On a ship of 5000 t displacement , a tank is partly full of DO of RD 0.88. If the moment of inertia of the tank about its centreline is 242m^{4} , find the FSC .
Solution:
W = 5000t, RD = 0.88, I = 242m4
We know that :
FSC = (i di /W )
= (242 x 0.88 ) /5000
= 0.042m.
If the tank in Question 1 was partly full of SW instead of DO, find the FSC.
Solution:
RD = 1.025
We know that:
FSC = (i di/W )
= (242 x1.025) /5000
= 0.049m.
On a ship of W 6000 t, KM 7.4m, KG 6.6m, a double bottom tank of i 1200m^{4} is partly full of FW . Find the GM fluid .
Solution:
W = 6000 t & KM = 7.4m,
KG = 6.6m, i = 1200m^{4}, RD = 1.00
GM (solid) = (KM – KG)
= (7.4 – 6.6)
= 0.8m
We know that:
FSC = (i di /W )
= (1200 x 1) /6000
= 0.2m
Fluid GM = GM (solid) – FSC
= (0.8 – 0.2)
= 0.6m
Given the following particulars, Find the GM fluid : W= 8800 t, tank of i = 1166 m4 is partly full of HFO of RD 0.95, KM 10.1m, KG 9.4m .
Solution:
W = 8800 t & i = 1166m^{4},
RD = 0.95,
KM = 10.1 m & KG = 9.0m.
GM (solid) = (KM – KG)
= (10.1 – 9.0)
= 1.1m
We know that:
FSC = (i di /W )
= (1166 x 0.95)/ 8800
= 0.126m
GM fluid = GM (solid) – FSC
= (1.1 – 0.126)
= 0.974m.
On a vessel of W 16000 t, NO.4 port DB tank 20m long and 8m wide is partly full of DW ballast of RD 1.010. Find the FSC.
Solution:
W = 16000t,
L = 20m & B = 8m,RD = 1.010
We know that:
FSC = (i di / W)
FSC =( LB^{3} X di )/( 12 x W)
= (20 x 8^{3} x 1.010)/(12 x 16000)
= 0.054m.
A vessel has a deep tank on the starboard side 12m long 9m wide which is partly full of coconut oil of RD 0.72 . If W = 1200 t , KM = 9m and KG = 8.5m , find the GM fluid .
Solution:
L = 12m, B = 9m, RD = 0.72
W = 12000 t, KM = 9m & KG = 8.5m
GM (solid) = (KM – KG)
= (9 – 8.5)
= 0.5m
We know that:
FSC = (i di /W )
= (LB^{3} x di ) / (12 x W )
= (12 x 9^{3} x 0.72) / (12 x 1200)
= 0.044m
Fluid GM = GM (solid) -FSC
= (0.5 – 0.044)
= 0.456m.
A vessel displacing 8000t , has a rectangular deep tank 10m long 8m wide and 9m deep full of SW . The KM is 7m and KG 6.2m. Find the GM when 1 / 3 of this tank as pumped out.
Solution:
W = 8000t.
L = 10m, B = 8m & D = 9m
RD = 1.025, KM = 7m, KG = 6.2m
Mass of the tank = (Volume x Density)
= (L x B x D) x 1.025
= (10 x 8 x 9) x (1.025)
= 738t
According to question, 1/3 of water pumped out
= (1/3 x738)
= 246t
After pumping out the water the KG of the tank = (6 + 1.5)
= 7.5m
Ship’s weight | KG | VM (t) |
8000t | 6.2m | 49600tm |
(-) 246t (discharge) | 7.5m | (-) 1845tm |
Final W = 7754 t Final VM = 47755tm
We know that :
Final KG = (Final VM)/ (Final W)
Final KG = (47755)/(7754)
= 6.159m
GM (solid) = (KM – KG)
= (7.0 – 6.159)
= 0.841m
We know that:
FSC = (i di / W)
= (LB^{3 }x di)/ (12 x W)
= (10 x 8^{3} x 1.025)/(12 x 7754)
= 0.05m
So, GM fluid = GM (solid) – FSC
= (0.841 – 0.055)
= 0.785m.
