001 Perhitungan Terjunan Tegak

PERHITUNGAN DIMENSI SALURAN Luas Areal Kebutuhan air di saluran Debit di saluran

A a Q

Lebar saluran Kemiringan saluran Kemiringan talud K"e#isien $e$asaran

b  m K

= 225 ha = 1.485 l/dt/ha = a xA = 0.33 0.33412 4125 5 m3/dt = 0.8 m = 0.000! = 1 = 50 %&asangan batu $ali'

A *

= b.h = b%2.h.%1m2'1/2

+

=

(ang dihitung adalah h dan )

) Q

b.h  m.h2 b2.h.%1m2'1/2 = K.+2/3 / 1/2 = 1.22 1.224,4 4,45 5 +2/3 = ).A +2/3

b h  m h2 bh  h2 b2.h.%1m2'1/2

= 1.224, 1.224,45 45 0.33 0.33 = 1.22 1.224,4 4,45 5

Persamaan akhir :

0

0.8h  h2

= 1.224,45

2/3

b h  h2

2/3

0.8 h  h 2

1/2

0.82.h.%2'

ntu$ menghitung h dila$u$an dengan ara iterasi &ada &ersamaan di atas  h hasil &ers se selisih h hasil &ers se selisih 0.400 -0.1!2 0.552 -0.002 0.004 0.!00 0.0!0 0.223 0.552 -0.001 0.001 0.500 -0.0!3 0.123 0.553 -0.001 0.000 0.550 -0.004 0.05 0.553 -0.001 0.000 0.5,5 0.02, 0.032 0.553 0.000 0.000 0.5!3 0.011 0.01! 0.553 0.000 0.000 0.55! 0.004 0.008 0.553 0.000 0.000 0.553 0.000 0.004 0.553 0.000 0.000 Dari Dari hasil hasil teras terasii dida&a dida&at$a t$an n nilai nilai h 

=

=

0.553 0.553

m

0.481 m/dt

adi Dimensi aluran en6adi  A = 225 Q = 0.334125  = 0.481 b = 0.8 h = 0.553 m = 1 i = 0.000! 7 = 0.4 $ = 50

ha m3/dt m/dt m m

m

7 = 0.4 1 1 h = 0.553

b = 0.8

- 0.334

PERHITUNGAN DIMENSI SALURAN IRIGASI Jenis Saluran

:

!ode Saluran

:

Tersier

Pilih utk diisi di kotak hiau :

L A ;<+ Kehila Kehilanga ngan n air di salura saluran n e > Kebutuhan air di saluran a Debit di saluran Q Luas Areal

Lebar saluran Kemiringan saluran Kemiringan talud K"e#isien $e$asaran K"ns K"nstr tru$ u$si si mema mema$a $aii lant lantai ai ? (ang ang dihi dihitu tung ng adal adalah ah h dan dan )

b  m K

A * +

= 25 = 11 11.83 = 0.81 = 0.000 0.000 = rr522 = 1 = rr522 = axA

*rimer9 e$under9 :ersier atau Kuarter

m ha l/dt/ha m3/dt

l/dt/ha

rr522 m3/dt 0.38 m b minimal = 0.3 b ma$simal = 1 0.02  minimal = 0.02  ma$simal = 0.02 0.50 1 m %h'  1 %)' rr522 Err:"## Err:"## 53.82 %&asangan batu' Pilih utk diisi di kotak hiau tida tida$ $ %isi %isi  @a@ @a@ atau atau @ tida tida$@ $@'' %&as %&asan anga gan n batu batu'' ;B  &asa &asang ngan an tana tanah h tida tida$ $ dila dilant ntai ai %&as %&asan anga gan n bet" bet"n' n' = b.h  m.h 2 %&asangan tanah' tanah'

= = = = =

= b%2.h.%1m2'1/2 = b.h  m.h 2

Jenis Tanah : %isi 6enis tanah sesuai dengan tabel disam&ing'

b2.h.%1m2'1/2 ) Q

Butir-butir dari maam lanau - brang$al

= K.+  = ,.!1123 ,.!11232 2 +2/3 = ).A 2/3

1/2

+2/3

= ,.!1123 ,.!11232 2 CCC =

%b h  m h 2'

bh  mh2

,.!1 ,.!11232 1232

2/3

%b h  mh 2'

2/3

%b h  mh 2' - rr522

b2.h.%1m2'1/2 Persamaan akhir :

0

=

bh  mh2

,.!11232

b2.h.%1m2'1/2 ntu$ menghitung h dila$u$an dengan ara iterasi &ada &ersamaan di atas  h hasil &ers 0.38 0.381 1 rr rr52 522 2 rr522 $$$ Dari Dari hasil hasil teras terasii dida& dida&at$ at$an an nilai nilai h 
= =

=

1.34 m/dt 1.00

0.381 0.381

m

sesuai den%an kriteria saluran &asan%an 'atu (! : karena enis aliran su'kritis sehin%%a tidak da&at ter'entuk lon)atan

adi Dimensi aluran en6adi  A = 11.83 ha Q  b h m i 7 $

= = = = = = = =

r rr522 1.34 0.38 0.38 0.50 0.02 rr522 53.8154

m3/dt m/dt m m

rr522

1

7 = rr522 0.50 1 h = 0.38

m

b = 0.38

S*arat '+h rasio : b/h =

rr522

Jari-ari minimum untuk len%kun%an : +

=

rr522

m

rr522

Nilai I R ./" = 0.005! ,atatan : +uas saluran di ba7ah ini minimal mem&unai nilai  +0.5

= 0.005!

