Irigasi Dan Bangunan Air
Irigasi Dan Bangunan Air
Irigasi Dan Bangunan Air
Oleh :
Kandiyas Sakta
DAFTAR ISI
HALAMAN JUDUL
KATA PENGANTAR .......................................................................................................
iii
B.
.........................................
C.
11
...............................................................................
15
F.
................................................................................ 21
................................................................................ 24
Kandiyas Sakta
30
BB 1
SP 1
+ 96.7
SS 1
BB 2
SP 2
Gorong-gorong
+ 96.4
+ 95.9
SS 2
Kandiyas Sakta
Kandiyas Sakta
: 29 m
: 0,7
: 0,0004
: 35 m/det
: 105 m3/det
Debit Normal
: 65 m3/det
Debit pengambilan
: 4,0 m3/det
Penyelesaian :
Perumusan debit
Q = AxV
Q = (A) x (Ks x R2/3 x )
Q = (A) x (1/n x (A/P)2/3 x )
Mencari Luasan Bendung
A = (b + m.h)h
= (29 + 0,7.h)h
= 29h + 0,7h2
Mencari keliling basah saluran
2
P = b + 2h
= 29 + 2h
= 29 + 2,441h
] 2/3 x
Kandiyas Sakta
1/0.03 x ( [
] 2/3 x
-
= 1/0.03 x ( [
= 1,205 m/det
Check Kontrol
Q = AxV
= 87,149 m2 x 1,205 m/det
= 105,014 m3/det 105 m3/det
(OK)
x LBendung
x 34 m
= 3,4 m 3,5 m
dibulatkan ke atas
Luas Pembilas
= 3,5 m
= 3
Tebal Pilar
= 1,5 m
Kandiyas Sakta
Bangunan
Pembilas
1.25
1.00
3.50
1.25
1.50
9.00
1.50
34.00
30.50
9.00
1.50
8.00
= 34 [ 3,5 m + (3 x 1,5 m) ]
= 26 m
Kp
Ka
= 26 ( (0,05 + 0,10) x H1 )
= 26 0,15 H1
Kandiyas Sakta
) x Leff x H11,5
)
(
= 1,438 m
o
= 0,9179 m
= 1,052 m
1.052
0.9885
Kandiyas Sakta
Check Kontrol
Cd = C0 x C1 x C2
= 1,3 x 0,9885 x 1
= 1,28505
1,285 = 1,285
o
(OK)
K = 2,000
N = 1,850
Y = (1,438) x
x.
= 0,367
Untuk pertemuan dengan garing kemiringan hilir 1:1, maka persamaannya :
= 1
= 1 = (0,367) x (1,850)
Kandiyas Sakta
= (0,367)
= 3,253 m
= 3,254 m
0,40
0,80
1,20
1,60
2,00
2,40
2,80
3,20
3,253
Y (m)
0,067
0,243
0,514
0,876
1,323
1,854
2,466
3,156
3,254
Kandiyas Sakta
C. Peredam Energi
Menghitung tinggi jatuh
Z = Elv. Crest Bendung Elv. dasar hilir
= 97,32 m 95,5 m
= 1.82 m
Menentukan kecepatan awal loncatan
V1 = ( ( )) (
= ( (
))
)
((
) (
= (
)
)
= 7,109 m/det
Menentukan kedalaman awal loncatan
-
=
= 4,074 m2/det
Y1
=
= 0,573 m
= 4,653
/ x .
(
)/ x .
