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STABILITY PROBLEMS NO. 2

Solve marine engineering problems including tank load calculations, vessel displacement, and draft adjustments with detailed solutions. Answer questions accurately for ship design and cargo loading.

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STABILITY PROBLEMS NO. 2

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  1. STABILITY PROBLEMS NO. 2

  2. 1. A rectangular tank 10 m long 8 m wide and 16 m deep is loaded with diesel oil of 0.85 relative density. If the ullage is 6.7 m, what is the outage? • ANSWER : 536 m³ • Solution: • Outage = L x W x Ullage • = 10m x 8m x 6.7m • Outage = 536 m³

  3. 2. A ship 80 m long, 18 m wide and 12 m deep is floating in fresh water. The length and breadth of the waterplane at a draught of 4.5 m is 78 m and 16 m respectively. What is the vessel’s displacement if block coefficient is 0.82 ? • Answer: 4,605.12 tons

  4. 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

  5. 3. A box-shaped vessel 65 m long, 10 m wide and 8 m deep is floating at an even keel of 4.62 m. If the displacement is 3,027 tons, what is the relative density of the water where the vessel is in? • Answer: 1.008

  6. 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 = 1.008

  7. 4. A box-shaped vessel is approaching her berth at a speed of 3 knots. Calculate the increase of draft due to squat. • Answer: 18 cms

  8. 4. Solution: • For Enclosed Water (metric) • Squat = ( Cb x Speed² ) x 2 100 • Squat = ( 1 x 3² ) x 2 100 Squat = 0.09 x 2 Squat = 18 cms.

  9. 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.

  10. 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 ft

  11. 6. Your vessel’s available cargo capacity is 950 tons and the remaining cubic capacity is 29,000 ft.³ You are to load steel with SF 18 and cotton with SF 52. If you are to load FULL AND DOWN, how much of each cargo should be loaded? • Answer: 600 t steel, 350 t cotton

  12. 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) Wt of Steel = 950 tons – 350 tons Wt of Steel = 600 tons

  13. 7 .Your vessel’s summer draft is 7.65 m. Calculate her Tropical draft. • Answer: 7.81mtrs

  14. 7. Solution: • Tropical Draft is 1/48 above Summer Draft therefore: • 1 / 48 = 0.02083 ( 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

  15. 8. If the vessel’s summer draft is 6.70 m, moulded depth is 12.3 m, what is her summer freeboard? • Answer: 5.6 m • Solution: • Summer Freeboard = MD – SD • = 12.3m – 6.7m • Summer Freeboard = 5.6 m

  16. 9. If the vessel’s summer draft is 6.70 m, moulded depth is 12.3 m, what is her tropical freeboard? • Answer: 5.46 m

  17. 9. Solution: • Summer Freeboard = MD – SD • = 12.3m – 6.7m • Summer Freeboard = 5.6 m • 1 / 48 = 0.02083 ( 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

  18. 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: 86,400 • 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 = 86,400

  19. 11. Find the approximate calculated squat if your vessel is proceeding to a channel not enclosed with a width of 90 meters deep and dredge surrounding depths of 20 feet. Your vessel's draft is 11 meters and beam of 27 meters, block coefficient is 0.75 , speed 7 knots. Answer: 0.3675 cm

  20. 11. Solution: • For Not Enclosed Water (metric) • Squat = ( Cb x Speed² ) 100 • Squat = ( 0.75 x 7² ) 100 Squat = 36.75 100 Squat = 0.3675 cms.

  21. 12. At the commencement of loading at 0800H, draft fwd was 4.30 m, aft 4.50 m. The stevedores worked continuously till 1800H, at which time drafts were read as follows: fwd 4.65 m, aft 4.75 m. If the TPC at this draft is 25, what is the rate of loading per hour? • Answer: 75 tons per hour

  22. 12. Solution: • 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

  23. 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.

  24. 13. A tank containing olive oil of rel.density 0.87 is 12m long x 10m wide x 14 m deep. At the start of discharging operation, the ullage was one meter. After an hour, the same tank had an ullage of 1.9 m. How much oil was discharged? • Answer: 93.96 tons

  25. 13. Solution: • Old ∆ = L x B x ( Depth – Ullage ) x R.D. • = 12m x 10m x ( 14 – 1 ) x 0.87 • Old ∆ = 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 tons • Old ∆ = 1,357.20 tons ( - ) • Disch. = 93.96 tons

  26. 14.You are to load lead, SF 18 and cotton, SF 78. The available deadweight capacity is 1,600 tons of cargo and cubic capacity is 58,800 cu.ft. Disregarding broken stowage,how much of each cargo should be loaded to make her full and down? • Answer: 1,100 t lead, 500 t cotton

  27. 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) Wt of Lead = 1,600 tons – 500 tons Wt. of lead = 1,100 tons

  28. 15. A barge 70 m long, 12 m wide with a depth of 8 m has an amidship compartment 15 m long, filled with cargo whose permeability is 35 %. She is on even keel at 6.10 m. Calculate the draft if this compartment is bilged. • Answer: 6.60 m

  29. 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 m

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