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CHS PROBLEM SOLVING 1 - 13

CHS PROBLEM SOLVING 1 - 13. 1. A ship has a displacement of 2,400 tons and KG 10.8 meters. Find the new KG if a weight of 50 tons mass already on board is raised 12 meters vertically . Answer: 11.05 meters. THE POS. OF K IS FIXED. THE POS OF M VARIES. M. W/ THE DFT OF VSL. G. G.

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CHS PROBLEM SOLVING 1 - 13

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  1. CHS PROBLEM SOLVING 1 - 13

  2. 1. A ship has a displacement of 2,400 tons and KG 10.8 meters. Find the new KG if a weight of 50 tons mass already on board is raised 12 meters vertically. • Answer: 11.05 meters

  3. THE POS. OF K IS FIXED • THE POS OF M VARIES M W/ THE DFT OF VSL. G G G • G MOVES TOWARDS G WT. LOADED G G • G MOVES AWAY FROM WT. DISCHARGE B • G MOVES PARALLEL TO THE DIR. OF THE WT BEING SHIFTED K • B MOVES TOWARDS THE LOW SIDE OF INCLINED V/L

  4. M G K

  5. 1. Solution: • GG’ = Wt. x Dist. Δ • = 50 tons x 12m 2,400 tons • GG’ = 0.25 m • Old KG = 10.80 m ( + ) • New KG = 11.05 m

  6. 2. A ship has displacement of 2,000 tons and KG 10.5 meters. Find the new KG if a weight of 40 tons mass already on board is shifted from the ‘tween deck to lower hold trough a distance of 4.5 meters vertically. • Ans. 10.41 m

  7. M G K

  8. 2. Solution: • GG’ = Wt x Dist Δ • = 40 tons x 4.5 m 2,000 tons • GG’ = 0.09 m • Old KG = 10.50 m ( - ) • New KG = 10.41 m

  9. 3. A ship of 2,000 tons displacement has KG 4.5 meters. A heavy lift of 20 tons mass is in the lower hold and has KG 2 meters. The weight is then raised 0.5 meter clear of the tank top by a derrick whose head is 14 meters above the keel. Find the new KG of the ship? Ans. 4.62 m

  10. M G K

  11. 3. Solution: • Height of Derrick fr. Keel = 14 m • KG. of Cargo = 2.0 m ( - ) • Distance = 12.0 m • GG’ = Wt x Dist. • Δ • = 20 tons x 12.0 m • 2,000 tons • GG’ = 0.12 m • Old KG = 4.50 m ( + ) • New KG = 4.62 m

  12. 4. A ship has a displacement of 7,000 tons and KG 6 meters. A heavy lift in the lower hold has KG 3 meters and mass 40 tons. Find the new KG when this weight is raised through 4.5 meters vertically and is suspended by a derrick whose head is 17 meters above the keel. Ans. 6.08 m

  13. 4. Solution: • Ht of Derrick = 17.0 m • KG of Cargo = 3.0 m ( - ) • Distance = 14.0 m • GG’ = Wt. x Dist. Δ • = 40 tons x 14.0 m 7,000 tons • GG’ = 0.08 m • Old KG = 6.00 m ( + ) • New KG = 6.08 m

  14. 5. Find the ship in the center of gravity of a ship of 1,500 tons displacement when a weight of 25 tons mass is shifted from the starboard side of the lower hold to the port side on deck through a distance of 15 meters. Ans. 0.25 m

  15. 5. Solution: • GG’ = Wt. x Dist. Δ • = 25 tons x 15 m 1,500 tons • GG’ = 0.25 m

  16. 6. A tank holds 120 tons when full of fresh water. Find how many tons of oil of relative density 0.84 it will hold, allowing 2% of the volume of the tank for expansion in the oil. Ans. 98.78 tons

  17. 6. Solution: • Capacity = Wt. of FW x Expansion • = 120 tons x 2 % or ( .02 ) • Capacity = 2.4 • (-) 120.0 • Capacity = 117.6 tons • Wt. of Oil = Capacity x R.D. of Oil • 117.6 tons x 0.84 • Wt. of oil = 98.78 tons

  18. 7. A tank when full will hold 130 tons of salt water. Find how many tons of oil relative density 0.909 it will hold allowing 1% of the volume of the tank for expansion. Ans. 114.13 tons

