stability problems 2
TRANSCRIPT
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1. A rectangu lar tank 10 m long 8 m
w ide and 16 m deep is loaded w ith d iesel
o i l o f 0.85 relat ive dens ity . If the ullage is6.7 m, what is the ou tage?
ANSWER : 536 m Solut ion:
Outage = L x W x Ullage
= 10m x 8m x 6.7m
Outage = 536 m
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2. A sh ip 80 m long, 18 m w ide and 12 m
deep is f loat ing in fresh water. The leng th
and breadth of the waterplane at a draught
of 4.5 m is 78 m and 16 m respect ively.
What is the vessels displacement if block
coeff icient is 0.82 ?
Answer: 4,605.12 tons
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2. Solu t ion :
= 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
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3. A box -shaped vessel 65 m long, 10 m
w ide and 8 m deep is f loating at an even
keel of 4.62 m . If the disp lacement is
3,027 tons, what is the relat ive density of
the water where the vessel is in?
Answer: 1.008
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3. Solu t ion :
= 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. Dens ity = 1.008
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4. A box-shaped vessel is app roaching
her berth at a speed o f 3 kno ts. Calcu late
the inc rease of d raft due to squat.
Answer: 18 cms
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4. Solu t ion :
For Enc losed Water (metric)
Squat = ( Cb x Speed ) x 2
100
Squat = ( 1 x 3 ) x 2100
Squat = 0.09 x 2
Squat = 18 cm s.
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5. A reefer vessel of 13,000 long tons
disp lacement is approach ing her berth
at a speed o f 3 kno ts. Its b lock
coeff icient is 0.79. Calcu late the value o f
squat.
Answer: 0.474 ft.
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5. For Engl ish System jus t change the
cons tant 100 to 30
For Enclosed Water (English) Squat = ( Cb x Speed ) x 2
30
Squat = ( 0.79 x 3 ) x 2
30.
Squat = 0.237 x 2
Squat = 0.474 ft
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6. Your vessels available cargo
capaci ty is 950 tons and the remaining
cub ic capacity is 29,000 ft. You are to
load steel w ith SF 18 and co tton w ith SF
52. If you are to load FULL AND DOWN,
how much o f each cargo shou ld beloaded?
Answer: 600 t s teel, 350 t co tton
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6 Solut ion :
WLF = Weight of cargo having the Large
Stowage FactorWLF = Cu. Ft.(Cargo Wt. x Small SF)
( Difference in SF )
WLF = 29,000 ft - ( 950 tons x 18 )5218
WLF = 29,00017,100
34
WLF = 350 tons (Weigh t of Cotton)
Wt of Steel = 950 tons350 tons
Wt of Steel = 600 tons
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7 .Your vessels summer draft is 7.65 m.
Calcu late her Trop ical draft.
Answer: 7.81mtrs
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7. Solu t ion :
Trop ical Draft is 1/48 above Summer Draft
therefore:
1 / 48 = 0.02083 ( Mu ltip l ier )
Tropical Draft = 0.02083 x Summer Draft
Tropical Draft = 0.02083 x 7.65 m
= 0.16 m
(+) 7.65 m
Trop ical Draft = 7.81 m
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8. If the vessels summer draft is 6.70 m,
mou lded depth is 12.3 m, what is her
summer freeboard?
Answer: 5.6 m
Solut ion:
Summer Freeboard = MDSD
= 12.3m6.7m
Summer Freeboard = 5.6 m
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9. If the vessels summer draft is 6.70 m,
mou lded depth is 12.3 m, what is her
trop ical freeboard?
Answer: 5.46 m
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9. Solu t ion :
Summer Freeboard = MDSD
= 12.3m6.7m
Summer Freeboard = 5.6 m
1 / 48 = 0.02083 ( Mu ltip l ier ) Tropical FB = 0.02083 x Summer Draft
Tropical FB = 0.02083 x 6.70 m
= 0.14 m (-) 5.60 m (Summer FB)
Trop ical FB = 5.46 m
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10. A vessel w i l l load 20 p i les o f wood .
Each p ile is 6 feet h igh , b read th is 6 feet
and leng th is 10 feet. This is equal to_________ board feet.
Answer: 86,400
Solut ion: 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
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11. Find the app roximate calcu lated
squat if your vessel is proceeding to a
channel no t enclosed w i th a width of 90meters deep and dredge surround ing
depths o f 20 feet. You r vessel 's d raft is
11 meters and beam of 27 meters , b lockcoeff icient is 0.75 , speed 7 kno ts .
Answer: 0.3675 cm
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11. Solu t ion :
For Not Enclosed Water (metric)
Squat = ( Cb x Speed )
100
Squat = ( 0.75 x 7 )100
Squat = 36.75
100
Squat = 0.3675 cms.
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12. A t the commencement of loading at
0800H, draft fwd was 4.30 m , aft 4.50 m .
The stevedores worked cont inuously t i l l1800H, at wh ich t ime drafts were read as
fo l lows: fwd 4.65 m , aft 4.75 m . If the TPC
at this d raft is 25, what is the rate o floading per hour?
Answer: 75 tons per hour
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12. Solu t ion :
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
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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/h r.
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13. A tank containing ol ive oi l of
rel .dens ity 0.87 is 12m long x 10m w idex 14 m deep. A t the start of d ischarging
operat ion , the u l lage was one meter.
A f ter an hou r, the same tank had an
ul lage of 1.9 m . How much o i l was
discharged?
Answer: 93.96 ton s
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13. Solu t ion :
Old = L x B x ( Depth Ullage ) x R.D.
= 12m x 10m x ( 141 ) x 0.87 Old = 1357.2 tons
New = L x B x ( Depth Ullage ) x R.D.
= 12m x 10m x ( 141.9 ) x 0.87
New = 1,263.24 tons
Old = 1,357.20 tons ( - )
Disch . = 93.96 tons
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14.You are to load lead, SF 18 and
co tton , SF 78. The available deadweigh t
capacity is 1,600 ton s o f cargo andcub ic capacity is 58,800 cu .ft .
Disregarding broken stowage,how
much of each cargo should be loaded
to make her fu l l and down?
Answ er: 1,100 t lead , 500 t co tton
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14. Solu t ion :
WLF = Weight of cargo having the Large
Stowage FactorWLF = Cu. Ft.(Cargo Wt. x Small SF)
( Difference in SF )
WLF = 58,800 ft - ( 1,600 t x 18 )
7818
WLF = 58,80028,800
60
WLF = 500 tons (Weigh t of Cotton)
Wt of Lead = 1,600 tons500 tons
Wt. o f lead = 1,100 tons
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15. A barge 70 m long, 12 m w ide w ith a
depth of 8 m has an amidship
compartment 15 m long , f i lled w i thcargo whose permeabi l i ty is 35 %. She
is on even keel at 6.10 m . Calcu late the
draf t if this compartment is b i lged.
Answer: 6.60 m
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15. Solu t ion :
Increase in Draft = V
A - a = 0.35 x 15 x 12 x 6.10
70 x 120.35 x 15 x 12
= 384.30 84063
= 384.3 / 777
Increase in Draft = 0.50 m Old Draft = 6.10 m ( + )
New Draft = 6.60 m