MATERI - 2 Saluran Komposite & Gabungan

8/13/2019 MATERI - 2 Saluran Komposite & Gabungan

1/14

HYDRAULICS 1

HIDROLIKA

SALURAN KOMPOSIT GANDA,PENAMPANG SALURAN PALINGEKONOMIS

SALURAN TERBUKAMATERI - 2


2/14

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


3/14

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


4/14

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


5/14

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


6/14

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


7/14

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


8/14

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.


9/14

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


10/14

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


11/14

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.


12/14

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


13/14

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


14/14

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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MATERI - 2 Saluran Komposite & Gabungan