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MENENTUKAN DEBIT RENCANA DAN DIMENSI SALURAN
SALURAN P0-P1
1.3
1 m
0.7
Vr = 1.5
A =1.3 + 0.7
x 1 = 12
Qr = A x Vr= 1 x 1.5= 1.5 (debit banjir rencana)
Dik: Q = 1.5 n manning = 0.0225L = 150 m
Ls =Elev. Patok P0 - Elev. Patok P1
150
=36.650 - 36.515 7
150= 0.0009
m =0.3
= 0.30 m1
B = 1 m
PENYELESAIAN:TAMPANG ABCDJIKA:A = B + m . H H
= BH + 0.3
P = B + 2 H 1 += B + 2 H 1 + 0.3 2
= B + 2 H 1.09= B + 2.09 H
mH B mH
H
B
SYARAT TAMPANG EKONOMIS
R =AP
JIKA:
R =A
=BH + 0.3 H 2
P B + 2.09 H
FORMULA MANNING
V =I
x R2/'3
x S1/'2
n
V =1
xBH + 0.3 H 2 2/'3
x 0.00091/2
0.0225 B + 2.09 H
V = 1.333 xBH + 0.3 H 2 2/'3
B + 2.09 H
m³/det
m²
m³/det
m³/det
H²
m²
Q = A . V
Q = BH + 0.3 . H2
x 1.333 xBH + 0.3 H 2 2/'3
B + 2.09 H
Q = 1.333BH + 0.3 H 2 5/'3
B + 2.09 H 2/'3
H = 1.30 (DICOBA)
1.5 = 1.333BH + 0.3 H 2 5/'3
B + 2.09 H 2/'3
1.5 = 1.3331 x 1 + 0.3 X 1 2 5/'3
1 + 2.09 x 1 2/'3
1.5 = 1.3331.3 + 0.5 5/'3
3.71 2/'3
1.5 = 1.5 OK
A = BH + 0.3 H 2
= 1.00 . 1.30 + 0.3 . 1.30 2
= 1.81
P = B + 2.09 H= 1.00 + 2.09 x 1.30= 3.71 m
R =A
=1.81
= 0.49 mP 3.71
W = 0.25 x H + 0.49= 0.25 x 1.30 + 0.49= 0.82 m dipakai W = 0.5 m
V = 1.333 xBH + 0.3 H 2 2/'3
B + 2.09 H
= 1.333 x1 . 1.30 + 0.3 . 1.30 2 2/'3
1 + 2.09 . 1.30
= 1.333 x1.8 2/'3
3.71= 1.333 x 0.619= 0.825
KONTROLQ = A x V
= 1.8 x 0.825= 1.5 OK W = 0.50 m
KESIMPULANQ = 1.5A = 1.81 H = 1.30 mP = 3.71 mR = 0.49 mW = 0.50 mB = 1.00 mH = 1.30 m ~ 1 m B = 1.00 mTHALUD = 0.30 : 1
m³/det
m³/det
m³/det
m³/det m³/det
m²
m³/det
m³/detm²
MENENTUKAN DEBIT RENCANA DAN DIMENSI SALURAN
SALURAN P1-P2
1.3
0.30
1 m
0.7
Vr = 1.5
A =1.3 + 0.7
x 1 = 12
Qr = A x Vr= 1 x 1.5= 1.5 (debit banjir rencana)
Dik: Q = 1.5 n manning = 0.0225L = 150 m
Ls =Elev. Patok P1 - Elev. Patok P2
150
=36.515 - 36.380 7
150= 0.0009
m =0.3
= 0.30 m1
B = 1 m
PENYELESAIAN:TAMPANG ABCDJIKA:A = B + m . H H
= BH + 0.3
P = B + 2 H 1 += B + 2 H 1 + 0.3 2
