Contoh Soal Pelabuhan

Oleh Alex Joe

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EOC 6430, Coastal Structures
Homework 1
1.
Construct the wind field for Hurricane Andrew at the instrument height and plot the
results using the following NOAA data:
Central Pressure:

922 mb (27.74 inHg) at 18:00 hrs, 23 Aug 1992
930 mb at land fall, 18:00 hrs, 24 Aug 1992
Ambient Pressure:
1014 mb (30 inHg)
Radius to maximum wind: 24 km (15 mi.)
Air Temperature:
22C (72oF)
Forward speed at land fall: 25 km/hr




Assume land fall at Latitude 25.5N Æ
f = 2ω sin λ = 2(0.286 radians/hr )sin (25.5) = 0.226 hr -1
Assume direction of motion is due west
Tair = 22C Æ ρair ~ 1.0 kg/m3

SPM Method
fr
f 2r 2 1
R
+
+ (p a − p e ) e −R / r
2
4
ρa
r
(add the square root term in the Northern Hemisphere)
Vg = −

assume the Line of Maximum Wind is at 115 degrees from the direction of forward
motion Æ θom = 180-115 = 65 degrees from the x-axis (direction of motion is
assumed to be due west or 180 degrees)
Calculate V(r ) = Vg (r ) + 0.5Vf cos θ , where θ is measured from the Line of Maximum

Wind, 65 degrees from the x-axis Æ at 65 degrees, V(r ) = Vg (r ) + 0.5Vf

at the instrument height (10 m) Æ V10 = 0.865Vg Æ V10 (r ) = 0.865Vg (r ) + 0.5Vf cos θ
units: R and r in km; f in rad/hr, ρa in kg/m3, pa and pe in mb (1 mb = 100 Pa)
−1

2

2

km
1 100 kg ⋅ m 1
km
 km 
 km 
m
Æ Vg = −
Vg = −
+ 
+ 
 +
 + 100  Æ
3
2
2
hr
kg / m
s
m
hr
 hr 
 hr 
s
2

2

km
 km 
 3600 m 
Vg = −
+ 
 + 100
 Æ
hr
 hr 
 1000 s 
Vg = −

fr
f 2 r 2 1296
(pa − pe ) R e −R / r
+
+
2
4
ρa
r

CEM Method
1/ 2

 rf  2 B(p − p )  R  B
  R  B  
n
c
Vgr =   +
  exp −    
ρa
r
 2 
  r   



rf
2

from Figure II-2-16 B~1.3 for p = 930 mb
assume the Line of Maximum Wind is at 115 degrees from the direction of forward
motion Æ θom = 180-115 = 65 degrees from the x-axis (direction of motion is
assumed to be due west or 180 degrees)
Calculate V(r ) = Vg (r ) + Vf cos θ (CEM says use direct vector addition), where θ is
measured from the Line of Maximum Wind, 65 degrees from the x-axis
Correction to instrument height should be done with Figure II-2-13 coefficient,
that V(r ) =

V10
, such
Vg

V10
Vg (r ) + Vf cos θ but requires a sea temperature. Æ assume
Vg

V10
≈ 0.865 Æ V(r ) = 0.865Vg (r ) + Vf cos θ
Vg

Note: CEM does not recommend using a constant (0.865) to get the 10-m wind
speed.

Land Fall Plots
Hurricane Andrew Wind Field (SPM Method, wind speeds in km/hr)

10
0

60

80

↑ NORTH
← line of max wind

150
10 0

12 0

80

10 0

20
114 0

12 0

12 0

80

0

16 0
14 0

50

24 Aug 1992
1800 EST
Land Fall

16 0

14 0

100

60

y distance from hurricane center (km)

80

200

-50

po = 930 mb

pamb = 1014 mb
R = 24 km

10
0

V f = 25 km/hr

← hurricane direction

-100

80

60

-150
60
40

-200
-200

-150 -100 -50
0
50
100 150
x distance from hurricane center (km)

200

Hurricane Andrew Wind Field (CEM Method, wind speeds in km/hr)
0

↑ NORTH8

200
60

← line of max wind

100
12 0

po = 930 mb

80

12 0

pamb = 1014 mb

60

60

-100

16 0

14
0

0
-50

24 Aug 1992
1800 EST
Land Fall

14
0

80

10 0

20 0
16
0
112
4 00

0
16

50

80

12 0

0
10

100

40

y distance from hurricane center (km)

