2 Analysis Procedure
The basic steps 1 through 6 below are described in this
section.
2.1
Step 1: Vessel design
The arrangement and dimensions of the example barge are as shown
in figure A1 (Barge arrangement).
For clarity purposes, a simple arrangement has been selected which
does not meet all MARPOL 73/78 requirements. However, for actual designs
submitted for approval as an alternative to double hull, the vessel
must meet all applicable regulations of Annex
I of MARPOL 73/78.
2.2
Step 2: Establishing of the full load
condition
An intact load condition shall be developed with the vessel
at its maximum assigned load line with zero trim and heel. Departure
quantities of constants and consumables (fuel oil, diesel oil, fresh
water, lube oil, etc.) should be assumed. Capacities of cargo oil
tanks should be based on actual permeabilities for these compartments.
All cargo oil tanks shall be assumed to be filled to 98% of their
capacities. All cargo oil shall be taken at a homogeneous density.
For this example, it is assumed that the permeability of
the cargo oil tanks is 0.99 and 0.95 for the double bottom/wing tank
ballast spaces. The 100% capacity of the cargo oil tanks CO1 and CO2
is:
| CO1:
|
9,623
m3
|
| CO2:
|
28,868
m3
|
| Total:
|
38,491
m3
|
Cargo tank capacity at 98% filling: C = 0.98
x 38,491 = 37,721 m3.
For this barge, for simplicity reasons, zero weight for
the constants and consumables has been assumed. At the 9.0 m assigned
load line the following values for the cargo oil mass (W) and
density (ρc) are obtained:
2.3
Step 3: Assembling of the damage cases
In this step the damage cases have to be developed. This
involves applying the probability density distribution functions for
side damage (figures 1 and 2) and the probability density distribution
functions for bottom damage (figures
3 and 4). Each unique
grouping of damaged compartments is determined together with its associated
probability. The sum of the probabilities should equal 1.0 for both
the side and the bottom damage evaluations.
There are different methods available for developing the
compartment groupings and probabilities, each of which should converge
on the same results.
In this example, the compartment groupings and the use of
the probability density functions is shown by a "step-wise" evaluation
method. This method involves stepping through each damage location
and extent at a sufficiently fine increment. For instance, it is assumed
(for the side damage) to step through the functions as follows: longitudinal
location = 100 steps, longitudinal extent = 100 steps, transverse
penetration = 100 steps, vertical location = 10 steps, and vertical
extent = 100 steps. You will then be developing 109 damage
incidents. The probability of each step is equal to the area under
the probability density distribution curve over that increment. The
probability for each damage incident is the product of the probabilities
of the five functions. There are many redundant incidents which damage
identical compartments. These are combined by summing their probabilities.
For a typical double-hull tanker, the 109 damage incidents
reduce down to 100 to 400 unique groupings of compartments.
-
2.3.1
Side damage evaluation
-
The damage density distribution functions provide independent
statistics for location, length, and penetration. For side damage,
the probability of a given damage longitudinal location, longitudinal
extent, transverse penetration, vertical location and vertical extent
is the product of the probabilities of these five damage characteristics.
-
To maintain the example at a manageable size, fairly coarse
increments have been assumed:
| Longitudinal location at 10
steps
|
=
|
L/10
|
=
|
0.10L per step
|
| Longitudinal extent at 3
steps
|
=
|
0.3L/3
|
=
|
0.10L per step
|
| Transverse penetration at 6
steps
|
=
|
0.3B/6
|
=
|
0.05B per step
|
-
To further simplify the evaluation, each damage is assumed
to extend vertically without limit. Therefore, the probabilities of
vertical location and vertical extent are taken as 1.0 for each damage
case. This is a reasonable assumption as the double bottom height
is only 10% of the depth. Taking the area under the density distribution
function for vertical location up to 0.1D (see figure 2, function f
S5) yields a value of 0.005. This means that the probability of
the centre of damage location falling within the double bottom region
is 1/200.
-
Figure A2 (Side
damage definition) shows the steps for longitudinal location, longitudinal
extent and transverse penetration in relation to the barge. Table
A1 (Increments for step-wise side damage evaluation) gives the range
for each step, the mean or average value over the step, and the probability
of occurrence of that particular step. For instance, Z
1 covers the range of transverse penetration beginning at the
side shell and extending inboard 5% of the breadth. The average penetration
is 0.025B or 2.5% of the breadth. The probability of
occurrence is the likelihood that the penetration will fall within
the range of 0% to 5% of the breadth. The probability equals 0.749,
which is the area under the density distribution function for transverse
penetration (figure 1, function f
S3) between
0.0B and 0.05B. The area under each probability
density function is 1.0, and therefore the sum of the probabilities
for all increments for each function is 1.0.
-
A total of ten longitudinal locations, three longitudinal
extents and six transverse penetrations will be evaluated. All combinations
of damages must be considered for a total of (10)\times(3)\times(6)
= 180 separate incidents. The damaged compartments are found by overlaying
each combination of location/extent/penetration onto the barge. These
damage boundaries define a rectangular box. Any compartment which
extends into this damage zone is considered damaged. Each of the 180
incidents results in damage to one or more compartments. Incidents
with identical damaged compartments are collected into a single damage
case by summing the probabilities of the individual damage incidents.