A ship of W 5000t has a tank 16m long, 10m wide and 4m deep which is empty. KM is 7.2m and KG 7.0m. Find the GM fluid if 400t of oil of RD 0.95 are received in it.
Solution:
W = 5000 t
Volume of tank = (L x B x D)
= (16 x 10 x 4) m^{3}
KM =7.2m & KG = 7.0m
W = 400 t, RD = 0.95
We know that, dimension of tank are given change of KG of tank due to change of sounding has to be considered
Mass = (Volume x Density)
400t = (Volume) x 0.95
Volume = (400/0.95)
= 421.05m^{3}
Now, Depth of the oil = (volume / area)
(421.052/ (16 x 10)
= 2.631m
Again, we know that:
KG = (depth/2)
= (2.631/2)
= 1.315m
Ship’s weight | KG | VM |
5000t | 7.0m | 35000tm |
(+) 400 (load) | 1.315m | (+) 526tm |
Final W = 5400t Final VM= 35526tm
We know that:
Final KG = (Final VM) / (Final W)
Final KG = (35526 / 5400)
= 6.578m
GM Solid = (KM – KG)
= (7.2 – 6.578)
= 0.622m
We know that:
FSC = (i di/ W)
= (LB^{3 }x di)/ (12 x W )
= (16 x 10^{3} x 0.95)/(12 x 5400)
= 0.234m
GM (fluid )= GM (solid) – FSC
= (0.622 – 0.234)
= 0.388m.
A vessel has two deep tanks port and starboard, each 12m long , 5m wide and 8m deep . The port side is full of SW while the starboard side is empty. W = 9840 t, KM = 8.5m KG = 8.0m . Calculate the GM if fluid if SW is transferred from P to S until each tank has equal quantity of ballast .
Solution:
Volume of tank = (L x B x D )
= (12m x 5m x 8m)
W = 9840t , KM = 8.5m and KG = 8.0m
We can calculate Mass :
Mass of the tank = (Volume x Density)
= (12 x 5 x 8 x 1.025)
= 492t
According to question, 1/2 of water transferred to stbd
= (492/2)
= 246t
KG, of the port side tank after shifting
= (4 + 2)m
= 6m
Now, Stbd side tank = (d/2)
= (4/2)m
= 2m
Again, d = ( KG of port tank – KG of stbd tank)
= (6 – 2)m
= 4m
Ship’s wt | KG | VM |
9840t | 8.0m | 78720tm |
246t ( shift) | 4m | (-) 984tm |
Final W = 9840t Final VM = 77736tm
We know that:
Final KG = (Final VM/Final W)
Final KG = (77736/9840)
= 7.9m
GM (Solid) = (8.5 – 7.9) m
= 0.6m
FSC when tanks one not divided ;
FSC = (i di/ W)
= (LB^{3 }x di )/ (12 x W )
= (12 x 10^{3 }x 1.025)/(12x 9840)
= 0.104m
Total FSC = (1/n2)
= (1/2^{2}) x (0.104 )
= 0.026m
GM fluid = (GM solid – FSC)
= (0.6 – 0.026)
= 0.574m
A ship displacing 10000t has KM 9. 9m. The following is her present condition:
Tank | KG (m) | 1(m4) | contents | RD | Remarks |
FP Tank | 6.3 | 10 | SW | 1.025 | Full |
No 1 DBT | 1.15 | 420 | HFO | 0.95 | Slock |
No 2 P or S | 0.65 | 720 | HFO | 0.95 | Port slock Stb empty |
No 3p or s | 0.65 | 240 | SW | 1.025 | Pord full Stbd slack |
No 3c | 0.60 | 1200 | HFO | 0.95 | Full |
No 4p | 0.70 | 300 | FW | 1.00 | Both Slack |
NO 5p | 0.85 | 180 | DO | 0.88 | Slack |
NO 5s | 0.85 | 100 | HFO | 0.95 | Full |
AP Tank | 8.80 | 20 | SW | 1.025 | Empty |
If the final KG is 8.954 m, find the final GM fluid .
