Buoyancy test

This test checks the values for the buoyancy components, see Buoyancy for details on the theory.

1 Test description

This test uses a cylinder floating in water. For different waves and initial conditions, the Buoyancy loads and Buoyancy springs from the Total hydro loads sensor are benchmarked against analytical solutions.

2 Model

The model is illustrated in the figure below:



The model consists of a uniform circular cylinder. The diameter of the cylinder is 
$$D = 6\text{ m}$$
and the thickness is 
$$t = 0.060\text{ m}$$
. This gives an outer area 
$$A = \pi(D/2)^2=28.27\text{ m}^2$$
and a structural area 
$$A_S = \pi(D/2)^2 - \pi(D/2-t)^2=1.120\text{ m}^2$$
.
The cylinder is made of steel with a density 
$$d = 8500\text{ kg}\cdot\text{m}^{-3}$$
. The length of the cylinder is 
$$L = 30\text{ m}$$
, which gives it a mass 
$$m_S = A_S\cdot L\cdot d = 285.5\text{ t}$$
. In addition, a point mass of 
$$m_P = 195.18\text{ t}$$
is added at the lowest node of the structure, which gives a total mass of 
$$m = m_S+m_P=580.69\text{ t}$$
The water density is 
$$d_W = 1026.9\text{ kg}\cdot\text{m}^{-3}$$
.
The gravitational acceleration is 
$$g=9.80665\text{ m}\cdot\text{s}^{-2}$$

3 Analytical solution

3.1 Draft 20 m

For this test, the initial draft of the cylinder is 
$$l = 20\text{ m}$$
.
The mass of the displaced water in the initial position is thus 
$$M_W=l\cdot A\cdot d_W$$
and the buoyancy force will be 
$$F_B = g\cdot M_W=5694\text{ kN}$$
The simulation is run with Loads only, so the cylinder does not move.

3.2 Draft 20 m  with waves

This test is the same as the previous one with the addition of linear regular waves of period 
$$T = 10\text{ s}$$
and amplitude
$$A = 2\text{ m}$$
. The time-varying wave elevation at the center of the cylinder will thus be 
$$\zeta(t) = A\sin(2\pi t/T)$$
 . The buoyancy force will be 
$$F_B(t) = g\cdot A\cdot d_W \cdot\left(l+\zeta(t)\right)$$
The simulation is run with Loads only, so the cylinder does not move.

3.3 Draft 20 m with waves two cylinders

This test is the same as the previous one, but a second floating cylinder is placed at a distance 
$$D = 70\text{ m}$$
ahead of the first cylinder, as shown in the figure below.



The wave elevation at the second cylinder is thus 
$$\zeta_2(t) = A\sin(2\pi t/T+k\cdot D)$$
where 
$$k$$
is the wave number and is equal to 
$$\omega^2/g$$
and 
$$\omega = 2\pi/T$$
.

The buoyancy force will be the sum of the buoyancy forces on the two cylinders, equal to 
$$F_B(t) = g\cdot A\cdot d_W \cdot\left(2l+\zeta(t)+\zeta_2(t)\right)$$

3.4 Draft 21 m

A static simulation is run with the model from the previous load case with no waves and a draft of 21 m. This simulation will find the static equilibrium of the cylinder given the initial conditions. 
The total mass of the cylinder is 
$$m=580.69\text{ t}$$
, which gives a total weight 
$$P=-mg=-5694\text{ kN}$$
. This is equal to the buoyancy force of the load case with a draft of 20 m, which means that the equilibrium position of this cylinder happens when the draft is equal to 20 m. Therefore, for the present load case where the initial draft is
$$l=21\text{ m}$$
, we expect the cylinder to rise by 1 m.

The mass of displaced water in the initial position is thus 
$$M_W=l\cdot A\cdot dw$$
and the buoyancy force will be 
$$F_B=g\cdot M_W = 5978\text{ kN}$$
.

Note:  the buoyancy force will not vary with the motion of the floater, as it is computed with the initial position of the floater. It will only change if parts of the structure are subjected to changes in the water elevation due to waves.

Since the cylinder will move in the vertical direction, it will be subjected to loads from Buoyancy springs. As explained in the Buoyancysection, the load from buoyancy sprngs can be expressed as 
$$P_-=-A\cdot L\cdot g\cdot d_W$$
 , where 
$$L$$
is the vertical displacement of the cylinder and is expected to be 1 m. This yields 
$$P_- = -285 \text{ kN}$$
Note that 
$$P_-+P+F_B=0$$
. This is expected since the cylinder is in equilibrium.

4 Results

For the first two load cases, a simulation of 30 seconds is run. The test is considered passed if the last 20% of the time series produced by Ashes is within 0.5% of the analytical solution
For the static load case, the test is considered passed if the results produced by Ashes are within 0.5% of the analytical solution.

The reports for this test can be found on the following links:

https://www.simis.io/downloads/open/benchmarks/current/Buoyancy test.pdf
https://www.simis.io/downloads/open/benchmarks/current/Buoyancy test static.pdf