PERHITUNGAN DIMENSI SALURAN PEM0UANGAN Jenis Saluran !ode Saluran Luas Areal

: :

Main+utama n + >* h a D A Q

Kebutuhan air di saluran Debit di saluran Lebar saluran Kemiringan saluran Kemiringan talud K"e#isien $e$asaran K"ns K"nstr tru$ u$si si mema mema$a $aii lant lantai ai ? (ang dihitung adalah h dan )


ain/utama atau e$under

= 3 hari = 185 mm = 5 mm/da = 100 mm = 0.25 = 48.33 l/dt/ha = 50.00 ha = D x A m3/dt m

b  m K

= 2.42 = 3.43 = 0.0088!055, = 4.14 = 55.01 tida tida$ $

A

= b.h  m.h 2

%&asangan tanah'

* +

= b%2.h.%1m2'1/2 = b.h  m.h 2

Jenis Pasan%an Tanah : 1isi enis tanah2 *asir halus $"l"id

b minimal =  minimal = 4 m %h'  1 %)' %&asangan batu' %isi %isi  @a@ @a@ atau atau @ tida tida$@ $@''

1 b ma$simal = ! 1-08  ma$simal = 0.1 4.1401104 (! (! Pilih utk diisi di kotak hiau %&as %&asan anga gan n batu batu'' %&asangan bet"n'

b2.h.%1m2'1/2 ) Q

= K.+2/3 1/2 = 5.1,8 5.1,81 102 02 +2/3 = ).A +2/3 %b h  m h 2' bh  mh2

= 5.1,81 5.1,8102 02 2.42 2.42 = 5.1,8 5.1,81 102 02

2/3

%b h  mh 2'

2/3

%b h  mh 2' - 2.41!!,

2 1/2

b2.h.%1m ' Persamaan akhir :

0

=

5.1,8102

bh  mh2 2 1/2

b2.h.%1m '

ntu$ menghitung h dila$u$an dengan ara iterasi &ada &ersamaan di atas  h hasil &ers 1.144 -0.00! -0.00! (! Dari Dari hasi hasill ter teras asii dida dida&a &at$ t$an an nila nilaii h 
= =

=

2.001 m/dt 0.5,

1.14 1.144 4

m sesuai den%an kriteria saluran &asan%an 'atu tanah

adi Dimensi aluran en6adi  A = 3 Q = 2.41,  = 2.001 b = 3.43 h = 1.14 m = 4.14 i = 0.0088!1 7 = 0.2 $ = 55.01083

ha

0.5

3

m /dt m/dt m m

7 = 0.2 4.14 1 h = 1.14

m b = 3.43 3.43

S*arat '+h rasio : b/h =

3

=

3

Jari-ari minimum untuk len%kun%an : +

=

!4.54

m

e$"n"mis

Nilai I R ./" = 0.0033 ,atatan : +uas saluran di ba7ah ini minimal mem&unai nilai  +0.5

= 0.0033

DESAIN 0ANGUNAN TERJUN In&ut : = =

0.80 m



=

3.13 m3/dt/m

g h

=

.81 m/dt 2

E le)asi hulu le)asi hilir >1 L Gd FG u

2.00 m3/dt

Q b1

= = = =

1.00 1.50 1 0.00 8.50

m m m m

= =

3.44 4.43

= =

1.!, m 0.83 m

=

5.42 m/dt

""$ untu$ ter6un tega$

(u

= 0.58 m Ti&e 0an%unan Terun Ter&ilih :

Bangunan Terjun Tegak

0ANGUNAN TERJUN TEGA!

(1

%' = :er&ilih

h = 1.00 h1

(

:irai Lua&an Q

*enurunan tinggi energi FG = 0.83

*1 *"t"ngan 

FE = 1.50

Gd = 1.!,

(u = 0.58

n = 0.30

L = 4.43

0ANGUNAN TERJUN MIRING

h = 1.00

%H' = :ida$ :er&ilih

( FG = 0.83 Q 2

FE = 1.50

1 Gd = 1.!, (u = 0.58 n = 0.30 L& = 3.00

L = 4.43

0ANGUNAN TERJUN TEGA!

(1

G1 = 1.00 h1

(

:irai Lua&an Q

*enurunan tinggi energi FG = 1.00

*1 *"t"ngan 

FE = 1.00

(d = 1.0 Gd = 1.00

(u = 0.20 L& = 2.8, Data : G1 Gd FG Q g n

L6 = 5.4! Lb = 8.33

= = = = = =

n = 0.50

1.00 1.00 1.00 2.00

m m m m2/dt .80 m/dt2 0.50 m

FE I 1.50 m FE = 1.00 m >atatan  sesuai untu$ bangunan ter6un tega$ u = .80 m/dt (u = 0.20 m
(2 = 0.5

0ANGUNAN TERJUN MIRING

G1 = 0.,5

( FG = 1.00 Q 2

FE = 1.42

Gu = 2.13 1

(d = 1.08

(2 = 0.,8

Gd = 1.1, (u = 0.18 n = 0.30 L& = 2.85

L6 = 5.40

Ta'el Per'andin%an Tak 0erdimensi Untuk Lon)at an Air 1dari 0os3 Re&lo%e and ,lemens3 45672 FG/G1 (d/(u (u/G1 u2/2gG1 Gu/G1 (d/G1 d2/2gG1 Gd/G1 0.244! 0.2!88 0.23 0.318 0.34!5 0.3,40

3.0000 3.1000 3.2000 3.3000 3.4000 3.5000

0.3!! 0.35 0.3533 0.348 0.340 0.3351

1.100 1.143! 1.18,0 1.2308 1.2,4 1.314

1.4!,5 1.5035 1.5403 1.5,,, 1.!158 1.!545

1.100! 1.115, 1.1305 1.144 1.150 1.1,28

0.1223 0.110 0.115 0.1130 0.1103 0.10,,

1.222 1.234, 1.24!4 1.25, 1.2!3 1.2805

DATA : FG = G1 = n= g=

1.0000 0.,500 0.3000 .8000

0.4022

3.!000

0.325

1.3!43

1.!38

1.18!3

0.1053

1.21!