150

10 0

← hurricane direction
8
0

-150

60

40

-200
-200

-150 -100 -50
0
50
100 150
x distance from hurricane center (km)

40

200

R = 24 km

V f = 25 km/hr

Line of Maximum Wind
240
Vmax-cem = 216.2 km/hr

220

Rcem = 24 km

24 Aug 1992
1800 EST
Land Fall

200
Vmax-spm = 183.3 km/hr
Rspm = 24 km

180

po = 930 mb

wind speed (km/hr)

pamb = 1014 mb
R = 24 km

160

V f = 25 km/hr
140
120
100
80
60
40

0

50

100
150
distance from hurricane center (km)

200

250

Offshore Plots

200

↑ NORTH 0
10

← line of max wind

12 0

10 0

10 0

14 0

0
12

-50

24 Aug 1992
1800 EST
Offshore

12 0

20
0

80

0

14 16 0
0

50

0
16

10 0

14
0

100

po = 922 mb

pamb = 1014 mb
R = 24 km

80

60

y distance from hurricane center (km)

150

80

80

Hurricane Andrew Wind Field (SPM Method, wind speeds in km/hr)

Vf = 25 km/hr

← hurricane direction100

-100

80

-150

60
60

40

-200

-200

-150 -100 -50
0
50
100 150
x distance from hurricane center (km)

200

Hurricane Andrew Wind Field (CEM Method, wind speeds in km/hr)

↑ NORTH

80

80

60

200

← line of max wind

10
0

12
0

14
0

50

0
10

100

24 Aug 1992
1800 EST
Offshore

16 0

-50

80

10 0

12006 0
14
1

1
14 6 0
0
12
0

12 0

80

0

220

20 0

60

po = 922 mb

pamb = 1014 mb
R = 24 km

10
0

V f = 25 km/hr

← hurricane direction

60

-100

40

y distance from hurricane center (km)

150

80

-150

60
40

-200

40

-200

-150 -100 -50
0
50
100 150
x distance from hurricane center (km)

200

Line of Maximum Wind
240
Vmax-cem = 225.2 km/hr

Rcem = 24 km

220

200

Vmax-spm = 191.3 km/hr

24 Aug 1992
1800 EST
Offshore

Rspm = 24 km

180

po = 922 mb

wind speed (km/hr)

pamb = 1014 mb
R = 24 km

160

Vf = 25 km/hr
140

120

100

80
60

40

0

50

100
150
distance from hurricane center (km)

200

250

2.
Using the data from problem 1, estimate the deep water wave parameters using the SMB
model. Assume fetch is equal to the radius of maximum wind and duration is 10 hours. Is the
condition fetch limited or duration limited?
Assume

U = 204 km/hr (highest wind speed in above two cases)
R = 24 km
t = 10 hrs = 36000 sec

Using the SMB model
Determine if seas are Fetch Limited or Duration Limited
Assume F = R = 24 km
U = 204 km/hr(hr/3600 sec)(1000 m/km) = 56.7 m/s
1.23
U A = 0.71U 1.23 = 0.71(56.7 ) = 101.9 m/s

 gt 
gFt

= 0.015
2
UA
U A 

3/ 2

Æ

9.8 Ft

(101.9)2

 9.8(36000 ) 
= 0.015

 101.9 

3/ 2

solving for Ft gives: Ft = 3200 km Æ Fetch Limited
F
H s = 1.6 × 10 U A  
g

1/ 2

−3

= 1.6 × 10 −3 (101.9)

1/ 3

U A  gF 


Ts = 2.7 ×10
g  U A2 
−1

24000
= 8.1 m
9.8
1/ 3

 101.9  9.8 × 24000 
= 0.27


2

 9.8  101.9

= 7.9 sec

SMB results Æ Hs = 8.1 m, Ts = 7.9 sec

Compare this to the Hurricane Wave formulae (from SPM, 1984)
 0.29αVF 
 exp R∆p  = 9.1 m
H 1 / 3 = 5.031 +

U R 
 4700 

 0.145αVF 
 exp R∆p  = 11.6 sec
T1 / 3 = 8.61 +

U R 
 9400 

∆p = (1014 – 922) mb x 760 mm/1013 mb = 69 mm Hg
Vf = 25 km/hr = 6.9 m/s
UR = 56.7 m/s
α = 1 (assumed)

3.
Wave data was collected by a submerged pressure gage at 10 m water depth at a project
site. The annual maximum values are tabulated below. Perform an extreme value analysis and
determine the design wave height for the following:
a.
b.
c.
d.

