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OOIP, OGIP, Recoverable Volume and Reserves: The Difference

Trace rock volume to oil or gas in place, separate an assumed recovery fraction from reserves classification, and work a volumetric OGIP example with clear units.

By PetroCalcHub Team | Updated 2026-10-24

Original diagram Interactive example Worked practice 3 technical sources

Engineering context

Original oil in place (OOIP) and original gas in place (OGIP) estimate how much petroleum an accumulation originally contained on a defined volume basis. A recovered quantity answers a further question about how much a stated process or project might produce. Reserves adds project and commercial requirements. These terms describe different information, so a larger calculator result is not automatically a larger reserves estimate.

The worked example builds a gas volume from rock geometry, porosity, initial water saturation and a compatible gas formation volume factor. It then applies a plainly hypothetical recovery fraction as separate arithmetic. The interactive control changes only that assumption. It does not assess a development project or assign a formal resources class. Use the official PRMS reference for those classifications.

1 · Rock

348,480,000 ft³

Area × net thickness

2 · Pores

52,272,000 ft³

Rock volume × porosity

3 · Gas-filled pores

39,204,000 reservoir ft³

Pore volume × (1 − Swi)

÷ Bgi → OGIP = 7.8408 billion scf

A physical estimate on a stated standard-gas basis.

Separate recovery assumption

60% × OGIP = 4.70448 billion scf

Assumed recovered quantity.

Additional project assessment

Discovery, recovery project, commercial evidence and evaluation date.

No reserves class supplied by this diagram.

The physical volume chain ends at an in-place estimate. A chosen recovery fraction is a separate assumption; project classification needs evidence beyond this arithmetic. This is an original learning diagram, not a PRMS classification chart.
Try the idea

Change one thing and observe the result

Keep A = 160 acres, h = 50 ft, porosity = 0.15, Swi = 0.25 and Bgi = 0.005 reservoir ft³/scf. Change only a hypothetical recovery fraction; it supplies no project or commercial evidence.

Calculated original gas in place (fixed)
7.84080 billion scf
Assumed recovered gas
4.70448 billion scf
Arithmetic remainder
3.13632 billion scf

Resources/reserves classification: not assessed. The slider changes an arithmetic assumption, including at its 0% and 100% endpoints.

The full bar is the fixed 7.8408-billion-scf OGIP. Cyan represents the 60% assumed recovered portion; gray is its arithmetic complement. The bar does not assign a resources class.

Practical workflow

  1. 1

    State the reservoir geometry, pore fraction, initial saturation and compatible fluid volume factor.

  2. 2

    Calculate the original surface-equivalent quantity while retaining each reservoir-volume conversion.

  3. 3

    Keep a recovery assumption and its resulting volume on a separate line.

  4. 4

    Use the official PRMS for project classification instead of relabeling the calculator output.

Start with rock volume and identify the occupied pores

Area multiplied by net thickness gives the rock volume included in a simple uniform reservoir model. With area in acres and thickness in feet, multiply by 43,560 to obtain cubic feet. For 160 acres and 50 ft, bulk rock volume is 348,480,000 ft³. This number includes the solid rock framework as well as its pore space; it is not a petroleum volume.

Multiply by porosity 0.15 to obtain 52,272,000 ft³ of pore volume. Under this example's initial gas-and-water occupancy assumption, water saturation 0.25 leaves gas occupying 75% of those pores. The initial gas-filled pore volume is therefore 39,204,000 reservoir ft³. The calculation does not establish net pay, spatial variability or saturation from raw measurements; those are supplied interpretations.

Vrock=43560Ah,Vgas,res=43560Ahϕ(1−Swi)V_{\mathrm{rock}}=43560Ah,\qquad V_{\mathrm{gas,res}}=43560Ah\phi(1-S_{wi})

Convert reservoir gas volume to a stated surface basis

The initial gas formation volume factor Bgi relates gas volume at initial reservoir conditions to its volume at the stated standard conditions. Here Bgi is 0.005 reservoir ft³ per standard cubic foot. Dividing 39,204,000 reservoir ft³ by that factor gives 7,840,800,000 scf, or 7.8408 billion standard cubic feet. The reservoir-volume unit cancels, leaving a standard-condition gas volume.

Write the reference pressure and temperature behind the chosen standard cubic foot when preparing a real case. Also check the numerator units: a Bg supplied in reservoir barrels per scf cannot be entered unchanged into this cubic-foot form. The example treats the gas phase and its supplied Bgi consistently; it does not infer a gas composition or separator-product specification from a volumetric estimate.

G=43560Ahϕ(1−Swi)Bgi[Bgi:reservoir ft3scf]G=\frac{43560Ah\phi(1-S_{wi})}{B_{gi}}\quad\left[B_{gi}:\frac{\mathrm{reservoir\ ft^3}}{\mathrm{scf}}\right]

Keep oil and gas in-place bases distinct

OOIP is commonly reported as a stock-tank oil volume, while OGIP is commonly reported as gas volume at specified standard conditions. Their formation volume factors relate different reservoir and surface volumes. A quantity stated in reservoir barrels, stock-tank barrels or standard cubic feet needs its label retained throughout the worksheet. The words 'in place' describe the original quantity being estimated, not the physical conditions of the reported unit.

