Production: components, productivity and diagnosis

Water cut, oil rate and time basis

Here water cut is the volume fraction of water in total oil-plus-water liquid on a consistent measurement basis. Oil rate equals liquid rate multiplied by one minus water cut; water rate equals liquid rate multiplied by water cut. Gas is excluded from that denominator. A water mass fraction cannot replace a volume fraction without density and condition information.

Across periods, sum oil and water volumes before calculating the overall fraction. Averaging water-cut percentages arithmetically is generally wrong when liquid volumes differ. An operating-time rate also differs from a calendar average. Establish whether downtime is included, how long the measurement lasted and whether it represents the whole period.

Higher liquid production need not mean higher oil production. Increased water cut can reduce the oil component even as liquid rises. This describes production composition, not the cause of change. Diagnosis requires comparable tests, operating context, measurement quality and well-condition evidence.

Walkthrough

Liquid 100 standard m³/day and fw=0.40 gives oil 60 and water 40. At liquid 120 and fw=0.60, oil is only 48 despite greater liquid rate.

Independent task

Day A produces 100 standard m³ liquid at fw=0.20; day B produces 200 at fw=0.50. Find cumulative oil, water and overall water cut.

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Drawdown and a conditional productivity index

In this learning model, drawdown is representative reservoir pressure minus flowing bottomhole pressure, referenced to a consistent datum and pressure basis. Wellhead pressure is not a direct substitute for bottomhole pressure. Define J as consistent liquid rate divided by positive drawdown. Preserve volume and time bases in its units, such as standard m³/day/MPa.

The linear relation q=JΔp is an explicitly assumed learning approximation over a limited range. Do not automatically extend it to gas wells, multiphase inflow or changing phase conditions. One test does not establish constant J. Even two matching values cannot diagnose damage or equipment condition without additional evidence.

Compare stabilisation, pressure representativeness, composition, duration and calibration between tests. A changed index may reflect different inputs and conditions rather than reservoir deterioration. Use the calculated index to organise hypotheses, not to recommend changes to drawdown, pump operation or operating limits.

Walkthrough

At pr=20 MPa, pwf=16 MPa and q=80 standard m³/day: drawdown=4 MPa and J=20 standard m³/day/MPa. Both pressures use the same datum.

Independent task

Test A: pr=18 MPa, pwf=15 MPa, q=60. B: pr=17, pwf=13, q=80. Rates are standard m³/day. Calculate both indices. Does higher rate prove improved productivity?

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Trend diagnosis: symptom versus cause

Lower production is an observation rather than a diagnosis. Reservoir pressure, composition, equipment operation, surface restrictions, availability or measurement may have changed. Start with a timeline: when did the signal change, which events preceded it and which quantities were measured simultaneously? Comparing unsynchronised averages can create a false causal link.

For this learning analysis, separate oil volume into operating liquid rate, oil fraction and operating time. This accounting highlights factor contributions but does not identify their physical causes. If several factors change, the order of sequential substitution affects the attributed contribution. State the decomposition method rather than present it as a unique causal explanation.

List competing hypotheses with observations that would support or challenge each one. Check measurement status as well as trend shape. Actual investigation and operating changes require approved procedures and responsible specialists. The learning output is a justified evidence request and bounded conclusion, not a remote repair prescription or well operating instruction.

Walkthrough

Over 10 days: operating liquid rate=100 standard m³/day, fw=0.30, availability=0.90. Oil=100×0.70×10×0.90=630 standard m³. Each quantity needs its own definition.

Independent task

A: operating qL=100, fw=0.20, availability=0.90. B: qL=120, fw=0.40, availability=0.75. Both periods last 10 days. Compare oil volumes and propose three testable hypotheses without prescribing intervention.

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Capstone: liquid rises while oil falls

Two fictional 30-day periods. A: operating qL=100 standard m³/day, fw=0.20, availability=0.90. B: qL=120, fw=0.40, availability=0.80. A report calls increased liquid rate an oil-production improvement. Only wellhead pressure data are available.

Walkthrough

Audit components, operating time and whether J can be calculated.

Independent task

Calculate oil and water totals, oil change and missing diagnostic evidence. Can J be calculated from wellhead pressures alone?

Attempt in the interactive worksheet →