 L95L!5677+8 D95 D!5<7++8 H567+4 (65( 95( !57.7! 3 K57.6<.
S,%1c'#:
V 2
Q2
=
π
V 2
×
2
4
×
( 0.05)
2.546 m
=
2
2
0.330 m
=
2 g
0or .ecuacion.de.con#inuid Q 1
Q2
=
K 2
K 3
=
Q3
+
=
f
=
×
0.005 L
D f
=
×
+
Q3 ..
0.03
×
1
0.05
L
× L 0.03 × 75 = = 3 D 0.075 Ecuacion.de.$ernoulli .en#r
K 1
f
=
0
V 1
+
2 g
γ
0
2
61
=
=
V 2
γ
V 2
2
2 g
+
2 +
2 g
29
V 1
30 V 1
2
2 g
+
6
2
2 g
..........
Ecuacion.de.$ernoulli .en#r 0
V 1
+
2 g
γ
0 γ
= 61.15
V 3
2
2 g
+ 29
2
V 1
+
10
=
V 3
2
2 g
+
V 1
2 g
2
2 g
− 10.....................( III )
Igualando .ecuacion.( II ). y ( III )> se.#iene 61
V 2
2
2 g
= 61.15
61 × 0.33
V 3
2
2 g
= 61.15
− 10
V 3
2
2 g
− 10 de.donde...V 3 = 1.803 m / seg .
Re em!la"ando .en.( I ) V 1
= 1.933 m / seg .
Finalmen#e .reem!la"am os.los .valores.de.V 1 . yV 2 .en.( II ) 0 γ
0 γ
= 61
V 2
2
2 g
+ 29
V 1
2
2 g
=
( 2.546 ) 2 61 2 g
+ 29
(1.993)
2
2
2 g
= 25 .676 × 1000 = 25676.21+g / m 2 = 2.56 +g / cm 2 ↵↵ →
<. P// $% &'&$+/ $ 1b$/&4 +,&/, $# %/ ('01/ 4 c/%c1%/ %/ >,$#c'/ #$c$&/'/ $ %/ b,+b/ 4*$# CV) c,# %/ $('c'$#c'/ $ ,c=$#/ >, c'$#, 4 >// 21$ B5 < %&$0. C,#&'$$ L5967+ 4 D57.67+ 3 (57.79<.
S,%1c'#8 0or .ecuacion.de.con#inuida d . se.#iene Q A
= QC + Q , = 0.005 + QC como.los .diame#ros . soniguales
A(V A
+ V C ) = 0.005
(V A + V C ) = K =
f × L
V , = V ,
0.005 × 4 × (0.1) 2
π
=
0.025 × 210 0.1
D 0.005 × 4 π
= 0.637m / seg .......... ....( I )
× (0.1)
= 52.5
2
2
2 g
= 0.0207m
Ecuacion.de.$ernoulli..en#re. A. y. , ∆ H = 52.5 ∆ H = 52.5
V A
2
2 g V A
+ (52.5 + 1)
V ,
2
2 g
+ 17.70
2
2 g
+ 18.807.......... .......... .......... ( II )
Ecuacion.de.$ernoulli..en#re. A. y.C ∆ H = 3 + 52.5
V A
2
2 g
+ (52.5 + 1)
V C
2
2 g
.......... ........( III )
Igualamos.ecuacion.( II )1 ( III ) 52.5
V A
2
2 g
+ 18.807 = 3 + 52.5
V A
2
2 g
+ (52.5 + 1)
V C
2
2 g
De.donde V C = 2.407m / seg . Re em!la"ando.en( I ) V A = 3.044m / seg . ∆ H = 52.5
(3.044) 2 g
2
+ 18.807 = 43.601m
0or . tan #o.el .caudal .#o#al .es Q A = 0.005 +
2.407 × π × (0.1)
2
= 0.0239m 3 / seg .
