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PORT and Harbor Engineering Radianta Triatmadja Lecture note 2
Waterway Design Depth of W aterway aterway
Waterway Design The depth of waterway depends on the tonnage of the ship and the water. water. Following Archimedes law, the amount of water displaced by the ship equals the weight of the ship. Therefore the more is the tonnage, the deeper deeper is the ship’s draft. Ships’ draft also depends on the cross section of the ship.
Draft
draft 50 000 ton
draft 50 000 ton
Ships of the same total weight may have different drafts.
Draft • Box Coefficient
draft
draft
50 000 ton
50 000 ton
Cb =0.8 Cb =0.6
= Cb
Draft • Box Coefficient
30 000 ton
draft
draft 50 000 ton
Cb =0.5 Cb =0.67
= Cb
Tonnage Displac ement tonnage : The weight weight of • Displacement water displaced, the physical weight of the ship Displac ement : the weight weight of the • Loaded Displacement ships + loading displac ement : the weight weight of the • Light displacement ships only
Tonnage • Dead weight ton : The difference between loaded displacement and light displacement, displacement, or the weight weight of loading (please do not confuse conf use this tonnage with with the weight weight of the ships) ships ) weight of fresh water, tools, and fuel f uel • The weight are considered as loading
Tonnage Gross Registered Regis tered ton (use (use for f or Passenger • Gross ships) Volume of ships space in cubic feet f eet divided by 100 • 1 GRT ~ 2.83 m 3
Tonnage • Net Register Ton (use for Passenger ships) • Volume of ships (GRT) minus all non earning spaces (machine, bunker, tank, room for staff, staff , tool room, radio radio and map room, storage room )
Typical Typical ships full f ull load draft of Bulk carr c arrier ier 14
Typical tonnage and full load draft of bulk carriers (ore, (ore, coal, cement etc) etc ) 20 18 16 ) d e d a o l ( t f a r D
14 12 10 y = 3.965Ln(x) - 3.1657
8
R2 = 0.8525
6 4 2 0 0
50
100
Dead weight (1000 tons)
150
200
Passenger Ships 12
Basin depth for unknown full draft
10 8 y = 1.9623Ln(x) + 2.7951
6
R2 = 0.9546 4
y = 1.6041Ln(x) + 2.6762 R2 = 0.9807
2 0 0
5
10
15
20
25
30
X 1000 ton Depth required / draft = 1.14
35
General Cargo 16
Basin depth for unknown full draft
14 12 10 8
y = 2.3647Ln(x) + 4.308
6
R2 = 0.9807 0.9807
4
y = 2.1454Ln(x) + 3.6605 R2 = 0.9903
2 0 0
10 W ei eight (x1000) 1 2 5 8 10 15 30 40 50
20 draft 4.2 4.9 6.8 8 8.5 9.3 10.9 11.7 12.4
30 depth 5 5.5 7.5 9 10 11 12 13 14
40
50
60
Depth/draft =1.14 =1.14 Depth required 14% more than the draft
Container Ships 16
Basin depth for unknown full draft
14 12 10 8 6
y = 0.1x + 10
y = 0.08x + 9.1
R2 = 1
R2 = 0.9877
4 2 0 0
10
W eight (x1000) 20 30 40 50
20
draft 10.6 11.6 12.4 13
30
depth 12 13 14 15
40
1.132075 1.12069 1.129032 1.153846
50
60
Depth/draft= 1.13
Depth of waterway • The depth depth of wat water erwa way y shoul should d be deepe deeperr than than the basin. Provisions should be made for possible pitching, heaving and squat • Squat Z
Z 2.4
2
L pp p p
2
F
2
1 F
Additional depth due to • Vertical movement due to wave (heaving) • Vertical movement due to Squat • Vertical movement due to pitching
Required Water W ater depth depth
Draft (ship size) d e g d e r d e b o T
Keel = bottom
Expected vertical vertical movement Net underkeel clearance
Gross underkeel clearance
Sounding tolerance
Expected sediment accumulation accumulation between two two dredging campaigns campaigns dredging tolerance tolerance
How to design waterway depth? S hips weight weight and type • Define Maximum Ships • Calculate/predict Ships draft • Define minimum water level • Calculate depth Required (1.15 times Ships draft) (nominal depth) • Calculate expected sediment accumulation during two consecutive dredging works • Calculate total depth required to dredge
W hat is minimum minimum water water level ? f luctuating ing due to moon and • Sea level is fluctuat sun attraction forces and centrifugal forces of the earth during its revolution about its common c ommon axis with the moon • This fluctuation is called tides • Maximum fluctuation occur during spring tide, while minimum fluctuation occur during neap tide.
Tidal constants waves es are periodic due to the • Tidal wav driving forces. Yet so many conditions affect the periodicity of the tidal waves. A complete period of tide wav wav e is approximately 19 years. In order to design a harbor one need to know tidal constants in the design area. These tidal constants govern the major water water level fluctuations. f luctuations.
Tidal constants
2t A cos T n
i
i 1
i
i
An Example of Water level Fluctuation 40.Air Pelayaran Sebelah Barat Surabaya (Karang Jamuang)
Spring
Neap
24
20
m16 d i s a v 12 e l E
8
4
0 0
100
200
300
400
500
600 Time
700
hours
800
900
1000
1100
1200
Discussion • 2. 3. 4.
5. 6. 7.
A Harbor designed is given to be discussed. Topics to be discussed are : W hat hat is is the the max maxim imum um siz size of of ship ships s that that can can be be serv served ed The width of the waterway Duri During ng spri spring ng tide tide even even larg larger er ship ships s can can ent enter er the the har harbo bor. r. What What is the maximum size of the ships during spring tide? Discuss the possibility of allowing such ships to berth in the harbor If the the spee speed d of of your your ship ships s are are lim limit ited ed to 10 knot knots s alo along ng the the waterway, what is the maximum squat ? Do you you hav have e any any com comment ent oth other er than than the the top topic ics s dis discu cusse ssed d? W ithi ithin n the the nex nextt 10 year years s the the har harbo borr wil willl be be upg upgra rade ded d to serv serve e ships of maximum maximum draft draft 4 meters. W hat works are needed?
