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8/13/2019 MATERI - 2 Saluran Komposite & Gabungan
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HYDRAULICS 1
HIDROLIKA
SALURAN KOMPOSIT GANDA,PENAMPANG SALURAN PALINGEKONOMIS
SALURAN TERBUKAMATERI - 2
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HYDRAULICS 2
SALURAN KOMPOSIT
Aliran dalam saluran dengan kekasaran bervariasi untuk tiap bagian daripenampang
Horton dan
Einstein (1942)
213
2
Sn
RV43
23
23
S
VnRor
N
1iii
N
1ii PRRPorAA
N
1iiiPn
S
V
S
Vn23
43
23
43
23
23
3
2
23
P
nP
n
N
1iii
e
Lotter
N
1i i
ii
e
n
RP
PRn
3
5
35
213
2
SP
A
n
1V
213
5
213
5
Sn
RPS
n
PR N
1 i
ii
e
2
13
5
Sn
PRQ
PRA
213
2
SP
PR
n
PRQ
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HYDRAULICS 3
PENAMPANG SALURAN GANDA
Saluran dengan kekasaran bervariasi tetapi dengan batas yang jelas
antara daerah / area aliran
Q1Q2
Q3
321 QQQQ
213
2
213
2
213
2
SP
A
n
AS
P
A
n
AS
P
A
n
AQ
3
3
3
3
2
2
2
2
1
1
1
1
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HYDRAULICS 4
CONTOH SOAL 2-1
Problem:
A trapezoidal channel with side slopes 1:1 and bed slope 1:1.000 has a 3 m wide bedcomposed of sand (n = 0.02) and side of concrete (n = 0.014). Estimate thedischarge when the depth of flow is 2.0 m.
Solution:A1(=A3) = 2x2/2 =2.0 m
2 A2= 3x2 = 6.0 m2 A= 10.0 m2
P1(=P3) =(4+4)0.5= 2.828 m P2 = 3.0 m P = 8.656 m
R1(=r3) = 2/2.828 = 0.7072 m R2 = 6/3 = 2.0 m R = 10/8.656 =1.155 m
3.0 m
2.0 m
1
1
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HYDRAULICS 5
CONTOH SOAL 2-1 (continued)
Lotter
ne= 0.0157
Q = 22.17 m3/dt
N
1i i
ii
e
n
RP
PRn
3
5
3
5
02.0
2x3
014.0
7072.0828.22
155.1x656.8n
25
35
35
e
21
32
001.0x155.1x0157.0
10Q
3
2
2
3
P
nP
n
N
1iii
e
Horton - Einstein
3
2
23
23
656.8
02.0x3014.0x282.22ne
21
32
001.0x155.1x0162.0
10Q
ne= 0.0162
Q = 21.49 m3/dt
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HYDRAULICS 6
CONTOH SOAL 2-2
Problem:
The cross section of the flow in a river during a flood was shown in the followingfigure. The roughness coefficient for the side channel and the main channel are 0.04and 0.03 respectively. Bed slope 0.005. Estimate the discharge if the area of the mainchannel (bank full) 280 m2and wetted perimeter of main channel 54 m.
Solution:A1(=A3) = 76.125 m
2 A2= 280+60 = 340 m2 A= 492.25 m2
P1(=P3) = 52.14 m P2 = 54 m P = 158.24 m
R1(=r3) = 1.461 m R2 = 6.296 m R = 3,111 m
40 m
1,5 m
1
1
40 m40 m
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HYDRAULICS 7
CONTOH SOAL 2-2 (continued)
Q = 3,079 m3/dt
21
21
32
005.0x296.6x03.0
340005.0x461.1x
04.0
125.76x2Q
32
Solution of this problem by the equivalent roughness method of Horton willproduce large error in the computed discharge due the inherent assumptions.
However the Lotter method should produce a similar result.
