CHAPTER EFFECT OF DENSITY ON DRAFT & DISPLACEMENTshipwonders.com/download/i/mark_dl/u/4013086932/4629493561/Chapter 5... · CHAPTER – 5 EFFECT OF DENSITY ON DRAFT & DISPLACEMENT

CHAPTER – 5 EFFECT OF DENSITY ON DRAFT & DISPLACEMENT

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Chapter –5

Effect of Density on Draft & Displacement

Exercise -5

Answer 1

Given:-

1. Displacement = 16000t

2. TPC = 20t/cm

3. DraftSW = 8m

FWA = Displacement 40 * TPC FWA = 16000 40 * 20 FWA = 20cms / 0.2m Draft in SW + FWA = Draft in FW 8m + 0.2m = Draft in FW 8.2m = Draft in FW Draft in FW is 8.2m Answer 2 Given:-

1. RD of DW = 1.008 2. FWA = 180mm

RD of SW = 1.025 DWA = (Difference in Density) * FWA 0.025 DWA = (1.025 – 1.008) * 180 0.025 DWA = 122.4mm/ 12.24cm

Change in draft = 12.24cms


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The vessel will rise it will move from lesser density to higher density.

Answer 3

Given:-

1. RD of water = 1.01 2. FWA = 160mm

RD of FW = 1.000 DWA = Difference in Density * FWA 0.025 DWA = (1.010 – 1.000) * 160 0.025 DWA = 64mm/ 6.4cms Change in draft is 6.4 cms The vessel will sink as it is moving from greater density to lesser density.

Answer 4

Given:-

1. FWA = 175mm 2. RD1 = 1.006 3. RD2 = 1.018

DWA = Difference in density * FWA 0.025 DWA = (1.018 – 1.006) * 175 0.025 DWA = 0.012 * 175 0.025 DWA = 84mm/ 8.4cms Change in draft = 8.4cms The vessel will rise as it is moving from lesser density to higher density.


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Answer 5

Given:-

1. Displacement = 18000t 2. TPC = 25t/cm 3. RD1 = 1.018

RD of SW = 1.025 FWA = Displacement 40 * TPC FWA = 18000 40 * 25 FWA = 18cms/ 180mm DWA = Change in Density * FWA 0.025 DWA = (1.025-1.018) * 180 0.025 DWA = 0.007 * 180 0.025 DWA = 50.4mm/ 5.04cms Her loadline can be immersed by 5.04cms

Answer 6

Given:-

1. Dimensions of vessels = 20m * 4m * 2m 2. Draft in SW = 1.05m 3. RD of DW = 1.012

20m

4m

2m

1.05m

RD =1.012


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RD of SW = Displacement of vessel Underwater Volume RD of SW = Displacement of vessel L * B * draft in SW 1.025 = Displacement of vessel 20 * 4 * 1.05 1.025 * 20 * 4 * 1.05 = Displacement of vessel 86.1 t = Displacement of vessel DISPLACEMENT OF VESSEL REMAINS CONSTANT RD of DW = Displacement of vessel Underwater Volume RD of DW = Displacement of vessel L * B * Draft in DW 1.012 = 86.1 20 * 4 * Draft in DW 1.012 * 20 * 4 * draft in DW = 86.1 Draft in DW = 86.1 1.012 * 20 * 4 Draft in DW = 1.063m

Answer 7

Given:-

1. Dimensions of vessel = 18m * 5m * 2m 2. RD of DW = 1.000 3. Draft at RD1.000 = 1.4m

18m

5m

2m

1.4m

RD =1.000


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RD of SW = 1.025 RD of DW = Displacement of vessel Underwater Volume RD of DW = Displacement of vessel L * B * draftDW 1.000 = Displacement of vessel 18 * 5 * 1.4 1.000 * 18 * 5 * 1.4 = Displacement of vessel 126 t = Displacement of vessel WE KNOW THAT DISPLACEMENT OF THE VESSEL WILL REMAIN CONSTANT RD of SW = Displacement of vessel Underwater Volume RD of SW = Displacement of vessel L * B * DraftSW 1.025 = 126 18 * 5 * draftSW

DraftSW = 126 18 * 5 * 1.025 DraftSW = 1.366m Total height of vessel = draft + freeboard 2m = 1.366m + freeboard 0.634m = freeboard Total volume of the vessel = L * B * H Total volume of the vessel = 18 * 5 * 2 Total volume of the vessel = 180m3 Volume of vessel above water line = L * B * freeboard Volume of vessel above water line = 18 * 5 * 0.634 Volume of vessel above water line = 57.06 m3 %Reserve Buoyancy = Volume of vessel above water line * 100 Total Volume


