Question 3 of 5: Bit Selection by Minimum Drilling Cost per Foot

Question 3: Bit Selection by Minimum Drilling Cost per Foot (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. ROP: $dD/dt=200\,e^{-0.0002303D}$ starting at $D_{in}=10{,}000$ ft; bit X: life $30$ hr, cost $\$25{,}000$; bit Y: life $20$ hr, cost $\$5{,}000$; $C_R=\$1{,}000/\text{hr}$; $t_t=0.001\,D_{out}$ hr; connection $=2$ min per 30-ft single, i.e. per 30 ft drilled.

Find. Which bit gives the lower drilling cost per foot, assuming each bit is run to the end of its rated life.

Approach. Integrate the ROP equation to find how much footage each bit drills over its rated life (bit life fixes rotating time $t_d$), then build the total cost of that bit run (bit cost + rig time for drilling, tripping and connections) and divide by footage drilled to get cost per foot; the lower $C_f$ wins.

1. Footage drilled to the end of bit life. Separating variables in $dD/dt=200e^{-aD}$ ($a=0.0002303$) and integrating from $D_{in}$ to $D_{out}$ over the bit's rated life $t_d$: $t_d=\dfrac{1}{200a}\left[e^{aD_{out}}-e^{aD_{in}}\right]$, so $\boxed{e^{aD_{out}}=e^{aD_{in}}+200a\,t_d}$, solved for $D_{out}$ per bit.
2. Bit X ($t_d=30$ hr). $e^{aD_{in}}=e^{0.0002303(10{,}000)}=10.004$; $200a\,t_d=200(0.0002303)(30)=1.382$; $e^{aD_{out}}=11.386\Rightarrow D_{out}=\ln(11.386)/a$, so $\boxed{D_{out,X}=10{,}562\ \text{ft}}$, i.e. $\Delta D_X=562\ \text{ft}$ drilled.
3. Bit Y ($t_d=20$ hr). $200a\,t_d=0.921$; $e^{aD_{out}}=10.925\Rightarrow D_{out}=\ln(10.925)/a$, so $\boxed{D_{out,Y}=10{,}383\ \text{ft}}$, i.e. $\Delta D_Y=383\ \text{ft}$ drilled.
4. Non-drilling time per run. Trip time $t_t=0.001D_{out}$; connections $=\Delta D/30$ singles $\times2\ \text{min}$. Bit X: $t_t=10.56$ hr, connections $=562/30=18.7\Rightarrow t_c=0.62$ hr; total non-rotating $+$ rotating time $=30+10.56+0.62=41.19$ hr. Bit Y: $t_t=10.38$ hr, connections $=383/30=12.7\Rightarrow t_c=0.42$ hr; total time $=20+10.38+0.42=30.81$ hr.
5. Cost per foot. $C_f=\dfrac{C_b+C_R(t_d+t_t+t_c)}{\Delta D}$. Bit X: $C_f=\dfrac{25{,}000+1{,}000(41.19)}{562}=\dfrac{66{,}186}{562}$, so $\boxed{C_{f,X}=\$117.8/\text{ft}}$. Bit Y: $C_f=\dfrac{5{,}000+1{,}000(30.81)}{383}=\dfrac{35{,}808}{383}$, so $\boxed{C_{f,Y}=\$93.6/\text{ft}}$.

Selection: Bit Y gives the lower cost per foot ($93.6 vs. $117.8) — run bit Y from 10,000 ft downwards.