Subsea Lifts

How does drag affect a subsea lift?

Drag in a subsea lift is similar to drag on a car due to air resistance and varies with the square of the speed. How big the drag force will be depends on the area perpendicular to the motion as well as the drag coefficient (which we again know from cars competing to always have lower drag coefficients to increase range). The main difference from a car however is that the fluid is water and not air which have 1000 times bigger mass density, which is another factor in the drag force.

$F_D = \rho_w C_D A_\perp \dot z |\dot z|$

Where $C_D$ is the drag coefficient (more info can be found in DNV RP-N103, some examples shown below) and $\dot z$ is the payload vertical velocity.

Shape	$C_D$	$A_{\perp}$	Notes
Sphere	$C_D = 0.5$	$A_{\perp} = \pi r^2$
Horiontal Cylinder	$C_D = 1.2$	$A_{\perp} = 2 r L$
Vertical Cylinder	$\frac{L}{2r}=0.5 \Rightarrow C_D=1.1$ $\frac{L}{2r}=1 \Rightarrow C_D=0.9$ $\frac{L}{2r}=2 \Rightarrow C_D=0.9$ $\frac{L}{2r}=4 \Rightarrow C_D=0.9$ $\frac{L}{2r}=8 \Rightarrow C_D=1.0$	$A_{\perp} = \pi r^2$
Cube	$C_D = 1.05$	$A_{\perp} = a^2$	Face normal to flow.
Cone	$C_D = 0.50$	$A_{\perp} = \pi r^2$	Pointed tip aligned with flow; dependent on cone angle.
Rectangular Plate	$C_D = 1.1+0.02 (\frac{L}{W}+\frac{W}{L})$	$A_{\perp} = L W$	Normal to flow.

Let’s do a practical example to get a feeling about the relevance. Assume you are lifting a payload that is shaped like a rectangular plate with length 15 m and width 10 m. The wave period is 8 seconds and the wave height is 4 m, assume sinusoidal waves. How big will the force be?

First lets find the peak speed; the wave motion is given as

z = \zeta \cos(\omega t)

So we find the peak speed by finding the max of the derivate which is:

\dot{z}_{\text{max}} = \zeta \, \omega

Since $\omega = \frac{2 \pi}{T_p}$ we then have a peak speed of 1.57 m/s.

We can then calculate the drag coefficient as:
$C_D = 1.1 + 0.02 \left( \frac{15}{10} + \frac{10}{15} \right) = 1.14$

Our drag area is:

A_\perp = 10 \cdot 15 = 150 \,\mathrm{m^2}

We can then calculate the maximum drag force, which is:

F_D = 1025\cdot 1.14 \cdot 150\cdot 1.57^2= 44\,\mathrm{t}

How does added mass affect a subsea lift?

Added mass in a subsea lift is very important as it can cause big additional inertia, which in turn can cause big dynamic forces. It comes from the fact that when a payload oscillates in water due to heave motion it also has to accelerate surrounding water and mathematically it is given as:

$m_A = \rho_w C_A V_R$

Where $C_A$ is the added mass coefficient (can be found in DNV RP-N103, some examples below) and $V_R$ is the reference volume.

Shape	$C_A$	$V_R$	Notes
Sphere	$C_A = 0.5$	$V_R = \frac{4}{3}\pi r^3$	Constant in all directions.
Cylinder	$\frac{L}{2r}=1.25 \Rightarrow C_A=0.62$ $\frac{L}{2r}=2.5 \Rightarrow C_A=0.78$ $\frac{L}{2r}=5 \Rightarrow C_A=0.90$ $\frac{L}{2r}=9 \Rightarrow C_A=0.96$ $\frac{L}{2r}=\infty \Rightarrow C_A=1.00$	$V_R = \pi r^2 L$	Vertical motion along cylinder axis in infinite fluid.
Rectangular Plate	$\frac{L}{W}=1 \Rightarrow C_A=0.58$ $\frac{L}{W}=2 \Rightarrow C_A=0.76$ $\frac{L}{W}=4 \Rightarrow C_A=0.87$ $\frac{L}{W}=8 \Rightarrow C_A=0.93$ $\frac{L}{W}=\infty \Rightarrow C_A=1.00$	$V_R = \frac{\pi}{4}W^2 L$	Motion normal to surface.
Circular Disc	$C_A = \frac{2}{\pi}$	$V_R = \frac{4\pi}{3} r^3$	Motion normal to surface.
Square Prism	$\frac{L}{W}=1 \Rightarrow C_A=0.68$ $\frac{L}{W}=2 \Rightarrow C_A=0.36$ $\frac{L}{W}=4 \Rightarrow C_A=0.19$ $\frac{L}{W}=10 \Rightarrow C_A=0.08$	$V_R = W^2 L$	Prismatic body with square base.

Let’s do another practical example. Assume you are lifting a payload that is shaped like a rectangular plate with length 20 m and width 10 m. The wave period is 8 seconds and the wave height is 4 m, assume sinusoidal waves. How big will the force due to added mass be?

We need to find the maximum acceleration (derivative of speed as shown in drag example), which is given by:

\ddot{z}_{\text{max}} = \zeta \, \omega^2

Thus the maximum acceleration is:

\ddot{z}_{\text{max}}=2 \cdot \left(\frac{2\pi}{8}\right)^2 = 1.23 \,\mathrm{m/s^2}

Next the added mass coefficient is 0.36 and our reference volume is 2000 cubic meters. We can then calculate the force using:

F=m a=1025 \cdot 0.36 \cdot 2000 \cdot 1.23=98 \mathrm{t}

Note that for subsea lifts close to seabed, the added mass may increase due to confinement.