= B + 2 H 1.09= B + 2.09 H
mH B mH
H
B
SYARAT TAMPANG EKONOMIS
R =AP
JIKA:
R =A
=BH + 0.3 H 2
P B + 2.09 H
FORMULA MANNING
V =I
x R2/'3
x S1/'2
n
V =1
xBH + 0.3 H 2 2/'3
x 0.00091/2
0.0225 B + 2.09 H
V = 1.333 xBH + 0.3 H 2 2/'3
B + 2.09 H
m³/det
m²
m³/det
m³/det
H²
m²
Q = A . V
Q = BH + 0.3 . H2
x 9.938 xBH + 0.3 H 2 2/'3
B + 2.09 H
Q = 1.333BH + 0.3 H 2 5/'3
B + 2.09 H 2/'3
H = 1.30 (DICOBA)
1.5 = 1.333BH + 0.3 H 2 5/'3
B + 2.09 H 2/'3
1.5 = 1.3331 x 1 + 0.3 X 1 2 5/'3
1 + 2.09 x 1 2/'3
1.5 = 1.3331.3 + 0.5 5/'3
3.71 2/'3
1.5 = 1.5 OK
A = BH + 0.3 H 2
= 1.00 . 1.30 + 0.3 . 1.30 2
= 1.81
P = B + 2.09 H= 1.00 + 2.09 x 1.30= 3.71 m
R =A
=1.81
= 0.49 mP 3.71
W = 0.25 x H + 0.49= 0.25 x 1.30 + 0.49= 0.82 m dipakai W = 0.5 m
V = 1.333 xBH + 0.3 H 2 2/'3
B + 2.09 H
= 1.333 x1 . 1.30 + 0.3 . 1.30 2 2/'3
1 + 2.09 . 1.30
= 1.333 x1.8 2/'3
3.71= 1.333 x 0.619= 0.825
KONTROLQ = A x V
= 1.8 x 0.825= 1.5 OK W = 0.50 m
KESIMPULANQ = 1.5A = 1.81 H = 1.30 mP = 3.71 mR = 0.49 mW = 0.50 mB = 1.00 mH = 1.30 m ~ 1 m B = 1.00 mTHALUD = 0.30 : 1
m³/det
m³/det
m³/det
m³/det m³/det
m²
m³/det
m³/detm²
MENENTUKAN DEBIT RENCANA DAN DIMENSI SALURAN
SALURAN P2-P3
1.3
0.30
1 m
0.7
Vr = 1.5
A =1.3 + 0.7
x 1 = 12
Qr = A x Vr= 1 x 1.5= 1.5 (debit banjir rencana)
Dik: Q = 1.5 n manning = 0.0225L = 150 m
Ls =Elev. Patok P2 - Elev. Patok P3
150
=36.380 - 36.245 7
150= 0.0009
m =0.3
= 0.30 m1
B = 1 m
PENYELESAIAN:TAMPANG ABCDJIKA:A = B + m . H H
= BH + 0.3
P = B + 2 H 1 += B + 2 H 1 + 0.3 2
= B + 2 H 1.09= B + 2.09 H
mH B mH
H
B
SYARAT TAMPANG EKONOMIS
R =AP
JIKA:
R =A
=BH + 0.3 H 2
P B + 2.09 H
FORMULA MANNING
V =I
x R2/'3
x S1/'2
n
V =1
xBH + 0.3 H 2 2/'3
x 0.00091/2
0.0225 B + 2.09 H
V = 1.333 xBH + 0.3 H 2 2/'3
B + 2.09 H
m³/det
m²
m³/det
m³/det
H²
m²
Q = A . V
Q = BH + 0.3 . H2
x 9.938 xBH + 0.3 H 2 2/'3
B + 2.09 H
Q = 1.333BH + 0.3 H 2 5/'3
B + 2.09 H 2/'3
H = 1.30 (DICOBA)
1.5 = 1.333BH + 0.3 H 2 5/'3
