SOME BASIC FORMULAS

Area of Waterplane = L x B x C_W

…. L = Length of vessel

…. B = Breadth of vessel

...C_W = Co-efficient of Waterplane

Volume of Displacement = L x B x d x C_B

…. d = depth of vessel

….C_B = Block co-efficient

Volume (V) = L x B x d

Displacement (W) = L x B x d x R.D

... R.D = Relative density of water

TRANSVERSE STABILITY

Rectangular Waterplanes

BM = I . where I = LB³

V 12

…. V = Volume of vessel

Depth of centre of buoyancy below water line:

= 1 ( d + V )

3 2 A

LONGITUDINAL STABILITY

a) Rectangular Waterplanes

I_L = L³B and BM_L = I_L .

12 V

b) Box Shapes

BM_L = L²

12d

LIST

GG_{1(Horizontal)} = w x d

W

….d = distance moved horizontal

….w = weight

….W = Final Displacement

GG_1(Vertical) = w x d

W

….d = distance moved vertical

….w = weight

….W = Final Displacement

TANq = GG₁

GM

….GM = Metacentric height

TANq = Listing Moments

W x GM ….GM = Use Fluid GM

…. W = Final Displacement

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DRY DOCKING

P = trim x MCTC

l

...l = Distance of COF from where vessel touches blocks first

...P = Upward force acts on ship where block first touches

P = COT x MCTC

l …..In case of declivity of Dock

Virtual loss of GM = P x KM

W

Virtual loss of GM = P x KM

(W - P) if P - force is very small

After taking the blocks (F & A):

P = Change in TMD ( cms ) x TPC or

P = Reduction in water level x TPC

….TMD = True mean draft

Change in Draft (rise) (cms) = P …always subtract from draft

TPC

TMD = Draft Aft – ( LCF x Trim )

LBP

….subtract if vessel is by the stern

….add if vessel is by the head

DRY DOCKING

HYDROSTATIC TABLES AND

VESSEL ‘A’ TYPE PROBLEMS

Proceed as follows :

1. Find mean draft from the present given drafts.

2. From this mean draft, look in tables for LCF

3. Using that LCF, calculate TMD

4. From the TMD, look in tables and find

MCTC, LCF and DISPLACEMENT

5. Calculate now P-Force

6. For Displacement (W) at Critical Instant, find W-P

7. From this new (W), look in tables for KM_T

8. Now find Virtual loss of GM and use new KM_T but old Displacement (W)

9. Find now initial GM, using the new KM_T

10. Apply Virtual loss of GM in it and find the

EFFECTIVE GM.

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FREE SURFACE EFFECT / MOMENT

FSE = l.b³.R.D

12W

FSM = l.b³.R.D

12

….R.D = Density of liquid in tank

FSE = FSM

W

Corrected FSM = Tabulated FSM x Actual R.D

Assumed R.D

New FSM = Original FSM x 1

n²

….n = number of tanks which are subdivided

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DYNAMICAL STABILITY

Dynamical Stability = W x Area under the curve

STATICAL STABILITY

Statical Stability = W x GZ

KN CURVES

GZ = KN – KG.SINq

INCLINING EXPERIMENT

GM = w x d x Length of Plumbline

W Deflection

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RIGHTING MOMENT

SMALL ANGLES OF HEEL (UPTO 10^O HEEL):

GZ = GM x SINq

LARGE ANGLES OF HEEL (WALL SIDED FORMULA):

GZ = SINq ( GM + 1.BM.TAN²q )

2

WIND HEELING MOMENT:

Total Wind heeling moment = F.A.d

1000

GZ (at angle of heel) = F.A.d

1000W

….d = Distance of centre of buoyancy to centre of windage area

….F = Steady wind force of 48.5 kg/m²

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SIMPSON’S RULES

SIMPSON’S FIRST RULE:

Area = h x ( a + 4b + 2c + 4d + 2e + 4f + g )

3

Remember : 1 4 1

SIMPSON’S SECOND RULE:

1 3 3 1

h h h

Area = 3 x h x sum of products

8

Remember : 1 3 3 1

SIMPSON’S THIRD RULE:

5 8 –1

h h

Area = h x ( 5a + 8b – c )

12

Remember : 5 8 –1

NB:

Divide the value of ‘ h (in degrees) ’ by ‘ 57.3 ’ while calculating the area.