(d/(u =

5.,!8 0.2410

0.4312

3.,000

0.3242

1.405

1.,33,

1.15

0.1030

1.3025

(u/G1 =

0.4!0

3.8000

0.311

1.4551

1.,,42

1.2125

0.1008

1.3133

u2/2gG1 =

2.5,!4

0.412

3.000

0.3142

1.500

1.8151

1.2253

0.08,

1.323

Gu/G1 =

2.8352 1.48!!

0.5222

4.0000

0.304

1.54,2

1.85!!

1.23,8

0.0!,

1.3345

(d/G1 =

0.58!1

4.2000

0.3005

1.!40,

1.412

1.2!21

0.030

1.3551

d2/2gG1 =

0.0!,1 1.5!55

0.!525 0.,211

4.4000 4.!000

0.222 0.2844

1.,355 1.8315

2.02,! 2.115

1.2855 1.3083

0.08! 0.08!!

1.3,52 1.348

Gd/G1 =

0.,20

4.8000

0.2,,1

1.28

2.20!0

1.3303

0.083,

1.4140

(u =

0.1808

0.8!51

5.0000

0.2,03

2.02,4

2.2,,

1.351!

0.0811

1.432,

(d =

1.0804

0.400

5.2000

0.2!3

2.12,1

2.310

1.3,23

0.0,8,

1.4510

u =

!.1541 0.28

1.01!

5.4000

0.25,

2.22,

2.4858

1.325

0.0,!4

1.4!8

d =

1.05,

5.!000

0.2521

2.32

2.5821

1.4121

0.0,43

1.48!4

Gu =

2.12!4

1.5035 1.5203 1.53!, 1.552 1.5!8, 1.5843 1.55 1.!145 1.!23 1.!43, 1.!580 1.!,20 1.!858 1.!4 1.,12, 1.,25 1.,38 1.,51, 1.,!43 1.,,!8 1.,81 1.8012 1.830 1.854 1.88,5 1.14! 1.411 1.!!, 1.1, 2.01,8 2.0401 2.0!35 2.08!3 2.108, 2.1305 2.1520 2.1,31 2.13! 2.2141 2.233 2.2534 2.2,2,

Gd =

1.1,41

1.1,!3 1.2585 1.342 1.4280 1.5150 1.!035 1.!3, 1.,851 1.8,,8 1.,20 2.0!,4 2.1!41 2.2!20 2.3!13 2.4!15 2.5!30 2.!!5! 2.,!4 2.8,41 2.801 3.08! 3.14 3.4!1 3.,41 4.0351 4.32!, 4.!233 4.252 5.2323 5.5424 5.8!05 !.1813 !.!50! !.83!3 ,.1,02 ,.5081 ,.848 8.158 8.5438 8.8085 .255, .!1!0

5.8000 !.0000 !.2000 !.4000 !.!000 !.8000 ,.0000 ,.2000 ,.4000 ,.!000 ,.8000 8.0000 8.2000 8.4000 8.!000 8.8000 .0000 .2000 .4000 .!000 .8000 10.0000 10.5000 11.0000 11.5000 12.0000 12.5000 13.0000 13.5000 14.0000 14.5000 15.0000 15.5000 1!.0000 1!.5000 1,.0000 1,.5000 18.0000 18.5000 1.0000 1.5000 20.0000

0.24!, 0.241, 0.23!, 0.2321 0.22,, 0.2235 0.215 0.215, 0.2121 0.2085 0.2051 0.201 0.188 0.158 0.12 0.101 0.18,4 0.184 0.1823 0.1, 0.1,,5 0.1,53 0.1! 0.1!4 0.1!03 0.15!0 0.1520 0.1482 0.144, 0.1413 0.1381 0.1351 0.1323 0.12, 0.12,1 0.124, 0.1223 0.1201 0.1180 0.115 0.1140 0.1122

2.4331 2.53,2 2.!42 2.,488 2.85!0 2.!43 3.0,3, 3.183 3.250 3.40,2 3.4,23 3.!343 3.,40 3.8!4 3.814 4.088 4.21,1 4.33!3 4.45!1 4.5,,0 4.!85 4.8208 5.1300 5.443, 5.,!23 !.0853 !.4124 !.,43, ,.0,4 ,.418 ,.,!25 8.10! 8.4!05 8.8153 .1,3! .5354 .005 10.2!3 10.!35 11.01!4 11.351 11.,,!5

2.!,8 2.,,8 2.8,! 2.80 3.083, 3.18,8 3.232 3.3! 3.50,1 3.!15, 3.,354 3.83!1 3.4,8 4.0!0, 4.1,43 4.288 4.4045 4.5211 4.!385 4.,5! 4.8,!0 4.!1 5.2 5.!08, 5.22, !.2413 !.5!44 !.81 ,.2241 ,.5!02 ,.00! 8.244, 8.52 8.450 .300, .!!01 10.022 10.384 10.,5,5 11.120 11.501 11.888,

1.4312 1.44 1.4!, 1.4858 1.5032 1.5202 1.52!8 1.5531 1.5!1 1.584, 1.!001 1.!152 1.!301 1.!44! 1.!58 1.!,30 1.!8! 1.,005 1.,13 1.,2,1 1.,402 1.,530 1.,843 1.814! 1.843 1.8,23 1.000 1.2!8 1.52 1., 2.0032 2.02,4 2.0511 2.0,42 2.0!8 2.110 2.140, 2.1!1 2.1830 2.2033 2.2234 2.2432

0.0,23 0.0,05 0.0!8, 0.0!,1 0.0!55 0.0!41 0.0!2, 0.0!14 0.0!02 0.050 0.05, 0.05!8 0.055, 0.0548 0.0538 0.052 0.0521 0.0512 0.0504 0.04, 0.048 0.0482 0.04!5 0.0450 0.043! 0.0423 0.0411 0.03 0.038 0.03, 0.03! 0.03!1 0.0352 0.0345 0.033, 0.0330 0.0323 0.031, 0.0311 0.0305 0.0300 0.025

m m m m/dt 2

DESAIN G(T MIRING INPUT DATA :

Tam&an% Atas

Qd  b h le)asi di hulu le)asi di hilir E >d L :inggi end sill

= = = = = = = = = =

0.0!4 0.0008 1.00 0.28 51., 40 11., 0.85 3 0.1

m3/dt

Bagian *emasu$an

m m m m m

1.00

=

0.21 b1

=

0.30 m

M1

=

0.28 e$ bilangan erderni$") e$ bilangan "nt"uri

0.0

0.1

T*&e I : 52.1! 51.,

!8.