B + 2.09 H 2/'3
1.5 = 1.3331 x 1 + 0.3 X 1 2 5/'3
1 + 2.09 x 1 2/'3
1.5 = 1.3331.3 + 0.5 5/'3
3.71 2/'3
1.5 = 1.5 OK
A = BH + 0.3 H 2
= 1.00 . 1.30 + 0.3 . 1.30 2
= 1.81
P = B + 2.09 H= 1.00 + 2.09 x 1.30= 3.71 m
R =A
=1.81
= 0.49 mP 3.71
W = 0.25 x H + 0.49= 0.25 x 1.30 + 0.49= 0.82 m dipakai W = 0.5 m
V = 1.333 xBH + 0.3 H 2 2/'3
B + 2.09 H
= 1.333 x1 . 1.30 + 0.3 . 1.30 2 2/'3
1 + 2.09 . 1.30
= 1.333 x1.8 2/'3
3.71= 1.333 x 0.619= 0.825
KONTROLQ = A x V
= 1.8 x 0.825= 1.5 OK W = 0.50 m
KESIMPULANQ = 1.5A = 1.81 H = 1.30 mP = 3.71 mR = 0.49 mW = 0.50 mB = 1.00 mH = 1.30 m ~ 1 m B = 1.00 mTHALUD = 0.30 : 1
m³/det
m³/det
m³/det
m³/det m³/det
m²
m³/det
m³/detm²
MENENTUKAN DEBIT RENCANA DAN DIMENSI SALURAN
SALURAN P3-P4
1.3
0.30
1 m
0.7
Vr = 1.5
A =1.3 + 0.7
x 1 = 12
Qr = A x Vr= 1 x 1.5= 1.5 (debit banjir rencana)
Dik: Q = 1.5 n manning = 0.0225L = 150 m
Ls =Elev. Patok P3 - Elev. Patok P4
150
=36.245 - 36.110 7
150= 0.0009
m =0.3
= 0.30 m1
B = 1 m
PENYELESAIAN:TAMPANG ABCDJIKA:A = B + m . H H
= BH + 0.3
P = B + 2 H 1 += B + 2 H 1 + 0.3 2
= B + 2 H 1.09= B + 2.09 H
mH B mH
H
B
SYARAT TAMPANG EKONOMIS
R =AP
JIKA:
R =A
=BH + 0.3 H 2
P B + 2.09 H
FORMULA MANNING
V =I
x R2/'3
x S1/'2
n
V =1
xBH + 0.3 H 2 2/'3
x 0.00091/2
0.0225 B + 2.09 H
V = 1.333 xBH + 0.3 H 2 2/'3
B + 2.09 H
m³/det
m²
m³/det
m³/det
H²
m²
Q = A . V
Q = BH + 0.3 . H2
x 9.938 xBH + 0.3 H 2 2/'3
B + 2.09 H
Q = 1.333BH + 0.3 H 2 5/'3
B + 2.09 H 2/'3
H = 1.30 (DICOBA)
1.5 = 1.333BH + 0.3 H 2 5/'3
B + 2.09 H 2/'3
1.5 = 1.3331 x 1 + 0.3 X 1 2 5/'3
1 + 2.09 x 1 2/'3
1.5 = 1.3331.3 + 0.5 5/'3
3.71 2/'3
1.5 = 1.5 OK
A = BH + 0.3 H 2
= 1.00 . 1.30 + 0.3 . 1.30 2
= 1.81
P = B + 2.09 H= 1.00 + 2.09 x 1.30= 3.71 m
R =A
=1.81
= 0.49 mP 3.71
W = 0.25 x H + 0.49= 0.25 x 1.30 + 0.49= 0.82 m dipakai W = 0.5 m
V = 1.333 xBH + 0.3 H 2 2/'3
B + 2.09 H
= 1.333 x1 . 1.30 + 0.3 . 1.30 2 2/'3
1 + 2.09 . 1.30
= 1.333 x1.8 2/'3
