Question 4 of 5: Drill String Design

Question 4: Drill String Design (25 marks)

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.

Find. The DC/DP composition — lengths, weights, and grade(s) — needed to drill safely to 10,000 ft.

Approach. Size the drill-collar length from the weight-on-bit and design factor, then check whether the cheaper E-75 pipe alone can carry the remaining length with the required overpull margin; if not, taper the top of the string in the stronger G-105 grade.

1. Buoyancy factor. $BF=1-MW/65.5=1-10/65.5$, so $\boxed{BF=0.8473}$.
2. Drill-collar length for WOB. $L_{DC}=\dfrac{DF\times WOB}{w_{DC}\,BF}=\dfrac{1.2(30{,}000)}{72(0.8473)}$, so $\boxed{L_{DC}=590\ \text{ft}}$. Remaining drillpipe length: $L_{DP}=10{,}000-590=\boxed{9{,}410\ \text{ft}}$.
3. Class 3 wear derating. With no wear table supplied on this open-book exam, Class 3 wear is taken as 70% of nominal wall remaining (a standard tubular-inspection convention). New wall $t=(OD-ID)/2$: E-75 pipe $t=0.337$ in., worn to $0.236$ in. ($ID_{worn}=4.028$ in.); G-105 pipe $t=0.430$ in., worn to $0.301$ in. ($ID_{worn}=3.898$ in.).
4. Worn-pipe tensile capacity, less MOP. $F_{allow}=\sigma_y\,A_{worn}-MOP$, $A_{worn}=\pi(OD^2-ID_{worn}^2)/4$. E-75: $A_{worn}=3.160$ in$^2$, tensile $=75{,}000(3.160)=237{,}010$ lbf, $\boxed{F_{allow,E75}=137{,}010\ \text{lbf}}$. G-105: $A_{worn}=3.971$ in$^2$, tensile $=105{,}000(3.971)=416{,}919$ lbf, $\boxed{F_{allow,G105}=316{,}919\ \text{lbf}}$.
5. Check E-75 for the full 9,410 ft. Buoyed weight: E-75 $=16.6(0.8473)=14.07$ lb/ft; DC $=72(0.8473)=61.0$ lb/ft. Tension at the top of an all-E-75 string $=L_{DC}(61.0)+L_{DP}(14.07)=590(61.0)+9{,}410(14.07)$, giving $168{,}356$ lbf, which exceeds $F_{allow,E75}=137{,}010$ lbf: $\boxed{\text{E-75 alone is insufficient}}$ — the string must be tapered.
6. Size the E-75 (bottom) section. Set the tension at the TOP of the E-75 run equal to its own allowable: $L_{DC}(61.0)+L_{E75}(14.07)=137{,}010 \Rightarrow L_{E75}=[137{,}010-590(61.0)]/14.07$, so $\boxed{L_{E75}=7{,}181\ \text{ft}}$. The remaining $9{,}410-7{,}181=\boxed{2{,}229\ \text{ft}}$ at the top is run in G-105.
7. Check surface tension against G-105's own capacity. G-105 buoyed weight $=20(0.8473)=16.95$ lb/ft. Full-string surface tension $=590(61.0)+7{,}181(14.07)+2{,}229(16.95)=174{,}777$ lbf, well under $F_{allow,G105}=316{,}919$ lbf, so the top-section length can be trimmed further on a real job; $2{,}229$ ft is the length needed for E-75's own limit to just be satisfied at its own top, and G-105 is comfortably safe for the rest of the string above that.