NB:

In the 3^rd rule of Simpson, we are only looking for a particular piece between the area i.e., from one co-ordinate to other and this is mainly used by surveyors for calculating sludge in bunker tank etc. Also for knowing the full area, we use Simpson’s first rule.

GM CONDITIONS

GM AT LOLL:

GM = 2(Initial GM)

COSq ….answer will be –ive but write +ive sign

WHEN GM IS NEGATIVE:

WHEN GM IS NIL:

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TURNING CIRCLE

TAN(Heel) = v²BG

gGMr

….v = velocity of ship(m/s)

….r = radius of turning circle

….g = Acceleration due to Gravity

(9.81 m/s)

….T = Period of Rolls (seconds)

….K = Radius of Gyration

….p = 3.142857143 (constant)

….I = Weight Moment of Inertia about

Rolling axis (tonne - metres²)

Hence we get,

Actual New Draft = [ Initial draft + B Tanq ] Cosq

2

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AIR DRAFT

CALCULATING LENGTH OF THE IMMAGINARY MAST

WHICH IS EXACTLY ABOVE THE ‘CF’:

Correction to Aft Mast

= Dist. of center mast from Aft Mast x Diff. of ht between masts

Dist. between the two masts

….subtract this value from the ht of Aft mast

or

Correction to Fwd Mast

= Dist. of center mast from Fwd Mast x Diff. of ht between masts

Dist. between the two masts

….add this value from the ht of Fwd mast

FOR FINDING DRAFT FWD AND AFT

Trim between masts = Trim of vessel

Distance between masts LBP

…..(from this, calculate ‘trim of vessel’ and proceed as follows)

Trim Effect Aft = l_a x Trim

L

Trim Effect Fwd = l_f x Trim

L

GRAIN

Weight of Grain = Volume

S.F

Weight of H.M = Volumetric H.M

S.F

Approx. Angle of heel = Total H.M x 12^o

Max.H.M

…. Max.H.M can be found in the Tables of Maximum permissible

Grain heeling moment against ‘W’ and KG

GG₁ ( lo) = w x d

W

….w = weight of Grain liable to shift while rolling

….d = horizontal distance of Grain shift

lo = Total volumetric H.M (in m⁴)

S.F x W

l₄₀ = GG₁(lo) x 0.80 ….80% of lo (GG₁)

NB:

If value for cargo is given for centroid then follow as normal

but if value given for ‘Kg’ of cargo then,

Multiply H.M value for fully filled compartment by 1.06 and

Multiply H.M value for partially filled compartment by 1.12

TRIM

HYDROSTATIC TABLES AND

VESSEL ‘A’ TYPE PROBLEMS

Proceed as follows :

1. Find mean draft from the present given drafts.

2. From this mean draft, look in tables for LCF

3. Using that LCF, calculate TMD

4. From the TMD, look in tables and find

MCTC, LCB and DISPLACEMENT

5. Calculate now INITIAL LCG

6. Now Calculate FINAL ‘W’ and FINAL LCG by MOMENTS

7. With this FINAL ‘W’, go in tables and look find TMD, LCB, LCF and MCTC

8. Calculate TRIM

9. After this calculate TRIM EFFECTS ( F & A )

10. Now apply this TRIM EFFECT to find FINAL DRAFTS.

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TRIM

Trimming Moment = w x d ( d = distance from COF )

Area of Waterplane = L x B x C_w

Volume of Displacement = L x B x D x C_B

TPC_sw = 1.025A

100

FWA = W .