3.71= 1.333 x 0.619= 0.825
KONTROLQ = A x V
= 1.8 x 0.825= 1.5 OK W = 0.50 m
KESIMPULANQ = 1.5A = 1.81 H = 1.30 mP = 3.71 mR = 0.49 mW = 0.50 mB = 1.00 mH = 1.30 m ~ 1 m B = 1.00 mTHALUD = 0.30 : 1
m³/det
m³/det
m³/det
m³/det m³/det
m²
m³/det
m³/detm²
MENENTUKAN DEBIT RENCANA DAN DIMENSI SALURAN
SALURAN P4-P5
1.3
0.30
1 m
0.7
Vr = 1.5
A =1.3 + 0.7
x 1 = 12
Qr = A x Vr= 1 x 1.5= 1.5 (debit banjir rencana)
Dik: Q = 1.5 n manning = 0.0225L = 150 m
Ls =Elev. Patok P4 - Elev. Patok P5
150
=36.110 - 35.975 7
150= 0.0009
m =0.3
= 0.30 m1
B = 1 m
PENYELESAIAN:TAMPANG ABCDJIKA:A = B + m . H H
= BH + 0.3
P = B + 2 H 1 += B + 2 H 1 + 0.3 2
= B + 2 H 1.09= B + 2.09 H
mH B mH
H
B
SYARAT TAMPANG EKONOMIS
R =AP
JIKA:
R =A
=BH + 0.3 H 2
P B + 2.09 H
FORMULA MANNING
V =I
x R2/'3
x S1/'2
n
V =1
xBH + 0.3 H 2 2/'3
x 0.00091/2
0.0225 B + 2.09 H
V = 1.333 xBH + 0.3 H 2 2/'3
B + 2.09 H
m³/det
m²
m³/det
m³/det
H²
m²
Q = A . V
Q = BH + 0.3 . H2
x 9.938 xBH + 0.3 H 2 2/'3
B + 2.09 H
Q = 1.333BH + 0.3 H 2 5/'3
B + 2.09 H 2/'3
H = 1.30 (DICOBA)
1.5 = 1.333BH + 0.3 H 2 5/'3
B + 2.09 H 2/'3
1.5 = 1.3331 x 1 + 0.3 X 1 2 5/'3
1 + 2.09 x 1 2/'3
1.5 = 1.3331.3 + 0.5 5/'3
3.71 2/'3
1.5 = 1.5 OK
A = BH + 0.3 H 2
= 1.00 . 1.30 + 0.3 . 1.30 2
= 1.81
P = B + 2.09 H= 1.00 + 2.09 x 1.30= 3.71 m
R =A
=1.81
= 0.49 mP 3.71
W = 0.25 x H + 0.49= 0.25 x 1.30 + 0.49= 0.82 m dipakai W = 0.5 m
V = 1.333 xBH + 0.3 H 2 2/'3
B + 2.09 H
= 1.333 x1 . 1.30 + 0.3 . 1.30 2 2/'3
1 + 2.09 . 1.30
= 1.333 x1.8 2/'3
3.71= 1.333 x 0.619= 0.825
KONTROLQ = A x V
= 1.8 x 0.825= 1.5 OK W = 0.50 m
KESIMPULANQ = 1.5A = 1.81 H = 1.30 mP = 3.71 mR = 0.49 mW = 0.50 mB = 1.00 mH = 1.30 m ~ 1 m B = 1.00 mTHALUD = 0.30 : 1
m³/det
m³/det
m³/det
m³/det m³/det
m²
m³/det
m³/detm²
MENENTUKAN DEBIT RENCANA DAN DIMENSI SALURAN
SALURAN P5-P6
1.6
0.30
1 m
1
Vr = 2
A =1.6 + 1
x 1 = 1.32
Qr = A x Vr= 1.3 x 2= 2.6 (debit banjir rencana)
Dik: Q = 2.6 n manning = 0.0225L = 125 m