Design wave for 100 year and 500 year return periods
Design wave for a risk of encounter = 50% , assume structure life = 50 years
Design wave for a risk of encounter = 20%
Design wave for a risk of encounter = 80%
Year
1982
1983
1984
1985
1986
1987

H1/3 (m)
3.21
2.61
2.94
2.70
3.76
2.54

Ts (sec)
8.6
8.0
8.5
8.4
8.4
8.2

Year
1988
1989
1990
1991
1992

H1/3 (m)
2.85
2.2
2.56
2.45
2.42

Ts (sec)
8.0
7.9
8.2
8.0
7.9

Using the Type I Asymptotic Distribution Æ F ( y n ) = exp[− exp(− α( y n − µ ))]
Where A =

6
6
var[y n ] =
σ and B = E[y n ] − 0.577 A
π
π

Let extreme wave height be H1/10 = 1.8Hs
2

σ
µ = E[yn]
A
B

Hs
0.188
2.554
0.338
2.749

H1/10
0.610
4.597
0.609
4.948

1
, where T is
1 − exp[− exp(− (y n − B) A )]
return period in years Æ solving for yn given T Æ

Return Period, observations are annual Æ T =


1 

y n = B − A ln − ln1 − 
 T 

a.
T
100 years
500 years

Hs
4.1 m
4.7 m

H1/10
7.40 m
8.38 m

Risk of Encounter:

b.
c.
d.

R
50 %
20 %
80 %

Td = −Te ln (1 − R ) Æ Te =

T
72 years
224 years
31 years

Hs
4.00 m
4.38 m
3.71 m

− 50
ln (1 − R )

H1/10
7.20 m
7.89 m
6.68 m

4.
Download the USGS daily discharge data for the Apalachicola River on the Coastal
Structures web site. Estimate the discharge for a 100 year and a 500 year return period.


A histogram distribution plot is generated from the data. The frequency data is then
converted into cumulative frequency data via
n

∑ fi

F ( xn ) = i =1 , where fi is the frequency of occurrence for each bin and N is the
N
total number of observations



This is then plotted in the Return Interval Plot as n =

1
and the plot
1 − F (xn )

constructed as ln(n) vs. ln(xn).


The Return Interval Plot can then be extended to determine the discharge at the
desired return interval.
Two fit equations can be used to extrapolate the data:
1. ln (n ) = 11.8 ln(q / 1000 ) − 52.2
2. ln (n ) = 4.4 ln (q / 1000 ) − 15
Solving gives:

Return Interval
T (years)
n (days)
100
36500
500
182500

Equation 1
ln(q/1000)
q/1000 cfs
5.31
203
5.45
233

Equation 2
ln(q/1000)
q/1000 cfs
5.80
329
6.16
475

Discharge Distribution
1000
900
800
700

frequency

600
500
400
300
200
100

173.9

161.8

149.6

145.6

139.5

133.4

129.4

125.3

121.3

117.2

109.1

105.1

97.0

101.0

92.9

88.9

84.8

80.8

76.7

72.7

68.6

64.6

60.5

56.4

52.4

48.3

44.3

40.2

36.2

32.1

28.1

24.0

20.0

15.9

7.8

11.9

0.0

0

q/1000 cfs

Return Interval Plot
2E+01

1E+01

y = 11.797x - 52.218

ln(n), n in days

1E+01

1E+01

8E+00

6E+00

y = 4.3852x - 14.998
4E+00

2E+00

0E+00
0

1

2

3

ln(q/1000), q in cfs

4

5

6

Judul: Contoh Soal Pelabuhan

Oleh: Alex Joe


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