40 TPC

DWA = FWA (1.025 – R.D)

0.025

MCTC = WGM_L

100L

TPC_DW = R.D x TPC_SW

1.025

MCTC_DW = R.D x MCTC_SW

1.025

Displacement_(DW) = RD x Displacement_(sw)

1.025

Sinkage (cms) = w .

TPC

COT = Trimming Moments

MCTC

COD Aft = l_a x COT

L

COD Fwd = COT – COD Aft

WHEN THE VESSEL IS EVEN KEEL

LCG = LCB

FOR A BOXED SHAPED VESSEL

BM = B²

12d

KB = draft

2

FOR A BOX SHAPED VESSEL WHEN DISPLACEMENT CONSTANT

New Draft = Old Density

Old Draft New Density

FOR A SHIP SHAPED VESSEL WHEN DRAFT CONSTANT

New Displacement = New Density

Old Displacement Old Density

TO KEEP THE AFT DRAFT CONSTANT

d = L x MCTC

l_a x TPC ….keeping the aft draft constant

d = L x MCTC

l_f x TPC ….keeping the fwd draft constant

d = Distance from the CF

l_a = Distance from the AP

l_f = Distance from the FP

TO PRODUCE A REQUIRED TRIM

Change in Draft (cms) = ( l. x w x d ) ± w .

L MCTC TPC

( - ive for Draft Aft)

( + ive for Draft Fwd)

( l_a for aft and l_f for fwd)

Trim (cms) = W (LCB - LCG)

MCTC

(Values for LCB, LCG and MCTC should be final)

COT WITH CHANGE OF DENSITY

COT = W(RD₁ – RD₂)(LCF – LCB)

RD₁ x MCTC₂

LCG_INITIAL = LCB ± ( Trim (cms) x MCTC )

W

….( - ive for stern trim )

….( + ive for head trim )

TRIM EFFECT AFT = l_a x Trim

L

TRIM EFFECT FWD = l_f x Trim

L

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BILGING

WHEN HEIGHT OF COMPARTMENT IS GIVEN AND ABOVE WATER LEVEL

CALCULATE SINKAGE BY RECOVERABLE BUOYANCY METHOD:

Sinkage = Buoyancy still to be recover

L x B

Buoyancy still to be recover = Lost buoyancy – Recoverable

Buoyancy

Volume of Lost Buoyancy = l x b x draft

Recoverable Buoyancy = ( L – l ) x B x ( Depth – Draft )

To find the Final Draft, add the Sinkage to Tank’s height

WHEN IN QUESTION PERMEABILITY OF THE CARGO IS GIVEN

CALCULATE THE EFFECTIVE LENGTH OF THE TANK:

Permeability ( m ) = Broken Stowage

Stowage Factor

Broken Stowage = Actual Stowage – Solid Stowage

Solid Stowage = 1 .

R.D of liquid in tank

Effective Length = Tank’s length _ORIGINAL x Permeability ( m )

NB

After calculating ‘Effective length’ always use this length for tank’s length.

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BILGING

MIDSHIP COMPARTMENT

NON WATER TIGHT WATER TIGHT

Sinkage = v .

A - a …If NON WATER TIGHT

Sinkage = v .

A …If WATER TIGHT

BM = LB³

12V …If WATER TIGHT

BM = (L – l)B³

12V …If NON WATER TIGHT

BILGING

SIDE COMPARTMENT

PLAN VIEW OF A SHIP

Sinkage = v .

A - a …If NON WATER TIGHT

Sinkage = v .

A …If WATER TIGHT

TANq = BB₁

GM ….q = List

BB₁ = a x d

Final A ….d = Distance from center of tank to

ship’s center line

….Final A = A - a

BM = I_OZ

V

I_OZ = I_AB - Ad² ….d = B + BB₁

2

….A = A - a

I_AB = LB³ - lb³

3 3

BILGING

END COMPARTMENT

AFT COMP. BILGED FWD COMP. BILGED

NON WATER TIGHT NON WATER TIGHT

Sinkage = v .