An in-place estimate can also come from material balance when production, pressure and fluid data support the chosen model. The linked general oil material-balance calculator rearranges an oil balance using compatible withdrawal, influx and expansion terms. It estimates N on a stock-tank oil basis. That is a different calculation route from this gas geometry example, and it still does not independently determine a project's commercial recoverable quantity.

Apply a hypothetical recovery fraction on a separate line

For a teaching assumption of 60% recovery, multiply the calculated OGIP by 0.60. The result is 4.70448 billion scf of assumed recovered gas on the same volume basis. The remaining arithmetic complement is 3.13632 billion scf. Neither number is a forecast derived from a reservoir drive model; the recovery fraction is an input selected for this exercise.

Changing that fraction must leave the in-place estimate unchanged. The explorer keeps geometry, saturation and Bgi fixed so the distinction is visible. A recovery model for an actual accumulation would need additional information about its process, pressure behavior, displacement, wells and operating constraints. A convenient round percentage is not evidence that those conditions support that recovered volume.

Hypothetical recovery sensitivity at fixed OGIP = 7.8408 billion scf
Assumed recoveryAssumed recovered (billion scf)Arithmetic remainder (billion scf)
40%3.136324.70448
60%4.704483.13632
80%6.272641.56816

Recognize the additional project question behind reserves

PRMS treats reserves as remaining quantities expected to be commercially recovered through defined development projects in known accumulations under stated conditions and at an evaluation date. Classification therefore calls for information about discovery, recoverability, commerciality and the project. The official document provides the detailed criteria. A geometric quantity and an assumed recovery percentage provide only part of that information.

For this synthetic case, no development project, costs, production schedule or commercial assessment has been supplied. The honest output label is 'assumed recovered volume at 60%,' followed by its gas basis. Do not attach a reserves category to the result or interpret the slider's low and high settings as proved, probable or possible volumes. The control explores arithmetic sensitivity rather than uncertainty categories or project maturity.

Make the boundary of the calculation easy to audit

Keep a short ledger that separates supplied reservoir inputs, calculated in-place volume and the separately assumed recovery fraction. Record the date and basis of each input interpretation. If a later model changes porosity, net thickness or Bgi, recalculate the in-place line before comparing recovered quantities. If only the assumed fraction changes, the original geometric estimate should remain the same.

Use a different worksheet for any project-based forecast and formal classification assessment. State whether a reported quantity concerns original or remaining petroleum, gas at standard conditions, or stock-tank liquid. This keeps a helpful educational calculation from acquiring an unsupported meaning as it moves into a slide deck or report. The linked calculators provide transparent physical arithmetic; their outputs need the surrounding case information to answer a project question.

Worked example

From 160 acres to OGIP, then a separate recovery assumption

Use A = 160 acres, h = 50 ft, phi = 0.15, Swi = 0.25 and Bgi = 0.005 reservoir ft³/scf. Separately assume 60% recovery for illustration.

  1. 1. Calculate bulk rock volume
    43,560 × 160 × 50 = 348,480,000 ft³.

    Convert acre-feet to the reservoir cubic-foot basis used by this equation.

  2. 2. Calculate initial gas-filled pore volume
    348,480,000 × 0.15 × (1 − 0.25) = 39,204,000 reservoir ft³.

    Apply porosity, then the gas share of pore occupancy for the stated gas-water case.

  3. 3. Convert to OGIP
    39,204,000 / 0.005 = 7,840,800,000 scf.

    Bgi removes the reservoir-volume unit and leaves the stated standard gas volume.

  4. 4. Apply the separate assumption
    7.8408 × 0.60 = 4.70448 billion scf assumed recovered.

    The 60% multiplier is a teaching input; it does not classify the quantity as reserves.

Result

Calculated OGIP is 7.8408 billion scf. The separate 60% assumption gives 4.70448 billion scf as an assumed recovered volume.

Interpretation

The in-place line describes storage on the stated gas basis. Project recovery and classification require information beyond these inputs.

Test your understanding

Change the recovery assumption to 40% while holding all reservoir inputs fixed. Which result changes?

Show the worked answer

OGIP remains 7.8408 billion scf. Assumed recovered volume becomes 3.13632 billion scf, and the arithmetic remainder becomes 4.70448 billion scf. No resources class is assigned.

Method selection guide

Match the calculation method to the physical question and the evidence available.

ConditionUseWhy
You have geometry, initial saturation and gas BgiCalculate volumetric OGIPThose inputs support the stated initial gas-volume account.
You want to explore an assumed recovery fractionKeep the multiplier and result explicitly hypotheticalThe calculation does not establish the recovery mechanism or project evidence.
You need a formal resources or reserves classificationApply the governing project-based evaluation frameworkPRMS classification requires additional information beyond in-place arithmetic.

Before using the result

  • Use net reservoir geometry and state its interpretation.
  • Enter porosity and initial saturation as fractions.
  • Match Bgi's reservoir-volume numerator to the equation.
  • Record the standard-condition gas or stock-tank oil basis.
  • Label recovery fractions as supplied assumptions unless supported by a separate model.
  • Keep original, recovered, remaining and formally classified quantities distinct.

Start with these calculators

These links keep the guide close to the working calculator flow.

2 calculators

Continue learning

Formula references in this workflow

Original Gas in Place - Volumetric Method

Original Oil in Place from General Oil Material Balance

Verified sources used for this guide