Solution:
Given:
W = 10000 t, KM = 9.9m & KG = 8.954m
FSM NO.1 = ( i x di )
= (420 x 0.95)
= 399
NO.2P = ( i x di )
= (720 x 0.95)
= 684
NO.3S = ( i x di)
= (240 x 1.025)
= 246
NO.4P = ( i x di )
= (300 x 1 )
= 300
NO.4S = ( i x di)
= (300 x 1)
= 300
NO.5P = ( i x di )
= ( 180 x 0.88 )
= 158.4
Total FSM = 2087.4 tm
FSC = (FSM / W)
= ( 2087.4 /10,000 )
= 0.208m
GM solid = (KM – KG)
=( 9.9 – 8.954 )
= 0.946m
GM fluid = (0.946 – 0.208)
= 0.738m.
Vessel is same condition as in question 10, transfers some HFO from NO.1 DBT to NO 2P such that NO 2 becomes full, while NO1 remains partly full. Find the resultant GM fluid if the final KG is 8.950m.
NOTE – 1 : It is not necessary to rework the entire problem . Just make the necessary change to the final part of the working of the answer to question 10 , Thus:
FSM obtain finally in question 10 | 2087.4tm |
NO2 DBT is now not slack | – 684tm |
Final FSM for the question | 1403.4tm |
NOTE -2 : Where a tank was originally slack but has now become full or empty , Its FSM has to be subtract from the final FSM obtained In question 10.
NOTE -3 : When a tank was originally empty or full but has now become slack , its FSM must be added to the final FSM obtain in question 10.
Solution:
Given:
Total FSM = 2087.4 tm
FSM deducted from 2P = 684.0 tm
New FSM = 1403.4
FSC =(FSM/W)
= (1403.4 /1000)
= 0.14
GM solid = (KM – KG)
= (9.9 – 8.950)
= 0.95m
We know that:
GM fluid = (GM solid – FSC)
= (0.95 – 0.14)
= 0.81m.
Vessel in same condition as in question 10, transfers SW ballast from FP IN TO NO 3s and in to AP, such that NO3s becomes full while FP and AP remain partly full . Find the GM fluid if final KG is 8.880m.
Solution:
According to question,
Now, No 35 becomes full, so FSM deducted
Again, New FSM created by FP and AP
We know that:
FP = (10 x 1.025)
= 10.25 tm.
AP = (20 x 1.025)
= 20.5 tm.
So, FSM = (10.25 + 20.50) tm
= 30.75 tm
New, final FSM = (2087.4 – 246.0 + 30.75)
= 1872.15 tm
FSC = (1872.15 /10000)
= 0.187
New KG = 8.880m (given)
GM (Solid) = (KM – KG)
= (9.9 – 8.880)
= 1.02m
GM (Fluid) = GM (solid) – FSC
= (1.02 – 0.187)
= 0.833m.
- Vessel in same condition as in question 10, consume the following while on passage :
- All the HFO (200t) from NO1 DB .
- Half the HFO (i.e 100t) from NO3c .
- All the FW from NO4 P and S (total 200t ) .
Find the fluid the GM on the arrival at the next port .
NOTE 4 : Find KG on the arrival by taking moments about keel , as explained in chapter 7 , and add an extra column for the changes in FSM , as shown below . Bear in mind notes 1,2 and 3 , under question 11.
Remarks
Weight (t) KG Moment About FSM
L D M Keel TM TM
L D
Ship in
Q 10 10000 – 8.954 89540 – 2027. 4
NO1 DB – 200 1.15 – 230 (-) 399
and so on .
Note – 5: When dealing with several tank’s in a ship calculation such as this the change of KG of tank due to change in its sounding may be ignored. This is the actual practice at sea .
FOR EXAMPLE: When half of the HFO (i.e. 100t from NO3 C is pumped out , the moment of the discharge weight about keel would 100X 0.6 = 60 tm .