A - a …If NON WATER TIGHT

Sinkage = v .

A …If WATER TIGHT

If ‘KG’ is not given, then GM_L = BM_L

BM = L³B

12V …If WATER TIGHT

BM = (L – l)³B

12V …If NON WATER TIGHT

COT = w x d

MCTC ….w = l x b x dft x R.D

….d = L ..(Non water tight case)

2

….d = tank’s center to CF

..(Water tight case)

MCTC = WGM_L

100L

COD Aft = l_a x COT

L ….l_a = ( L - l ) + tank’s length

2

(For measuring the CF from AP) ..(Non water tight case)

…. l_a = L

2

(CF hasn’t changed and is amidships) ..(Water tight case)

When Fwd compartment is bilged (and non water tight), then just use

….l_a = ( L - l )

2

(Again for measuring the CF from AP) ..(Non water tight case)

IN CASE OF WATER TIGHT COMPARTMENT BELOW WATER LINE AND BELOW THE TANK THERE IS AN EMPTY COMPARTMENT

a) Deal as normal water tight case

b) Use volume of the tank only which is filled with water but not the portion beneath it.

c) But for KB of tank, use from K to center of tank

NB

IN WATER TIGHT CASE

· BM remains the same before and after

· KB is different before and after bilging

KB₁ is half of Original Draft

KB₂ is found by moments

IN NON WATER TIGHT CASE

· BM is different before and after bilging

BM₁ is LB³ and BM₂ is (L - l)B³

12V 12V

· KB is different before and after bilging

KB₁ is half of Initial Draft

KB₂ is half of New Draft

PLEASE NOTE THE FOLLOWING CONDITIONS

WATER TIGHT CASE NON WATER TIGHT CASE

Calculate:

a) Sinkage by non w/t method

b) KB₂ by Moments

NB:

In all cases of WATER TIGHT COMPARTMENT, calculate KB by the MOMENTS METHOD& use ‘New Draft’ in calculating this KB when calculating volume.

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TANKER CALCULATIONS

TOTAL OBSERVED VOLUME (T.O.V.):

The Total Observed Volume of all Petroleum Liquids and Free Water at observed temperature.

GROSS OBSERVED VOLUME (G.O.V.):

The Total Volume of all Petroleum Liquids, excluding Free Water at observed temperature.

G.O.V. = T.O.V. - Vfw (at observed temperature)

GROSS STANDARD VOLUME (G.S.V.):

The Total Volume of all Petroleum Liquids, excluding Free Water, corrected by appropriate Volume Correction Factor for the observed temperature and API Gravity 60º F, Relative Density 60º F / 60º F or Density 15º C.

G.S.V. = G.O.V. X V.C.F.

FREE WATER (Vfw):

The volume of water present in a tank which is not in suspension in the contained liquid at observed temperature.

ONBOARD QUANTITY (O.B.Q):

Quantity of water, oil, slops, residue, sludge or sediment, remaining in the tanks prior to loading.

TOTAL CALCULATED VOLUME (T.C.V.):

It is the Gross Standard Volume plus Free Water.

T.C.V. = G.S.V. + Vfw

TOTAL RECEIVED VOLUME (T.R.V.):

Is equal to the Total calculated Volume minus O.B.Q.

Weight Correction Factor (W.C.F.) is applied to this Volume to obtain Weight in Metric Tons or Long Tons.

Shore Gross B/L figure is to be compared with this figure.

Whenever Free Water is found in Cargo:

T.R.V. = T.C.V. - O.B.Q.

LOADED OIL WEIGHT:

Is equal to the Gross Standard Volume minus O.B.Q.

Weight Correction Factor is applied to this Volume to obtain weight in Metric Tons or Long Tons.

Shore Gross B/L figure is to be compared with this figure when no Free Water is found in Cargo.

