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1. A rectangu lar tank 10 m long 8 m

w ide and 16 m deep is loaded w ith d iesel

o i l o f 0.85 relat ive dens ity . If the ullage is6.7 m, what is the ou tage?

ANSWER : 536 m Solut ion:

Outage = L x W x Ullage

= 10m x 8m x 6.7m

Outage = 536 m

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2. A sh ip 80 m long, 18 m w ide and 12 m

deep is f loat ing in fresh water. The leng th

and breadth of the waterplane at a draught

of 4.5 m is 78 m and 16 m respect ively.

What is the vessels displacement if block

coeff icient is 0.82 ?

Answer: 4,605.12 tons

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2. Solu t ion :

= L x W x Dr. x Cb x Rel. Density

= 78m x 16m x 4.5m x 0.82 x 1.0

= 4,605.12 tons

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3. A box -shaped vessel 65 m long, 10 m

w ide and 8 m deep is f loating at an even

keel of 4.62 m . If the disp lacement is

3,027 tons, what is the relat ive density of

the water where the vessel is in?

Answer: 1.008

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3. Solu t ion :

= L x W x Dr. x Cb x Rel. Den.

3,027 tons = 65m x 10m x 4.62m x Rel.Den.

Rel. Density = 3,027 tons

65m x 10m x 4.62m

Rel. Density = 3,027

3,003

Rel. Dens ity = 1.008

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4. A box-shaped vessel is app roaching

her berth at a speed o f 3 kno ts. Calcu late

the inc rease of d raft due to squat.

Answer: 18 cms

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4. Solu t ion :

For Enc losed Water (metric)

Squat = ( Cb x Speed ) x 2

100

Squat = ( 1 x 3 ) x 2100

Squat = 0.09 x 2

Squat = 18 cm s.

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5. A reefer vessel of 13,000 long tons

disp lacement is approach ing her berth

at a speed o f 3 kno ts. Its b lock

coeff icient is 0.79. Calcu late the value o f

squat.

Answer: 0.474 ft.

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5. For Engl ish System jus t change the

cons tant 100 to 30

For Enclosed Water (English) Squat = ( Cb x Speed ) x 2

30

Squat = ( 0.79 x 3 ) x 2

30.

Squat = 0.237 x 2

Squat = 0.474 ft

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6. Your vessels available cargo

capaci ty is 950 tons and the remaining

cub ic capacity is 29,000 ft. You are to

load steel w ith SF 18 and co tton w ith SF

52. If you are to load FULL AND DOWN,

how much o f each cargo shou ld beloaded?

Answer: 600 t s teel, 350 t co tton

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6 Solut ion :

WLF = Weight of cargo having the Large

Stowage FactorWLF = Cu. Ft.(Cargo Wt. x Small SF)

( Difference in SF )

WLF = 29,000 ft - ( 950 tons x 18 )5218

WLF = 29,00017,100

34

WLF = 350 tons (Weigh t of Cotton)

Wt of Steel = 950 tons350 tons

Wt of Steel = 600 tons

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7 .Your vessels summer draft is 7.65 m.

Calcu late her Trop ical draft.

Answer: 7.81mtrs

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7. Solu t ion :

Trop ical Draft is 1/48 above Summer Draft

therefore:

1 / 48 = 0.02083 ( Mu ltip l ier )

Tropical Draft = 0.02083 x Summer Draft

Tropical Draft = 0.02083 x 7.65 m

= 0.16 m

(+) 7.65 m

Trop ical Draft = 7.81 m

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8. If the vessels summer draft is 6.70 m,

mou lded depth is 12.3 m, what is her

summer freeboard?

Answer: 5.6 m

Solut ion:

Summer Freeboard = MDSD

= 12.3m6.7m

Summer Freeboard = 5.6 m

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9. If the vessels summer draft is 6.70 m,

mou lded depth is 12.3 m, what is her

trop ical freeboard?