Solution:
Ship’s wt | KG | VM | FSM |
10000 | 8.954 | 89540tm | 2087.4 |
(- ) 200 (dish) | 1.15 | (-) 230tm | (- )399.0 |
(- ) 100 (dish) | 0.60 | (-)60 tm | ( +)1140.0 |
(- ) 200(dish) | 0.70 | (- )140 tm | (-) 600.0 |
Final W = 9500 Final VM=89110tm Final FSM= 2228.4tn
We know that:
Final KG = (Final VM/ Final W)
Final KG = (89110/9500)
Final KG = 9.38m
Again, FSC = (FSM /W)
= (2228.4 /9500)
= 0.234m
GM Solid = (KM –KG)
= (9.9 – 9.38)
= 0.52m
GM fluid = GM (solid) – FSC
= (0.52 – 0.234)
= 0.286m.
Vessel in same condition as in question 10 consume’s the following on a passage :
- All HFO (i.e 150t) from NO2 port .
- Part HFO (i.e 50t ) from NO.3c .
- All FW (i.e. 100 t ) from NO4 S.
Find the GM fluid on arrival port .
Solution:
Ship’s wt | KG | VM | FSM |
10000 | 8.954 | 89540tm | 2087.4 |
(-) 150t | 0.65 | (-) 97.5tm | (-) 684.0 |
(-) 50t | 0.60 | (-) 30tm | (+) 1140.0 |
(-) 100t | 0.70 | (-)70 tm | (-)300.0 |
Final W =9700 t FVM = 89342.5 tm FSM= 2243.5
We know that :
Final KG = (Final VM/ Final W)
= (89342.5/9700)
Final KG = 9.21m
Again, FSC = (FSM / Final W)
= (2243.5 /9700)
= 0.231m
GM (Solid) = KM – KG
= (9.9 – 9.21)
= 0.69m
GM fluid = GM (solid) – FSC
= (0.69 – 0.231)
= 0.459m.
Vessel in same condition as in question 10, loads 1000t cargo in NO2 LH KG 4m ; 2000t cargo NO4 LH KG 5m . Find the final GM fluid, given that the final KM is 10.0m .
Solution:
Ship’s wt | KG | VM |
10000t | 8.954m | 89540 tm |
(+ ) 1000 | 4m | (+) 4000 tm |
(+) 2000 | 5m | (+)10000 tm |
Final W = 13000 Final VM = 103540tm
We know that:
Final KG = (Final VM/ Final W)
Final KG = (103540 / 13000 )
= 7.964m
KM = 10m ( given)
Again, GM solid = ( KM – KG)
= (10 – 7.964) m
= 2.036m
Now, FSC =( FSM/ W)
= (2087.4 /13000)
= 0.16m
GM fluid = GM (solid) – FSC
= (2.036 – 0.16)
= 7.876m.
A ship of 5000t displacement has a DB tank 18m long and 12m wide , partly full of SW . Find the FSC in the following cases:
- If the tank is undivided .
- If the tank is divided into identical P and S watertight divisions and
- Both side are slack.
- only one side is slack.
- If the tank is divided in to P, S and C identical watertight division and
- All three of these are slack .
- Only two of these are slack .
- Only one of these are slack .
- If the tank is divided into four identical watertight division – Port inner , port outer ,Stbd inner , Stbd outer, – and
- All four of these are slack .
- Any three of these are slack .
- Any two of these are slack .
- Any one of these is slack .
NOTE : The use of 1/n2 would be a very quick method of solving this question .
Solution :
a). if the tank is undivided
We know that
FSC = (LB 3 x i )/ (12 x W)
= ( 123x 18 x 1.025)/ (12 x 5000)
So, FSC = 0.531m
b). If Divided into identical P and S
1 ). Both sides are slack
FSC = (1/n2)
= (1/22 ) x 0.531
= 0.133m
2.) If only one side is slack
We can calculate
FSC = (0.133 /2)
= 0.066m
C). If the tank is divided in to P , S and C identical water tight division and
1.) If all three of these are slack .