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% Reserve Buoyancy = 57.06 * 100 180 %Reserve Buoyancy = 31.7% Percentage of reserve buoyancy is 31.7% Answer 8 Given:-

1. DraftSW = 5m 2. Displacement = 3000t 3. Draft at RD 1.018 = 5m

RD of SW = 1.025 RD of SW = Displacement of vessel Underwater Volume 1.025 = 3000 Underwater Volume Underwater Volume = 3000 1.025 Underwater Volume =2926.829m3 We know that at same draft, the underwater volume remains constant RD of DW = Displacement of vessel Underwater Volume 1.018 = Displacement of vessel 2926.829 2979.5t = Displacement of vessel Displacement of vessel when afloat at 5m draft in water of RD 1.018 is 2979.5t. Answer 9 Given:-

1. Displacement of vessel at FW = 4500t 2. RD of DW = 1.020

RD of FW = Displacement of vessel Underwater Volume


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1.000 = 4500 Underwater Volume Underwater Volume = 4500m3 We know that at same draft underwater volume remains constant RD of DW = Displacement of vessel Underwater Volume 1.020 = Displacement of vessel 4500 4590t = Displacement of vessel Displacement of the vessel in RD 1.020 at the same draft as in FW is 4590t. Answer 10 Given:-

1. Length of vessel = 100m 2. Breath of vessel = 20m 3. Block Coefficient (CB) = 0.8 4. Draft SW = 8.0m

RD of FW = 1.000 Underwater Volume = L * B * DraftSW * CB Underwater Volume = 100 * 20 * 8 * 0.8 Underwater Volume = 12800m3 RD of SW = Displacement of vesselSW Underwater Volume 1.025 = Displacement of vesselSW

12800 13120t = Displacement of vesselSW We know that underwater volume at same draft remains constant RD of FW = Displacement of vesselFW Underwater Volume 1.000 = Displacement of vesselFW 12800


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1.000 * 12800 = Displacement of vessel FW

12800t = Displacement of vessel FW Difference in Displacement = DisplacementSW – Displacement FW Difference in Displacement = 13120t – 12800t Difference in Displacement = 320t Difference in displacement when floating at the same draft in FW is 320t. Answer 11 Given:-

1. Displacement = 14500t 2. Till winter load line 3. RD of DW = 1.010

RD of SW = Displacement of vessel SW Underwater Volume 1.025 = 14500t Underwater Volume Underwater Volume = 14500 1.025 Underwater Volume = 14146.341m3 We know that, at same draft the underwater volume remains constant. RD of DW = Displacement of vessel DW

Underwater Volume 1.01 = Displacement of vessel DW 14146.341 1.01 * 14146.341= Displacement of vessel DW

14287.8 t = Displacement of vessel DW

Max cargo that can be loaded = Displacement of vessel SW – Displacement of vessel DW

Max cargo that can be loaded = 14500t – 14287.8t Max cargo that can be loaded = 212.2 She can load 212.2 t of cargo so that she would float at winter load line in SW.


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Answer 12 Given:-

1. Displacement of vesselSW = 12000t 2. Draft SW = 10.0m 3. DW RD = 1.012

RD of SW = 1.025 RD of SW = Displacement of vessel Underwater Volume 1.025 = 12000 Underwater Volume Underwater Volume = 12000 1.025 Underwater volume = 11707.317m3 We know that at same draft, underwater volume remains constant. RD of DW = Displacement of vessel DW Underwater Volume 1.012 = Displacement of vessel DW 11707.317 1.012 * 11707.317 = Displacement of vessel DW Displacement of vessel DW = 11847.8t Cargo to discharge = Displacement of vessel SW – Displacement of vessel DW Cargo to discharge = 12000t – 11847.8t Cargo to discharge = 152.2t She should discharge 152.2 t so that her draft in an upriver port of RD 1.012 would be 10.0m Answer 13 Given:-

1. RD of DW = 1.005 2. Upper edge of summer loadline in water line to stbd 3. Upper edge of summer loadline 50mm above water line to port 4. FWA = 180mm 5. TPC = 24t/cm


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Arithmetic mean draft = 50mm+ 0mm 2 Arithmetic mean draft = 50mm 2 Arithmetic mean draft = 25 mm above water line Summer load line is 25mm above water line DWA = (RDSW – RDDW) * FWA 0.025 DWA = (1.025 – 1.005) * 180 0.025 DWA = 0.02 * 180 0.025 DWA = 144mm/14.4cms Max permissible sinkage = Arithmetic Mean Draft + DWA Max permissible sinkage = 2.5cms + 14.4cms Max permissible sinkage = 16.9cms Cargo to load = Max permissible sinkage * TPC Cargo to load = 16.9 * 24 Cargo to load = 405.6t Cargo that can be loaded in the vessel to bring her to her permissible draft is 405.6t.

