Rapida-claseResuelto

DISEÑO DE RAIDA

-! $i$ $i$t# t#ma ma mo$ mo$tr tra" a"o o #n !a &/ &/ra ra on on$t $taa "# /n an ana! a! tra trap# p#oi"a oi"a!! "# on onr# r#to to "on "on"# "# $# $# "#$a "#$arr rro! o!!a !a /n /n /o /o / "# $#in $#in trap# trap#oi"a! oi"a! a r#t r#tan&/! an&/!ar ar m#"ian m#"iant# t# /na tran$ii tran$iin n !a /a! on#ta on#ta a !a !a rápi"a rápi"a "# $#in $#in r#tan r#tan m3+$ m3+$ b1= 1.75m 1.75m 1= 0.75 0.75 S1= S1= 0.0003 0.0003 b3=1 b3=1.2 .2 m ota1= ota1= 1129 m$nm m$nm ota6 ota6 = 1120 1120 m$nm m$nm  /# #xi$ #xi$t# t# p !a tran$iin. a) :im#n$ionar !a !on&it/" "# !a tran$iin b) a!/!ar !a ota "# on"o on"o "# !a $#in 2 ) ;S# orma #$a!to #$a!to i"rá/!io> i"rá/!io> $i $# ormara ormara "#t#rminar "#t#rminar !o$ ?rant#$ on/&a"o$ on/&a"o$ "#! #$a!to #$a!to @i"rá/!io @i"rá/!io  "# !a rápi"a. ") a!/! a!/!ar ar !a ota ota "#! o!n o!n "i$ip "i$ipa"or a"or "# ta! ta! mo"o mo"o /# $# orm# orm# /n r#$a!t r#$a!to o !aro !aro #n #n #! mi$mo mi$mo on$i"# on$i"#rr $#in 6 ?#n# !a mi$ma $#in  p#n"i#nt# /# #! "#! prim#r tramo. #) Araar #! p#r! "# /o

DISEÑO DE TRA!SI"IO! DE E!TRADA

Longitud Transicion

yn1=?

Lt = 4.46 4.4698 98061 0611 1

Manning

Q

 A R h

2 3

.

Lt =

S e

n

Datos

b (m)= z=

S= n=  (mB+$)=

1.75 0.75 0.0003 0.014 2

%annin&' y1 (m)=

C (mD)= (m)= R (m)= T (m)=

 (mB+$)= * (m+$)= D (m)=

,=

y2 = yc=?

4.5

0.95457621 propon#r 2.35392017 4.13644053 0.56906902 3.18186432 1.99986626 omprobar 0.84958967 0.73979276 0.31536949 Subcríco

α B

b 2

 L 

 B 2

t 

α < =12.5°. máximo 30°

#$u%o "rico :ato$ & (m+$D)= b (m)= =  (mB+$)=



 g 

9.81 1.2 0 2

 (mB+$)=

&'()((*(2+ 0.78797553 1.2 1.99992161

* (m+$)=

2.53805042

yc (m)=

C (mD)= Tc (m)=

3

Q2

 Ac T 

iorm# #! ana! ambia &/!ar. Sabi#n"o /#  =2 r"i"a "# ar&a "#bi"o a

# $# orma a&/a$ abao n"o /# #! ana! "# !a

b



 

"i$#Ear on #$t# Fa!or ara ontro! "# on"a$ a !a #ntra"a

,ALA!"E DE E!ER-IA E!TRE SE""IO! TRAE.OIDAL / RE"TA!-0LAR

S#in'

1

 

1  m3+$

b

1m

z

1

S#in'

2

y

1m

 A

1 m2

V

1 m/s

2   1.75 b 0.75 @F2 z 9.81 0.3283488739 & 0.9545762106 y 2.353920175 @F1   A 0.8496464839 0.0367940442 V

