Unfilled-Hole Limits and Tubular Displacement

A full hole is part of the barrier

Removing tubulars increases the volume available for fluid in the well. If that volume is not filled, the fluid level falls and hydrostatic support is reduced. In a simple vertical column, pressure lost equals density × 0.052 × vertical level drop. Continued loss can remove the static overbalance and permit influx from a permeable formation. The calculation of how much pipe could be pulled before that happens is a training limit, not permission to omit scheduled hole filling.

The effect depends on whether the tubular is open and draining or closed-ended and wet. Open-pipe displacement is the volume of steel removed. Closed-end displacement includes the internal capacity that leaves with the tubular. In a vertical cased interval, the effective area available at the fluid surface determines the level drop. When the surface intersects pipe, that area is casing capacity minus the relevant displacement. State these assumptions before using the formula.

Dry pipe: pressure loss per foot pulled

Let casing capacity be 0.0800 bbl/ft and dry-pipe steel displacement be 0.0070 bbl/ft. For each foot of pipe pulled without fill, 0.0070 bbl of additional space must be provided by a falling fluid level. In the simplified geometry, the level drop per foot pulled is 0.0070 ÷ (0.0800 − 0.0070) = 0.09589 ft. With 10.0 ppg mud, gradient is 0.520 psi/ft, so pressure loss per foot pulled is 0.049863 psi.

For 500 ft pulled, level drop is 47.945 ft and pressure loss is approximately 24.932 psi. The hole-fill shortfall is 500 × 0.0070 = 3.5 bbl. Check the two routes: 3.5 ÷ 0.0730 gives the same level drop. If geometry or the fluid surface location changes, the effective capacity and formula may also change.

Wet pipe and a smaller allowable length

Suppose the closed-end displacement is 0.0248 bbl/ft. The level drop per foot pulled becomes 0.0248 ÷ (0.0800 − 0.0248) = 0.449275 ft. Pressure loss per foot is 0.520 × 0.449275 = 0.233623 psi. The same 500 ft would lose approximately 116.812 psi of head, much more than the dry-pipe case.

If initial static overbalance is 150 psi, the level drop that removes it is 150 ÷ 0.520 = 288.462 ft. The allowable unfilled pulled length in this simplified model is:

With dry displacement, L = 150 × 0.0730 ÷ (0.520 × 0.0070) = 3,008.24 ft. With closed-end displacement, L = 150 × 0.0552 ÷ (0.520 × 0.0248) = 642.06 ft. The corresponding missing fill volumes are 21.058 bbl and 15.923 bbl. Neither result includes swab pressure, uncertain formation pressure, losses or measurement error; actual operating controls require a conservative approved plan and continued filling.

Recognise a change in the calculation boundary

When drill collars are entirely below the fluid surface, a specified collar displacement removed over a length may be divided by the casing or riser capacity available at that surface. For a given training case with 100 ft of collar, 0.0300 bbl/ft displacement and 0.1000 bbl/ft surface capacity, the missing 3.0 bbl causes a 30 ft level drop. Do not substitute this special geometry into the previous pipe-at-surface formula. Identify what occupies the surface interval and whether the tubular interior contributes fluid to the well.

Deviation adds another distinction. Volume capacities are per measured foot, while hydrostatic loss requires vertical level change. A fluid surface dropping through an inclined interval cannot be treated as a vertical interval of the same measured length without trajectory information. The calculation should use the known fluid-level position and the well geometry. An apparent precision of two decimals does not compensate for an unknown boundary.

Use the trip sheet to find the discrepancy early

Record each relevant tubular type, cumulative length and the correct predicted displacement. Compare incremental and cumulative actual fill, allowing only documented transfers and equipment effects. A change from dry to wet pipe or failure of a draining arrangement changes the baseline. Inform the supervisor and recalculate the expectation instead of attributing every change to instrumentation.

If actual fill is below prediction, stop tripping and assess for influx using the approved procedure; if above prediction, assess losses and keep the hole filled with the approved fluid. An identified influx requires the appropriate containment and control plan. A trip tank must be small enough and instrumented sensitively enough to detect meaningful differences, with its circulation pump and valve line-up understood. The aim is to recognise a departure long before the computed overbalance is consumed. These relationships follow the API-unit IWCF formula sheet, with the geometry made explicit.

Dry and wet example comparison