LOADED OIL WEIGHT = (G.S.V. - O.B.Q.) W.C.F.

VESSEL’S EXPERIENCE FACTOR (V.E.F.):

Is equal to the Total of Gross B/L figures divided by the Total of Ship’s figures over the last 10 voyages.

For the purpose of calculating V.E.F., 10 TO 20 Voyages may be taken. However all voyages must ‘qualify’. A minimum of 5 ‘qualified’ voyages is needed for some level of V.E.F.

The defination of a “qualified voyage is one that meets the following criteria:

· Any voyage that is within +/- 0.0030 of then average ratio of all voyages listed. (eg. If the average listed is 1.00105, then all voyages within the range 0.99805 through 1.00405 would qualify)

· Excludes all voyage prior to any structural modification which affected the vessel’s cargo capacity.

· Excludes load or discharge data where shore measurements were not available.

This Factor is not to be applied to ship’s figure for assessing Ship / Shore difference.

The Factor may be applied to Ship’s figure to obtain an approximate B/L figure, only as a counter check where:

SHIP’S FIGURE X V.E.F = APPROXIMATE B/L FIGURE.

TABLES, VOLUME AND DENSITY:

API = AMERICAN PETROLEUM INDEX

ASTM = AMERICAN STANDARD OF TESTING MATERIALS

Previously there were 3 versions of ASTM Tables:

· USA Version - Giving API at 60º F

· UK Version - Giving Specific Gravity at 60º / 60º F (Ratio of density of Oil at 60º F to density of Water at 60º F = Specific Gravity)

· Metric Version - Giving Density at 15º C (eg. 0.865 kg/m3)

IN PRACTICE:

Volume at observed temperature is calculated by taking ullages.

Density at 15º C is given by Shore authorities.

Now Volume at 15º C = Volume at observed temperature X V.C.F.

And Volume at 15º C X Density at 15º C = Weight at 15º C (IN VACUUM)

But we want Weight IN AIR, therefore we apply the W.C.F.

W.C.F. = DENSITY AT 15º C - 0.0011

NOTES:

Now all 3 Versions are combined together and made into total 14 Volumes which contain all calculations regarding CRUDE, LUBE OIL, DIESEL OIL and all kinds of fluids / liquids.

A particular ship may have selected Volumes only, for the trade on which she is being run.

HYDROMETER - TO MEASEURE DENSITY OF WATER

PICNOMETER - TO MEASURE DENSITY OF ANY LIQUID OTHER THAN WATER

SOLVED NUMERICALS:

1. Volume at observed temperature = 10000 m3

Density at 15º C. Use observed temperature = VCF

Volume at 15º C = 9000 m3 (Volume at observed temperature X VCF)

Therefore Weight in Vacuum = 9000 X 0.8 = 7200

Density at 15º C - 0.0011 = WCF

0.8 - 0.0011 = 0.7989

Therefore W = Volume at 15º C X 0.7989

OBQ = ON Board Quantity

ROB = Remaining On Board

TOV = Vo + Vfw at observed temperature

GOV = TOV - Vfw

GSV = GOV x VCF

TCV = GSV + Vfw

TRV = TCV - obq

TRU x WCF is the figure used to compare B/L figure.

(GSV – OBQ) x WCF = Weight of Oil Loaded

API at 60º F = 141.5 - 131.5

SG 60º / 60º F

API 10 is for fresh water.

Higher the API, the lighter the product

1m3 = 6.28981 barrels

6A A - stands for Crude Oil

6B B - stands for Product Oil

2. On commencement of discharge of No. 3 © tank at 1324, EK draft of 9.00 m, ullage of 0.20 m with waterdip 15 cm.

On completion of bulk discharge at 1800, sounding of 3 © was 20 cm.

The tank is box shaped with dimensions L = 14, B = 12, D = 10, density at 15º C = 0.8937

Average Cargo temperature = 26.0º C. Find the rate of discharge.