P

1m

4.1364405265

P

2m

2.51329254

R

1m

0.5690690244

R

2m

0.3135232

T

1m

T

2m

D

1m

D

2m

cota

1m

cota

2m



1

3.1818643159 0.7397927571 1129   0.315390584 Subcríco



2

1.2 0.65664627 1128.91891 1.0000392

!T

1m

1129.9913703

!T

2m

1129.99137

&

m2+$

-G1H-G2H=0

2 m3+$

2m

2 1.2 0 9.81 0.65664627

2 m2

0.78797553

2 m/s

2.5381499

2m 2

0

α  Pérdida por convergencia: C i Δhv

i =

B

0.3

IK

IF = 0.29155483 i JIF

b

0.08746645

E1 = E2  "i 345 b/$ar @p (p#r"i"a) 0.08746645

R6sa$to 7idrau$ico 6n S6cci8n r6ctangu$ar yn9 = y*= ?

7a$$ando rant6 con%ugado :ayor

Manning Q

 A R h

2 3

.

S e

z=

S= n=  (mB+$)=

y3 (m)=

1.2 0 0.022 0.014 2

%annin&' y3 (m)=

C (mD)= (m)= R (m)= T (m)=

 (mB+$)=

2

  

2 1  8F  4

Longitud

  1  

* Schoklits

:ato$'

n

Datos

b (m)=

 y5 

 y4  

 (mB+$)= b (m)= & (m+$D)= C1 (mD)= *1 (m+$)= T1 (m)

0.40523345 propon#r D1 (m) 0.48628014 ,4= 2.01046691 y4 (m)= 0.24187423 1.2 1.99999808 omprobar

0.40523345 2 1.2 9.81

 F 

V  

 gD

0.48628014 4.1129 1.2 0.40523345 2.0628 Su;6rcríco 0.9968

A= P=

a= L1 =  0S,R L2+2= L2 =

Lprom = Lon& o!on

by"zy#2 b"2y(z#2"1)#0.5

G=b2 * (m+$)=

O.L1=

4.11285163 0.40523345 2.06279417 Su;6rcríco MJ30N 0.28637286 m O.L2=

@3=

0.69160632

O.L4= @4= @4=

0.5 m 0.90523345 0.9 -:PQ:-C:P

D (m)=

,=

MJ40N 0.26265851 m

@3=

0.7 *a!or on$tr/?Fo

O.L5= @5= @5=

1.5

OL6 OL1 OL2 b!3

0.28637286 0.26265851 0.7

Oor"# !ibr#

0.5 m 1.4968 -:PQ:-C:P

MJ30N 0.286372863

O!6

0.28637286

  6sa$to h

 L





5 a 6  y 2



y1 

5 2.95774414

4.48 4.46558462 m 3.71166438 m  

"i"ipa"or= 4

m

5o6 #$ S-ARQ -L ,P: %CMP

 

aor

0.286373

,ALA!"E DE E!ER-IA E!TRE SE""IO! TRAE.OIDAL / RE"TA!-0LAR S#in'   5  m3+$

)

S#in'

y

5m

 A

5 m2

V

5 m/s

2   1.2 b 0@F2 z 9.81 0.03679404 & 0.99678228 y 1.19613874 @F1 A 1.67204684 0.14249443 V

P

5m

3.19356456

R

5m

T

5m

D

5m

b

5m

z

5

&

m2+$

cota 5 m



5

!T 5 m

$m

2 1.75 0.75 9.81 0.95457621

$ m2

2.35392017

$ m/s

0.84964648

P

$m

4.13644053

0.3745466

R

$m

0.56906902

1.2

T

$m

3.18186432

D

$m m

0.73979276 1120 0.31539058

!T $ m

1121.04422

0.99678228 1119.90494 ropon#r 0.53470443 Subcríco 1121.04422

T m3+$

(

$m $

  cota  $



$

-G1H-G2H=0

0

Gomar @F1 $o!o /an"o #$ maor p=

0.0528501911 Δ

=

DISEÑO DE TRA!SI"IO! G5=

b

α B

 Pérdida por divergencia: Co Δhv o = 0.5

IK IF = 0.10570038 o JIF

E) = E(  "o 345

E SALIDA 1.2

G6= 3.18186432

Longitud Transicion Lt = 4.46980611 Lt = 4.5

 L 

 B 2

α < =12.5°.

 b tg 

 