Ls =Elev. Patok P5 - Elev. Patok P6
125
=35.975 - 35.785 7
125= 0.0015
m =0.3
= 0.30 m1
B = 1.2 m
PENYELESAIAN:TAMPANG ABCDJIKA:A = B + m . H H
= BH + 0.3
P = B + 2 H 1 += B + 2 H 1 + 0.3 2
= B + 2 H 1.09= B + 2.09 H
mH B mH
H
B
SYARAT TAMPANG EKONOMIS
R =AP
JIKA:
R =A
=BH + 0.3 H 2
P B + 2.09 H
FORMULA MANNING
V =I
x R2/'3
x S1/'2
n
V =1
xBH + 0.3 H 2 2/'3
x 0.00151/2
0.0225 B + 2.09 H
V = 1.733 xBH + 0.3 H 2 2/'3
B + 2.09 H
m³/det
m²
m³/det
m³/det
H²
m²
Q = A . V
Q = BH + 0.3 . H2
x 9.938 xBH + 0.3 H 2 2/'3
B + 2.09 H
Q = 1.733BH + 0.3 H 2 5/'3
B + 2.09 H 2/'3
H = 1.40 (DICOBA)
2.6 = 1.733BH + 0.3 H 2 5/'3
B + 2.09 H 2/'3
2.6 = 1.7331 x 1 + 0.3 X 1 2 5/'3
1 + 2.09 x 1 2/'3
2.6 = 1.7331.68 + 0.6 5/'3
4.12 2/'3
2.6 = 2.6 OK
A = BH + 0.3 H 2
= 1.20 . 1.40 + 0.3 . 1.40 2
= 2.27
P = B + 2.09 H= 1.20 + 2.09 x 1.40= 4.12 m
R =A
=2.27
= 0.55 mP 4.12
W = 0.25 x H + 0.49= 0.25 x 1.40 + 0.49= 0.84 m dipakai W = 0.5 m
V = 1.733 xBH + 0.3 H 2 2/'3
B + 2.09 H
= 1.733 x1.2 . 1.40 + 0.3 . 1.40 2 2/'3
1.2 + 2.09 . 1.40
= 1.733 x2.3 2/'3
4.12= 1.733 x 0.671= 1.163
KONTROLQ = A x V
= 2.3 x 1.163= 2.6 OK W = 0.50 m
KESIMPULANQ = 2.6A = 2.27 H = 1.40 mP = 4.12 mR = 0.55 mW = 0.50 mB = 1.20 mH = 1.40 m ~ 1 m B = 1.20 mTHALUD = 0.30 : 1
m³/det
m³/det
m³/det
m³/det m³/det
m²
m³/det
m³/detm²
MENENTUKAN DEBIT RENCANA DAN DIMENSI SALURAN
SALURAN P6-P7
1.6
0.30
1 m
1
Vr = 2
A =1.6 + 1
x 1 = 1.32
Qr = A x Vr= 1.3 x 2= 2.6 (debit banjir rencana)
Dik: Q = 2.6 n manning = 0.0225L = 125 m
Ls =Elev. Patok P6 - Elev. Patok P7
125
=35.785 - 35.595 7
125= 0.0015
m =0.3
= 0.30 m1
B = 1.2 m
PENYELESAIAN:TAMPANG ABCDJIKA:A = B + m . H H
= BH + 0.3
P = B + 2 H 1 += B + 2 H 1 + 0.3 2
= B + 2 H 1.09= B + 2.09 H
mH B mH
H
B
SYARAT TAMPANG EKONOMIS
R =AP
JIKA:
R =A
=BH + 0.3 H 2
P B + 2.09 H
FORMULA MANNING
V =I
x R2/'3
x S1/'2
n
V =1
xBH + 0.3 H 2 2/'3
x 0.00151/2
0.0225 B + 2.09 H
V = 1.733 xBH + 0.3 H 2 2/'3
