PRELIMINARY CAPACITY PLAN

The allocation and dimensions of main spaces like length of cargo tanks, width of

double skin and height of double bottom etc of double hull tankers are determined by the

regulation 13 F MARPOL 73/78 [ref6] for the construction of new tankers. All new tankers of

dead weight above 5000 t are to have either a double hull or mid deck or an alternative

arrangement for the prevention of oil out flow in case of a damage to the hull due to collision or

grounding.

The Mid Deck arrangement makes use of a horizontal subdivision (mid deck) of the

cargo spaces so that the oil pressure is reduced to a level less than the hydrostatic pressure. As a

result of this even if the hull is damaged the oil out flow will be considerably reduced.

Double hull construction makes use of wing tanks and double bottom spaces

throughout the cargo region, so that even if the outer hull is damaged, oil out flow will not occur.

Double hull construction is the modern trend.

2.4.1 Ballast Tanks or Spaces

According to regulations 13F and 13G of MARPOL 73/78 [ref6], the entire cargo

length should be protected by ballast tanks or spaces other than cargo and fuel oil tanks.

a) Wing Tanks or Spaces

Wing tanks or spaces should extend for the full length of ships side, from the top of

the double bottom to the upper most deck, disregarding a rounded gunwale where fitted. They

should be arranged such that the cargo tanks are located in board of the molded line of side shell

plating nowhere less than the distance W at any cross section is measured at right angles to the

side shell, as specified below.

W = 0.5 + Dwt / 20000 m

= 0.5 + 200,000 / 20 000

= 10.5 m

OR W = 2m, which ever is the lesser.

The minimum value of W is 1m. Therefore W = 2 m

b) Double Bottom Tanks or Spaces

At any cross section the depth of each double bottom tank or space is such that the

distance h between the bottom of the cargo tanks and the molded line of the bottom shell plating

measured at right angles to the bottom shell plating is not less than specified below:

h = B /15 = 3.6m

OR h = 2 m, whichever is lesser

The minimum value of h is 1.0m

Therefore h = 2 m.

2.4.2 Size and Arrangement of Cargo Tanks

The length of each cargo tank shall not exceed 10m or one of the following values,

whichever is the greatest

a) When a longitudinal bulkhead is provided inside the cargo tanks

(0.25 bi/B +0.15) LBP m,

where bi = 2 m, the width of wing tank

L = length of cargo tank

B = 54m, breadth of ship

b) When two or more longitudinal bulkheads are provided inside the cargo tanks

i) for wing cargo tanks 0.2LBP

ii) for center cargo tanks

-if bi/B>l/5 0.2LBP

- if bi /B < 1/5 0.2LBP

(0.5 bi /B + 0.1) LBP m, where no centerline bulkhead is provided

(0.25 bi /B + 0.15) LBP m, where a centerline bulkhead is provided

c) where no longitudinal bulkhead is provided inside the cargo tanks

(0.5 bi /B + 0.1) L m, but not to exceed 0.2LBP m

One centerline longitudinal bulkhead is fitted in this ship. So case (a) applies

Length of cargo tank, L= (0.25 bi/B +0.15) LBP

bi = 2m

L = 47.22 m

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2.6.CAPACITY CHECK

The total capacity of the ship is the volume required for cargo plus the minimum

volume required for ballast.

2.6.1. Volume of hold

VHD = (VDD + VSH + VCA + VHT + VHS)-(VFP + VAP + VER + VDB

+ VTA + VSS + VCOF)

where:

VHD = volume of hold

VDD = volume up to upper deck

VSH = volume of sheer

VCA = volume of camber.

VHT = volume of hatchway trunks

VHS = volume of holds in superstructure

VFP = volume of forepeak tank

VAP = volume of aft peak tank

VER = volume of engine

VDB = volume of double bottom

VTA = volume of tank in the hold

VSS = volume of side tanks

(1) VSH = VHT = VHS = VTA = 0

(2) VDD = L.B.T.CBD

CBD = CB + (0.25/T) (D-T)(1- CB)= 0.869

CB = 0.847

VDD = 366368.53 m3

(3) VAP = KAP (LAP/LBP)2 L.B.D.CBD

where KAP = 2.16 (2-K)

K = 3.33 AB/L –0.667

AB = 0.523 L when CB > 0.72

KAP = 1.805

LAP = 0.05 LBP = 15 m

CBD = block coefficient at uppermost deck.

