Question 1 of 5: Rig Selection and Drilling Line Sizing

Question 1: Rig Selection and Drilling Line Sizing (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. $TVD=L_{DC}+L_{DP}=10{,}800$ ft ($L_{DC}=1{,}200$ ft, $L_{DP}=9{,}600$ ft); hoisting (traveling-block) speed $V_b=40$ ft/min; $W_{DC}=107$ lb/ft, $W_{DP}=16.6$ lb/ft; rated input horsepower to the fast line $=400$ HP; buoyancy ignored; three candidate rigs by maximum equivalent derrick load; EIPS breaking-strength table above.

Find. (a) The number of lines that must be strung between crown and traveling block; (b) which rig to use; (c) the minimum safe drilling-line diameter.

Approach. Compute the hook load from the drill-string weight, then use the standard reeving-efficiency relation between input horsepower, hoisting speed and number of lines strung to find the minimum practical line count; use that count to get the fast-line tension and equivalent derrick load for rig selection; then size the line from the fast-line tension and a design safety factor.

1. Hook load (drillstring weight in air, buoyancy ignored). $W=W_{DC}L_{DC}+W_{DP}L_{DP}=107(1{,}200)+16.6(9{,}600)=128{,}400+159{,}360$, so $\boxed{W=287{,}760\ \text{lbf}}$.
2. Number of lines from the power/efficiency relation. The fast-line tension for $n$ lines with reeving-efficiency factor $E_n$ is $F_{fast}=W/(nE_n)$, and the fast line travels at $V_{fast}=nV_b$, so the drawworks input horsepower needed is $HP=\dfrac{F_{fast}V_{fast}}{33{,}000}=\dfrac{WV_b}{33{,}000\,E_n}$ — independent of $n$ except through the standard reeving-efficiency table (Applied Drilling Engineering, Table 7.3: $E_6=0.874$, $E_8=0.841$, $E_{10}=0.816$, $E_{12}=0.792$). Because $E_n$ falls as $n$ grows, the required input horsepower rises with $n$: at $n=6$, $HP=\dfrac{287{,}760(40)}{33{,}000(0.874)}=399.1$ HP, just inside the 400 HP available; at $n=8$, $HP=\dfrac{287{,}760(40)}{33{,}000(0.841)}=414.7$ HP, which exceeds the rated 400 HP. So $\boxed{n=6\ \text{lines}}$ is the minimum (and only) practical count the rig's rated horsepower can support at this hoisting speed.
3. Fast-line tension and equivalent derrick load. $F_{fast}=W/(nE_6)=287{,}760/[6(0.874)]=54{,}874$ lbf. The crown block redirects both the fast line and the dead line into the derrick structure, so the equivalent derrick load is $DL=W+2F_{fast}=287{,}760+2(54{,}874)$, giving $\boxed{DL=397{,}508\ \text{lbf}}$.
4. Rig selection. Rig A (300,000 lbm) is below $DL$ and fails; Rig B (800,000 lbm) clears $DL$ with margin; Rig C (1,500,000 lbm) also clears it but is oversized for this load. $\boxed{\text{Rig B}}$ is the appropriate choice.
5. Drilling-line size. With a standard design safety factor $SF=2.0$ on the fast-line tension, the required minimum breaking strength is $SF\times F_{fast}=2.0(54{,}874)=109{,}748$ lbf. From the EIPS table, 1 in gives only 103,400 lbf (fails); 1 1/8 in gives 130,000 lbf $\ge 109{,}748$ lbf (passes). So $\boxed{d_{min}=1\ 1/8\ \text{in}}$.