TOV = 14 x 12 x 9.8 = 1646.4 m3

Vfw = 14 x 12 x 0.15 = 25.2 m3

GOV = 1646.4 – 25.2 = 1621.2 m3

GSV = GOV x VCF = 1608.026 m3

By Interpolation:

0.8937

0.890 0.895

26º C 0.9918 0.9919

therefore VCF = 0.991874

WCF = 0.8937 - 0.0011 = 0.8926

Therefore Weight of oil on arrival = 1608.026 x 0.8926 = 1435.324 Tons

Volume ROB (0.2 x 14 x 12) x VCF x WCF = 29.747 Tons

Therefore Rate of Discharge in MT = (1435.324 - 29.747)

TIME

WEDGE FORMULA:

It is applicable to Center Tanks only and when the ship is upright and trimmed.

Refer to Figure 1.

Let Breadth of the tank = ‘b’, trim = 0º, dist. of ullage port from aft b/head = ‘d’, height of the tank = ‘h’.

In triangle DGB, angle DGB = 90º

DB = DG Cosec 0 [ DG = Sounding = Pm = Pmiddle ]

= Pm Cosec 0

EC = Dist of ullage port from aft b/head = d

Now, BE = EC + BD – CD

Therefore BE = (d + Pm Cosec 0 – h Tan 0) ………………………. (A)

BE is < / = AE [Wedge is formed]

EF = BE Tan 0

= (d + Pm Cosec 0 - h Tan 0) x Tan 0

Therefore Volume of Wedge = [1/2 x BE x EF] x b

Vwedge = [1/2 x BE x P] x b (B)

Now, Trim = 0

Therefore Tan 0 = Trim = T ………….(1)

Length L

Also from Figure 1

Tan 0 = EF (2)

BE

Comparing (1) and (2)

T = EF

L BE

Therefore BE = EF x L = P x L ………(3)

T T

Putting value of BE (3) in (B),

Vwedge = ½ x (P x L) x P x b

T

Vwedge = L x b x P2 Where P = EF = Ht. Of liquid at aft b/head

2T T = Trim

See Figure 2. below

IN CASE WHEN TRAPEZIUM IS FORMED INSTEAD OF WEDGE

Refer to Figure 3 as above.

Corrected Sounding (P) = {Obs. Sounding (P’) + [d – (h – T/L)] x T/L}

Area of Trapezium = [ (Pmax + z) + z] x l = [ 2P – Pmax] x l

1 2

Where P = Corrected sounding at aft b/head

P’ = Sounding observed from ullage port

d = Dist of ullage port from aft b/head

h = Height of the tank

T = Trim, L = Length of the ship, l = Length of the tank, b = Breadth of the tank

Pmax Max EF due to wedge

Pmax = T/L x l ………………………………….(1)

Pcorr. sdg = P’ + [ d – (h x T/L) ] x T/L ……….(2)

If (1) < (2) ie. Pmax < Pcorr. sdg

The instead of wedge, a trapezium is formed of volume…

Volume = [ Pmax + 2z ] x l x b

2

MORE NUMERICALS:

1. Given tank ullage = 37 cm

Observed temperature of Cargo = 19.3º C

Density of Cargo in Vacuum at 15º C = 0.857 t/m3

Water Content = 165 m3

Calculate amount of Cargo in the tank in MT.

TOV = 3014.9 m3 at 19.3º C

GOV = TOV - Vfw

GOV = 3041.9 – 165 = 2876.9 m3

GSV = 2876.9 x VCF

= 2876.9 x 0.9966

= 2867.1185 m3

WCF = 0.8570 – 0.0011 = 0.8559

Therefore amount of Cargo in MT = 2867.1185 x 0.8559

= 2453.967 MT

2. LBP = 150 M, Trim = 1.30 m

P’ = 58 cm = 0.58 m

d = 1.500 m, l = 12, b = 15, h = 19

R.D. = 0.89

Pmax = T/L x l = 1.3/150 x 12 = 0.104 m

Pcorr.sdg = P’ + [ d – (h x T/L) ] x T/L

= 0.58 + [ 1.5 – (19 x 1.3/150) ] x 1.3/150 = 0.592 m

Since Pmax < Pcorr.sgd, therefore a trapezoid if formed.

