Stability Problems

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1 | Page Stability Problems Solved For more details please visit: www.captyashpal.blogspot.com  Problem: A vessel afloat to complete loading in a summer zone is expected to enter winter zone after steaming from the load port for 10 days. On passage fuel consumption is expected to be 30 MT/day and water consumption 15 MT/day. The ship is at present floating in water of R.D. 1.013 at a draft of 9.00 m. If summer load draft is 9.475 m, FWA = 203 mm and TPC (sw) = 30, then a. Determine maximum permissible draft on completion of loading. b. Determine amount of cargo vessel can load. Solution: The vessel has to be at a draft of not more than winter draft on entering the w inter zone. Going backwards, the vessel can load 450 MT (30 x 10 + 15 x 10) more than permit ted by winter draft, at load port. This extra 450 MT will be consumed during the passage and vessel will be at winter draft on entering the winter zone. Now, FWA = 203 mm DW density = 1.013 SW density = 1.025 Difference = 1.025 – 1.013 = 0.012 DWA = (0.012 x 203)/0.025 = 97.44 mm = 0.097 m TPC (sw) = 30 TPC (DW) = (30 x 1. 013)/1.025 = 29.65 Sinkage for 450 MT = 450/29.65 = 15.177 cm = 0.152 m Summer draft = 9.475 mm Winter draft = 9.475 – (9.475/48) = 9.278 m Present draft = 9.00 m Sinkage available to winter draft = 9.278 – 9.00 = 0.278 m Total sinkage available = 0.097 + 0.152 + 0.278 = 0.527 m Thus, maximum permissible draft on completion of loading = 9.00 + 0.527 = 9.527 m (Ans. a) Total sink age available = 52.721 cm; TPC ( DW) = 29.65 Thus, amount of cargo vessel can load = 52.721 x 29.65 = 1563.18 (Ans. b)

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8/3/2019 Stability Problems

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 Problem: A vessel afloat to complete loading in a summer zone is expected to enter winter zone

after steaming from the load port for 10 days. On passage fuel consumption is expected to be 30

MT/day and water consumption 15 MT/day.

The ship is at present floating in water of R.D. 1.013 at a draft of 9.00 m. If summer load draft is

9.475 m, FWA = 203 mm and TPC (sw) = 30, then

a.  Determine maximum permissible draft on completion of loading.b.  Determine amount of cargo vessel can load.

Solution: The vessel has to be at a draft of not more than winter draft on entering the winter zone.

Going backwards, the vessel can load 450 MT (30 x 10 + 15 x 10) more than permitted by winter

draft, at load port. This extra 450 MT will be consumed during the passage and vessel will be at

winter draft on entering the winter zone. Now,

FWA = 203 mm

DW density = 1.013

SW density = 1.025

Difference = 1.025 – 1.013 = 0.012DWA = (0.012 x 203)/0.025 = 97.44 mm = 0.097 m

TPC (sw) = 30

TPC (DW) = (30 x 1.013)/1.025 = 29.65

Sinkage for 450 MT = 450/29.65 = 15.177 cm = 0.152 m

Summer draft = 9.475 mm

Winter draft = 9.475 – (9.475/48) = 9.278 m

Present draft = 9.00 m

Sinkage available to winter draft = 9.278 – 9.00 = 0.278 m

Total sinkage available = 0.097 + 0.152 + 0.278 = 0.527 m

Thus, maximum permissible draft on completion of loading = 9.00 + 0.527 = 9.527 m (Ans. a)

Total sink age available = 52.721 cm; TPC (DW) = 29.65

Thus, amount of cargo vessel can load = 52.721 x 29.65 = 1563.18 (Ans. b)