hydrostatics and stability (deepwater)
DESCRIPTION
Stability DeepwaterTRANSCRIPT
Hydrostatics & stabilityNomenclature
Talking with Naval Architects
Coordinates
Heel= static roll Trim = static pitch
BowStern
Keel(K)
Nomenclature
• The bow is pointing forward,• The stern is facing aft (backwards)• The starboard side is on your right as you
face forward,• The port side is on your left as you face
forward
Archimedes Principle
• The buoyancy of an object is equal to the weight of displaced fluid
Buoyancy ForceStatic Pressure
P=ρgz B
T
F = P*A = [ρgT]*BL = ρg*TBL = ρg*(Volume)
z
Tip: All fluid forces of importance for offshore structures are pressure forces!
Displacement
LBT=∇L
BT
BLBTC=∇L
TB
L = LengthB = Beam (width)T = DraftCB = Block coefficient∇= Displacement (Volume) ∆= Displacement (Buoyancy Force)γw = Density of fluid = ρg
∆ = γwLBT
∆ = γwLBTCB
30 m
100m
4000 t
T ?γw =1.025 t/m3
Seawater
∆= γwLBT = 4000 t
T = 4000 t / [γwLB] = 4000 t / [1.025 t/m3 * 100 m * 30 m] =
1.3 m Draft
Example 1: Compute draft (T) for prismatic barge, 100 m long, 30 m wide, weight 4000 t
Hydrostatic Stiffness - HeaveW
W + δW
δTT
K = δW/δT = γwAwp
Awp = LB (Waterplane Area)
Hydrostatic Stiffness – Roll/Pitch
Pontoon Cross Section10 m
5 m
60 m typ
10 m Dia typ
25 m
Example 2: Semi-Submersible Weighs 3000 t. How much payload can it carry? What is its vertical stiffness? Seawater is 1.025 t/m3.
Three Rules of Floating Platform Design
First Rule of Floaters: Buoyancy must equal weight plus any external vertical forces.
Weight includes:
1. Hull Steel Weight
2. Hull Outfitting
3. Topsides Payload (fixed and variable)
4. Topsides Structure
5. Ballast in Hull (fixed and variable)
External Vertical Forces Include
1. Component of Mooring Load
2. Component of Riser Load
3. Suspended weights (e.g. from crane)
2nd Rule of Floaters: The weight shall be positioned such that the hull will not tip over!
Typhoon TLP after Hurricane Rita
3rd Rule of Floaters: There should be enough Reserve Buoyancy to maintain balance and stability even with tanks flooded!
P-36 After Explosion in Column – Not Enough Reserve Buoyancy
Thunderhorse Semi after Hurricane Dennis – Reserve Buoyancy in Deck saved it from sinking!
Transverse Stability
W B B
W
Case 1: Positively stable Case 2: Negatively stable or unstable
Stability = Tendency to return to a previous condition when perturbed!
The metacenter is the point of intersection between the action of the buoyant force and the centerline of
the vessel.
W
L
W1 L1
G
B B1
M
K
z
Lost Buoyancy
Gained Buoyancy
Metacentric Height is the distance from the center of gravity to the
metacenter.
W
L
W1 L1
G
B B1
M
K
z
K – keel pointG – point of action of weight, i.e. center of gravityB – point of action of buoyancy, i.e. center of buoyancy; The position of B
shifts with the amount of heel (B to B1)M – Point of intersection of line of buoyancy and centerline, i.e. MetacenterGM – Distance between G and M, i.e. metacentric height.KB - Distance from Keel Point to BKG- Distance from Keel Point to G
“GM” is the metacentric height
and must be positive!
Metacentric Height, GMW
L
W1 L1
G
B B1
M
K
z
KGBMKBGM −+=
BM
KB
KG
B
M
TB
LBTLBIBM xx
1212/ 23
==∇
=
L
BT
Different “B”
“BM” is a function of the waterplane inertia (moment of square of the distance from the axis)
Metacentric Height (GM) and Restoring Moment
W
L
W1 L1
GB B1
M
K
z
θ
Μ θ = GZ*B ~ GM*B* sin(θ) (small angles)
M θ
GZ is the “righting arm”
Example Restoring Moment Curve
• Plot GZ (righting arm) vs. angle of heel– Ship’s G does not change as angle changes (not
true if tanks are partially full)– Ship’s B always at center of underwater portion of
hull– Ship’s underwater portion of hull changes as heel
angle changes– GZ changes as angle changes
B
G
M
K
T
B
GM=T/2 + B2/12T - KGStable if GM > 0
KG<T/2 +B2/12T
Substitute T = W/(γwLB)
WLB
LBWKG w
w 122
3γγ
+≤
0
10
20
30
40
50
60
70
0 5000 10000 15000 20000
Weight, t
Dra
ft (T
), K
Gm
ax -
m
T
Kgmax
L=100 mB = 30 m
Max KG for Payload is required in operating manual for vessels.