11.35

40.00

N L4 8 20.00 T*&e II :

0.1 m

=

0.0! m2

=

1.15 m/dt

L0

=

1. m

J b1/M1

=

10

= #/ 0a%ian Peralihan N = T*&e I : "t N =
1.!0

52.1! 0.28

0.1

51.,

ti da$ ter6 adi &enambahan )"l ume ai r 40.23

N

11.35

e$"n"mis I=

25.00

K

11.,

3.2! 0.8, 0

= I= I=

2.3 1.00 0.5

40.81

r

K K K

0.,1 L4 8 96/#"

40.00

=

20.00 m

L1L2

=

3.00 m =

3

K

=

0.1

I=

0.5

K

r "t N
= = =

0.,5 m 3.2! 0.!

= I=

3.24 1.00

K K

=

0.00

=

20.0!

O"

=

14.0!

%OLO"'/2

= =

1,.0! 5 m

= =

0.038 m 3.2!

=

2.3

K

0.8,

I=

1.00

K

L9 8 4.28

T*&e III :

0.28 I=

0.5

0.1

11.35 41.53 O"

=

0.0 m

L3

=

4.28 m

L0

=

2.14 m

11.,

0.,1 L4 8 97

OL h )

Lt 8 "

0.3

0.1 ,ross Se)tion Got Mirin%

ANALISA HIDROLIS le)asi K"lam la$ = le)asi *elim&ah = DE = E1 =

40.000 m 51.,0 m 11.,0 m

d1 =

0.038 m 1.15 m/dt

Kee&atan Aliran %' = 2/2g =

0.0, m .81 m/dt2 !0 %*asangan batu'

g= $=

Pilih utk diisi di kotak hiau %*asangan batu' %Bet"n' %Ba6a' %Ba6a bet"n'

Hasil Analisa Q

b

q

d2

db

(m3/dt)

(m)

(m3/dt/m)

(m)

(m)

./.;7

./9.

./#47

./.47

./64.

n

Lp

Vb2/2g

(m) ./.7;

7/#6

Vb

8

4#/."

,ek 0ilan%an >ederniko@ 1>2 dan 0ilan%an Mountouri 1M2: > = 0.2! tidak ada aliran %etar 0.01

> ta'el utk M # 8 ./.4 *aitu : 1.50 d+P 8 0.28 tan ? 8 0.31 d+P ta'el utk tan ? 8 ./94 *aitu : 0.1

tidak ada aliran %etar

hB

(m/dt) 44/65.

4"/#9

d1h)1E1 = d2h)2h# E2

,ek Rumus 0ernoulli :

40.00 L9 8 4.28

TIPE G(T MIRING TERPILIH : Ti&e I

=

40.81

K

0.38 m

b3

2.14

52.1! 51.,

K

1,.0!

0.10

! 8 1tan L - tan .2 # h@ )os# . + Lt

N =

0.81

r 8 ./"

1.00 m

L1

M#

2.14

! 8 @# +% r )os 

"

1,.0!

0.81

0.10

L2 = 1.00 L9 8 4.28

m

=

4#/."

40.81 0.,1

11.,

M1

"t N

0.30

h# harus dihitung

A 1

Lt h)

Bagian K"lam la$

1. Tam&an% Sam&in%

m m

0.0!4 b1

T*&e II : K

Bagian *eralihan L2

titi$ A

HASIL PERHITUNGAN : 4/ 0a%ian Pemasukan 3/2 Q = >d 1., b h 1

%2/3' h1

Bagian *eralihan L1

J

(! < 0( ===

./4#

iB

./..77

0.10 2.14

0.81

DESAIN 0(C 0AGI+ SADAP INPUT DATA : !ode 0o 0a%i+Sada& : !ondisi Areal Iri%asi 4 : !ondisi Areal Iri%asi # : !ondisi Areal Iri%asi 9 : Areal 1  A1 = !0 ha aluran = ;<+1 = e1 = > a1

> a2

50 ha e$under

e$under9 :ersier atau Kuarter



0.,2 1 1.13 l/dt/ha 30 ha e$under 0.81 l/dt/ha 0.,2

> a3

=

1

Q"ut3 =


0.81 l/dt/ha

=

Q"ut1 = Q"ut2 =


1.13 l/dt/ha

e3

= bin = Qin =


0.,2 1

= =

= Areal 3  A3 = aluran = ;<+3 =

%&ilih ut$ diisi$an &ada $"ta$ hi6au  datar atau miring' %&ilih ut$ diisi$an &ada $"ta$ hi6au  datar atau miring' %&ilih ut$ diisi$an &ada $"ta$ hi6au  datar atau miring'

0.81 l/dt/ha

=

= Areal 2  A2 = aluran = ;<+2 = e2

e$under

datar datar miring

B Q"ut1 =

1.13 l/dt/ha 0.8

m

1.20

0.15,5

m3/dt

0.0!,5

3

m /dt

0.05!25

m3/dt

0.033,5

3

>d =

1.03

n =

0.3

0.40 Qin = 0.15,5 A

m /dt

0.8

m

1.!0

Q"ut3 =

1.40

h/b

0.30 arat

h/b

tatus

D :i&e Aliran

1.2

m

1.10

=

1

K

diambil =

1.2

m

0.14

1

m

1.30

=

1

K

diambil =

1

m

0.14

m b1h11.5

2.20

=

0

K

diambil =

0.!

m

0.14

b3 =

0.!

