Question 5 of 5: Well Control — Engineer's Method

Question 5: Well Control — Engineer's Method (a=15, b=5, c=5)

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. Shut-in kick at $D=12{,}000$ ft (casing shoe at $8{,}000$ ft, $L_{DC}=1{,}000$ ft); $MW=11$ ppg, $SIDPP=190$ psig, $SICP=400$ psi, pit gain $=15$ bbl; three annulus capacity zones and two internal (drillstring) capacities above; safety factor $100$ psi, kill-mud safety margin $0.2$ ppg, reduced pump pressure $900$ psi at kill rate $5$ bbl/min. Engineer's (wait-and-weight) Method: kill-weight mud is pumped from the start of circulation.

Find. (a) Surface casing pressure when the top of the kick reaches surface; (b) mud volume pumped at that moment; (c) the drill-pipe pressure schedule while kill mud is displacing the drillstring.

Approach. Bottom-hole pressure equals formation pressure throughout a properly controlled kill and is held constant at $SIDPP+0.052\,MW\,D$. Treat the kick as a constant-volume slug (its length changes only with the local annulus capacity as it migrates, the standard assumption for this problem class), back its own hydrostatic gradient out of the shut-in data, then track it and the kill-mud front position as circulation proceeds with kill-weight mud from the very start.

1. Formation (bottom-hole) pressure — held constant throughout. $BHP=SIDPP+0.052\,MW\,D=190+0.052(11)(12{,}000)$, so $\boxed{BHP=7{,}054\ \text{psi}}$.
2. Back out the kick's own hydrostatic gradient from the shut-in data. At shut-in the kick sits at the bottom, in the DC–open-hole annulus: length $L_0=15/0.0272=551.5$ ft, so its top is at $12{,}000-551.5=11{,}448.5$ ft, with 11 ppg mud above it: $P_{mud\,above}=0.052(11)(11{,}448.5)=6{,}548.6$ psi. From $BHP=SICP+P_{mud\,above}+g_k L_0$: $g_k=(7{,}054-400-6{,}548.6)/551.5$, so $\boxed{g_k=0.191\ \text{psi/ft}}$ (higher than a pure-methane rule-of-thumb, but this is what the exam's own shut-in readings imply, and it is what keeps the rest of the solution internally consistent).
3. Kill mud weight. $KMW=MW+\dfrac{SIDPP}{0.052\,D}+SM=11+\dfrac{190}{0.052(12{,}000)}+0.2=11+0.305+0.2$, so $\boxed{KMW=11.505\ \text{ppg}}$.
4. (a) Casing pressure when the top of the kick reaches surface. In the Engineer's Method, kill mud is pumped continuously from the start, so by the time the kick's top reaches surface, everything below the kick is already kill-weight mud (plug-flow assumption — no old mud gap remains). Once entirely within the cased-hole annulus, the kick's length is $L_{top}=15/0.059=254.2$ ft (well inside the $8{,}000$ ft cased section), leaving a kill-mud column of $12{,}000-254.2=11{,}745.8$ ft below it. $$P_c=BHP-0.052\,KMW(12{,}000-L_{top})-g_k\,L_{top}=7{,}054-0.052(11.505)(11{,}745.8)-0.191(254.2)$$ $$P_c=7{,}054-7{,}026.7-48.6\approx-21\ \text{psi}$$ so $\boxed{P_c\approx0\ \text{psi}}$: the calculation comes out slightly negative, meaning the well is already comfortably over-balanced with the choke essentially wide open by the time the kick reaches surface — the $0.2$ ppg kill-mud safety margin, applied over nearly the entire $12{,}000$ ft column by this point, more than offsets the light gas column still left at the top.
5. (b) Mud volume pumped at that moment. Internal (drillstring) volume needed to bring kill mud to the bit: $V_{DP,int}=(12{,}000-1{,}000)(0.014)=154.0$ bbl, $V_{DC,int}=1{,}000(0.0087)=8.7$ bbl, total $162.7$ bbl. The kill-mud front in the annulus has by then advanced up to the kick's current bottom ($254.2$ ft below surface), sweeping the full DC–OH zone ($1{,}000$ ft $\times0.0272=27.2$ bbl), the full DP–OH zone ($3{,}000$ ft $\times0.0425=127.5$ bbl), and $7{,}745.8$ ft of the DP–casing zone ($\times0.059=457.0$ bbl) — a total annular kill-mud volume of $611.7$ bbl (check: $611.7+15$ bbl kick $=626.7$ bbl $=$ the well's full annulus volume, confirming nothing is missed). $$V_{pumped}=162.7+611.7$$ so $\boxed{V_{pumped}=774.4\ \text{bbl}}$ (at the $5$ bbl/min kill rate, about $155$ minutes of pumping).
6. (c) Drill-pipe pressure schedule while kill mud goes down. $ICP=SIDPP+P_{reduced}+SF=190+900+100=1{,}190$ psi; $FCP=P_{reduced}\left(KMW/MW\right)=900(11.505/11)=941.3$ psi. While kill mud fills the drillstring (the first $162.7$ bbl pumped), drill-pipe pressure ramps down LINEARLY with volume pumped: $DPP(V)=ICP-(ICP-FCP)\,V/162.7$ for $0\le V\le162.7$ bbl; once kill mud reaches the bit, $\boxed{DPP=FCP=941.3\ \text{psi, held constant}}$ for the remainder of the job (through the kick reaching surface and beyond).