-
Let us begin at the aft end of the barge and proceed forward.
The first damage location X
1 is centred 0.05L forward of the transom. The first damage extent Y
1 has an average length of 0.05L. The average value
for the first transverse penetration Z
1 is
0.025B. The resulting damage box lies entirely within the WB1 compartment
and therefore damages that compartment only. The probability of this
incident is:
-
If we step through the transverse penetrations Z
2 through Z
6, we find that only the
WB1 compartment is damaged for each of these cases. The probabilities
for these cases are 0.01074, 0.00216, 0.00216, 0.00216, 0.00216, and
0.00216 respectively. The combined probability for the six cases at
longitudinal damage location X
1 is:
-
Next, we move to damage extent Y2. The damage
box X
1
Y
2
Z
1 once again falls within the WB1 compartment. Likewise, transverse
penetrations Z
2 through Z
6 fall
within this compartment. We compute the probability for these cases
and find that: P
121-6 (X
1
Y
2
Z
1-6) = 0.01925.
-
Similarly, the damage boxes defined by X
1
Y
3
Z
1-6 lie within the WB1
compartment and have a combined probability P
131-6(X
1
Y
3
Z
1-6)
= 0.00350.
-
We now move to the next longitudinal location, X
2. With longitudinal extent Y
1, the
damage stays within the WB1 compartment. The combined probability
is P
211-6(X
2
Y
1
Z
1-6) = 0.07725.
-
The forward bound of the damage box X
2
Y
2
Z
1 extends forward of
the transverse bulkhead located 20.0 m from the transom, damaging
compartments both fore and aft of this bulkhead. Transverse penetration Z
1 extends to a point just outboard of the longitudinal
bulkhead. Therefore, this combination damages both the WB1 and WB2S
compartments. The probability is P
221(X
2
Y
2
Z
1) = 0.01442.
-
We find that the damage box X
2
Y
2
Z
2 extends inboard of the longitudinal
bulkhead, damaging compartments WB1, WB2S and CO1. A cargo oil tank
has been damaged and oil outflow will occur. Similarly, damage penetrations Z
3 through Z
6 result in
breaching of the three compartments. The combined probability for
these five incidents is:
-
By stepping through the barge for all 180 incidents and
combining unique damage compartment groupings, we obtain the compartment
grouping and probability values shown in table A2 (Probability values
for side damage). Each compartment group represents a unique set of
compartments. The associated probability is the probability that each
particular group of compartments will be damaged in a collision which
breaches the hull. For instance, the probability of damaging the WB1
compartment is 0.17725. This means there is approximately a 17.7%
likelihood that only this compartment will be damaged. Likewise, the
probability of concurrently damaging the WB1 and WB2S compartments
is 0.03408 or about 3.4%. Note that the cumulative probability of
occurrence for all groups equals 1.0.
-
2.3.2
Bottom damage evaluation
-
For bottom damage, the probability of a given damage longitudinal
location, longitudinal extent, vertical penetration, transverse location
and transverse extent is, analogously to the side damage evaluation,
the product of the probabilities of these five damage characteristics.
-
The following increments are assumed for the bottom damage
evaluation:
| Longitudinal location at 10
steps
|
=
|
L/10
|
=
|
0.10L per step
|
| Longitudinal extent at 8
steps
|
=
|
0.8L/8
|
=
|
0.10L per step
|
| Vertical penetration at 6
steps
|
=
|
0.3D/6
|
=
|
0.05D per step
|
-
To further simplify the evaluation, all damage is assumed
to extend transversely without limit. Therefore, the probabilities
of transverse extent and transverse location are taken as 1.0 for
each damage case.
-
Compartment groupings are developed using the same process
as previously described for side damage.
-
Analogously, a total of ten longitudinal locations, eight
longitudinal extents and six vertical penetrations need to be evaluated.
The damage incidents to be taken into account for groundings sum up
to a total of (10) x (8) x (6) = 480 separate incidents.
-
Figure A3 (Bottom
damage definition) shows the steps for longitudinal location, longitudinal
extent and vertical penetration in relation to the barge. Table A3
(Increments for step-wise bottom damage definition) gives the range
for each step, the mean or average value over the step, and the probability
of occurrence of that particular step.
-
Again, putting the aftmost compartment WB1 together in terms
of damage increments, the following probabilities have to be summed
up:
-
Therefore the likelihood of damaging the WB1 compartment
sums up to:
-
By addressing each of the 480 incidents to the relevant
compartment (or groups of compartments) the likelihood of a damage
to these resulting from a grounding is obtained. This is shown in
table A4 (Probability values for bottom damage).
2.4
Step 4: Computation of the equilibrium
condition for each damage case
This example describes the concept analysis only. Damage
stability analyses to determine the equilibrium conditions are only
required for the final shipyard design, in accordance with paragraph
5.1.5.10 of the Guidelines.