Q"ut1 =

1.,!

0.0!,5 = Q"ut2 =

0.0!,

0.05!3 = Q"ut3 =

0.05!

1.,! 1.,!

b2h11.5 b3h11.5

0.0338 = 0.033 Q"ut1  Q"ut2  Q"ut3 = ,ross Se)tion A-0 : Q"ut1 = 0.0!

>d = L1 = G1 =

Q"ut2 = 0.0! >

0.!0

HASIL PERHITUNGAN :

&1 = h1 =

1.00

0.40

>atatan = in&ut data (! : in 8 total out h1n&1 = 1.50 m = 0.10 m ------- K

b1 = b2 =

0.0,

1.20

0.15! m3/dt m

I= "ut1

0.1 m 0.1,!08 0.10

m

0.12

m

m3/dt

0.0, =

0.54

K

simulasi selesai

m/dt

K 0.10

A

0.10

B

1.20

G1/L1 = /%h1&1' = 1

1.1!

I=

1.5

0.08

I=

0.35

h1&1 = P aliran = Q"ut1 =

1.30 100 0.0,

m3/dt

m P 3

m /dt

0.30

1.30

1 1

0.10 0

,ross Se)tion A-, : Q"ut2 = 0.05 &1 = h2 =

1.20

m3/dt m

0.1 m >d = 0.1,!08 L2 = 0.10 m G2 = 0.12 m G2/L2 = 1.1! I= /%h2&2' = 2 h2&2 =

0.08 1.30 100

P aliran = Q"ut1 = 0.0! ,ross Se)tion A-D : Q"ut3 = 0.03 &1 = h3 =

1.20

I=

P aliran = Q"ut1 =

0.08

0.0! =

m3/dt

0.54

K

simulasi selesai

m/dt

K 0.10 0.10

A

>

1.20 1.5

0.30

1 1.30

1

0.10

0.35

0

m P m3/dt m3/dt m

0.1 m >d = 0.1,!08 L3 = 0.10 m G3 = 0.12 m G3/L3 = 1.1! I= /%h3&3' = 3 h3&3 =

I= "ut2

I=

1.30 100

m

0.03

m /dt

P 3

I= "ut3

0.03 =

m3/dt

0.54

K

simulasi selesai

m/dt

K 0.10

A

0.10

D

1.20 1.5 0.35

0.30

1 1.30

1

0.10 0

,ek arat I= I= I=

r=

r=

r=

DESAIN 0(C 0AGI+ SADAP INPUT DATA : !ode 0o 0a%i+Sada& : !ondisi Areal Iri%asi 4 : !ondisi Areal Iri%asi # : !ondisi Areal Iri%asi 9 : Areal 1  A1 = 500 ha aluran = ;<+1 = e1 = > a1

=

= Areal 2  A2 = aluran = ;<+2 = e2 > a2

= =

= Areal 3  A3 = aluran = ;<+3 =

e$under







0.,2 1 1.13 l/dt/ha 300 ha e$under 0.81 l/dt/ha 0.,2 1 1.13 l/dt/ha 200 ha e$under 0.81 l/dt/ha

=

0.,2

> a3

=

1

B Q"ut1 =

0.5!

1.13 l/dt/ha 0.8

m

0.!0

1.125

m3/dt

0.5!25

3

m /dt

0.33,5

m3/dt

Q"ut3 =

0.225

3

>d =

1.03

Q"ut1 = Q"ut2 =


0.81 l/dt/ha

e3

= bin = Qin =

datar datar miring

0.!1 Qin = 1.125 A

m /dt

0.8

0.30

n = 0.52844,5 m

0.20

HASIL PERHITUNGAN : h1 = 0.!,482! m b1 = 0.5,513! m

0.22

h1/b

2.02

D

arat

h1/b

tatus

1.20

=

1

K

diambil =

0.!

m

0.33

1.0

=

1

K

diambil =

0.3

m

0.33

2.0

=

0

K

diambil =

0.2

m

0.33

:i&e Aliran

0.5! 1.,!

b2h11.5

0.34 1.,!

b3h11.5

0.225 = 0.22 Q"ut1  Q"ut2  Q"ut3 = ,ross Se)tion A-0 : Q"ut1 = 0.55

1.125 m3/dt

K =  = a =

0.,4 0.53 0.!8

m

g = h1/a = h2 =

.80

m/dt2

1.00

J =

0.00

P aliran = Q"ut1 =

Q"ut3 =

b2 = 0.34!5,2 m b3 = 0.230424, m Q"ut1 = b1h11.5 1.,!

0.33,5 = Q"ut3 =

2.02

0.!1

>atatan = in&ut data (! : in 8 total out h1  n = 1.20 m = 0.!8 m ------- K

0.5!25 = Q"ut2 =

Q"ut2 = 0.34 >

0.!8 100 0.5!

I= "ut1

m3/dt m3/dt

0.5! =

1.42

K

simulasi selesai

m/dt

K

A

B

0.!, J

m

0.53

0.!8 0.,8

"

P 3

m /dt

h2/a

=

3.023401,,

h2/a

=

1.00325,2



=

0.0000002

3

0.!8

1 1

0 2

K



-3.4014,

K



%h1/a'!



-0.00002

%h1/a'5



,ross Se)tion A-, : Q"ut2 = 0.34 K  a g h1/a

= = = =

0.,! 0.53 0.!8 .80

= h2 =

1.00

J =

0.00

P aliran = Q"ut2 =

0.!8

m3/dt

,ross Se)tion A-D : Q"ut3 = 0.22

m/dt

K

A

>

0.!,

m

0.53 P

3

m /dt

m3/dt

m

.80

m/dt2

1.00

J =

0.00 0.23

1.45

simulasi selesai

0.!8

I= "ut3

=

3.02338455

h2/a

=

1.002014



=

0.0000002

m3/dt

0.23 =

h2/a

0.,8

1.41

K

simulasi selesai

m/dt

K

K

3

%h1/a'!

