1. a rectangular tank 10 m long 8 m wide and 16 m deep is loaded with diesel oil of 0.85 relative...

• 1. A rectangular tank 10 m long 8 m 1. A rectangular tank 10 m long 8 m wide and 16 m deep is loaded with diesel wide and 16 m deep is loaded with diesel oil of 0.85 relative density. If the ullage is oil of 0.85 relative density. If the ullage is 6.7 m, 6.7 m, what is the outage?what is the outage?

• ANSWER : 536 m³

• Solution:

• Outage = L x W x Ullage

• = 10m x 8m x 6.7m

• Outage = 536 m³

• 2. A ship 80 m long, 18 m wide and 12 m 2. A ship 80 m long, 18 m wide and 12 m deep is floating in fresh water. The length deep is floating in fresh water. The length and breadth of the waterplane at a draught and breadth of the waterplane at a draught of 4.5 m is 78 m and 16 m respectively. of 4.5 m is 78 m and 16 m respectively. What is the vessel’s displacement if block What is the vessel’s displacement if block coefficient is 0.82 ?coefficient is 0.82 ?

• Answer: 4,605.12 tonsAnswer: 4,605.12 tons

• 2. Solution:2. Solution:

• ∆ = 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= 4,605.12 tons

• 3. A box-shaped vessel 65 m long, 10 m 3. A box-shaped vessel 65 m long, 10 m wide and 8 m deep is floating at an even wide and 8 m deep is floating at an even keel of 4.62 m. If the displacement is keel of 4.62 m. If the displacement is 3,027 tons, what is the relative density of 3,027 tons, what is the relative density of the water where the vessel is in?the water where the vessel is in?

• Answer: 1.008Answer: 1.008

• 3. Solution:3. Solution:

• ∆ = 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. Density = 1.008Rel. Density = 1.008

• 4. A box-shaped vessel is approaching 4. A box-shaped vessel is approaching her berth at a speed of 3 knots. Calculate her berth at a speed of 3 knots. Calculate the increase of draft due to squat.the increase of draft due to squat.

• Answer: 18 cms Answer: 18 cms

• 4. Solution4. Solution:

• For Enclosed Water For Enclosed Water (metric)(metric)

• Squat = ( Cb x Speed² ) x 2

100• Squat = ( 1 x 3² ) x 2

100

Squat = 0.09 x 2

Squat = 18 cms.Squat = 18 cms.

• 5. A reefer vessel of 13,000 long tons displacement is approaching her berth at a speed of 3 knots. Its block coefficient is 0.79. Calculate the value of squat.

• Answer: 0.474 ft.Answer: 0.474 ft.

• 5. For English System just change the constant 100 to 30

• For Enclosed Water (English)

• Squat = ( Cb x Speed² ) x 2

30

• Squat = ( 0.79 x 3² ) x 2

30.

Squat = 0.237 x 2

Squat = 0.474 ftSquat = 0.474 ft

• 6. Your vessel’s available cargo 6. Your vessel’s available cargo capacity is 950 tons and the remaining capacity is 950 tons and the remaining cubic capacity is 29,000 ft.cubic capacity is 29,000 ft.³³ You are to You are to load steel with SF 18 and cotton with SF load steel with SF 18 and cotton with SF 52. If you are to load FULL AND DOWN, 52. If you are to load FULL AND DOWN, how much of each cargo should be how much of each cargo should be loaded?loaded?

• Answer: 600 t steel, 350 t cottonAnswer: 600 t steel, 350 t cotton

• 6 Solution:6 Solution:• WLF = Weight of cargo having the Large

Stowage Factor

WLF = Cu. Ft. – (Cargo Wt. x Small SF)

( Difference in SF )

WLF = 29,000 ft³ - ( 950 tons x 18 )

52 – 18

WLF = 29,000 – 17,100

34

WLF = 350 tons (Weight of Cotton)WLF = 350 tons (Weight of Cotton)

Wt of Steel = 950 tons – 350 tons

Wt of Steel = 600 tons Wt of Steel = 600 tons

• 7 .Your vessel’s summer draft is 7.65 m. 7 .Your vessel’s summer draft is 7.65 m. Calculate her Tropical draft.Calculate her Tropical draft.

• Answer: 7.81mtrs Answer: 7.81mtrs

• 7. Solution:7. Solution:

• Tropical Draft is 1/48 above Summer Draft Tropical Draft is 1/48 above Summer Draft therefore:therefore:

• 1 / 48 = 0.02083 ( Multiplier )( Multiplier )

• Tropical Draft = 0.02083 x Summer Draft

• Tropical Draft = 0.02083 x 7.65 m

• = 0.16 m

• (+) 7.65 m

• Tropical Draft = 7.81 m

• 8. If the vessel’s summer draft is 6.70 m, 8. If the vessel’s summer draft is 6.70 m, moulded depth is 12.3 m, what is her moulded depth is 12.3 m, what is her summer freeboard?summer freeboard?

• Answer: 5.6 mAnswer: 5.6 m

• Solution:Solution:

• Summer Freeboard = MD – SD

• = 12.3m – 6.7m

• Summer Freeboard = 5.6 m Summer Freeboard = 5.6 m

• 9. If the vessel’s summer draft is 6.70 m, 9. If the vessel’s summer draft is 6.70 m, moulded depth is 12.3 m, what is her moulded depth is 12.3 m, what is her tropical freeboard?tropical freeboard?