4 γ × Q A × ∆ H 1000 × 0.0239 × 43.601 = 0 (CV ) = n × 75 0.80 × 75 0 (CV ) = 17.37CV ↵↵.......... .......... .......... ....davis
". D$&$ 1# $>,&',4 c13, #'K$% c,#&/#$ &$ +/#'$#$ / %/ $%$K/c'# $ 97 +. >/$ 1# c,#1c, $c, $ 677 + $ %,#0'1. E&$ $&$+b,c/ / %/ +'/ $ 1# c,#1c, =,',#/% >$>$#'c1%/ /% >'+$, c,# $% c1/% &$
1#$ $# (,+/ $ T. E% &$01#, c,#1c, $&$+b,c/4 $# c// $Q$+,4 / 1# /#21$ c13, #'K$% &$+/#'$#$ / %/ $%$K/c'# $ <.77 +. L/ '&/#c'/ $&$ %/ T / c// /#21$ $& $ <7 +. D$$+'#/ %,& '/+$,& $ %,& c,#1c,& &' &$ $&$/ ,b$#$ - $# c// $Q$+, $% &$01#, c,#1c, 1# 0/&, $4 >, %, +$#,&4 9< %&$0. L,& c,#1c,& &, $ ('$, (1#', 3 %,& '/+$,& *c,+$c'/%$&) K/'/# $ 67 $# 67++ .
S,%1c'#. P// /01/ / 1#/ $+>$/1/ $ 6<c. Ѵ56.6967-" P// %/ 1b$'/ 6. S1>,#0/+,& 1# '/+$, $ 697 ++. E#,#c$&:
V 1 =
Q1
0.05*4
→ V1 =
2
= 4.42m / seg.
A1 π *(0.12 ) V1 * D1 4.42 * 0.12 6 R E = 10 = 4.64 *10 5 = 1.142 ν 0.0238 P// ('$, (1#', D$% DIAGRAMA A 6: f 1 = ♣ B$#,1%%' $#$ A 3 D
!A1 γ
2
+
VA
20 = ( f
2 g
l1 D1
+ %A+ =
+ 1)
V12 2 g
+
!D γ
+
!D γ
VD
2
2 g
+ %D + h fA− D
.....................I
P$,8
f
l 1 D1
= 0.0238
100
= 19.83..........II
0.12 R$$+>%//#, II $# I &$ '$#$. 2 (4.42) !D + 20 = (19.83 + 1) γ 2 g !D = 19.75m..............III
γ CORRIGIENDO: EN LA ECUACION I
l1 V 12 l1 V 12 20 = 19.756 + ( f ) → 0.25 = ( f ) D1 2 g D1 2 g V 1 5 7."< m / seg V * D1 0.465* 0.12 6 R E = 1 10 = 5.09 *10 5 = 1.142 ν P// ('$, (1#', D$% DIAGRAMA A 6: → f 2 = 0.0255 Q1 0.05*4 → D 12 = 0.465* → D1 = 0.37m = 370mm. A1 = π V 1
P, c,#'#1'/.
Q1 = Q2 + Q3 >$, Q2 = Q3 → Q1 = 2Q2 V *A Q2 = Q3 = 1 1 → V2 A2 = 0.025m3 / seg. 2 ♣ B$#,1%%' $#$ D 3 B.
!D γ
+
VD 2
+ %D + =
2 g (0.465)2
19.75 +
14.76 = K 2
2 g
V 22 2 g
!, γ
+
= 5 + K 2
V, 2
2 g V 2 2
+ %, + h fD− ,
2 g
→ 14.76 = f
l V 2 2 D
2 g
2
5.7854 = f
V
2
D 2 g
............................IV
P// %/ 1b$'/ 9. S1>,#0/+,& 1# '/+$, $ 697 ++.
V2
=
Q2 * 4 π *(0.12
& E = P//
V *D
=
2
= 2.21m / seg.
) 2.21* 0.12
= 2.23*105 1.142 ν ('$, (1#', D$% DIAGRAMA A 6: → f 2 = 0.0245 106
f 2 EN LA ECUACIÓN IV4 SE TIENE. V2 = 5.32m / seg. R$$+>%//#,
Y CON ESTA NUEVA VELOCIDAD SE CALCULA EL
& E =
V *D ν
=
5.32 * 0.12 1.142
106
= 5.59 *10 5
P// ('$, (1#', D$% DIAGRAMA A 6: E% c1/% $&
%$ & E
→ f 2 = 0.0238
f 2 />,Q'+/, >, $'//c'#.
Y c,# $&$ K/%,4 c/%c1%/+,& $% K/%, +/& />,Q'+/, $ V 2 4 $$+>%/#,
f 2 $# %/ $c1/c'# IV. Q2 * 4 2 → D2 = 0.0767 m. D2 = π * V 2 $% K/%, $
L1$0, %,& '/+$,& c,$&>,#'$#$& $#$+,&: >// c// /+,
D2 = D3 = 80mm. D1 = 370mm.