RL

TF

F T

S

W

WNA

STBD SIDE

L R

TF

F T

S

W

WNA

PORT SIDE

50mm


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Answer 14 Given:-

1. Draft of vessel = 7.8m 2. RD of DW = 1.010 3. TPC = 18t/cm 4. FWA = 250mm 5. Max permissible draft SW =8m

DWA = (RDSW – RDDW) * FWA 0.025 DWA = (1.025 – 1.010) * 250 0.025 DWA = 0.015 * 250 0.025 DWA = 150mm/15.0cms Change in draft = Max permissible draft – initial draft Change in draft = 8m – 7.8m Change in draft = 0.2m Permissible sinkage = Change in draft + DWA Permissible sinakge = 0.2m + 0.15m Permissible sinakge = 0.35m/35cms TPCDW = TPCSW * RDDW

RDSW TPCDW = 18 * 1.01 1.025 TPCDW = 17.737t/cm DWT Available = Permissible sinkage * TPC DW DWT Available = 35 * 17.737 DWT Available = 620.8t DWT Available is 620.8t


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Answer 15 Given:-

1. Statutory freeboard = 2m 2. RD of DW = 1.015 3. Freeboard = 2.1m 4. TPCSW = 24t/cm 5. FWA = 200mm

TPCDW = TPCSW * RDDW

RDSW TPCDW = 24 * 1.015 1.025 TPCDW = 23.768t/cm DWA = (RDSW – RDDW) * FWA 0.025 DWA = (1.025 – 1.015) * 200 0.025 DWA = 0.010 * 200 0.025 DWA = 80mm/8.0cms Change in draft required = Freeboard – Statutory freeboard Change in draft required = 2.1m – 2m Change in draft required = 0.1m/10cms Permissible sinkage = change in draft + DWA Permissible sinkage = 10cms + 8 cms Permissible sinkage = 18cms DWT available = Permissible sinkage * TPCDW DWT available = 18 * 23.768 DWT available = 427.8t DWT available is 427.8t


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Answer 16 Given:-

1. RD of river birth = 1.01 2. Summer load line 20mm above water on stbd side 3. Summer load line 50mm above water on port side 4. Summer displacement = 15000t 5. TPC = 25t/cm

Arithmetic mean draft = 20mm+ 50mm 2 Arithmetic mean draft = 70mm 2 Arithmetic mean draft = 35 mm above water line Summer load line is 35mm above water line FWA = Displacement 40 * TPC FWA = 15000 40 * 25 FWA = 15cms DWA = (RDSW – RDDW) * FWA 0.025 DWA = (1.025 – 1.01) * 150 0.025

RL

TF

F T

S

W

WNA

STBD SIDE

L R

TF

F T

S

W

WNA

PORT SIDE

50mm20mm


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DWA = 0.015 * 150 0.025 DWA = 90mm/9cms TPCDW = TPCSW * RDDW

RDSW TPCDW = 25 * 1.01 1.025 TPCDW = 24.634t/cm Max permissible sinkage = Arithmetic Mean Draft + DWA Max permissible sinkage = 3.5cms + 9cms Max permissible sinkage = 12.5cms Cargo to load = Max permissible sinkage * TPC Cargo to load = 12.5 * 24.634 Cargo to load = 307.9t Cargo that can be loaded in the vessel to bring her to her permissible draft is 307.9t.


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Answer 17 Given:-

1. RD of DW = 1.005 2. Stbd WNA 30mm below water line 3. Port WNA 60 mm below line 4. Summer water draft = 8.4m 5. TPC = 30t/cm 6. FWA = 160mm

Arithmetic mean draft = 30mm+ 60mm 2 Arithmetic mean draft = 90mm 2 Arithmetic mean draft = 45 mm below water line Winter North Atlantic load line is 45mm below water line Distance between WNA and winter load line = 50mm Winter load line = 50mm – 45mm above water line Winter load line = 5mm above water line Distance between summer load line and winter load line = 1 * Summer draft 48 Distance between summer load line and winter load line = 1 * 8.4m 48

RL

TF

F T

S

W

WNA

STBD SIDE

L R

TF

F T

S

W

WNA

PORT SIDE

60mm30mm


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Distance between summer load line and winter load line =0.175m/17.5cms DWA = (RDSW – RDDW) * FWA 0.025 DWA = (1.025 – 1.005) * 160 0.025 DWA = 0.020 * 160 0.025 DWA = 128mm/12.8cms Max permissible sinkage = DWA + Distance between Summer load line and winter load line + Distance between Winter load line and water line Max permissible sinkage = 12.8cms + 17.5cms + 0.5cms Max permissible sinkage = 30.8cms TPCDW = TPCSW * RDDW

RDSW TPCDW = 30 * 1.005 1.025 TPCDW = 29.415t/cm Cargo to load = Max permissible sinkage * TPC Cargo to load = 30.8 * 29.415 Cargo to load = 906t Cargo that can be loaded in the vessel to bring her to her permissible draft is 906 t.


