Determination of principal particulars of ship

1. Requirement:
Ship type : General Cargo
Dead Weight : 20,000 tonnes
Route :
Distances : Chittagong Port to Keppel Harbor Port, Singapore 1863 Nautical mile
Distance from Bangladesh to China 3231
Distance from Bangladesh to Hong kong 3038
Distance from Bangladesh to Japan 4284
Distance from Bangladesh to Malaysia 2021
Average distance : 3500 Nautical mile
Speed : 17 knots
Endurance : 10 days
There is some restriction to design our ship. Which are given below:
1. The Maximum permissible draft for entering Chittagong Port is 9.15 m.
2. The Maximum permissible length for entering Chittagong Port is 186 m.
3. The Maximum permissible entry length for night navigation is 153 m.
4. The Maximum permissible draft for Main Jetty areas i.e. Jetty No. 2 to Jetty
No.13 is 8.55 m.
The entry permissible draft would however depend on the day’s maximum
permissible draft

2. Basis Ship Data:
Ship Type : General Cargo
Length, LOA : 142 m
Length, LBP : 132 m
Breadth, BMLD : 22.23 m
Depth, D : 11.30 m
Draught, T : 8.25 m
Dead weight, DWT : 15000 tonnes
Speed : 13.8 Knots
Complement : 18 persons
Power : 4,900KW
Principal Particulars Calculation:
Displacement of basis ship, Δ = 132 × 22.23 × 8.25 × 1.025 × 0.775
= 10106.53 tonnes

3. Dead weight coefficient, Cd =
15000
10106.53
= 0.78
L/B = 5.93
B/T = 2.69
T/D = 0.72
Taking same Ac for basis sip & model ship :
∇1
2
3×V13
𝑝1
=
∇2
2
3×V23
𝑝2
P2= 11097 kw
Deadweightofnew ship:
 Weight of crew: 75 kg/person
Number of crew = 18
Weight = 75 × 18 = 1350 kg
= 1.35 tonnes
 Stores & Provisions: 5 kg/person/day
Weight = 5 × 18 × 10 × 1.20 kg
= 1080 kg
= 1.08 tonnes
 Fresh Water:
1. Consumption for crew = 20 kg/person/day
= 20 × 18 × 10 × 1.20 kg

4. = 4320 kg
= 4.32 tonnes
2. Bath & Laundry: 200 kg/person/day
= 200 × 18 × 10× 1.20 kg
= 43200 kg
= 43.2 tonnes
3. Cooking: 4 kg/person/day
= 4 × 18 × 10× 1.20 kg
= 862.85 kg = 863 kg
= 0.863 tonnes
4. Main engine Cooling: 7 g/ kwh
= 7 × 7300×10× 24 × 1.20 g
= 15207kg
= 15.3 tonnes
5. Auxiliary engine Cooling: 0.20 × Main engine cooling water
= 0.20 × 15.3 tonnes
= 3 tonnes
Total fresh water needed : 4.32+43.2 +0.863 +15.3 +3 = 66.7 tonnes
 Fuel Oil:
Engine: B32:40V12P
Engine power: 7300 kw
SFC : 184 g/kwh
Fuel needed : 184 × 6000× 10 × 24 × 1.20

5. = 318 tonnes
 Lubricating Oil:
Specific lubricating oil consumption = 0.8 g/BHPh
Weight = 0.8 × 7300 ×10× 24 × 1.20 g
= 1560 kg
= 1.23 tonnes
 Generator Fuel:
Model : 6090HF484
SFC : 59.1 l/h( 15.6g/h)
Generator power : 2x250 kw
Fuel needed : 59.1 × 2 × 10× 18 × 1.20 g
= 25528.5 kg
= 25.52 tonnes
[Source :
https://www.deere.com/en_US/products/engines_and_drivetrain/marine/marine_di
esel_engines.page]
Summary:
DWT : 20000 tonnes
Fresh Water : 66.7 tonnes
Engine Fuel : 318 tonnes
Generator Fuel : 25.52 tonnes
Lubricating Oil : 1.23 tonnes