/
(
= (0,2865 m) x (12,198 m)
= 3,495 m
Kandiyas Sakta
10
No
DEFINISI
BLOK MUKA
INISIAL
RUMUS
HASIL (m)
INISIAL
BLOK HALANG
RUMUS
HASIL (m)
2,7 x (Y2)
=
=
2.7 x (3.495)
9.437
2,7 x (Y2)
=
=
2.7 x (3.495)
9.437
0.82 x (Y2)
=
=
0.82 x (3.495)
2.866
0.82 x (Y2)
=
=
0.82 x (3.495)
2.866
Yu = Y1
0.573
Y1 x ( 4 + fr1 )
6
=
=
0.573 x ( 4 + 4.653)
6
0.826
3 Tinggi blok
Yu
h3
h3
4 Lebar blok
Yu
Yu = Y1
5 Tebal blok
0.5 x (Y1)
Y1 x ( 18 + fr1 )
18
=
=
=
=
Kandiyas Sakta
0.573
0.5 x (0.573)
0.2865
h3
0.75 x (h3)
=
=
0.75 x (0.826)
0.620
h3
0.2 x (h3)
=
=
0.2 x (0.826)
0.165
0.375 x (h3)
=
=
0.375 x (0.826)
0.310
0.573 x ( 18 + 4.653))
18
0.721
11
D. Perencanaan Intake
Perencanaan Intake
Perumusan Debit
Q = AxV
Menentukan luasan
A = (b + mh) h
= 29 h + 0,7 h2
Menentukan luasan basah saluran
P = b+2h
= 29 + 2 h
= 29 + 2,441 h
Menentukan kecepatan aliran
V = Ks x
= 35 x . /
h = 1,84866 m
P = 33,983 m2
A = 62,119 m2
V = 1,047 m/det
Check control
Q =AxV
= 62,119 m2 x 1,047 m/det
= 65,039 65 m3/det
Mencari H1
o
Kandiyas Sakta
12
Persamaan Debitnya
Q = Cd x
x Leff x H11,5
Hd = H1 = 1,0882 -
(
(
)
)
= 1,032 m
Menentukan nilai Co, C1 & C2
o
Nilai C1
- P = Elv. Crest Bendung Elv.dasar hulu
= 97,32 m 96,00 m
= 1,32 m
=
= 1,2791 m
=
= 1,054 m
Kandiyas Sakta
13
1.047
0.995
Dari grafik C1 di dapatkan = 0,995
Nilai C2
Kemiringan hulu 1 : 0,67
=
= 1,213 m
1.005
1.213
Dari grafik C2 di dapatkan = 1,005
Check control
Q = Co x C1 x C2
= 1,3 x 0,995 x 1,005
= 1,2999 1,3 (ok)
Kandiyas Sakta
14
-a
-b
-z
Menentukan Q rencana
Diasumsikan tebal pilar = 1 meter
-
- Q
= mxaxbx
(
)(
4,8
= 0,7 x 1,132 x b x
4,8
= 0,7973 x b x 1,981
Kandiyas Sakta
15
1.50
1.00
1.50
Qn = 4,0 m3/det
W = 0,004 m/det
T
= 1 minggu
Kandiyas Sakta
16
B syarat =
=
= 11,18 m
(B) > ( B syarat = 11,18 m)
L syarat =
=
= 89,45 m
(L) > (L syarat = 89,445 m)
di ambil L = 89,69 m
L/B =
= 8,044 > 8
( OK )
Koefisien Strickler
60
70
55
35 - 45
85
Kandiyas Sakta
17
hn =
=
= 0,897 m
= ( b + mh )h
10,0 = ( b + ( 2 x (
) x 0,897 m) x 0,897 m
0.897
3
0.4
9.00
= 11,153 m
Kandiyas Sakta
18
=
`
= 0,897 m
Kandiyas Sakta
19
= 0,356 m
)
(
= 0,329 m
/
(
=
=
= 0,470 < 1
(OK)
Kandiyas Sakta
20
0.897
+ 97.094
+ 97.09
In
0.356
In
+ 97.094
0.178
Is
54.415
Pintu Intake
+ 97.22
+ 97.12
Saluran Pengantar
+ 97.1
Kantong Lumpur
+ 97.094
+ 97.09
+ 96.738
+ 96.00
Kandiyas Sakta
SP 1
