18 heave compensation
Post on 05-Sep-2015
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Heave compensationPassiveActiveCombined passive / active
Single versus two degrees of freedom system
Modes of operationConstant tension(drill strings, risers, guidelines)Motion control (landing on bottom)Slack prevention (lift off)Snap load prevention (towing)
Heave_compensation.PRZ 1
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Passive compensator
Combined passive / active compensator
Heave_compensation.PRZ 2
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Dynamic model
Load
Compensator mass
Compensator stiffness and damping
Load damping
Wire stiffness
Crane top motion
Load motion
Compensator motion
Heave_compensation.PRZ 3
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Dynamic model (SDOF)
y Ignore compensator massy Linear passive systemy No excitation on load
M $$3 + c $ 3 + k( 3 3T ) = F3M (t) = 0
Total Complex stiffness
k = 1kw +1
kc + i* c c1
j kc + i*cc for kw >> kcHarmonic oscillation of top:
3T = 3Ta cos (* t)
Dynamic response of load (general solution) :
33T2 =
1 + **02 cc2kc M
1 **02 2 + **0
2 (cc + c) 2k c M
Solution at resonnance: 33T * = *0
2 = kcM + cc2(cc + c)2
Solution at high frequencies :
33T * >> * 02
* 0*4 + cc* M
2
Heave_compensation.PRZ 4
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Dynamic model (2DOF)
y Include compensator massy Linear passive systemy No excitation on loadmc
$$3P +cc $3P + kc3P + kw(3P 3) = F3t(t)M $$3 +c $3 + kw(3 3P ) = F3M(t) = 0
Excitation on compensator mass:
F3t(t) = tkc + $t cc
On matrix form we may write:
,M $$ +C $ +k = F
M = mc 00 M
C = cc 00 c
k = kc +kw kwkw kw
F = tkc + $t cc F3M T
Heave_compensation.PRZ 5
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Time domain simulations (2DOF):Use "state space" representation:
Solves for the derivative of the state vector by:
where:
Heave_compensation.PRZ 6
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In general (N DOF) we may write:
Forward Euler:
Heave_compensation.PRZ 7
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Example, two connected masses, one is given an initial displacement
0 20 40 60 80 100 120-2
-1.5
-1
-0.5
0
0.5
1
1.5
2Motion of masses in 2 dof system (Euler)
Time (s)
Dis
plac
emen
t (m
)
Mass nr. 1Mass nr. 2
m1 = 10[kg]m2 = 20[kg]k1= 2 [N/m]k2= 2 [N/m]c1 = c2 =0
Heave_compensation.PRZ 8
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Pneumatic cylinders
y Adiabatic compression:
. pV = constant , l 1.4
kc = dFdz = Adpdz
kc = A ddzp0 V0V
= Ap0V0
Addz
1z
= Ap0V0
A ()z(1)
= p0V0
V Az
Small displacements: . . V lV0 = Az0 lAz
kc l p0A 1z0 = F0 AV0
I.e Stiffness increases with load level
Heave_compensation.PRZ 9
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