= CB + 0.25/T (D-T) (1-CB)

= 0..869

VAP = 889.04 m3

(4) VFP = KFP (LFP/LBP) 2

.L.B.T.CBD

where KFP = 1.7 K.b

b = 1.4 (with bulbous bow)

KFP = 2.5575

LFP = 0.08 L= 24 m

VFP = 4512.75 m3

(5) VER = B.D.LER K ((KERA+KERF)/2)

Where LER = 0.15 L =45 m

KERA = 5.4 XERA /L +0.11

KERA = 0.3

KERF = 5.4 XERF/L +0.11

KERF = 1.08

VER = 42911.98 m3

(6) VCA = (2/3) (L-LAP- LER - LFP – LCOF ) B/50 B C3

where

C3 = 0.76CB + 0.273 = 0.916

VCA = 6590.16 m3

In segregated ballast tankers the ballast water is carried in the wing tanks and the

double bottom tanks. Therefore the volume required for ballast water must be subtracted from

the volume of hold, to get the actual volume available for the carriage of cargo.

2.6.2. Volume of Required Minimum Segregated Ballast Water

The minimum volume of ballast water that the vessel should carry is given by the

MARPOL regulations.

Draft at aft, da = 0.7T (for full propeller immersion)

= 11.9 m

Mean draft, dm = 2+0.02L

= 8 m

Maximum trim by stern, tm = 0.015L

= 4.5 m

Draft at fore, d f = da–tm = 7.4 m

Mean draft, dmean > dm

dmean = (da +df)/2 =9.65 m

Ballast displacement, B = (dmean /T ) (Cw/Cb)

B = 129516.62 t

Mass of ballast water = B-LS

= 90962.75 t

Volume of ballast water = 90962.75 /1.025= 88744.15m3

2.6.3. Volume of Stores

Range = 8000 nm

Speed = 15 kn

Hours of travel, H = 533Hrs

Hours in port = 72Hrs

1) Volume of heavy fuel oil (VHFO)

Specific fuel consumption, SFC = 185 g / kWhr.

(assumed for a slow speed large bore diesel engine)

Brake power, PB = 1.02 PD =19670.9 kW

Mass of heavy fuel oil, MHFO = SFC PB H / 1000000 +20%

(Allowance)

= 1942.44 t

Volume of HFO, VHFO = MHFO /0.95 = 2044.7 m3

2) Volume of diesel oil (VDO)

SFC = 220 g /kWhr

Power of auxiliary machinery, Paux

= (1554+38.4 X1.–0.269 X2 + 0.046X12+16.21 X2

2

2.31X1.X2) 0.76

where X1 = 0.001 dwt = 200

X2 = 0.001 PB = 19.7

Paux = 6259.9 kW

Mass of diesel oil, MDO = SFC Paux (H+60)/1000000

= 874.4 t

Volume of diesel oil, VDO = MDO/0.835

= 1028 m3

3) Volume of lubricating oil (VLO)

Mass of lube oil, MLO = 0.03 (MHFO + MDO )

= 84.5 t

4) Volume of fresh water, (VFW)

Consumption of fresh water = 20 litres / person / day

Mass of fresh water, M FW = 16.1 t

Volume of fresh water, VFW = 16.1m3

5) Volume of washing water (VWW)

Consumption 120 litres /person/ day for officers

60 litres /person/ day for crew

Mass of washing water, MWW = 62.1 t

Volume of washing water, VWW = 62.1 m3

6) Mass of crew and effects

Assume 150 kg per officers and 120 kg per crew

Mcrew = 4.77 t

Mass of stores = MHFO + MDFO +MLO + MFW + MWW

= 186.405 t

2.6.4. Mass of Cargo

Mass of cargo, MCR = dwt- Total mass of stores - Mcrew

= 196765.96t

2.6.5. Volume of Cargo

Volume of cargo = VCG = (VDD+ VCA )– VBALLAST - VSTORES

VBALLAST = 88744.15 m3

VSTORES = 3499.9 m3

Giving an allowance of 2% for oil expansion,

Volume of cargo, VCG = 232400.85 m3

Stowage factor = VCG/ MCG = 1.181 m3 / t

Conclusion: The vessel has sufficient capacity.