Area of Trapezium ABCF,

= ( 0.592 + 0.488 ) x 12 = 6.48 m2

2

Volume of the trapezoid = 6.48 x 15

Therefore amount of Oil in the tank = 6.48 x 15 x 0.89 = 86.508 MT

2. The vessel is on even keel

(A) Ullage = 0.43 m, Water dip = 0.12 m

Density at 15º C = 0.8572

Observed temperature = 33.5º C

For ullage 0.43 m,

Tov = 1955.0 m3

Vfw = 14.8 m3

Therefore GOV = TOV – Vfw = 1940.2 m3

VCF = 0.985388

GSV = GOV x VCF = 1940.2 x 0.985388 = 1911.8498

WCF = 0.8572 – 0.0011 = 0.8561

Therefore the amount of Cargo = 1911.8498 x 0.8561

(B) At the end of discharge,

Ullage = 10.3 m, Sdg = 0.20 m = P’

Trim = 2 m, l x b x h = 20 x 8 x 10.5

d = 1.000 m, LBP = 150

Pmax = T/L x l = 2/150 x 20 = 0.266

Pcorr.sdg = P’ + [ d – (h x T/L) ] x T/L

= 0.20 + [ 1.0 – (10.5 x 2/150) ] x 2/150 = 0.21466

Since Pmax > Pcorr. sdg, therefore a wed ge is formed

P = BE Tan 0 = [ d + Pm Cosec 0 – h Tan 0 ] x Tan 0.

(Tan 0 = 2/150. Therefore 0 = 0.1458. Cosec 0 = 81.404 )

P = [ 1.0 + 0.2Cosec 0 – 10.5 x 2/150] x 2/150

= [ 1.0 + 0.2 x 81.404 – 10.5 x 2/ 150 ] x 2/150 = 0.22854

Volume of wedge = L x b x P2 = 150 x 8 x (0.228544 x 0.228544)

2T 2 x 2

Therefore Volume of wedge = 15.6697 m3

3. l x b x h = 40 x 20 x 20 m3

Ullage of oil = 1.24 m Trim = 3 m

Height of ullage point = 1.10 m LBP = 200 m

Depth of free space above oil = 0.14 m d = 1.6 m

Depth of oil = P’ = 19.86 m

Water dip = 21.1 – 20.94 = 0.61 m

Pmax = T/L x l = 3/20 x 40 = 0.60 m

Pcorr.sdg = { [P’] + [ (d – h x T/L) ] x T/L }

= [ 19.86 + ( 1.6 – 20 x 3 / 200) x 3 / 200 ] = 19.8795

Since Pmax < Pcorr.sdg, a trapezium is formed.

Volume of trapezium = ( 19.2795 – 19.8795) x 40 x 20 = 15663.8 m3

2

Now considering the water dip,

Pmax = EF /AE = T/L

Pmax = 3/200 x 40 = 0.60 m

Present P’ = ( d + Pm Cosec 0 – h Tan 0) x Tan 0

= ( 1.6 + 0.16 x 66.901 – 20 x 0.015 ) x 0.015 = 0.180 m

Since Pmax > P a wedge is formed

Volume of wedge EBF = L x b x P2/ 2T = 200 x 20 x 0.18 x 0.18/ 2 x 3 = 21.6 m3

TOV = 15663.8 m3

Vfw = 21.6 m3

GOV = 15642.2 m3

VCF = x 0.980784

GSV = 15341.619 m3

WCF = x 0.8135

Therefore amount of oil in Tonnes = 12480.407 MT