+ 97.02
+ 96.556
54.415 m
21
+ 96.556
a) Saluran Primer 1
Diketahui
: Q = 4,0 m3/det
Ks = 55 (pasangan batu kali)
L = 300 m
Q ( m3 / det)
b/h
0.00 0.15
0.25 0.30
1:1
0.15 0.30
0.30 0.35
1:1
0.30 0.40
1.5
0.35 0.40
1:1
0.40 0.50
1.5
0.40 0.45
1:1
0.50- 0.75
0.45 0.50
1:1
0.75 1.50
0.50 0.55
1:1
1.50 3.00
2.5
SP 2
0.55 0.60
1 : 1,5
SP 2
3.00 4.50
SP 1
0.60 0.65
1: 1,5
SP 1
4.50 6.00
3.5
0.65 0.70
1 : 1,5
6.00 7.50
0.70
1 : 1,5
7.50 9.00
4.5
0.70
1 : 1,5
9.00 11.00
0.70
1 : 1,5
11.00 15.00
0.70
1:2
15.00 25.00
0.75
1:2
25.00 40.00
10
0.80
1:2
40.00 60.00
12
1.00 1.50
0.6
0.00 0.30
0.3
1.50 2.50
0.75
0.30 0.50
0.4
> 2.50
1.0
0.50 1.00
0.5
Q (m3/det)
Kandiyas Sakta
22
V = 0,65
= 3 b=3h
m = 1 : 1,5
= 1
A = (b + m h) h
= (3 h + 1,5 h) h
= 3 h2 + 1,5 h2
= 4,5 h2 meter
P = b+2h
= 3h+2h
= 3 h + 3,61 h
= 6,61 h meter
Perumusan Debit
Q
= AxV
= 2,925 h2
= 1,169 m
=
= 0,796 m
Kandiyas Sakta
23
Mencari kemiringan
V = Ks x R2/3 x
I
= .
= .
/
(
h = I x L
= (0,0001893) x (300)
= 0,0568 m 0,057 m
1.00
9.985
3.00
+ 99.189
+ 99.189
0.30
1.00
+ 98.189
1
1.5
1.169
0.30
1.169
+ 97.02
0.60
0.30
0.50
3.507
0.50
Kandiyas Sakta
24
1.00
9.985
3.00
+ 99.132
+ 99.132
0.30
1.00
+ 98.132
1
1.5
1.169
1.169
+ 96.963
0.30
0.30
0.50
3.507
0.50
b) Saluran Primer 2
: Q = 3,0 m3/det
Diketahui
- m = 1,5
- = 1
Untuk saluran bentuk trapesium
A = (b + m h) h
= (2,5 h + 1,5 h) h
= 2,5 h2 + 1,5 h2
= 6 h2 meter
P = b+2h
= 2,5 h + 2 h
= 2,5 h + 3,61 h
= 6,11 h meter
Perumusan Debit
Q
Kandiyas Sakta
= AxV
25
= 0,913 m
= 2,5 (0,913)
= 2,283 m
= 6 (0,9132)
= 5,001 m2
R=
= 0,897 m
Mencari kemiringan
V = Ks x R2/3 x
I
= .
= .
/
(
h = I x L
= (0,000137) x (400)
= 0,0548 m
Kandiyas Sakta
26
1.00
7.996
3.00
+ 98.726
+ 98.726
0.30
1.00
1.00
+ 97.726
1
1.5
0.913
0.913
+ 96.813
0.30
0.60
0.50
2.283
0.50
1.00
7.996
3.00
+ 98.573
+ 98.573
0.30
1.00
1.00
+ 97.573
1
1.5
0.913
0.913
+ 96.66
0.30
0.50
0.60
2.283
0.50
Kandiyas Sakta
27
Q gorong-gorong = 3 m3/det
Direncanakan :
-
Penyelesaian
Persamaan Debit :
Kandiyas Sakta
28
)
(
I = 0,00116
Fr =
= IxL
= 0.00116 x 10
= 0,0116 m
= 0,2 [
= 0,00826 m
= 0,4 [
= 0,0165 m
Kandiyas Sakta
29
Kandiyas Sakta
30
Kandiyas Sakta
31
Kandiyas Sakta
32
Kandiyas Sakta
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Kandiyas Sakta
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Kandiyas Sakta
35
Kandiyas Sakta
36