Minimum Drilling Line Size: 24-Pet-A4 Oil and Gas Well Drilling and Completion, May 2013, Q1

Question 1: Minimum Drilling Line Size (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $n=10$ lines strung; max depth $D=10{,}000$ ft; drill pipe $4.5$ in.–$20$ lb/ft; drill collars $7$ in.–$120$ lb/ft; $L_{DC}=0.10\,L_{total}$; safety factor $SF=2.0$; buoyancy ignored; nominal breaking strengths of galvanised wire (table above, extra-improved plow steel, EIPS).

Find. The smallest drilling-line nominal diameter that safely carries the fast-line load at $D=10{,}000$ ft.

Approach. Split the string into drill-collar and drill-pipe lengths from the 10% rule, sum their weights to get the hook load, divide by the number of lines strung to get the fast-line tension (frictionless sheave assumption, consistent with the buoyancy-ignored simplification), apply the safety factor, then read the smallest wire size whose EIPS breaking strength clears that value.

String geometry. $L_{DC}=0.10(10{,}000)=1{,}000$ ft, so $\boxed{L_{DP}=9{,}000\ \text{ft}}$.
Hook load (drillstring weight in air, buoyancy ignored). $W_{DP}=9{,}000(20)=180{,}000$ lbf, $W_{DC}=1{,}000(120)=120{,}000$ lbf. $W=W_{DP}+W_{DC}=180{,}000+120{,}000$, so $\boxed{W=300{,}000\ \text{lbf}}$.
Fast-line tension. With $n=10$ lines strung and no sheave-friction efficiency given, $T_{fast}=W/n=300{,}000/10$, so $\boxed{T_{fast}=30{,}000\ \text{lbf}}$.
Required minimum breaking strength. $T_{req}=SF\times T_{fast}=2.0(30{,}000)$, so $\boxed{T_{req}=60{,}000\ \text{lbf}}$.
Select the line. From the table, $3/4$ in gives only $58{,}800$ lbf $<60{,}000$ lbf (fails); $7/8$ in gives $79{,}600$ lbf $\ge 60{,}000$ lbf. So $\boxed{d_{min}=7/8\ \text{in}}$.

Check: assumes a frictionless sheave system (fast-line tension = hook load / number of lines strung, no efficiency-factor derating) since no sheave-efficiency table is supplied with this exam — consistent with the paper's own instruction to ignore buoyancy, i.e. use the simplest static load path. A real rig-load calc would also derate by the sheave efficiency $K_e$ for $n=10$ lines (typically 0.87–0.90), which would push the requirement toward the next larger line size.

Quantity	Value
Drill collar length / drill pipe length	1,000 ft / 9,000 ft
Total string weight (hook load), $W$	300,000 lbf
Fast-line tension, $T_{fast}$	30,000 lbf
Required minimum breaking strength	60,000 lbf
Minimum drilling line size	7/8 in (79,600 lbf EIPS)

Topic: Rotary rig hoisting-system load path — hook load, fast-line tension and drilling-line selection.

Key relations: $L_{DC}=0.10\,L_{total}$; $W=w_{DP}L_{DP}+w_{DC}L_{DC}$; $T_{fast}=W/(K_e\,n)$ (here $K_e=1$); required breaking strength $=SF\times T_{fast}$.

Why this works: the traveling block is suspended by $n$ parallel line segments strung between the crown and traveling blocks; in a frictionless idealization each segment carries an equal share of the hook load, so the fast line (the segment that spools onto the drawworks drum, carrying the highest tension of the reeving) sees $W/n$. The line must be rated well above this working tension because it experiences millions of bending-fatigue cycles over the drum and sheaves, hence the design safety factor.

Common pitfall: forgetting the drill-collar/drill-pipe split (using a single uniform weight per foot for the whole string), or dividing by the wrong line count (deadline tension differs from fast-line tension when sheave friction is included — fast line always governs since it carries the highest load).

Source: Bourgoyne et al., Applied Drilling Engineering, Ch. 7 (Rig Selection and Rig Load Design), §7.1–7.2.

Question 1 of 5: Minimum Drilling Line Size

Question 1: Minimum Drilling Line Size (equal value)