Example: Maximum KG as function of total weight for the 100 x 30 m barge.
KGBMKBGM −+=
A. A rectangular barge is 80 m long by 20 m wide. It weighs 5000 tonnes with its cargo. What is it’s draft if the density of water is 1.025 t/m3?
B. What is the maximum value of KG for this barge for positive stability, i.e. GM>0?
C. Man is in rowboat on lake. There is an anchor in the boat. He throws the anchor overboard. Does the level of the lake:
• Raise• Go down• Stay the same?
D. A glass of water has an ice cube. When the ice melts is the level of water.
• Higher• Lower• The Same?
Hint: The density of water is greater than the density of ice.
E. A glass of Scotch has an ice cube. When the ice melts, is the level• Higher• Lower• The Same?
Hint: The density of Scotch is less than the density of ice.
hVw
Vboat = (Wboat + Wanchor)/γw
h = (Vboat + Vw)/(WL) = 1/(WL)*(Wboat/ γw+Wanchor/γw+Vw)
W
hVw
Vboat = Wboat /γw
h2 = (Vboat + Vanchor +Vw)/(WL) = 1/(WL)*(Wboat/ γw+Wanchor/γa+Wanchor/γw+Vw)
W
Vanchor = Wanchor/γa
h1 = (Vboat + Vw)/(WL) = 1/(WL)*(Wboat/ γw+Wanchor/γw+Vw)
Since γa>γw
h2 < h1…. The level goes down!
h Vw
Vice = Wice /γw
W
h Vw
Vicewater = Wice /γw
W
The ice cube and the melted ice cube weigh the same!.... H is the same…
Ice cube in water
h1 Vw
Vice = Wice /γice
W
h2 Vw
Vicewater = Wice /γw
W
Since γice < γw
Vice > Vicewater
h1 > h2
Ice cube in scotch
Class Rules for Stability
Overturning Moment
6 m
Wind Force
Reaction force from:
a) Centroid of Drag or
b) Moorings (use worst)
Wind Load Calculations
Wind Force
1
23
4
5
Select different block areas based on elevation, shape
z2
z1
z4
z5
Tip: For wind calcluations use the waterline as z=0 .. Convert to the keel reference for stability!!!
6 m3.6 m
z3
Force in normalized by Ur2
Centroid based on 1st moment of forces
Zc_wind
Wind Force Coefficient
• Cw = Wind Force Divided by Velocity (at 10 m) Squared
• Units: – tonne/(m/sec)2
– Kips/(ft/sec)2
– kN/(m/sec)2
Overturning Moment
6 m
Wind Force = 898*Ur2 (N)
Reaction force from:
Moorings at keel (assume worst)
2.4 + 16.5 = 18.9 Moment Arm
2.4 m draft
• Watertight Integrity– Compartments and fittings designed for
hydrostatic head (minimum 20 ft) of surrounding structure.
• Weathertight Integrity– Under design sea conditions water will not
enter the structure (e.g. typically limited by vent openings)
0
2000
4000
6000
8000
10000
12000
14000
16000
18000
0 10 20 30 40 50 60 70 80
Theta (deg)
Mom
ent (
t-m)
Righting Moment KG=5Overturning Moment
Min extent of weathertight integrity (assumed when
deck imerses!)
1st Intercept
2nd InterceptB
A
C
KG = 5 GM = 9.9
KG = 10 GM = 4.9
Intact Stability Example
KG 5 10Area A 88500 20000Area B 72750 67500Area C 10500 22500A+B 161250 87500B+C 83250 90000Ratio (A+B)/(B+C) 1.94 0.97Min Reqd 1.4 1.4
OK NO!!
Even with positive GM the barge does not meet Class Rules for Stabiltiy:
Reduce KG or Reduce Weight.. Or? Add fixed ballast..
Increases Weight but lowers KG
0
1000
2000
3000
4000
5000
6000
7000
8000
9000
0 5 10 15 20 25 30 35 40
Theta (deg)
Mom
ent (
t-m)
Righting Moment KG=10Overturning Moment
A
B
C
Intact Stability Rules for Offshore Floating Platforms
• Must have a positive GM for all conditions• Survive overturning moment from 100 kt wind
(51.5 m/sec)– Waves not considered in class rules– Reaction from drag (free floating) or mooring,
whichever is worse• Inclining test is required for first unit of series• Righting moment curves and overturning
moment curves are required for all operating drafts
Damage Stability Rules
• Consider accidental floating of any compartment with external connections to the sea, or adjacent to the sea
• Mean heeling angle plus overturning due to 50 kt wind should not exceed limits of weathertight integrity.
Wind Tunnel Testing
Photos from Force Technology (www.force.dk)