Question

An underwater riser with an outer diameter of 250 mm and wall thickness of 20 mm is subjected to tension and pressure. The effective tension is $1200$ kN wherein the internal and external pressures of the riser are $25$ MPa and $6$ MPa, respectively. The true wall tension in the riser is $\times 10^{6}$ N (rounded to two decimal places).

Hint:

For risers and pipelines, convert effective tension to true wall tension by adding the internal end force ($p_iA_i$) and subtracting the external end force ($p_oA_o$).

Answer 1

Step 1: Geometry.
Outer radius $r_o=\dfrac{250}{2}\text{ mm}=125\text{ mm}=0.125$ m, \; thickness $t=20$ mm $\Rightarrow$ inner radius $r_i=r_o-t=105$ mm $=0.105$ m.
Areas: $A_o=\pi r_o^2=\pi(0.125)^2=0.049087\ \text{m}^2$, \; $A_i=\pi r_i^2=\pi(0.105)^2=0.034641\ \text{m}^2$.
Step 2: Relation between effective and true wall tension.
For a pressurized pipe/riser, \[ T_{\text{true}}=T_{\text{eff}}+p_i A_i - p_o A_o, \] where $p_i$ and $p_o$ are internal and external pressures (end-pressure forces).
Step 3: Substitute numbers.
$T_{\text{eff}}=1.200\times 10^{6}\ \text{N}$, \; $p_iA_i=25\times10^{6}\times0.034641=0.866\times10^{6}\ \text{N}$, \; $p_oA_o=6\times10^{6}\times0.049087=0.295\times10^{6}\ \text{N}$.
\[ T_{\text{true}}=1.200+0.866-0.295=1.771\ \times 10^{6}\ \text{N} \approx 1.77\times10^{6}\ \text{N}. \]