,4' en la red5 motrada en la fi#ura7 e pide calcular lo diametro teorico de la tu"eria ! de manera .ue :B,/ 4, l! B3/-(l! N/-0 5 Poeny7 y la car#a de preion minima en la decar#a ean de por lo meno %, m de columna de a#ua! lo tu"o on de fierro #alvani*ada'
&('((m -,'((m
%
L/0(( D/ ,
L/,(( D/ -
L/3(( D/ B/4, lt
L/4((( D/ &
L/&(( D/ B/-( lt %-'((m
%('((m 3
&,%1c'#:
,
?/%, m Se pide calcular:
6. C/%c1%/#, Caudal aumido:
9. C/%c1%/#, Caudal aumido: Con eto dato "ucamo en la ta"la el
!. C/%c1%/#, Aumimo
Caudal ucamo en la ta"la el
. C/%c1%/#, Aumimo
Caudal ucamo en la ta"la el
<. C/%c1%/#, Aumimo
R>/
ucamo en la ta"la el
<. en la red motrada e re.uiere calcular B4!B& y B3 en ete cao B/<3', lte#! ?3/%( m! L%/0( m! D%/4(( mm! f%/('(4%! L/%(( mm y f/('(4,
L! D! f L/,( m D/%(( mm f/('(4,
L/,( m D/%(( mm f/('(4, B4 /O
?3 /% ( m
B& /O
?/O 3
&
4
G- 5O" %- : &$0
L/0( m D/4(( mm
L/0( m D/4(( mm f/('(4%
f/('(4%
L/0( m D/4(( mm f/('(4%
S,%1c'#. Calculando
con la formula de darcy y Rei"ac)
Si B total 9eempla*amo:
Calculando C con la formula de ?a*en Rilliam
9eempla*ando
Como:
Caudal upueto Entonce calculamo el la velocidad:
9eempla*amo con la formula darcy y Rei"ac)
entonce
entonce:
Calculamo el caudal con la ta"la de )acenRilliam:
O"tenemo el
entonce:
C/%c1%/#,
C/%c1%/#, H: P, B$#,1%%' :
R>/
R>/ ".) L/ ('01/ +1$&/ $% >,3$c, $% &'&$+/ $ 1b,& >// c,+b/' '#c$#',& $# 1#/ '#&/%/c'# '#1&'/%. E# %,& >1#,& 64 94! 3 &$ $21'$$# '#&/%/ ='/#$& >// /b/&$c$ 0/&,& $ 6<4 !74"7 3 6< %&$0.
R$&>$c'K/+$#$. D$$+'#/ $% 0/&, $# %,& 1b,& $% &'&$+/. *U'%'c$ %/ (,+1%/ $ H/$#-W'%%'/+&4 C H = 95.) c,#&'$/#, 21$ %/ $%$K/c'# $ ,,& %,& #1,& $& 7.77 +. C/%c1%/ %/ /%1/ $ %/& c/0/& $ >$&'#4 $# c// #1,.
Q A −1
= 15l# + 30l# + 60l# + 15l# Q A−1 = 120l# %'7 Suponiendo #ato iniciale en la tu"er$a: Q1− 2 Q 2 −3 Q 3− 4 Q 4 −1
= 53 L# . = 23 L# . = 37 L# . = 25 L# .
