Calculation method |
Definition of N |
Symbol |
Other calculation conditions |
Damaged tank method through application of hypothetical sub-compartments |
The number of longitudinal sub-compartments |
◆ |
- |
Damage scenario method denoted in the Interim Guidelines [2] refered to in regulation 19.5 |
The number of steps for longitudinal location |
■ |
Longitudinal extent at 3 steps Transverse extent at 6 steps |
|
|
▲ |
Longitudinal extent at 6 steps Transverse extent at 6 steps |
|
|
● |
Longitudinal extent at 6 steps Transverse extent at 12 steps |
|
BLG8/18 ANNEX 4 DRAFT MEPC RESOLUTION
“EXPLANATORY NOTES ON MATTERS RELATED TO THE ACCIDENTAL OIL
OUTFLOW PERFORMANCE FOR MARPOL REGULATION I/21”からの抜粋
PART C - EXAMPLES
1 Tank barge example
1.1 General
1.1.1 The application of the Accidental Oil Outflow Performance regulation is shown in the following worked example illustrating the calculation procedure for a tank barge.
1.1.2 The arrangement and dimensions of the sample barge are as shown figure 26. For clarity purposes, a simple arrangement has been selected which does not comply with all MARPOL requirements. However, for actual designs, the vessel must satisfy all applicable regulations of MARPOL Annex I.
Figure 26 - Barge Arrangement
1.2 Establishing the nominal cargo oil density
1.2.1 The deadweight (DW) equals the displacement at the summer load line draft measured in seawater with a density of 1.025 t/m3 minus the lightship. No deduction is taken for consumables.
DW = 36,900 - 2,951 = 33,949t
1.2.2 The cargo volume C equals the total cargo volume at 98% filling. In accordance with paragraph 4.5 of regulation 21, the capacity of cargo tanks are calculated based on a permeability of 0.99.
|
100% Capacity
(m3) |
98% Filling
(m3) |
CO1
CO2 |
9,623
28,868 |
9,430
28,291 |
|
C=37,721 |
|
1.2.3 In accordance with paragraph 4.4 of regulation 21 , the nominal density is calculated as follows:
ρn = 1000 (DW)/C (kg/m3) = 1000 (33,949)/37,721 = 900 kg/m3 (1.2.3)
1.3 Calculating the probabilities of side damage
1.3.1 The first step is to determine the values for the dimensions and clearances Xa, Xf, Zl, Zu and y as defined in paragraph 8.2 of regulation 21:
Tank |
Xa
m-AP |
Xf
m-AP |
Zl
m-BL |
Zu
m-BL |
y
m |
CO1 |
20.000 |
35.000 |
2.000 |
20.000 |
2.000 |
CO2 |
35.000 |
80.000 |
2.000 |
20.000 |
2.000 |
|
1.3.2 From the ratios Xa/L, Xf/L, Z/Bs, Zl/Ds, Zu/Ds, Yl/Ds, and y, the probabilities associated with these subdivision locations are interpolated from the table of probabilities for side damage provided in Paragraph 8.3 of regulation 21. For instance, for compartment CO1 , the forward boundary Xf is at 35.0 m from the A.P, and Xf/L = 0.35. From the table, we find that Psf = 0.617. The probabilities for CO1 and CO2 are as follows:
Tank |
Xa/L |
PSa |
Xf/L |
PSf |
Zl/DS |
PSl |
Zu/Ds |
Psu |
y/Bs |
Psy |
CO1
CO2 |
0.2000
0.3500 |
0.1670
0.3170 |
0.3500
0.8000 |
0.6170
0.1670 |
0.1000
0.1000 |
0.0010
0.0010 |
1.0000
1.0000 |
0.0000
0.0000 |
0.0500
0.0500 |
0.7490
0.7490 |
|
1.3.3 In accordance with paragraph 8 of regulation 21 , the probability factors are then combined to find the probability, PS, of breaching a compartment from side damage.
For tank CO1:
PSL=(1 - Psf - Psa) = (1 - 0.617 - 0.167) = 0.216
PSV = (1 - Psu - Psl) = (1 - 0.000 - 0.001) = 0.999
PST = (1 - Psy) = (1 - 0.749) = 0.251
Ps = PSL PSV PST = (0.216)(0.999)(0.251) = 0.0542
For tank CO2:
PSL = (1 - Psf - Psa) = (1 - 0.167 - 0.317) = 0.516
PSV = (1 - Psu - Psl) = (1 - 0.000 - 0.001) = 0.999
PST = (1 - Psy) = (1 - 0.749) = 0.251
Ps = PSL PSV PST = (0.216)(0.999)(0.251) = 0.1294
1.3.4 Given a collision that penetrates the outer hull, Ps is the probability that the damage will extend into a particular cargo tank. As shown above, the probability of breaching the CO2 tank from side damage is 0.1294, or about 12.9%.
1.4 Calculating the mean outflow from side damage
1.4.1 For side damage, the total content of the tank is assumed to outflow into the sea when the tank is penetrated. Thus, the mean outflow is calculated by summing the products of the cargo tank volumes at 98% filling and the associated probabilities, in accordance with the formula given in paragraph 6 of regulation 21:
1.4.2 C3 = 0.77 for ships having two longitudinal bulkheads inside the cargo tanks extending over the length of the cargo block, and 1.0 for all other ships. In this case, there are no longitudinal bulkheads within the cargo tanks, and C3 = 1.0.
The mean oil outflow from side damage is therefore:
OMS = (1.0)(0.0542)(9,430) + (1.0)(0.1294)(28,291) = 4,172 m3
1.5 Calculating the probabilities of bottom damage
1.5.1 The first step is to determine the values for the dimensions and clearances Xa, Xf, Yp, Ys and z. Xa and Xf are as previously specified for side damage. Yp, Ys and z are defined in paragraph 9.2 of regulation 21:
Tank |
YP
m |
YS
m |
z
m |
CO1
CO2 |
38.000
38.000 |
2.000
2.000 |
2.000
2.000 |
|
1.5.2 From the ratios Xa/L, Xf/L, Yp/BB, Ys/BB, and z, the probabilities associated with these subdivision locations are interpolated from the table of probabilities for bottom damage provided in Paragraph 9.3 of regulation 21.
Tank |
Xa/L |
PBa |
Xf/L |
PBf |
Yp/BB |
PBp |
Ys/BB |
PBs |
Z/Ds |
PBz |
CO1
CO2 |
0.2000
0.3500 |
0.0290
0.0760 |
0.3500
0.8000 |
0.8100
0.2520 |
0.9500
0.9500 |
0.0090
0.0090 |
0.0500
0.0500 |
0.0090
0.0090 |
0.1000
0.1000 |
0.7800
0.7800 |
|
1.5.3 In accordance with paragraph 8 of regulation 21, the probability factors are then combined to find the probability, PB, Of breaching a compartment from bottom damage.
Fort ank CO1:
PBL = (1 - PBf - PBa) = (1 - 0.810 - 0.029) = 0.161
PBT = (1 - PBp - PBs) = (1 - 0.009 - 0.009) = 0.982
PBV = (1 - PBZ) = (1 - 0.780) = 0.220
PB = PBL PBT PBV = (0.161)(0.982)(0.220) = 0.0348
For tank CO2:
PBL = (1 - PBf - PBa) = (1 - 0.252 - 0.076) = 0.672
PBT = (1 - PBp - PBS) = (1 - 0.009 - 0.009) = 0.982
PBV = (1 - PBZ) = (1 - 0.780) = 0.220
PB = PBL PBT PBV = (0.161)(0.982)(0.220) = 0.1452
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