B + 2.09 H
m³/det
m²
m³/det
m³/det
H²
m²
Q = A . V
Q = BH + 0.3 . H2
x 9.938 xBH + 0.3 H 2 2/'3
B + 2.09 H
Q = 1.733BH + 0.3 H 2 5/'3
B + 2.09 H 2/'3
H = 1.40 (DICOBA)
2.6 = 1.733BH + 0.3 H 2 5/'3
B + 2.09 H 2/'3
2.6 = 1.7331 x 1 + 0.3 X 1 2 5/'3
1 + 2.09 x 1 2/'3
2.6 = 1.7331.68 + 0.6 5/'3
4.12 2/'3
2.6 = 2.6 OK
A = BH + 0.3 H 2
= 1.20 . 1.40 + 0.3 . 1.40 2
= 2.27
P = B + 2.09 H= 1.20 + 2.09 x 1.40= 4.12 m
R =A
=2.27
= 0.55 mP 4.12
W = 0.25 x H + 0.49= 0.25 x 1.40 + 0.49= 0.84 m dipakai W = 0.5 m
V = 1.733 xBH + 0.3 H 2 2/'3
B + 2.09 H
= 1.733 x1.2 . 1.40 + 0.3 . 1.40 2 2/'3
1.2 + 2.09 . 1.40
= 1.733 x2.3 2/'3
4.12= 1.733 x 1.000= 1.733
KONTROLQ = A x V
= 2.3 x 1.733= 3.9 OK W = 0.50 m
KESIMPULANQ = 3.9A = 2.27 H = 1.40 mP = 4.12 mR = 0.55 mW = 0.50 mB = 1.20 mH = 1.40 m ~ 1 m B = 1.20 mTHALUD = 0.30 : 1
m³/det
m³/det
m³/det
m³/det m³/det
m²
m³/det
m³/detm²
REKAPITULASI DIMENSI SALURAN
NO PATOKDIMENSI SALURAN
V (m3/det) A (m2) B (m) H (m) T (m) W (m) Ls m
1 P0-P1 150 1.5 0.825 1.81 1.00 1.30 1.30 0.50 0.0009 0.30
2 P1-P2 150 1.5 0.825 1.81 1.00 1.30 1.3 0.50 0.0009 0.30
3 P2-P3 150 1.5 0.825 1.81 1.00 1.30 1.3 0.50 0.0009 0.30
4 P3-P4 150 1.5 0.825 1.81 1.00 1.30 1.3 0.50 0.0009 0.30
5 P4-P5 150 1.5 0.825 1.81 1.00 1.30 1.3 0.50 0.0009 0.30
6 P5-P6 125 2.6 1.163 2.27 1.20 1.40 1.6 0.50 0.0015 0.30
7 P5-P6 125 3.9 1.733 2.27 1.20 1.40 1.6 0.50 0.0015 0.30
JARAK (M) Q
(m3/det)
PERHITUNGAN DIMENSI BOX CULVERT 1
DIK: Q = 1.5B = 1.00 mM = 0.30S = 0.0009H = 1.30 mV = 0.825L = 12 mZ = 0.15 m
COBA KECEPATAN ALIRAN BOX CULVERT, V = 2.000
A =QV
=1.5
= 0.75 m 2
2
Y
BB = 1.00 M
BY =QV
Y =Q
V.B
Y =1.5
2 x 1.00= 0.75
LUAS BASAH BOX CALVERTA = B x Y
= 1.00 x 0.75= 0.75
V =QA
=1.5
= 20.75
KEHILANGAN ENERGI (HEADLOS) Z = + +
KEHILANGAN ENERGI DI INLET
Δhin = SinV - Vu 2
2 g
PAKAI TRANSISI RECTANGULAR:Sin = 0.5 tabel 6.2
MAKA
Δhin = 0.52 - 0.82 2
2 . 9.81= 0.0352 m
KEHILANGAN ENERGI DIOUTLET: Sout = 1 TABEL 6.2