A

1

m

0.53

-3.40141

K



-0.00002

%h1/a'5

P 3

m /dt

h2/a

=

3.023401,,

h2/a

=

1.00325,2



=

0.0000002

K

%h1/a'!

 

D 0.!8 0.,8

" 3

0 2



0.!, J

0.!8

1

"

g = h1/a = h2 =

100

K

J

0.,4 0.53 0.!8

P aliran = Q"ut3 =

=

m/dt2

K =  = a =

0.!8

m3/dt

0.34

m

100 0.34

I= "ut2

0.!8

1 1

0 2



-3.4014,

K



-0.00002

%h1/a'5

 


r=

1.25!153 -0.0002

r=

1.25!14, -0.0002

r=

1.25!153 -0.0002


=

= Areal 2  A2 = aluran = ;<+2 = e2

= =

> a2

= Areal 3  A3 = aluran = ;<+3 =

e$under








0.81 l/dt/ha 0.,2 1 1.13 l/dt/ha 300 ha e$under 0.81 l/dt/ha 0.,2 1 1.13 l/dt/ha 200 ha e$under 0.81 l/dt/ha

e3

=

0.,2

> a3

=

1

= bin = Qin =

datar datar miring

B Q"ut1 =

0.5!

1.13 l/dt/ha 0.8

m

0.!0

1.125

m3/dt

Q"ut1 = Q"ut2 =

0.5!25

3

m /dt

0.33,5

m3/dt

Q"ut3 =

0.225

3

>d =

1.03

0.!1 Qin = 1.125 A

m /dt

0.8

0.30

n = 0.52844,5 m

0.30


1.,!

h1/b

2.02

D

arat

h1/b

tatus

1.20

=

1

K

diambil =

0.!

m

0.32

1.0

=

1

K

diambil =

0.3

m

0.32

2.0

=

0

K

diambil =

0.3

m

0.32

:i&e Aliran

b2h11.5

0.34 1.,!

,ross Se)tion A-0 : Q"ut1 = 0.53 = = = =

0.,3 0.53 0.!! .80

= h2 = J =

1.02

P aliran = Q"ut1 =

0.22

0.5!

b3h11.5

0.225 = 0.22 Q"ut1  Q"ut2  Q"ut3 =

K  a g h1/a

Q"ut3 =

b2 = 0.34!5,2 m b3 = 0.230424, m Q"ut1 = b1h11.5 1.,!

0.33,5 = Q"ut3 =

2.02

0.!1

>atatan = in&ut data (! : in 8 total out h1  n = 1.20 m = 0.!, m ------- K

0.5!25 = Q"ut2 =

Q"ut2 = 0.34 >

0.!, 0.00 5 0.53

1.125 m3/dt

m m/dt2

I= "ut1

m3/dt m3/dt

0.5! =

1.40

K

simulasi selesai

m/dt

K

A

B 0.!, J

m

0.53

0.!! 0.,8

"

P

h2/a

=

3.01202422

3

h2/a

=

1.015!10!!



=

0.0000002

m /dt

3

0.!,

1 1

0 2

K



-3.351,

K



%h1/a'!



-0.00002

%h1/a'5



,ross Se)tion A-, : Q"ut2 = 0.32 K  a g h1/a

= = = =

0.,4 0.53 0.!5 .80

= h2 =

1.04

J =

0.00

P aliran = Q"ut2 =

0.!,

0.32

g = h1/a = h2 = J = P aliran = Q"ut3 =

0.,4 0.53 0.!5 .80

I= "ut2

m3/dt

0.34 =

1.42

K

simulasi selesai

m/dt

K

A

m m/dt

> 0.!,

2

J m

0.53 P

3

m /dt

m3/dt

I= "ut3

=

3.00!1,4!31

h2/a

=

1.0321!028



=

0.0000002

m3/dt

0.23 =

h2/a

1.40

K

simulasi selesai

m/dt

K

0.!, 5 0.21

1

m/dt



-3.32208

K

%h1/a'!



-0.00002

%h1/a'5

m

0.53

m /dt



0.!, 0.!5 0.,8

"

P



D

2

J

3

0 2

K

A

m

0.!,

1

0.,8

3

1.04 0.00

0.!5

"

5

,ross Se)tion A-D : Q"ut3 = 0.21 K =  = a =

m3/dt

h2/a

=

3.00!1,4!31

h2/a

=

1.0321!028



=

0.0000002

3

0.!,

1 1

0 2

K



-3.32208

K

%h1/a'!



-0.00002

%h1/a'5

 


r=

1.250,1 -0.0002

r=

1.24,2, -0.0002

r=

1.24,2, -0.0002


=

= Areal 2  A2 = aluran = ;<+2 = e2 > a2

= =

= Areal 3  A3 = aluran = ;<+3 =

e$under








=

0.,2

> a3

=

1

B Q"ut1 =

0.5!

1.13 l/dt/ha 0.8

m

0.!0

1.125

m3/dt

0.5!25

m3/dt

0.33,5

m3/dt

Q"ut3 =

0.225

3

>d =

1.03

Q"ut1 = Q"ut2 =


0.81 l/dt/ha

e3

= bin = Qin =

datar datar miring

0.!1 Qin = 1.125 A

m /dt

0.8

0.30

n = 0.52844,5 m

0.20


1.,!

h1/b

2.02

D

arat

h1/b

tatus

1.20

=

1

K

diambil =

0.!

m

0.33

1.0

=

1

K

diambil =

0.3

m

0.33

2.0

=

0

K

diambil =

0.2

m

0.33

:i&e Aliran

b2h11.5

0.34 1.,!

,ross Se)tion A-0 : Q"ut1 = 0.5!

b3h11.5 1.125 m3/dt

K =  = a =

0.,4 0.53 0.!8

m

g = h1/a = h2 =

.80

m/dt2

P aliran = Q"ut1 =

0.22

0.5!