0241.0
03.0296.6x54
04.0461.112.522
111.3x24.158n
25
35
35
e
dt/m077,3005.0x111.3x0241.0
25.492Q 32
132
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HYDRAULICS 8
PENAMPANG SALURAN PALING EKONOMIS
Bagian saluran dianggap paling ekonomis ketika dapat
melewatkan debit maksimum pada luas penampang, koefisien
hambatan dan kemiringan dasar
Berdasarkan persamaan kontinuitas, jelas bahwa untuk luaspenampang yang konstan, debit maksimum saat kecepatan
aliran maksimum.
Dari rumus Chezy atau Manning dapat dilihat bahwa untuk nilai
tertentu dari kemiringan dan kekasaran, kecepatan aliran akan
maksimum jika R (radius hidrolik) maksimum. Untuk luas penampang konstan, R (radius hidrolik) adalah
maksimal jika P (perimeter basah) adalah minimum.
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HYDRAULICS 9
BhA
SALURAN SEGIEMPATYANG PALING EKONOMIS
B
h
h
AB
h2h
AP
02h
A
dh
dP2
Bhh2A 2
2
BhORh2B
2
h
h2h2
h2R
2
r
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HYDRAULICS 10
SALURAN TRAPESIUMYANG PALING EKONOMIS
B
hr
m
1
h)mhB(A
1mh2BP 2
1mh2PB 2
22
mhh1mh2PA
222 mh1mh2PhA
0mh21mh4Pdh
dA 2 mh21m4P 2
0h21m
m2
h42
1
dm
dP
2
31
31m
3h23h3
23h
3
8P 3h
3
23h
3
43h2B
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HYDRAULICS 11
SALURAN SEGITIGAYANG PALING EKONOMIS
r
h
mm
11
tanhA 2
tan
Ah
sech2P
sectan
A2P
0
tan2
sec
tan
tansecA2
d
dP
2
3
3
0sec-tansec 22
0sec-2tan 22 sectan2 = 45o, or m = 1.
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HYDRAULICS 12
CONTOH SOAL 2-3
Problem:
Saluran Trapesium mempunya kemiringan dasar 1:5.000, and koefisien
Manning 0,012. Tentukan penampang paling ekonomis untuk debit 10 m3/det.
2
hR
3hA
3h2P
2
Solution:
213
2
S2
h
n
1x3hQ 2
21
32
000,5
1
2
h
012.0
1x3h10 2
By trial and error method, it is found
h = 2.16 m
3h3
2B B = 2.49 m
B=2,49 m
h=2,16 mr
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HYDRAULICS 13
PROBLEMS
1. Design the most economical trapezoidal channel composed of concrete
(n=0,02) to bring the discharge of 25 m3/s. The land slope is 0,1%.
2. Do the problem No. 1 for a rectangular channel composed of silt having
manningn = 0,025, and the channel would be built in land of 0,05%
gradient.3. Determine the discharge in the channel shown below. The slope is 0.0002.
Mannings n is 0.025 for the part below EL. 72, and 0.05 for the part above
El. 72.
EL. 74 m
EL. 73 m
25 m25 m
5 m
EL. 72 m
EL. 70 m
Q1Q2
Q3
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HYDRAULICS 14
PROBLEMS
5. A rectangular has a bed slope 1:1.250 and 5 m bottom wide composed
of gravel (n = 0.03) and side of masonry (n = 0.02). Estimate thedischarge when the depth of flow is 2.5 m.
6. A trapezoidal channel with side slopes 1:1.5 and bed slope 1:1.200 hasa 5 m bottom wide composed of sand (n = 0.025) and side of concrete(n = 0.015). Estimate the discharge when the depth of flow is 2.5 m.
4. A river is diverted around the city throuh a dyked channel; the cross-
section is shown below. In the central portion Mannings n is 0.024, and
in the flood plain Mannings n is 0.05. The slope is 0.0001. Calculate
the discharge when the water level is 9.14 m above the channel bottom.
122 m
30 m
7,61 m
9,14 m
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