Mud Level Drop While Tripping Dry and the Onset of a Kick: 24-Pet-A4 Oil and Gas Well Drilling and Completion, May 2013, Q2

Question 2: Mud Level Drop While Tripping Dry and the Onset of a Kick (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. See table above. No fill-up mud is added as pipe is pulled (open-ended drill pipe, U-tube connected to the annulus).

Find. (a) mud-level drop after 15 stands pulled dry; (b) the resulting reduction in bottomhole pressure; (c) the number of stands pulled dry before the well kicks.

Approach. As dry pipe is pulled, the steel volume it displaced is removed from the wellbore; with no replacement mud added, the existing mud must drop in level to fill the space, spread over the casing–drillpipe annular area near the top of the well (where the remaining string's top sits). Convert the level drop to a hydrostatic-pressure loss, then find how much level drop (and how many stands) exhausts the pore-pressure overbalance margin.

Fig. 1 — Mud level in the casing–drillpipe annulus drops by Δh once dry pipe is pulled without replacing the vacated steel volume with mud; the drop is spread over the annular area (casing ID minus drillpipe OD) since that is where the remaining string's top sits.

Steel volume vacated per foot of drill pipe pulled. Drill-pipe steel cross-section factor $=OD^2-ID^2=5^2-4.276^2=25-18.284$, so $6.716\ \text{in}^2$ (as a bbl/ft displacement factor, divide by 1029.4). Length pulled: $15$ stands $\times 90\ \text{ft/stand}=\boxed{1{,}350\ \text{ft}}$.
Annular area receiving the drop. Casing ID $9.00$ in, drill pipe OD $5$ in: area factor $=9.00^2-5^2=81-25=\boxed{56\ \text{in}^2}$ (this is the annulus around the remaining string, which still reaches the surface after 15 stands are pulled since only $1{,}350$ ft of the $5{,}600$ ft of drill pipe in the hole is removed).
(a) Mud level drop. Volume vacated must equal level drop $\times$ annular area: $\Delta h = L_{pulled}\times\dfrac{OD^2-ID^2}{D_{csg}^2-OD^2}=1{,}350\times\dfrac{6.716}{56}$, so $\boxed{\Delta h = 161.9\ \text{ft}}$.
(b) Bottomhole pressure reduction. $\Delta P = 0.052\times MW\times\Delta h = 0.052(10)(161.9)$, so $\boxed{\Delta P = 84.2\ \text{psi}}$.
Overbalance margin before a kick. Static (full) bottomhole pressure $=0.052(10)(6{,}500)=3{,}380$ psi against a pore pressure of $3{,}250$ psi, so the margin is $3{,}380-3{,}250$, giving $\boxed{130\ \text{psi}}$ of overbalance to lose before underbalance (a kick) begins.
(c) Stands pulled before a kick. The allowable level drop is $\Delta h_{kick}=130/(0.052\times10)=250\ \text{ft}$. Each stand pulled drops the level by $90\times(6.716/56)=10.79$ ft. After 23 stands the cumulative drop is $23(10.79)=248.2$ ft (still overbalanced); after 24 stands it is $24(10.79)=259.0\ \text{ft} > 250\ \text{ft}$, so the well goes underbalanced during the pulling of the 24th stand. $\boxed{N_{kick}=24\ \text{stands}}$.

Check: assumes the drill pipe is open-ended with no float valve, so the pipe bore stays in free (U-tube) communication with the annulus and the "missing" volume is exactly the removed steel — if a float valve were run, the pipe would trip dry inside and the vacated volume (and hence the level drop) would instead equal the pipe's full internal capacity, not just its steel cross-section. Also assumes the remaining string's top stays above the drill-collar interval throughout (true here: only 1,350–1,470 ft of the 5,600 ft drill-pipe interval is ever pulled in parts a–c), so the annular area used ($D_{csg}^2-OD_{DP}^2$) does not change.

Quantity	Value
(a) Mud level drop after 15 stands	161.9 ft
(b) Bottomhole pressure reduction	84.2 psi
Static overbalance margin (before any pulling)	130 psi
(c) Stands pulled dry before a kick	24 stands (drop reaches 259.0 ft, vs. 250 ft allowable)

Topic: Swabbing / trip-margin effect — mud-level drop and bottomhole underbalance from pulling dry pipe without filling the hole.

Key relations: $\Delta h = \dfrac{L_{pulled}(OD_p^2-ID_p^2)}{D_{csg}^2-OD_p^2}$; $\Delta P_{BH}=0.052\,MW\,\Delta h$; kick onset when $P_{BH}(\text{static})-\Delta P_{BH}\le P_{pore}$.

Why this works: the annulus and the open-ended drill-pipe bore form one connected fluid column (a U-tube) that was full to surface before tripping. Removing pipe steel from that column without replacing it leaves a volume deficit that can only be satisfied by the free surface dropping; because the drop occurs where the mud surface actually sits (in the annulus around the remaining string), the relevant cross-section is the casing–drillpipe annular area, not the full casing bore.

Common pitfall: dividing the vacated volume by the full casing cross-sectional area instead of the annular area (ignoring that the remaining drill pipe still occupies part of that cross-section), or using the pipe's full bore capacity instead of just its steel displacement for an open-ended (no float) string.

Source: Bourgoyne et al., Applied Drilling Engineering, Ch. 4 (well control fundamentals, trip-margin and swabbing calculations), §4.4.

Question 2 of 5: Mud Level Drop While Tripping Dry and the Onset of a Kick

Question 2: Mud Level Drop While Tripping Dry and the Onset of a Kick (equal value)