Contoh Soal Pelabuhan Kelas 1 E

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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

(

) (

)

-1

• Assume direction of motion is due west • Tair = 22C Æ ρair ~ 1.0 kg/m3

SPM Method

(

)

R/r

(add the square root term in the Northern Hemisphere)

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)

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CEM Method

(

)

2 rf r

R exp r

R p p B 2 rf V

2 / 1 B B

a c n 2

gr −

    

  

    

  

      − 

     ρ

− +

      =

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 =V_g

( )

r +V_fcosθ (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,

g 10

V V

, such that

( )

= V

( )

r +V cosθ

V V

r _g _f

g 10

V but requires a sea temperature. Æ assume 865

. 0 V V

g

10 ≈ Æ _V

( )

_r =₀_.₈₆₅_V

( )

_r +_V _cosθ f g

(3)

Land Fall Plots

x distance from hurricane center (km)

y

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

40

← hurricane direction ↑ NORTH

24 Aug 1992 1800 EST Land Fall

p_o = 930 mb

y

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

4₀

← hurricane direction

↑ NORTH

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0 50 100 150 200 250 40

60 80 100 120 140 160 180 200 220 240

distance from hurricane center (km)

w

in

d s

pe

ed (

k

m

/hr

)

Line of Maximum Wind

V_max-cem = 216.2 km/hr

R_cem = 24 km

V_max-spm = 183.3 km/hr

R_spm = 24 km

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Offshore Plots

y

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

40

24 Aug 1992 1800 EST Offshore

y

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

40

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0 50 100 150 200 250 40

60 80 100 120 140 160 180 200 220 240

distance from hurricane center (km)

w

ind s

peed

(k

m

/hr

)

Line of Maximum Wind

V_max-cem = 225.2 km/hr

R_cem = 24 km

V_max-spm = 191.3 km/hr

R_spm = 24 km

p_o = 922 mb p_amb = 1014 mb

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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

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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. Design wave for 100 year and 500 year return periods

b. Design wave for a risk of encounter = 50% , assume structure life = 50 years c. Design wave for a risk of encounter = 20%

d. Design wave for a risk of encounter = 80%

Year H1/3 (m) Ts (sec) Year H1/3 (m) Ts (sec)

1982 3.21 8.6 1988 2.85 8.0 1983 2.61 8.0 1989 2.2 7.9 1984 2.94 8.5 1990 2.56 8.2 1985 2.70 8.4 1991 2.45 8.0 1986 3.76 8.4 1992 2.42 7.9 1987 2.54 8.2

Using the Type I Asymptotic Distribution Æ F(y_n)=exp

[

−exp

(

−α

(

y_n −µ

)

]

Where

[ ]

σ

π = π

= 6 vary 6

A _n and B=E

[ ]

y_n −0.577A

Let extreme wave height be H1/10 = 1.8Hs

Hs H1/10

σ2

0.188 0.610 µ = E[yn] 2.554 4.597

A 0.338 0.609

B 2.749 4.948

Return Period, observations are annual Æ

(

)

(

[

exp y B A

)

]

exp 1

1

n −

− − − =

T , where T is

return period in years Æ solving for yn given T Æ

   

 

      − − −

=

T 1 1 ln ln A B y_n

a.

T Hs H1/10

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Risk of Encounter: T_d =−T_eln

(

1−R

)

Æ

(

R

)

e ₋

T = −

1 ln

50

R T Hs H1/10

b. 50 % 72 years 4.00 m 7.20 m c. 20 % 224 years 4.38 m 7.89 m d. 80 % 31 years 3.71 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 f x

F

n

i i

n ∑ = =1

)

( , where fi is the frequency of occurrence for each bin and N is the

total number of observations

• This is then plotted in the Return Interval Plot as

_{( )}

n x F n

− =

1 1

and the plot 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 =11.8ln

(

q/1000

)

−52.2

2. ln

( )

n =4.4ln

(

q/1000

)

−15 Solving gives:

Return Interval Equation 1 Equation 2

T (years) n (days) ln(q/1000) q/1000 cfs ln(q/1000) q/1000 cfs

100 36500 5.31 203 5.80 329

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Discharge Distribution

0 100 200 300 400 500 600 700 800 900 1000

0.0 7.8

11.9 15.9 20.0 24.0 28.1 32.1 36.2 40.2 44.3 48.3 52.4 56.4 60.5 64.6 68.6 72.7 76.7 80.8 84.8 88.9 92.9 97.0 _101.0 _105.1 _109.1 _117.2 _121.3 _125.3 _129.4 _133.4 _139.5 _145.6 _149.6 _161.8 _173.9

q/1000 cfs

frequency

Return Interval Plot

y = 4.3852x - 14.998 y = 11.797x - 52.218

0E+00 2E+00 4E+00 6E+00 8E+00 1E+01 1E+01 1E+01 2E+01

0 1 2 3 4 5 6

ln(q/1000), q in cfs