Question 14.2: A submarine pipeline of internal diameter 275 mm rests on a ......
A submarine pipeline of internal diameter 275 mm rests on a rocky sea bed at a depth of 20 m. The pipe is made of 16 mm thick steel with a 100 mm thick concrete coating. Waves of 6m height and 9 s period approach the shore with the crests at an angle of 60° with the pipeline. The prevailing current of 0.5 m s ^{-1} is along the same direction as that of the waves. Check whether the pipe is stable against rolling.
Assuming the following: C_{ D }=1.1 ; C_{ M }=3.1 ; C_{ L }=1.2 ; density of sea water = 1030 kg m ^{-3} ; density of concrete = 2700 kg m ^{-3} ; density of steel = 7800 kg m ^{-3} . The coefficient of friction for the rock–pipe interface is 0.3.
The external diameter of the pipe is 275 + 32 + 200 = 507 mm. The weight of concrete is 1/4π(0.507² – 0.307²) × 2700 × 9.81 = 3.39 × 10³ N m ^{-1} . The weight of steel is 1/4π(0.307² – 0.275²) × 7800 × 9.81 = 1.13 × 10³ N m ^{-1} . The weight of the pipe in air is 4.52 × 10³ N m ^{-1} . The buoyancy is 1/4π × 0.507² × 1030 × 9.81 = 2.04 × 10³ N m ^{-1} . The submerged weight of the pipe is (4.52 – 2.04) 10³ = 2.48 × 10³ N m ^{-1} .
From Fig. 14.4, for d = 20 m, T = 9 s and L = 105 m,
From equation (14.16)
u=\frac{a \sigma \cosh [k(y+d)]}{\sinh (k d)} \sin (k x-\sigma t), (14.16)
u=-\frac{6}{2} \times \frac{2 \pi}{9} \frac{\cosh [0.06(-19.746+20)]}{\sinh (0.06 \times 20)} \sin \sigma t
= – 1.4 sin σt.
MORISON EQUATION (14.65)
F_{ i }=C_{ D } \frac{\rho}{2}|u| u D+C_{ M } \frac{\rho \pi D^2}{4} \frac{ d u}{ d t} . (14.65)
In determining the wave force, the component of velocity normal to the pipeline is considered. The algebraic sum of the velocity of the current and the particle velocity due to the wave is
u=(0.5-1.4 \sin \sigma t) \cos 60^{\circ}DRAG TERM
INERTIA TERM
The inertia force is
-3.1 \times 1030 \times \frac{\pi}{4} \times 0.507^2 \cos 60^{\circ} \cos \sigma t=-322.3 \cos \sigma t\left( Nm ^{-1}\right)The in-line force on the pipe per unit length is
71.8|0.5-1.4 \sin \sigma t|(0.5-1.4 \sin \sigma t)-322.3 \cos \sigma t\left( Nm ^{-1}\right)The lift force per unit length is
\begin{aligned} & 1.2 \times \frac{1030}{2} \times(0.5-1.4 \sin \sigma t)^2 \times 0.507 \cos ^2 60^{\circ} \\ & =78.3(0.5-1.4 \sin \sigma t)^2\left( Nm ^{-1}\right) . \end{aligned}Table 1
The frictional force = 0.3 × (submerged weight – lift). For stability, the frictional force must be greater than the in-line force. The pipeline is found to be stable.
Table 1
| t | Drag | Inertia force | In line | Lift | Frictional force |
| 0 | 18.00 | -322.3 | -304.3 | 19.6 | 738 |
| 1 | -11.5 | -246.9 | -258.4 | 12.5 | 740 |
| 2 | -55.4 | -56.0 | -111.4 | 60.5 | 726 |
| 3 | -36.4 | 161.1 | 124.7 | 39.8 | -732 |
| 4 | 0.0 | 302.9 | 302.9 | 0.0 | -744 |
| 5 | 68.8 | 302.9 | 371.7 | 75.0 | -722 |
| 6 | 210.5 | 161.2 | 371.7 | 229.7 | -675 |
| 6.5 | 253.4 | 56.0 | 309.4 | 276.5 | -661 |
| 7 | 253.4 | -56.0 | 197.5 | 276.5 | -661 |
| 8 | 140.7 | -246.9 | -106.2 | 153.2 | 698 |
| 9 | 18.0 | -322.3 | -304.3 | 19.6 | 738 |