Lue#o reempla*ando en la formula de ?a*en Tilliam 5N/%'0,%7 para cada tramo:
hf =
L (0.279.C H D
∆Q = −
O
2.63
)
1.851
Q 1.85
∑ hf hf N ∑ ( ) Q
8rimera interacción: @9A; D L B ) 5m7 5m7 5m-7 f 5m7 B %4 (! 4 ( (
)f
∆
B 5m-7
%-
('
('
-
((
!(-<
(! 4-
4,
((
4,
'(44
((
!(-<
(! &% 4, SU;A @O9IA
(%4342 ( % 3& (' '&<&& '%(&( (%4342 ( 4& (' ('0004 '((3, (%4342 ( %4 (' 3'330 0'4-00 (%4342 44 ,'-<%& 2'<34-
0
(! &-
!<%(2
0 ((
Se#unda interacción: @9A; D O 5m7 5m7 ( %4 !(( ( 4!4, (( ( -& !4, (( ( &% !4, (( SU;A @O9IA @ercera interacción: @9A; D O 5m7 5m7 ( %4 !(( ( 4!4, (( ( -& !4, (( ( &% !4, (( SU;A @O9IA
!(,4
L
B 5m-7 4 ( '(3, 0 ( '(-3 4 ( '(4& 0 ( '(-2
L
) f 5m7
4
)f
% '(-< '0%, ('-22
'3%3
-'2%3& ('&4 ( '(202< 3'0%3
) f 5m7
(
(-,3 ('(4&-< ('(-2&
B 5m-7
('(((4-,< 2 ('(((4-,< %3 ('(((4-,< %( ('(((4-,< 44
)f
(' (3&0 (' (-,<3 (' (4-<3 (' (-0<3
∆
B %
('
%, '2,&
'-<<
(3,3
∆Q
B
B 5m-7
'&%-(
B 5m-7
%,
'(3, '(-< '2,& ('(((4 0 ( 2 '(-3 '-<<& '0%,('(((4 4 %3 ('(4& ('-200 '3%3& ('(((4 0 %( ('(-2 -'2%3& ('&4(('(((4 ( 44 '((24 3'0(3
(' (34'& (' (-%, (' (420 (' (&-0<
De la ltima interacción e o"tiene .ue:
= 0.0624m 3 = 62.4l# . Q2 −3 = 0.0315m 3 = 31.5 L# . Q3− 4 = 0.0298m 3 = 29.8 L# . Q4 −1 = 0.04387 m 3 = 43.9 L# . Q1− 2
9ta'
4'7 reempla*ando en la formula de ?a*en Tilliam dada anteriormente: hf A−1
= 0.397
9etando la altura total con la perdida de ener#$a de cada tramo e o"tiene la altura de la car#a de preión para cada nudo:
= 100 − 0.397 = 99.6m h2 = 99.6 − 1.017 = 98.57 m h3 = 98.57 − 3.377 = 95.14m h4 = 95.14 − 0.3958 = 95.59m h1
69.)
9ta'
E# %/ $ c$// +,&// $# %/ ('01/ &$ >'$ c/%c1%/ $% 0/&, 21$ &$ '$#$ $# c// 1#/ $ %/& 1b$/&4 &' $% 21$ &/%$ $ %/ >$&/ $& Q12 = 90l# / seg . E# c// ,+/ *!4 4<) $% 0/&, $b$ $ &$ $ !7%&$0. A 1#/ >$&'# +'#'+/ $ 67 +. $ c,%1+#/ $ /01/8 %/& 1b$/& &,# $ /c$, #1$K,4 &'# c,&1/&. C/%c1%/ /+b'# %/& $%$K/c',#$& $ %/& c/0/& >'$,+$'c/& $# '&'#,& #1,&.
Q12
= 90l#
De la formula de 6o*eny :
f =
2 g (8.86 !"g D + N ) 2
8ara acerro nuevo in cotura: N/-0! reempla*ando en cada tu"er$a:
f 12
=
f 25
=
2(9.81) (8.86 !"g(0.4) + 38) 2
f 12
= 1.65 × 10 −2
f 25
= 1.70 × 10 −2
2(9.81) (8.86 !"g(0.35) + 38) 2
f 54
=
f 43
=
2(9.81) (8.86 !"g(0.3) + 38) 2 2(9.81)
f 54
= 1.76 × 10 −2
(8.86 !"g(0.25) + 38) 2
f 43
= 1.84 × 10 −2
f 32
f 35
=
=
2(9.81) (8.86 !"g(0.15) + 38) 2
f 32
= 2.08 × 10 −2
2(9.81) (8.86 !"g(0.15) + 38) 2
f 35
= 2.08 × 10 −2
9eempla*ando en la formula de DarcyTei"ac) 5N/47 para reali*ar la interacción: hf =
8 fL π
2
gD
∆Q = −
5
Q2
∑ hf hf N ∑ ( ) Q
I @ 9A;O f 5m7 4 ( , '(%< !-, , ( '(4(0 !%, ( 4 '(4(0 !%, SU;A@ O9IA
II @ 9A;O f 5m7 , ( & '(%<3 !-( & ( '(%0& !4, (
D 5m7 ( ,(( ( ((( ( (((
L
D 5m7 ( ((( ( ((( (
L
B 5m-7
% (< 4 (%, % (4(
f 5m7 %
4 '23,< 0'(0%& (' % 3 ('%,0 <0'230 (! & 2'(,- ,4'3, % '(2< %,2'32,
B
) f 5m7
(! (4,
4
('((,
4
)f B
(!
