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

Passive compensator

Combined passive / active compensator

Heave_compensation.PRZ 2

Dynamic model

Load

Compensator mass

Compensator stiffness and damping

Load damping

Wire stiffness

Crane top motion

Load motion

Compensator motion

Heave_compensation.PRZ 3

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

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

Time domain simulations (2DOF):Use "state space" representation:

Solves for the derivative of the state vector by:

where:

Heave_compensation.PRZ 6

In general (N DOF) we may write:

Forward Euler:

Heave_compensation.PRZ 7

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

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