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Answer 18 Given:-

1. Stbd WNA 60mm above water line 2. Port WNA 20 mm below line 3. Summer water draft = 8.4m 4. TPC = 30t/cm

Arithmetic mean draft = 60mm - 20mm 2 Arithmetic mean draft = 40mm 2 Arithmetic mean draft = 20 mm above water line Winter North Atlantic load line is 20mm above water line Distance between summer load line and winter load line = 1 * Summer draft 48 Distance between summer load line and winter load line = 1 * 7.2m 48 Distance between summer load line and winter load line =0.15m/15cms Distance between tropical load line and summer load line = 1 * Summer draft 48 Distance between tropical load line and summer load line = 1 * 7.2m 48

RL

TF

F T

S

W

WNA

STBD SIDE

L R

TF

F T

S

W

WNA

PORT SIDE

20mm

60mm


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Distance between tropical load line and summer load line =0.15m/15cms Max permissible sinkage = Distance between Summer load line and winter load line + Distance between tropical load line and summer load line + Distance between Winter load line and water line Max permissible sinkage = 15cms + 15cms + 2cms Max permissible sinkage = 32cms Cargo to load = Max permissible sinkage * TPC Cargo to load = 32 * 20 Cargo to load = 640t Cargo that can be loaded in the vessel to bring her to her permissible draft is 640 t.

Answer 19 Given:-

1. Port freeboard = 3cms 2. Stbd freeboard = 2.9cms 3. RD of DW = 1.020 4. FWA = 200mm 5. TPC = 30t/cm 6. Statutory summer freeboard = 2.8m

2.8m2.9m

RL

TF

F T

S

W

WNA

STBD SIDE

L R

TF

F T

S

W

WNA

PORT SIDE

3m2.8m


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Arithmetic Mean freeboard = Port freeboard + Stbd freeboard 2 Arithmetic Mean freeboard = 3.0 + 2.9 2 Arithmetic Mean freeboard = 2.95m Change in draft required = Arithmetic mean freeboard – Statutory summer freeboard Change in draft required = 2.95m – 2.8m Change in draft required = 0.15m/15cms DWA = (RDSW – RDDW) * FWA 0.025 DWA = (1.025 – 1.020) * 200 0.025 DWA = 0.005 * 200 0.025 DWA = 40mm/4cms Max permissible sinkage = change in draft required + DWA Max permissible sinkage = 15cms + 4cms Max permissible sinkage = 19cms TPCDW = TPCSW * RDDW

RDSW TPCDW = 30 * 1.020 1.025 TPCDW = 29.854t/cm Deadweight available = TPCDW + Max permissible sinkage Deadweight available = 29.854 * 19 Deadweight available = 567.2t

DWT available is 567.2t


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Answer 20 Given:-

1. Port freeboard = 1.68m 2. Stbd freeboard = 1.79m 3. RD of DW = 1.017 4. Tropical SW freeboard = 1.63m 5. Tropical SW draft = 9.6m 6. FWA = 150mm 7. TPC = 20.4t/cm

Arithmetic Mean freeboard = Port freeboard + Stbd freeboard 2 Arithmetic Mean freeboard = 1.68 + 1.79 2 Arithmetic Mean freeboard = 1.735m Change in draft required = Arithmetic mean freeboard – Tropical SW freeboard Change in draft required = 1.735m – 1.63m Change in draft required = 0.105m/10.5cms DWA = (RDSW – RDDW) * FWA 0.025 DWA = (1.025 – 1.017) * 150 0.025

1.63m

1.79m

RL

TF

F T

S

W

WNA

STBD SIDE

L R

TF

F T

S

W

WNA

PORT SIDE

1.68m

1.63m


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DWA = 0.008 * 150 0.025 DWA = 48mm/4.8cms Max permissible sinkage = change in draft required + DWA Max permissible sinkage = 10.5cms + 4.8cms Max permissible sinkage = 15.3cms TPCDW = TPCSW * RDDW

RDSW TPCDW = 20.4 * 1.017 1.025 TPCDW = 20.241t/cm Deadweight available = TPCDW + Max permissible sinkage Deadweight available = 20.241 * 15.3 Deadweight available = 309.7t

DWT available is 309.7t

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CHAPTER EFFECT OF DENSITY ON DRAFT & DISPLACEMENTshipwonders.com/download/i/mark_dl/u/4013086932/4629493561/Chapter 5... · CHAPTER – 5 EFFECT OF DENSITY ON DRAFT & DISPLACEMENT