5m-7 %
)
)f
∆Q
B 5m-7 ( ('((%'(30< ( ('((%-('((< '((3< ('((%('(4%-
∆Q
B (
% '-<& &'23%4 % ('(<<0 ,',30 3
('((< ('((< ('((<
B 5m-7 ( '(-4 ( '((4
,
'(4(0 !%, SU;A@
(((
(!(%,
%('%0, <0'230 < 2'000 (2'&20
O9IA
5('((%-7
('((3<
Se#unda interacción: I @ 9A;O f 5m7 4 ( , '(%< !-, , ( '(4(0 !%, ( 4 '(4(0 !%, SU;A@ O9IA
II @ 9A;O f 5m7 , ( & '(%<3 !-( & ( '(%0& !4, ( , '(4(0 !%, SU;A@ O9IA
D 5m7 ( ,(( ( ((( ( (((
L
B
)
)
D 5m7 ( ((( ( ((( ( (((
L
B
)
)f
∆Q
B 5m-7 f 5m7 fB 5m-7 % (! % 4 ( (3< '2(22 <'30(('((-3 '(<4 (' 4 ('((-3 ( ((< '4%0 %3'0, 5('(((,7 '(% % & (!(4% 2'240 <,'40 ('((-3 ('(%< , 0 '0,%2'0%
∆Q
B 5m-7 f 5m7 B 5m-7 % (! ( 4 ( (-4 '2%24 0'<43 ('(((, '(-4, 4 3' ('((4 ('(%4, 44<('(((, ('((%, 4 % ('(((, (!((< %'%(2 ,0'&43 5('((-3 ('(% % ('4(4 2-'-0
O"tenindoe: Q2−3 Q2−5 Q5 − 3 Q5 − 4 Q4−3
hf 1− 2
= 0.017 m 3 = 17l# . = 0.073m 3 = 73 L# . = 0.01m 3 = 10 L# . = 0.0325m 3 = 32 L# = 0.0015m 3 = 1.5 L# .
9ta'
= 11m.
9etando la altura total con la perdida de ener#$a de cada tramo e o"tiene la altura de la car#a de preión para cada nudo:
h2 h3 h4 h5
= 44 − 11 = 32.89m = 32.89 − 0.928 = 31.96m = 31.96 − 0.0125 = 31.95m = 32.89 − 1.9099 = 30.98m
<4' Determinar la ditri"ución de #ato en la red! motrada! donde lo tu"o on de fierro fundido! vieo 5C) /%(( ?AEN TILLIANS7'
126. 1 !t/#eg
250 mm 25 1 0 m m 24 !t/#eg 25.2 6 250 mm
306 m
306 m
m 200 mm 305 4 4 2 m 2 m 0 0 2 25. 2 !t/#eg 250 mm 306 m 5
m 4 12.6 t!/#eg 4 2 m 3 m 0 0 2 4 63.2 !t/#e
126. 1 !t/#eg
250 mm
200 mm 305
306 m
12.6 t!/#eg m 3 4 4 2 g e m # / t ! m 5 0 2 . 7 2 0
40.1 !t/#eg 1 25 0 m m 24 4 m
70.1 !t/#eg
g e # / t ! 6 5
m 2 4 4 2 m g e m # / t ! 0 0 3 0 2
30.8 !t/#eg
6
250 mm
35.6 !t/#eg
5
306 m
250 mm
4 63.1 !t/#e
306 m
25. 2 !t/#eg 25.2 !t/#eg
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7.
30'Determinar la ditri"ución de #ato en la red! motrado en la fi#ura! donde lo tu"o on de fierro fundido! vieo 5C/%((7
/+,
D *C+ )
6 9
9 <
9 <
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96.<79< 97.66
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9. 97! 9. 97! 9. 97! 9. 97!
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<4 Una red de tu"o tiene la #eometr$a motrada en la fi#ura' @odo lo tu"o on de acero oldado! nuevo' Determinar el #ato en lo conducto del itema Utili*ando la formula de DarcyTei"ac) calcular la perdida de fricción D S L H H% H% # *C+ *LS0 *+6777 /+, *+) % *+) H% *+) *LS0) *LS0) ) ) ) 6 9 6 7.9 !7 9 7 77 77 7 7 9.97" .<
