Question 4 of 5: Well Control — Driller's Method

Question 4: Well Control — Driller's Method (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. $D=12{,}200$ ft; casing shoe at $4{,}000$ ft; $MW=12$ ppg; $L_{DP}=11{,}600$ ft, $L_{DC}=600$ ft; annular/string capacities as tabulated; pit gain $=15$ bbl (methane, assumed a non-expanding slug); $SICP=400$ psi, $SIDPP=200$ psi; kill-rate ($200$ gpm) parasitic loss $=750$ psi; safety margins $0.2$ ppg (kill mud) and $100$ psi (initial circulating pressure).

Find. The kick-zone length, equivalent shoe mud density, surface casing pressure and mud pumped at the instant the kick's top reaches the shoe (first, kick-removal circulation with original mud), and the drill-pipe pressure schedule while kill mud is pumped down (second circulation).

Approach. Use the constant bottomhole pressure (BHP) principle of the Driller's method: hold BHP equal to its shut-in value throughout by balancing the annulus hydrostatic column (mud + the migrating, constant-volume kick) against a calculated surface casing pressure; back out the kick's own hydrostatic gradient from the given shut-in data, then re-apply it once the kick reaches the shoe. Volume pumped follows from the annular capacities swept plus the drill-string volume; the second-circulation drill-pipe schedule follows the standard ICP→FCP straight-line method.

1. Bottomhole pressure (from the drill-pipe side, which stays filled with original mud throughout). $BHP=SIDPP+0.052\,MW\,D=200+0.052(12)(12{,}200)=200+7{,}612.8$, so $\boxed{BHP=7{,}812.8\ \text{psi}}$.
2. Kick zone length at the shoe (part a). The kick was originally only $15/0.035=428.6$ ft long at the bottom (drill-collar/open-hole annulus), spanning $11{,}771.4$–$12{,}200$ ft. Since the drill collars occupy only the bottom $600$ ft and the shoe ($4{,}000$ ft) is far above the drill-pipe/open-hole transition at $11{,}600$ ft, by the time the kick nears the shoe it is entirely within the drill-pipe/open-hole annulus (capacity $0.0775$ bbl/ft). Treating the $15$ bbl kick as a non-expanding slug, $L_k=15/0.0775$, so $\boxed{L_k=193.5\ \text{ft}}$ (occupying $4{,}000$–$4{,}193.5$ ft when its top is exactly at the shoe).
3. Back out the kick's own hydrostatic gradient from the initial shut-in data. At shut-in, the annulus above the kick ($0$–$11{,}771.4$ ft) is pure mud: $0.052(12)(11{,}771.4)=7{,}345.4$ psi. Balancing the annulus against $BHP$: $SICP+7{,}345.4+g_kL_{k,0}=BHP$, i.e. $400+7{,}345.4+428.6\,g_k=7{,}812.8$, giving $\boxed{g_k=0.157\ \text{psi/ft}}$ — a physically reasonable downhole methane gradient, derived directly from the exam's own shut-in data rather than an assumed textbook constant.
4. Casing pressure at the surface, kick at the shoe (part c). With the kick now at $4{,}000$–$4{,}193.5$ ft: mud above $=0.052(12)(4{,}000)=2{,}496.0$ psi; kick $=0.157(193.5)=30.5$ psi; mud below $=0.052(12)(12{,}200-4{,}193.5)=4{,}996.0$ psi. Total annulus hydrostatic $=2{,}496.0+30.5+4{,}996.0=7{,}522.5$ psi. $P_c=BHP-7{,}522.5=7{,}812.8-7{,}522.5$, so $\boxed{P_c=290.3\ \text{psi}}$ — lower than the original $400$ psi $SICP$ because the kick has migrated up under a shorter, capacity-thinned annular section, reducing the "missing" hydrostatic weight it displaces at this configuration.
5. Equivalent mud density at the shoe (part b). $EMW_{shoe}=\dfrac{P_c+(\text{mud column above shoe})}{0.052\,D_{shoe}}=\dfrac{290.3+2{,}496.0}{0.052(4{,}000)}=\dfrac{2{,}786.3}{208}$, so $\boxed{EMW_{shoe}=13.40\ \text{ppg}}$ — the density this casing shoe "feels" while the kick is being circulated past it, to be checked against the $17.6$-ppg-class fracture margin used elsewhere in this well.
6. Mud volume pumped (part d). The kick's bottom boundary moves from the bit ($12{,}200$ ft, where new mud enters the annulus) up to $4{,}193.5$ ft. Annular volume swept: drill-collar/open-hole leg ($12{,}200$ to $11{,}600$ ft, $600$ ft $\times0.035=21.0$ bbl) plus drill-pipe/open-hole leg ($11{,}600$ to $4{,}193.5$ ft, $7{,}406.5$ ft $\times0.0775=574.0$ bbl) $=595.0$ bbl. Before any of that displacement can occur, the pump must first fill the drill string: $V_{string}=L_{DP}(0.014)+L_{DC}(0.0087)=11{,}600(0.014)+600(0.0087)=162.4+5.2=167.6$ bbl. Total pumped $=167.6+595.0$, so $\boxed{V_{pumped}=762.6\ \text{bbl}}$.
7. Kill mud weight (needed for the second circulation, part e). $KMW=MW+\dfrac{SIDPP}{0.052D}+0.2=12+\dfrac{200}{0.052(12{,}200)}+0.2=12+0.315+0.2$, so $\boxed{KMW=12.52\ \text{ppg}}$.
8. Drill-pipe pressure schedule while kill mud is pumped down (part e). The second circulation starts at $ICP=SIDPP+P_{p,200}+\text{margin}=200+750+100=1{,}050$ psi (original mud still fills the string, plus the given $100$ psi safety margin) and, once kill mud completely displaces the string, ends at $FCP=P_{p,200}\times(KMW/MW)=750(12.52/12)=782$ psi. The time to pump the drill-string volume ($167.6$ bbl) at the $200$ gpm ($=4.762$ bbl/min) kill rate is $167.6/4.762=35.2$ min. So $\boxed{DP\ \text{pressure falls linearly from } 1{,}050\ \text{psi at } t=0 \text{ to } 782\ \text{psi at } t\approx35.2\ \text{min, then holds constant at } 782\ \text{psi}}$ for the remainder of the circulation, once kill mud fills the string and only the annulus (still returning original mud and the kick) remains to be displaced.