0.225 = 0.22 Q"ut1  Q"ut2  Q"ut3 =

J =

Q"ut3 =

b2 = 0.34!5,2 m b3 = 0.230424, m Q"ut1 = b1h11.5 1.,!

0.33,5 = Q"ut3 =

2.02

0.!1

>atatan = in&ut data (! : in 8 total out h1  n = 1.20 m = 0.!8 m ------- K

0.5!25 = Q"ut2 =

Q"ut2 = 0.34 >

I= "ut1

m3/dt m3/dt

0.5! =

1.44

K

simulasi selesai

m/dt

K

A

B 0.!,

1.00 0.!8 55.00

J m

0.53

0.!8 0.,8

"

100

P

h2/a

=

3.023401,,

0.5!

3

h2/a

=

1.00325,2



=

-0.000020022

m /dt

3

K

J2

0.!8

1 1



-3.4014,



-0.0018

0 2

K

 J



,ross Se)tion A-, : Q"ut2 = 0.34 K =  = a = g = h1/a = h2 = J = P aliran = Q"ut2 =

0.,! 0.53 0.!8 .80 0.!8 55.00

m3/dt

0.34 =

0.!8

K

simulasi selesai

m/dt

K

A

m m/dt

>

2

0.!, J

m

0.53 P

h2/a

=

3.02338455

0.34

3

h2/a

=

1.002014



=

-2.00215-05

0.,4 0.53 0.!8 .80 1.00

J =

55.00

0.!8

0.!8 0.,8

m /dt

m3/dt

I= "ut1

m3/dt

0.23 =

0.43

K

simulasi selesai

m/dt

K

K3 J2

1



-3.40141



-0.0018

A

m m/dt

0 K2

 J



D

2

0.!, J

m

0.53

0.!8 0.,8

"

100

P

h2/a

=

3.023401,,

0.23

3

h2/a

=

1.00325,2



=

-0.000020022

m /dt

0.!8

1

"

100

g = h1/a = h2 = P aliran = Q"ut3 =

I= "ut1

1.00

,ross Se)tion A-D : Q"ut3 = 0.22 K =  = a =

m3/dt

K3 J2

0.!8

1 1



-3.4014,



-0.0018

0 K2

 J




r=

1.25!153 0.,012

r=

1.25!14, 0.,0131

r=

1.25!153 0.,012


=

= Areal 2  A2 = aluran = ;<+2 = e2 > a2

= =

= Areal 3  A3 = aluran = ;<+3 =

e$under








=

0.,2

> a3

=

1

B Q"ut1 =

0.5!

1.13 l/dt/ha 0.8

m

0.!0

1.125

m3/dt

0.5!25

m3/dt

0.33,5

m3/dt

Q"ut3 =

0.225

3

>d =

1.03

Q"ut1 = Q"ut2 =


0.81 l/dt/ha

e3

= bin = Qin =

datar datar miring

0.!1 Qin = 1.125 A

m /dt

0.8

0.30

n = 0.52844,5 m

0.20


1.,!

h1/b

2.02

D

arat

h1/b

tatus

1.20

=

1

K

diambil =

0.!

m

0.24

1.0

=

1

K

diambil =

0.3

m

0.24

2.0

=

0

K

diambil =

0.2

m

0.24

:i&e Aliran

b2h11.5

0.34 1.,!

,ross Se)tion A-0 : Q"ut1 = 0.3

b3h11.5 1.125 m3/dt

K =  = a =

0.,4 0.!0 0.2

m

g = h1/a = h2 =

.80

m/dt2

P aliran = Q"ut1 =

0.22

0.5!

0.225 = 0.22 Q"ut1  Q"ut2  Q"ut3 =

J =

Q"ut3 =

b2 = 0.34!5,2 m b3 = 0.230424, m Q"ut1 = b1h11.5 1.,!

0.33,5 = Q"ut3 =

2.02

0.!1

>atatan = in&ut data (! : in 8 total out h1  n = 1.20 m = 0.5! m ------- K

0.5!25 = Q"ut2 =

Q"ut2 = 0.34 >

I= "ut1

m3/dt m3/dt

0.5! =

2.31

K

simulasi selesai

m/dt

;

A

B 0.!,

2.33 0.5! 55.00 ,0 0.3

J m

0.53

0.2 0.,8

"

P 3

m /dt

h2/a

=

3.1283303,45

h2/a

=

1.20045305



=

!.53,2!-0!

3

K

J2

0.5!

1 1



-4.558!



-0.003!82

0 2

K

 J



,ross Se)tion A-, : Q"ut2 = 0.24 K =  = a =

0.,! 0.!0 0.2

g = h1/a = h2 =

2.33

J =

55.00

P aliran = Q"ut2 =

.80 0.5!

0.24

0.,4 0.!0 0.2

g = h1/a = h2 =

2.33

J =

55.00

P aliran = Q"ut3 =

.80 0.5! ,0 0.1!

I= "ut1

m3/dt

0.34 =

1.12

K

simulasi selesai

m/dt

K

A

m m/dt

>

2

0.!, J

m

0.53

0.2

P 3

m /dt

m3/dt

I= "ut1

=

3.1283303,45

h2/a

=

1.143!05,2



=

!.53,2!-0!

m3/dt

0.23 =

h2/a

0.,8

0.,2

K

simulasi selesai

m/dt

K

K3 J2

1



-4.558!



-0.003!82

A

m m/dt m

0.53

m /dt

K2

 J



0.!, 0.2 0.,8

"

P

0

D

2

J

3

0.5!

1

"

,0

,ross Se)tion A-D : Q"ut3 = 0.1! K =  = a =

m3/dt

h2/a

=

3.1283303,45

h2/a

=

1.20045305



=

!.53,2!-0!

3

K

J2

0.5!

1 1



-4.558!



-0.003!82

0 2

K

 J




r=

1.123355 0.,,13!

r=

1.123355 0.,,13!

r=

1.123355 0.,,13!

