Bit Economics — Run the Used Bit or a New One?: 24-Pet-A4 Oil and Gas Well Drilling and Completion, December 2014, Q5

Question 5: Bit Economics — Run the Used Bit or a New One? (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. Average bit life $30$ hr; the pulled bit already has $15$ hr on it (used, free); both options resume drilling from $D_{in}=12{,}000$ ft; rig cost $=\$1{,}500$/hr; new bit cost $=\$10{,}000$; trip time $=0.0012\,D_{out}$ hr; connection time $=2$ min per $30$-ft single; ROP relation $D_{out}(t_d)$ as given.

Find. Whether continuing with the half-used bit or installing a new one gives the lower cost per foot for the next bit run.

Approach. For each bit, run it to the end of its own remaining life ($15$ hr for the used bit, a full $30$ hr for a new one), compute the footage each drills from the given ROP relation, then compare total cost (bit $+$ rig time for drilling, tripping and connections) divided by footage — the standard minimum-cost-per-foot bit-economics criterion.

Footage drilled by the used bit (remaining life $t_d=15$ hr). $D_{out}=\dfrac{\ln(0.025(15)+e^{0.00008(12{,}000)})}{0.00008}=\dfrac{\ln(0.375+2.6117)}{0.00008}=\dfrac{1.0937}{0.00008}$, so $\boxed{D_{out,used}=13{,}677\ \text{ft}}$, i.e. $1{,}677$ ft of new footage.
Footage drilled by a new bit (full life $t_d=30$ hr). $D_{out}=\dfrac{\ln(0.025(30)+e^{0.96})}{0.00008}=\dfrac{\ln(0.75+2.6117)}{0.00008}=\dfrac{1.2124}{0.00008}$, so $\boxed{D_{out,new}=15{,}156\ \text{ft}}$, i.e. $3{,}156$ ft of new footage.
Cost per foot, used bit. Trip time $=0.0012(13{,}677)=16.41$ hr; connection time $=(1{,}677/30)(2/60)=1.86$ hr; total time $=16.41+1.86+15=33.28$ hr; total cost (bit free) $=1{,}500(33.28)=\$49{,}914$. $\boxed{C_{f,used}=49{,}914/1{,}677=\$29.76/\text{ft}}$.
Cost per foot, new bit. Trip time $=0.0012(15{,}156)=18.19$ hr; connection time $=(3{,}156/30)(2/60)=3.51$ hr; total time $=18.19+3.51+30=51.69$ hr; total cost $=10{,}000+1{,}500(51.69)=10{,}000+77{,}539=\$87{,}539$. $\boxed{C_{f,new}=87{,}539/3{,}156=\$27.74/\text{ft}}$.
Decision. $27.74<29.76$, so despite the used bit costing nothing up front, $\boxed{\text{run the NEW bit}}$: its higher rate of penetration while sharp drills nearly twice the footage per rig-hour spent, more than repaying the $\$10{,}000$ purchase price once amortized over that extra footage.

Quantity	Used bit (15 hr remaining)	New bit (30 hr)
Footage drilled	1,677 ft	3,156 ft
Trip + connection + drilling time	33.28 hr	51.69 hr
Total cost	$49,914	$87,539
Cost per foot	$29.76/ft	$27.74/ft
Decision		Run the new bit

Topic: Minimum cost-per-foot bit economics under an exponential rate-of-penetration model.

Key relations: $C_f=\dfrac{C_b+C_r(t_t+t_c+t_d)}{\Delta D}$, with $t_t=0.0012D_{out}$, $t_c=(\Delta D/L_{single})(2/60)$, and footage from the given $D_{out}(t_d)$ relation for each bit's own remaining/full life.

Why this works: a dulling bit's ROP decays with cumulative footage/time (captured here by the exponential ROP law), so a "free" used bit is only economical if the extra footage a fresh bit would drill is not worth its purchase price. Cost-per-foot correctly weighs the sunk-cost bit price against the value of the additional footage, rather than comparing raw operating cost alone.

Common pitfall: concluding that the used bit must be cheaper simply because it "costs nothing," without checking that its lower remaining ROP-life footage can still make its cost per foot higher once trip and connection overhead are spread over less new hole.

Source: Bourgoyne et al., Applied Drilling Engineering, Ch. 6 (Bit Selection and Evaluation), §6.4–6.5.

Question 5 of 5: Bit Economics — Run the Used Bit or a New One?

Question 5: Bit Economics — Run the Used Bit or a New One? (equal value)