We know that
FSC = (1/n2)
= (1/32 ) x 0.531
= 0.059m
2.) If only two of these are slack then
FSC = (0.059 /2)
= 0.039m
3.) If only one of these is slack
= (0.059/3)
= 0.019m
d). If the tank is divided into four identical tanks
1). All four of these are slack
We can calculate:
FSC = (1/n2) x 0.531
= (1/42 ) x 0.531
= 0.0332m.
2). Any three of these are slack
FSC would be 0.0247 m
3). Any two of these are slack
FSC would be (0.0332/2) m
= 0.0165m
4). Any one of these is slack = (8.25/1000)
= .0082m.
A ship of 1000 t displacement has a fresh water DB tank which is 20m long and divided in to P ,S and C watertight divisions . The P and S division are each 4m wide , while the she division is 12m wide . Calculate the FSC in the following cases :
- Only the port side tank is slack .
- Both port and stbd tanks are slack .
- Only the centre tank is slack .
- All three tanks – PS AND C are slack .
- If all the three divisions were not separate but formed one undivided tank 20m broad.
NOTE: 1/n^{2} cannot be used here as the watertight division are not identical.
Solution:
a). Only the port side is slack
We can calculate
FSC = (LB^{3 }x di )/(12 x W)
= (4^{3} x 20)/( 12 x 10000)
= 0.016m
b). Both P and S tanks are slack
= (2 x 0.016)
= 0.0213m
C ). Only the centre tank is slack
= (LB^{3} x di )/( 12 x W)
= ( 12 ^{3} x 20) /( 12 x 10000)
= 0.288m.
d). All three tank P, S and C are slack
FSC = (0.213 + 0.288) m
= 0.309m.
e) If all the three division were not separate but forward are undivided tank 20 m broad .
We can calculate as
FSC == (LB^{3} x di )/( 12 x W)
= (20 x 20^{3}x 1) / (12 x 10000)
= 1.333m.
A ship has a very small GM it is decided to fill up SW ballast , one tank at a time , in six tanks whose particulars are :-
Tank | ‘i’ about tank’s centre line |
NO 1 DB | 280m4 |
NO 2 O or S | 600 |
NO 3 P or S | 350 |
NO 3 C | 650 |
State the order in which the tanks must be filled so as to keep FSC at a minimum at all times.
An unstable vessel is it her angle of loll. The following tanks are available for SW ballast :-
NO 1 DB | 400m4 |
NO 2 P | 700 |
NO 2 S | 700 |
NO 4 P | 250 |
NO 4 S | 250 |
NO 4 C | 800 |
If it is decided to ballast three tanks with SW, one at a time , state what order should be followed so as to keep FSE to a minimum .
A vessel of W 8000t, KM 7.9m, KG 7.0m,has a tank 15m long and 12m wide , partly full of HFO of RD 0.95 .
- Find her moment of her statical stability at 6degree heel
- If her BM is 4.9m, find her moment of statical stability at 20 degree heel assuming that she is wall- side.
Solution:
Given:
W = 8000t, KM = 7.9m & KG = 7.0m
Area of tank = (L X B)
= (15 x 12) m^{2}.
RD = 0.95, θ = 6degree
We know that:
FSC= (LB^{3} x di )/( 12 x W)
FSC = (15 x 12^{3} x 0.95)/(12 x 8000)
= 0.256m
We know that GM can be calculated as
GM = (KM – KG)
= (7.9 – 7.0) m
= 0.9m
Again, RM = (W x GM sinθ )
So, RM = (8000 x 9 x sin 6 ^{0} )
= 538.5tm
B). If the BM is 4.9m find the moment of statical stability at 20 degree heel assuming that the ship is wall sided .
Given:
BM = 4.9 m, θ = 20 degree,
RM = (W x Sinθ (GM + 1/2 BM tan^{2 }θ)
RM = (8000 x Sin 20 (9 + ½ x 4.9tan ^{2}20 ^{0})
Hence, RM = 2650 tm
GM fluid = (0 .9 – 0.256)
= 0.644 m.