Answer: 5.46 m

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9. Solu t ion :

Summer Freeboard = MDSD

= 12.3m6.7m

Summer Freeboard = 5.6 m

1 / 48 = 0.02083 ( Mu ltip l ier ) Tropical FB = 0.02083 x Summer Draft

Tropical FB = 0.02083 x 6.70 m

= 0.14 m (-) 5.60 m (Summer FB)

Trop ical FB = 5.46 m

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10. A vessel w i l l load 20 p i les o f wood .

Each p ile is 6 feet h igh , b read th is 6 feet

and leng th is 10 feet. This is equal to_________ board feet.

Answer: 86,400

Solut ion: Board Feet = L x B x H x 12

= 10 ft x 6 ft x 6 ft x 12

= 4,320 BF x 20 piles

Board Feet = 86,400

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11. Find the app roximate calcu lated

squat if your vessel is proceeding to a

channel no t enclosed w i th a width of 90meters deep and dredge surround ing

depths o f 20 feet. You r vessel 's d raft is

11 meters and beam of 27 meters , b lockcoeff icient is 0.75 , speed 7 kno ts .

Answer: 0.3675 cm

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11. Solu t ion :

For Not Enclosed Water (metric)

Squat = ( Cb x Speed )

100

Squat = ( 0.75 x 7 )100

Squat = 36.75

100

Squat = 0.3675 cms.

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12. A t the commencement of loading at

0800H, draft fwd was 4.30 m , aft 4.50 m .

The stevedores worked cont inuously t i l l1800H, at wh ich t ime drafts were read as

fo l lows: fwd 4.65 m , aft 4.75 m . If the TPC

at this d raft is 25, what is the rate o floading per hour?

Answer: 75 tons per hour

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12. Solu t ion :

0800H Fwd. = 4.30m

Aft = 4.50m (+)

Mean Draft = 8.80/2

Mean Draft = 4.40m

1800H Fwd. = 4.65m

Aft = 4.75m(+)

Mean Draft = 9.40m/2

Mean Draft = 4.70m

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MD 1800H = 4.70m

MD 0800H = 4.40m ( - )

CMD = 30 cm

TPC x 25

Total Load = 750 tons 10 Hrs.

Rate of Loading= 75 tons/h r.

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13. A tank containing ol ive oi l of

rel .dens ity 0.87 is 12m long x 10m w idex 14 m deep. A t the start of d ischarging

operat ion , the u l lage was one meter.

A f ter an hou r, the same tank had an

ul lage of 1.9 m . How much o i l was

discharged?

Answer: 93.96 ton s

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13. Solu t ion :

Old = L x B x ( Depth Ullage ) x R.D.

= 12m x 10m x ( 141 ) x 0.87 Old = 1357.2 tons

New = L x B x ( Depth Ullage ) x R.D.

= 12m x 10m x ( 141.9 ) x 0.87

New = 1,263.24 tons

Old = 1,357.20 tons ( - )

Disch . = 93.96 tons

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14.You are to load lead, SF 18 and

co tton , SF 78. The available deadweigh t

capacity is 1,600 ton s o f cargo andcub ic capacity is 58,800 cu .ft .

Disregarding broken stowage,how

much of each cargo should be loaded

to make her fu l l and down?

Answ er: 1,100 t lead , 500 t co tton

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14. Solu t ion :

WLF = Weight of cargo having the Large

Stowage FactorWLF = Cu. Ft.(Cargo Wt. x Small SF)

( Difference in SF )

WLF = 58,800 ft - ( 1,600 t x 18 )

7818

WLF = 58,80028,800

60

WLF = 500 tons (Weigh t of Cotton)

Wt of Lead = 1,600 tons500 tons

Wt. o f lead = 1,100 tons

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15. A barge 70 m long, 12 m w ide w ith a

depth of 8 m has an amidship

compartment 15 m long , f i lled w i thcargo whose permeabi l i ty is 35 %. She

is on even keel at 6.10 m . Calcu late the

draf t if this compartment is b i lged.

Answer: 6.60 m

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15. Solu t ion :

Increase in Draft = V

A - a = 0.35 x 15 x 12 x 6.10

70 x 120.35 x 15 x 12

= 384.30 84063

= 384.3 / 777

Increase in Draft = 0.50 m Old Draft = 6.10 m ( + )

New Draft = 6.60 m