  19. 7. Solution: • Capacity = Capacity of FW x Expansion • = 130 x 1% or .01 • Capacity = 1.3 • (-) 130.0 tons • Capacity = 128.7 tons • R.D. Oil = 0.909 ( x ) • Wt. of Oil = 116.98 tons • Actual Wt. = Weight of Oil / Density of SW • = 116.98 tons / 1.025 • Actual Wt. of Oil = 114.13 tons

  20. 8. A tank measuring 8m x 6m x 7m is being filled with oil of relative density 0.9. Find how many tons of oil in the tank when the ullage is 3 meters. Ans. 172.8 tons

  21. 8. Solution: • Wt. of Oil = L x W x (H – Ullage) x R.D of Oil • = 8m x 6m x ( 7m – 3m ) x 0.9 • = 8m x 6m x 4m x 0.9 • Wt. of Oil = 172.8 tons

  22. 9. Oil of relative density 0.75 is run into a tank measuring 6m x 4m x 8m until the ullage is 2 meters. Calculate the number of tons of oil the tank then contains. Ans. 108 tons

  23. 9. Solution: Weight of Oil = L x W x ( H – Ullage ) x R.D. Oil = 6m x 4m x (8m – 2m) x 0.75 = 6m x 4m x 6m x 0.75 Weight of Oil = 108 tons

  24. 10. A tank will hold 100 tons when full of fresh water. Find how many tons of oil of relative density 0.85 may be loaded if 2% of the volume of the oil loaded is to be allowed for expansion. Ans. 83.3 tons

  25. 10. Solution: • Capacity = Wt. of FW x Expansion • = 100 t x 2% or .02 • = 2 • (-)100 • Capacity = 98 tons • Weight of Oil = Capacity x R.D. of Oil • = 98 tons x 0.85 • Weight of Oil = 83.3 tons

  26. 11. A box-shaped vessel 120m x 6m x 2.5 m floats at a draft of 1.5 m in water of density 1,013 kgs/m³. Find the displacement in tons, and the height of the centre of buoyancy above the keel. Ans. 1094.04 tons / 0.75 m

  27. 11. Solution: • Δ = L x B x Draft x Cb x R. Density • = 20m x 6m x 1.5m x 1 x 1.013 tons/m³ • Δ = 1,094.04 tons • KB = ½ x Draft • = ½ x 1.5 m • KB = 0.75 m

  28. 12. A box-shaped barge 55m x 10m x 6m is floating in fresh water on an even keel at 1.5 m draft. If 1,800 tons of cargo is now loaded, find the difference in the height of the centre buoyancy above the keel? • Ans. 1.636 m

  29. 12. Solution: • Δ = L x W x Draft x Cb x R. Density • = 55m x 10m x 1.5m x 1 x 1.000ton/m³ • Δ = 825 tons • KB = ½ x Draft • KB = ½ x 1.5m • KB = 0.75m

  30. If 1,800 is now loaded • Δ = L x W x Draft x Cb x R. Density • 1,800 tons = 55m x 10m x Draft x 1 x 1.000 • Draft = 1800 tons • 55m x 10m x 1 x 1.000 ton/m³ • Draft = 1800 • 550 • Draft = 3.27m • KB = ½ x Draft • = ½ x 3.27m • KB = 1.635m

  31. 12. Solution: • New DraftNew Δ • Old Draft Old Δ • New Draft1,800 tons 1.5m 825 tons • New Draft = 1.5m x 1,800 tons 825 tons • New Draft = 3.27m • KB = ½ x Draft • KB = ½ x 3.27m • KB = 1.635m

  32. 13. A box-shaped barge 75m x 6m x 4m displaces 180 tons when light. If 360 tons of iron are loaded while the barge is floating in fresh water, find her final draft reserve buoyancy. Ans. 1.2m / 70%

  33. 13. Solution: Light Δ = L x W x Draft x Cb x R. Density 180 tons = 75m x 6 m x Draft x 1 x 1.0 Light Draft = 180 tons 75m x 6m x 1 x 1.000 Light Draft = 180 tons 450 Light Draft = 0.4m

  34. New DraftNew Δ Old Draft Old Δ • New Draft360 tons 0.4m 180 tons • New Draft = 0.4m x 360 tons 180 tons • New Draft = 0.8m • Light Draft = 0.4m ( + ) • Final Draft = 1.2m

  35. RESERVE BUOYANCY 4m Depth 1.2m Draft • Reserve Buoyancy = Final Draft/ Depth • = 1.2m / 4m • Intact Buoyancy = 0.3 or 30 % • ( - ) 100 % • Reserve Buoyancy = 70 %

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