6. TOTAL = 414.53 TONNES
CAPACITY = 20000 – 415 = 19585 TONNES
Crew : 1.35 tonnes
Store & Provisions : 1.08 tonnes
Displacement, Δ =
𝐷𝑊𝑇
Cd
=
20000
0.7733
= 25863.183 𝑡𝑜𝑛𝑛𝑒𝑠
Lightweight, LWT = 25863.183 – 20000 = 5863.183 tonnes
We know,
Δ = (L/B) × (B/T)2 × T3 × Cb
Or, T =√
Δ
𝐶𝑏×(
𝐿
𝐵
)×(
𝐵
𝑇
)2
3
= √
26692.105
0.795×(5.93)× 2.692
3
= 9.21 m
And,
B/T = 2.69
Or, B = 2.69 × 9.21 = 24.78 m

7. And,
L/B = 5.93
Or, L = 5.93 × 24.78 = 145.28 m
And,
T/D = 0.72
Or, D = 8.35 /.72= 12.79 m
Summary of Principal Particulars:
Displacement, Δ = 25863.183 tonnes = 26103 tonnes
CARGO CAPACITY = 20000 – 485 = 19515 TONNES
Dead weight, DWT = 20000tonnes
Light weight, LWT = 5863.183 tonnes
Length, LOA = 149.77 m
Length, LBP = 145.28 m
Breadth, BMLD = 24.78 m
Depth, D = 12.79 m
Draught, T = 9.21 m
Block coefficient, Cb =0 .775
Free Board Calculation:
[According to :ILLC 1966, Bulk Carrier: Type ‘B’ Vessel.]

8. Length Free board
243 3893
244 3934
245 3949
246 3965
247 3975
248 3978
289 3992
[source : Rules for Ships, January2001 ; Pt.3 Ch.5 Sec.3 – Page17]
By liniear inter pulation for length L = 145.28 FreeBoard Is 3953.48 mm
Sheer Calculation :
Minimum Bow Height:
As Length, L < 250m
=
56𝐿(1 −
𝐿
500
)1.36
𝐶 𝐵 + 0.68
𝑚𝑚
=
8135.7(1 −
145.28
500
)1.36
0.79 + 0.68
𝑚𝑚
= 5339.89 𝑚𝑚
AP
1
6
𝐿
1
3
𝐿
Midship
2
3
𝐿
5
6
𝐿
FP
25(
𝐿
3
+ 10) 11.1(
𝐿
3
+ 10) 2.8(
𝐿
3
+ 10)
0
5.6(
𝐿
3
+ 10) 22.2(
𝐿
3
+ 10) 25(
𝐿
3
+ 10)
2066 648.5 163.59 0 327.189 1297 1460.676.65

9. Wheel House Height:
[According to https: panama-canal-bridge-visibility-requirements]
[source : https://officerofthewatch.com/2012/10/10/panama-canal-bridge-
visibility-requirements/ ]
AC = 2× 145.28 m = 290560 mm
Bow height h = 5339.89 𝑚𝑚
Length of Poop Deck :
[According to: Ship Construction By David J.Eyres,George J. Bruce ;
page -238 ]
Poop deckcovers 15% of ships length & Pood Deck’s front should not
exceed 50% Of Length Amitship.
For Our ship Poop deckHeight =
36.25+2×145.28
2×145.28
× (5.3+ 12.79) m = 20 m
(maximum)

10. Dimention for Bulbusbow:
Length of Bulbusbow Will be calculated by subtracting at X distance forward per
pendicular .
𝑋 =
a
2
= 0.015L = 2.175 m
a = 4.35m (maximum)
Length of collision bulkhead should not be more than 7% of L
So 0.07×145.28 m = 10.15 m
[According to: Ship Construction By David J.Eyres,George J. Bruce ;
page -237 ]

11. Hc = Height Of Hatchways = 600mm
Calculation of Determining the Distance of Cargo holds
from Collision Bulkhead:
For Position 1 : Hatchway height Hc = 600 mm
𝐿 𝑓 = 5 × ( 𝐻𝑓 − 𝐻𝑐)0.5
𝑚 = 3.53 m
Hf = (Hc+0.5) mm = 600.5 mm
Navigation In Ice:
Angles Ranges ( °) For Our Ship
Α ( A1) 15 ˜ 55 45

12. Awf = Area of waterplane in Bow
B/4 = Breadth /4 = 6.133 m
Length Of Cargo Hold :
Length remaining = L - (7% of L For Forcastle ) –( 15% of L/2 For
Poop deck) – Lf
= 145.28 – ( .07×145.28+.15×(145.28/2)+3.53)m
= 145.28 – 25.59 m
ß (A2) 25 ˜ 90 48

13. =120.68 m
So Length of Cargo hold = 120/5 = 24 m (minimum)