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1 - All solved problems of ship stability chapter one
introduction to ship stabilty
Indian Maritime University
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Stability -I : Chapter 1
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- A rectangular tank measures 16mx15mx6m. How many tonnes of oil of RD 0 can it hold?
####### SOLUTION:
Given : L =16m , B = 15m, H = 6m, RD=0.
Volume of rectangle =(LxBxH) =16x15x =1440m
We know that : Density = Mass /Volume 0 =Mass/1440m
Hence, Mass =1123 tonnes
- A cylindrical tank of diameter 8m is 10m high of oil of RD 0 is poured in to it. Find the ullage, assuming π to be 3.
####### SOLUTION:
Given: Diameter = 8m,
Radius = (d/2)m =(8/2)m =4m Height =10m Mass = 400t RD =0.
We know that: Volume of the cylindrical tank = π h = 3 x 4 x 4 x 10 =502
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We know that: Density = (Mass/Volume) 0 = 400/ (Volume of the oil)
Hence, Volume of the oil = 400/0. =444
Depth of oil = volume/area Area of cylinder =( π r) =(3 x 4 x 4) =502656m
We can calculate Depth of oil = (volume of oil)/ ( area of cylinder) = (444. 44)/(50 ) =8 m
Hence, ullage = (10 – 8) =1
- A tank of 2400m volume and 12 depth, has vertical side and horizontal bottom. Find how many tonnes of oil of RD 0 it can hold, allowing 2% of the tank for the expansion .state the ullage of loading.
Solution :-
Given : Volume of the tank = 2400m Depth of tank =12m RD =0.
According to question 2% of the volume is allowed for expansion. As we know that
Density = Mass/ Volume Since volume =(L x B x H) = 2400m( given )
So, Area = (volume / depth) = (2400/12) =200m
Mass = ( volume x density) Mass = (2400×0) =1680 t
Since 2% of the volume of the tank allowed for expansion = (2/100) x 2400 = 48m
Volume of the oil = (volume of the tank – free space ) = ( 2400 -48) = 2352 m
Mass of the oil = (volume x density) =2352 x 0.
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Ullage = ( 10 – 9) = 0
- A rectangular tank measuring 25mx 12m x8m has an ullage pipe projecting 0 above the tank top. nd the mass of SW in the tank when the ullage is 3.
Solution : Volume of the rectangular tank = ( L x B x H) = (25 x 12 x 8 )
Pipe above the tank top is 0
Total ullage =3
Ullage inside the tank = (3 – 0) = 3m
Depth of the SW = (8 – 3) = 5m
Volume of the SW = (L x B x H ) = ( 25 x12 x5) = 1500 m
Mass of the SW = (volume x density) = (1500 x1) = 1537 t
- A rectangular tank measures 30mx 16mx 14m. It has an ullage pipe projecting 0 above its top. Oil of RD 0 is to be loaded. The pipe line leading from the renery to the ship is 10km long and 40cm in diameter. At the time of completion, all the oil in pipeline has to be taken. nd at what ullage the valve at the renery end must be shut so that the nal ullage in the ship’s tank would be 0. State also, the mass of oil loaded nally. (Assume π to be 3).
Given : The volume of the rectangular tank = (L x B x H) =(30 x16 x14) Ullage pipe projecting above its top = 0 RD of oil = 0. The length of the pipe line from tank to the ship = 10km = 10000m The diameter of the pipe = (d) = 40cm Hence, radius = (40 /2 ) = 20cm
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Volume of the tank = (L x B x H) = (30m x16m x14m) =6720m
Area = (L x B) = (30 x 16) = 480m
The total ullage =0
So ,ullage inside the tank = (0 – 0) = 0
If ullage inside the tank = 0,The depth of oil to be loaded is = (14 -0) m =13
Volume of the oil = (L x B x D) =( 30x 16x 13)m = 6585 m
Mass of the oil to be load = (volume x density) = (6585 x 0) = 5136 t
Now , volume of oil in the pipe = ( π rh) =(3 x 0 x 0 x 10000) =1256
Depth of oil when it poured in to the tank = (volume /area) = (1256 /480) = 2
Depth of the tank without the pipe line oil = (13 -2) = 11
Thus the ullage will be = (14 -11) = 2
Exact ullage where pump should be shut = (2 + 0)m = 3.
- A tank with a horizontal base and vertical sides is 10m deep and has rectangular trunkway 1m high. The volume of the tank alone is 8000m and that of the trunkway 500m. Find the ullage when 5320t of vegetable oil of RD 0 is loaded.
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Depth of the oil = (volume /area) = (400/600) = 0
Hence, Ullage = (1 -0) = 0 9. A rectangular tank has a total depth of 10 and volume 820 0m3, which include a trunkway of depth 0 m and volume 200m. nd the mass of oil of oil of RD 0 loaded and the ullage, if 2% of the volume of the tank is left for expansion.
Given : Depth of the tank is = 10 Volume of the tank = 8200m Depth of the trunkway = 0 Volume of the trunkway = 200m RD of the oil = 0.
According to question, 2% of the tank volume left for expansion. Volume of the tank alone = (8200-200) = 8000m
Height of the tank alone = (10 -0) = 10 m
2% of the volume of the tank left for expansion. = (2/100 x 8200) =164m
Volume of the oil to be load = (8200 -164) =8036 m
Mass of oil to be load = ( volume x density) = (8036 x 0) = 6428
Volume of the trunkway = (L x B x 0) = 200m ( given)
We can calculate Area as = (200 /0) =400m
Volume of the free space = ullage= (volume of the free space)/(area) = (164/ 400)m =0.
- A rectangular tank has a total depth of 21m and volume 10250 m which include a trunkway of trunkway of depth 1m and volume 250m3. Oil of RD 0 is to be loaded for expansion. nd the mass of oil to be loaded and the nal ullage.
Given :
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Total depth of the tank =21m
Total volume of the tank =10250m RD =0.
According to question , Space left for expansion is =(3% of the volume of the oil)
Let 8v9 be the volume of the oil 3 /100 x v = volume of free space (Volume of free space) + (volume of oil) = (volume of tank) ( 3/100 x v + v ) = 10250m (3v /100 + v) = 10250m (103v /100) = 10250m V = (10250×100 /103) = 9951
Volume of the free space = (10000 -9951) m = 48
Depth of the free space = (volume / area ) = (48 /500) = 0
Hence , nal Ullage =( 1 + 0) = 1.
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1 - All solved problems of ship stability chapter one
Course: introduction to ship stabilty
University: Indian Maritime University
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