= SoutV - Vu 2
2 g
m³/det
m³/det
m³/det
m²
m³/det
Dhin Dhf Dhout
Dhout
PAKAI TRANSISI RECTANGULARMAKA
= 1.02 - 0.82 2
2 . 9.81= 0.0704 m
KEHILANGAN AKIBAT GESEKAN
ΔHf =V 2 . LK 2 . R 3/'4
DIMANAP = B + 2 Y
= 1.00 + 2 . 0.74515= 2.49031 m
R =A
=0.75
P 2.49031= 0.299
UNTUK BETON, KOEFISIEN KEKERASAN K = 70
ΔHf =2 2 . 12
70 2 . 0.299 4/'3 ###
=48
980.66= 0.0489 m
MAKA TOTAL ENERGI YG DIPERLUKANZ = + +
= 0.0352 + 0.0704 + 0.0489= 0.15 < 0.15 OK
Dhout
Dhin Dhout DHf
PERHITUNGAN BANGUN TERJUN
DIK: Q = 1.5B = 1.00 mm = 0.30S = 0.0009h = 1.30 mV = 0.825
a HITUNG ENERGI UDIK H1
H1 = h1 +2
2 g
H1 = 1.30 +0.825 2
2 . 9.81= 1.335 m
b AMBIL PENAMPANG PENGONTROL ALIRAN SEGI EMPAT DENGAN LEBAR bc = 1.00 mMAKA DEBIT PER SATUAN LEBAR:
q =Q
=1.4903
= 1.4903 m³/detbc 1.00
c 1.5 mPERKIRAAN AWAL TINGGI TERJUNHd = 2.2 H1 = 2.2 x 1.335
= 2.9363 m= 1.5 + 2.9363 - 1.335= 3.102 m
d HITUNG KONYUGASI, yu DAN ydKECEPATAN, Vu = 2 g ( DZ ) 60.853
= 19.62 x 3.102= 7.309 m/det
yu =q
=1.4903
= 0.2039 mVu 7.309
BILANGAN Froude Fr
Fr =Vu
g . yu
=7.309
9.81 x 0.2039
=7.309
1.7012= 4.296
DARI GRAFIK 6.10 DIDAPATLp
= 2 MAKA Lp = 2 x 3.102 = 6.203 mDZyd
= 0.5 MAKA yd = 0.5 x 3.102 = 1.551 mDZ
e KONTROL KEDALAMAN TERJUNAN:
y2 = 1.5 M, MAKA y2=
1.5yu 0.20
= 7.357
m³/det
m³/det
V1
TINGGA TERJUN: DZ = DH + Hd - H1, DIMANA DH =
JADI DZ
UNTUK KEAMANAN LONCATAN AIR, MAKA yd ≤ n+y2, DAN TINGGI ENERGI Hd ≥ n+y2+(V2²/2g). KARENA DIMENSI SALURAN DI UDIK DAN DI HILIR BANGUNAN SAMA, MAKA TINGGI MUKA AIR DI HILIR BANGUNAN TERJUN
Fr = 4.296NILAI:
n/y2 =1
= 0.66671.5
n = 0.204 x 0.6667= 0.1359 m SEHINGGA
y2 + n = 1.5 + 0.1359= 1.6359
f Yd = 1.55 ≤ n + y2 = 1.6359 ............... OKE!
Hd = ≥ n + y2 +²
= 2.936 ≥ 1.6359 +0.825 2
2 g 2 x 9.81= 2.936 ≥ 1.671 ........................OKE!
PANJANG LONCATAN AIRLj = 5 n + y2
= 5 0.1359 + 1.5= 8.180 m ~ 8 m
Lp = 6.203 m ~ 6 m
PANJANG KOLAM OLAKLb = Lj + LpLb = 8 + 6
= 14 m
V2