• Answer: 5.46 mAnswer: 5.46 m

• 9. Solution:9. Solution:

• Summer Freeboard = MD – SD

• = 12.3m – 6.7m

• Summer Freeboard = 5.6 m Summer Freeboard = 5.6 m

• 1 / 48 = 0.02083 ( Multiplier )( Multiplier )

• Tropical FB = 0.02083 x Summer Draft

• Tropical FB = 0.02083 x 6.70 m

• = 0.14 m

• (-) 5.60 m (Summer FB)

• Tropical FB = 5.46 m

• 10. A vessel will load 20 piles of wood. Each pile is 6 feet high, breadth is 6 feet and length is 10 feet. This is equal to _________ board feet.

• Answer: 86,400Answer: 86,400

• Solution: Solution:

• 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 = 86,400Board Feet = 86,400

11. Find the approximate calculated 11. Find the approximate calculated squat if your vessel is proceeding to a squat if your vessel is proceeding to a channel not enclosed with a width of 90 channel not enclosed with a width of 90 meters deep and dredge surrounding meters deep and dredge surrounding depths of 20 feet. Your vessel's draft is depths of 20 feet. Your vessel's draft is 11 meters and beam of 27 meters, block 11 meters and beam of 27 meters, block coefficient is 0.75 , speed 7 knots.coefficient is 0.75 , speed 7 knots.

Answer: 0.3675 cmAnswer: 0.3675 cm

• 11. Solution11. Solution:

• For Not Enclosed Water For Not Enclosed Water (metric)(metric)

• Squat = ( Cb x Speed² )

100• Squat = ( 0.75 x 7² )

100

Squat = 36.75

100

Squat = 0.3675 cms.Squat = 0.3675 cms.

• 12. At the commencement of loading at 12. At the commencement of loading at 0800H, draft fwd was 4.30 m, aft 4.50 m. 0800H, draft fwd was 4.30 m, aft 4.50 m. The stevedores worked continuously till The stevedores worked continuously till 1800H, at which time drafts were read as 1800H, at which time drafts were read as follows: fwd 4.65 m, aft 4.75 m. If the TPC follows: fwd 4.65 m, aft 4.75 m. If the TPC at this draft is 25, what is the rate of at this draft is 25, what is the rate of loading per hour?loading per hour?

• Answer: 75 tons per hourAnswer: 75 tons per hour

• 12. Solution:12. Solution:

• 0800H Fwd. = 4.30m

• Aft = 4.50m (+)

• Mean Draft = 8.80/2

• Mean Draft = 4.40mMean Draft = 4.40m

• 1800H Fwd. = 4.65m

• Aft = 4.75m(+)

• Mean Draft = 9.40m/2

• Mean Draft = 4.70m Mean Draft = 4.70m

• 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/hr.Rate of Loading= 75 tons/hr.

• 13. A tank containing olive oil of 13. A tank containing olive oil of rel.density 0.87 is 12m long x 10m wide rel.density 0.87 is 12m long x 10m wide x 14 m deep. At the start of discharging x 14 m deep. At the start of discharging operation, the ullage was one meter. operation, the ullage was one meter. After an hour, the same tank had an After an hour, the same tank had an ullage of 1.9 m. How much oil was ullage of 1.9 m. How much oil was discharged?discharged?

• Answer: 93.96 tonsAnswer: 93.96 tons

• 13. Solution:13. Solution:

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

• = 12m x 10m x ( 14 – 1 ) x 0.87

• Old ∆ = 1357.2 tonsOld ∆ = 1357.2 tons

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

• = 12m x 10m x ( 14 – 1.9 ) x 0.87

• New ∆ = 1,263.24 tonsNew ∆ = 1,263.24 tons

• Old ∆ = 1,357.20 tons ( - )

• Disch. = 93.96 tons Disch. = 93.96 tons

• 14.You are to load lead, SF 18 and 14.You are to load lead, SF 18 and cotton, SF 78. The available deadweight cotton, SF 78. The available deadweight capacity is 1,600 tons of cargo and capacity is 1,600 tons of cargo and cubic capacity is 58,800 cu.ft. cubic capacity is 58,800 cu.ft. Disregarding broken stowage,how Disregarding broken stowage,how much of each cargo should be loaded much of each cargo should be loaded to make her full and down? to make her full and down?

• Answer: 1,100 t lead, 500 t cottonAnswer: 1,100 t lead, 500 t cotton

• 14. Solution:14. Solution:• WLF = Weight of cargo having the Large

Stowage Factor

WLF = Cu. Ft. – (Cargo Wt. x Small SF)

( Difference in SF )

WLF = 58,800 ft³ - ( 1,600 t x 18 )

78 – 18

WLF = 58,800 – 28,800

60

WLF = 500 tons (Weight of Cotton)WLF = 500 tons (Weight of Cotton)

Wt of Lead = 1,600 tons – 500 tons

Wt. of lead = 1,100 tons Wt. of lead = 1,100 tons

• 15. A barge 70 m long, 12 m wide with a 15. A barge 70 m long, 12 m wide with a depth of 8 m has an amidship depth of 8 m has an amidship compartment 15 m long, filled with compartment 15 m long, filled with cargo whose permeability is 35 %. She cargo whose permeability is 35 %. She is on even keel at 6.10 m. Calculate the is on even keel at 6.10 m. Calculate the draft if this compartment is bilged.draft if this compartment is bilged.

• Answer: 6.60 m Answer: 6.60 m

• 15. Solution:• Increase in Draft = µV• A - µa • = 0.35 x 15 x 12 x 6.10• 70 x 12 – 0.35 x 15 x 12• = 384.30• 840 – 63• = 384.3 / 777• Increase in Draft = 0.50 m• Old Draft = 6.10 m ( + )• New Draft = 6.60 mNew Draft = 6.60 m