Reservoir development: volumes, replacement and scenarios

PVT and volume basis: interpreting Bo

Oil volume depends on pressure, temperature and dissolved gas. Here Bo means oil with dissolved gas at specified reservoir conditions divided by the corresponding oil volume at specified standard conditions. Reservoir volume therefore equals surface volume multiplied by Bo; the reverse conversion uses division. Retain reservoir cubic metres and standard cubic metres in the notation, although the coefficient is a volume ratio.

Bo belongs to a particular fluid and state. Do not borrow a neighbouring well value without evidence of representativeness or hold it constant throughout production history without justification. Check sampling conditions, PVT study identification, pressure, temperature and the surface reference basis. A volume conversion does not replace a phase model when phase behaviour changes.

An initial reservoir oil volume can be converted to standard conditions using consistent initial properties. The result remains an estimate of initial oil quantity, not recoverable reserves. Keep geometric assumptions, saturation uncertainty and the absence of a recovery forecast explicit in the report.

Walkthrough

Fictional oil: 1200 reservoir m³ at Bo=1.20 reservoir m³/standard m³ corresponds to 1000 standard m³. Reverse conversion: 1000×1.20=1200.

Independent task

Initial oil pore volume is 1800 reservoir m³ and Bo=1.25. Find standard volume, then repeat at Bo=1.20 and 1.30 with fixed pore volume. Explain the limits.

Attempt in the interactive worksheet →

Withdrawal and injection on a common reservoir basis

Comparing surface injection and production volumes can mislead because oil, water and gas use different properties and measurement conditions. In this oil-and-water-only exercise, reservoir withdrawal equals surface oil multiplied by Bo plus surface water multiplied by Bw. Convert injected water using a consistent factor for that stream at the target conditions.

The injected-to-withdrawn ratio is a simplified replacement indicator over one common period and boundary. Free gas, gas-cap effects, aquifer support, compressibility and exchange with neighbouring blocks are excluded. This is not a complete reservoir material-balance equation. If those effects matter, extend the model rather than insert an arbitrary correction.

A ratio near one does not establish pressure stability or effective injection distribution. Streams may belong to different compartments, exhibit time lags or follow conductive channels. Connectivity, pressure and flow-distribution evidence are needed. One ratio cannot establish waterflood effectiveness or justify an operating change.

Walkthrough

Over one period: oil 100 standard m³ at Bo=1.20, water 50 standard m³ at Bw=1.00. Withdrawal=170 reservoir m³. Injection 160 standard m³ at Bi=1.00 gives 160/170≈0.941.

Independent task

Oil=200 standard m³, Bo=1.25; water=100 standard m³, Bw=1.02; water injection=340 standard m³, Bi=1.01. Calculate reservoir withdrawal, injection and ratio. Does this prove stable pressure?

Attempt in the interactive worksheet →

Scenario, history and evidence of effect

Matching measured history does not make a model unique. Different combinations of connectivity, fluid properties, permeability and boundary conditions may reproduce similar observations. Examine time-dependent residuals, pressure and independent evidence rather than cumulative production alone. Data used for calibration are not a fully independent forecast test.

A constant-rate cumulative calculation is a scenario, not a depletion model. Specify operating days, availability and the rate basis separately. Do not apply availability twice when a calendar-average rate already includes downtime. Compare alternatives over a common horizon and identify the assumptions that differ.

The difference between scenarios does not establish a causal intervention effect. Comparable conditions, concurrent changes and a justified counterfactual are needed. In a learning report, show the arithmetic, identify major uncertainties and propose useful additional observations. Do not turn an arithmetic difference into a promise of incremental production, revenue or project payback.

Walkthrough

Constant operating rate 10 standard m³/day, 30 calendar days, availability 0.90: 270 standard m³. If 10 is already a calendar average, the total is 300; applying 0.90 again would be wrong.

Independent task

A: operating rate 100 standard m³/day, availability 0.90. B: 110 and 0.80. Over 30 days compare volumes and explain why higher operating rate need not mean greater monthly production.

Attempt in the interactive worksheet →

Capstone: audit a waterflood improvement claim

One fictional block and common period: oil 300 standard m³, Bo=1.20; water 200 standard m³, Bw=1.00; injection 550 standard m³, Bi=1.00. A report says: “550/500=1.10, pressure is guaranteed to rise and recovery has improved.” Pressure history and injection distribution are absent.

Walkthrough

First audit stream bases, then calculate the ratio, then assess the claims.

Independent task

Correct the calculation, list simplified-model exclusions and request evidence to assess pressure and injection effect.

Attempt in the interactive worksheet →