2.5
Step 5: Computation of the oil outflow
for each damage case
In this step the oil outflow associated with each of the
compartment groupings is calculated for side and bottom damage as
outlined below.
-
2.5.1 Side damage evaluation
-
For side damage, 100% of the oil in a damaged cargo oil
tank is assumed to outflow into the sea. If we review the eleven compartment
groupings for side damage, we find that oil tank damage occurs in
three combinations: CO1 only, CO2 only, and concurrent damage to CO1
and CO2.
-
The oil outflow for these tanks is as follows:
| CO1 (98% full volume)
|
=
|
9,430 m3
|
| CO2 (98% full volume)
|
=
|
28,291 m3
|
| CO1 + CO2 (98% full
volume)
|
=
|
37,721 m3
|
-
2.5.2 Bottom damage evaluation
-
For bottom damage, a pressure balance calculation must be
carried out. The vessel is assumed to remain stranded on a shelf at
its original intact draught. For the concept analysis, zero trim and
zero heel are assumed. An inert gas overpressure of 5 kPa gauge is
assumed in accordance with paragraph 5.1.5.5 of the Guidelines. The
double bottom spaces located below the cargo oil tanks "capture" some
portion of the oil outflow. In accordance with paragraph 5.1.5.8 of
the Guidelines, the flooded volume of such spaces should be assumed
to contain 50% oil and 50% seawater by volume at equilibrium. When
calculating the oil volume captured in these spaces, no assumptions
are made on how the oil and seawater is distributed in these spaces.
-
The calculations are generally carried out for two tidal
conditions: 0.0 and 2.5 m fall in tide.
-
The actual oil volume lost from a cargo tank is calculated
for each of the two tidal conditions, assuming hydrostatic balance
as follows:
- where:
|
z
c
|
= |
height of remaining oil in the damaged tank (m) |
|
ρ
c
|
= |
cargo oil density (0.9 t/m 3 )
|
|
g
|
= |
gravitational
acceleration (9.81 m/s2)
|
|
Δp
|
= |
set
pressure of cargo tank pressure/vacuum valves (5 kPa gauge) |
|
z
s
|
= |
external seawater head above inner bottom (m) |
|
z
s
|
= |
T - 2 = 7.00 m |
|
ρ
s
|
= |
seawater density (1.025 t/m3)
|
-
See also figure A4.
-
From the above equation one obtains for the height of remaining
oil zc for the zero-tide condition:
-
Thus, the height of lost oil (h
1 =0.98 h
c - z
c) is:
-
The volume of lost oil (V
1) of cargo
tank CO1 is:
-
In this case the total volume (V
WO)
of oil and water in the water ballast tanks is:
- where:
|
z
wo
|
= |
0.5(z
c + z
s)
= 7.20 m.
|
-
If one assumes that 50% of V
WO is
occupied by captured oil, one obtains for the total oil outflow (Voutflow) of cargo tank CO1:
-
The oil outflow of cargo tank CO2 is:
-
and the total oil outflow of cargo tanks CO1 and CO2 is:
-
Step-wise application of the damage extents and assumed
increments results in fourteen compartment groupings for bottom damage.
Oil tank and double bottom damage occurs in three combinations. The
oil outflows for these tanks at 0.0 m and 2.5 m fall in tide are summarized
in the table below:
|
Tank combination
|
Oil outflow [m3] at
|
|
0.0 m tide
|
2.5 m fall in tide
|
| WB2S + WB2P + C01
|
2,373
|
3,862
|
| WB2S + WB2P + C02
|
13,322
|
17,244
|
| WB2S + WB2P + CO1 + C02
|
18,796
|
23,935
|
2.6
Step 6: Computation of the oil outflow
parameters
In this step the oil outflow parameters are computed in
accordance with paragraph 4.3 of the Guidelines. To facilitate calculation
of these parameters, place the damage groupings in a table in ascending
order as a function of oil outflow. A running sum of probabilities
is computed, beginning at the minimum outflow damage case and proceeding
to the maximum outflow damage case. Tables A5 and A6 (Cumulative probability
and oil outflow values) contain the outflow values for the side damage
and bottom damage for the two tide conditions.
Probability of zero oil outflow, P
O:
This parameter equals the cumulative probability for all damage cases
for which there is no oil outflow. From table A5, we see that the
probability of zero outflow for the side damage condition is 0.83798,
and the probability of zero outflow for the bottom damage (0.0 m tide)
condition is 0.84313.
Mean oil outflow parameter, O
M:
This is the weighted average of all cases, and is obtained by summing
the products of each damage case probability and the computed outflow
for that damage case.
Extreme oil outflow parameter, O
E:
This represents the weighted average of the damage cases falling within
the cumulative probability range between 0.9 and 1.0. It equals the
sum of the products of each damage case probability with a cumulative
probability between 0.90 and 1.0 and its corresponding oil outflow,
with the result multiplied by 10.
|