DESAIN 0ANGUNAN U!UR AM0ANG LE0AR
=

3

2.00 m /dt

K = 0.28

HASIL PERHITUNGAN : L = 1.12 m (min = 4.00 m <

=

1.20 m

*min

=

0.,0 m

*Rmin

= = = =

0.85 3.80 0.20 2.80

:

= =

1.12 m 0.8! m

hmax

=

0.5! m

BR r B smin

0.5! 5 1

m m m m

2

0.30

,R(SS ,E! : Q = 2.00 m3/dt G = 0.5!

1.12

4.00

0.30

0.20

K 2.80

=

0.14

K

=

0.55

K

=

0.14

K

2.80

1.12

0.8!

0.5! 1 1 0.,0

0.30 0.!0

3.80

0.30 0.!0

DESAIN 0ANGUNAN U!UR ,IP(LETTI
3

2 m /dt

<

= =

4.00 m 1.20 m

*min

=

0.,0 m

*Rmin

= = =

0.85 m 3.,0 m 2.,0 m

:

= =

1.00 m 0.80 m

hmax

=

0.54 m

BR B smin

1.20

K

HASIL PERHITUNGAN : (min

7 = 0.05

4 4

0.4!

5 1

0.0, 0.43 2

,R(SS ,E! : Q = 2.00 m3/dt G = 0.54

0.50

4.00

0.30

45 K 3.,0

=

0.14

K

=

0.5,

K

=

0.10

K

2.,0

1.00 0.3

1 4

0.54

0.80 1 1 0.,0

0.4 3.,0 0.002

0.0,

0.00, 0.0, 1 1 >+: *LA:

DESAIN TALANG INPUT DATA : Apakah lokasi rencana talang berada pada belokan sungai ? a7ab  (a esi!pulan tentang lokasi " esuai untu$ talang tan&a tiang
m/dt ,0 %Bet"n' 10.00 m

1

FGmasu$ = Fh$eluar = Q = B/h =

Pilih utk diisi di kotak hiau %*asangan batu'

0.25

%Bet"n'

0.5

%Ba6a' %Ba6a bet"n'

HASIL PERHITUNGAN : )a = 1.1,1 m/dt )a = i = = FG# =

0.5

1.1,1 m/dt 0.00020 I= 0.5 I= 2.501!5 m 53.5543



1

K

 0.002 0.5

1

K K K

0.002 m 0.0005 m 0.001 m 0.!0 m3/dt 2 e$"n"mis

0.50 10.00

DESAIN G(R(NG-G(R(NG INPUT DATA : Apakah rencana pe!buatan gorong#gorong un tuk saluran irigasi ? a7ab  (a Apakah rencana pe!buatan gorong#gorong un tuk saluran pe!buang ? a7ab  :ida$ >r"ss eti"n T"r"ng-T"r"ng esi!pulan tentang kecepatan aliran " C > =1.5 m/dt
=

1.00

=

1

)"ut

= = =

1

K L Smasu$

=

0.25

S$eluar

=

0.5

m

i FG#

m/dt

;B  H = tida$ di&a$ai

m/dt  = di&a$ai ,0 %Bet"n bertulang' Pilih utk diisi di kotak hiau 10.00 m tida$ di&erlu$an &erhitungan ang teliti

%*asangan batu' %Bet"n bertulang' %Ba6a'

= = = = =

1.5012, m/dt 0.00021 0.5 I= 3.32055 m 5,.2824

=

0.0020, m

=

0.00321 m

Fh$eluar

=

0.00!41 m

Q B/h

= =

1.50 1.!1

= bet"n bertulang

%Ba6a bet"n'

FGmasu$

h

2

HASIL PERHITUNGAN : )a = 1.5012, m/dt )a = 1.5012, m/dt )a

C



1

K



1

K

=

1.5

K

0.5

m3/dt e$"n"mis

K

0., 10.00

DESAIN SIPH(N

esi!pulan tentang lokasi $eru!nas Baler Bale Agung
0.3!

m2

0.5

m/dt

>r"ss eti"n i&h"n >

C

Ta!pang Sa!ping 0.42

0.!0

0.84

0.!0 0.!0

;B  H = tida$ di&a$ai

%Bet"n'

0.5

m/dt  = di&a$ai %Ba6a' ,0 %Bet"n' Pilih utk diisi di kotak hiau %Ba6a bet"n' 3 m tida$ &erlu di&asang lubang &eri$sa %manh"le' dan &intu &embuang N1

0.25 0.5

N2

&emasangan si&"n tida$ memerlu$an ahli me$ani$ dan hidr"li$ 3

HASIL PERHITUNGAN : )a = 2.,8154 m/dt )a = i FG#

= = = =

= FGmasu$ = Fh$eluar = FGb1 = FGb2 =

2.,8154 m/dt 0.00150 1.1 I= 1.8 m 53.530,

 I=

1 3

K K

 I=

1 3

K K

1.5

Ta!pang Atas J2 1.5

K

0.0045 m 1.0844! m

0.!0

2.1!82 m

2

0.10!58 m

0.10!58 m FGbt"tal = 0.2131! m Kehilangan energi &ada si$u / sudut )erti$al  " N1 = 45 N2 = Kb1 = Kb2 =

45 0.2,

"

m

0.2, m Kehilangan energi &ada ti$ungan / sudut h"riM"ntal  +b3 = 2 m +b4 = J1 =

1.5 30

"

J2 = +b3/D =

20

"

+b4/D = Kb3 =

3.33 2.50

0.04 Kb4 = 0.04 :"tal Kehilangan energi )in2/2g = 0.01 )a2/2g = Fh) = 7ater seal min = 7ater seal max = 7ater seal = B/h =

m

m m =

0.3

m

0.38 0.42 0.5, 0.42

m m m m

1

0.84 m

m

= 0.15 I= 0.45 e$"n"mis

K K

J1

001 Perhitungan Terjunan Tegak

Recommend Documents