OIL and gas in place anD reserves
Calculating oil or gas in place from petrophysical analysis results is a simple matter of calculating volumes from reservoir thickness, porosity, and water saturation. The area of the reservoir is usually contoured from maps of the reservoir properties. For single well analysis, a spacing unit is usually defined as 160 acres for oil wells and 640 acres (1 square mile) for gas wells. These dimensions are approximately 640,000 and 2,550,000 square meters in Metric units.

Reserves are defined as the amount of oil or gas that can be produced from a reservoir with current technology at current prices and current costs. Since these change on a daily basis, economic reserves can vary with time, increasing or decreasing with changes in the overall world economic conditions.

Decline curve analysis and material balance methods are used to calculate remaining reserves based on actual production and pressure data. Coupled with the volumetric analysis from petrophysical data, a reasonable solution can usually be found, although differences between the results from the three models may lead to a revised reservoir description. Additional data might be needed to resolve discrepancies, such as additional production history, new wells, better PVT data, or a new geological interpretation.

Calculate oil or gas in place
      1: Bo = 1.05 + 0.0005 * GOR
      2: OOIP = KV3 * HPV * AREA / Bo

Where:
  Bo = oil formation volume factor (fractional(
  HPV = hydrocarbon pore volume (feet or meters)
  GOR = gas / oil ratio (scf/bbl)
  KV3 = 7758 bbl for English units
  KV3 = 1.0 m3 for Metric units

      3: Bg =  (PS * (TF + KT2)) / (PF * (TS + KT2)) * ZF 
      4: OGIP = KV4 * (1 - Qnc) * HPV * AREA / Bg

Where:
  KT2 = 460 for English units
  KT2 = 273 for Metric units
  KV4 = 43.56 mcf for English units
  KV4 = 0.001 e3m3 for Metric units (e3m3 = 1000 cubic meters)
  Qnc = fraction of gas that is non-combustible (CO2, N2, etc)

NOTE: AREA is in acres for English units
           AREA is in m2 for Metric units

Spacing unit for oil is usually 160 acres (640 000 m2 approx)
                     for gas                640 acres (2 550 000 m2 approx)

In some parts of the world, oil is measured in metric tonnes instead of barrels or cubic meters.
       5: OOIPT  = KV6 * DENShy * OOIP  / 1000

Where:
  KV6 = 1.00 for Metric units (m3)
  KV6 = 0.159 for English units (bbl)
  DENShu = hydrocarbon density (kg/m3)
  OOIP = original oil in place (m3 or bbl)
  OOIPT = original oil in place (tonnes)

Oil and Gas Reserves
Calculate reserves
      6: RF = (Sxo - Sw) / (1 - Sw)   or from decline curve analysis  or analogy.
      7: Roil = RF * OOIP
      8: Rgas = RF * OGIP

Where:
  AREA = reservoir area (acres or m2)
  Bg = gas formation volume factor (fractional)
  Bo = oil formation volume factor (fractional(
  OGIP = original gas in place (mcf or m3)
  OOIP = original oil in place (bbl or m3)
  RF = recovery factor (fractional)
  Rgas = recoverable reserves of gas (mcf or m3)
  Roil = recoverable reserves of oil (bbl or m3)
  Sw = water saturation in un-invaded zone (fractional)
  Sxo = water saturation in invaded zone (fractional)

COMMENTS:
Recovery factor is difficult to estimate and is often known only after the pool, or an analogous pool, is depleted.

RECOMMENDED PARAMETERS:
Recovery factor can have a broad range for oil (0.01 to 0.95) and a narrower range for gas (0.50 to 0.95).

NUMERICAL EXAMPLE:
Using the data
  Sxo = 0.5
  Sw = 0.2
  HPV = 1.56 ft
  GOR = 500 scf/bbl
  AREA = 640 acres
  gas gravity = 0.85

  TF = 460 + 160 = 620 degrees R
  TS = 460 + 60 = 520 degrees R
  PF = 0.46 * 3000 = 1380 psi
  PS = 14.7 psi
  Bo = 1.05 + 0.0005 * 500 = 1.30
  Bg = 1 / 99.78
  RF = (0.5 - 0.2) / (1 - 0.2) = 0.37

If an oil well:
  OOIP = 7758 * 1.56 * 640 / 1.30 = 5.958 * 10^6 bbl/section
  Roil = 6 * 10^6 * 0.31 = 2.2 * 10^6 bbl/section

If gas:
  OGIP = 43.56 * 1.56 * 640 * 99.78 = 4.34 Bcf/section
  Rgas = 4.34 * 0.37 = 1.60 Bcf/section

Remember to round your answers to two or three significant digits, which you started with.

oil sand Reserves (Weight)
Tar or bitumen, and sometimes heavy oil, is measured by weight of tar in place as opposed to volume of oil in place. Some Former Soviet Union countries record conventional oil reserves in tonnes.

The following formulas are for use in areas where reserves are measured in metric tonnes, or as weight fraction (or weight percent).

Oil Weight Calculation
Tar weight.
      1: WTtar = PHIe * (1 - Sw) * DENShy / 1000
Shale weight.
      2: WTsh = Vsh * DENSSH / 1000
Sand weight.
      3: WTsnd = (1 - Vsh - PHIe) * DENSMA / 1000
Water weight.
      4: WTwtr = PHIe * Sw * DENSW / 1000
Total rock weight.
      5: WTrock = WTtar + WTsh + WTsnd + WTwtr
Tar mass fraction.
      6: TARfrac = WTtar / WTrock
Tar weight percent.
      7: TARwt% = 100 * WTtar / WTrock
Bitumen in place. (OOIP)
      8: OOIP = 0.001 * TARfrac * NetPay * DENSoil * AREA  (tonnes, AREA in m2)
Or 8A: OOIP = HPV * NetPay * AREA / Bo                           (m3, AREA in m2)
Or 8B:  OOIP = 7758 * HPV * NetPay * AREA / Bo                (bbl, AREA in acres)

Where:
  AREA = reservoir area (m2)
  Bo = oil volume factor        default = 1.0 for oil sands
  OOIP = bitumen in place  (tonnes, m3, bbl)
  DENSoil = hydrocarbon density (kg/m3)    default = 1000 kg/m3 for oil sands
  DENSMA = matrix density (kg/m3)
  DENSSH = shale density (kg/m3)
  DENSW = water density (kg/m3)
  NetPay = rock thickness (meters)
  PHIe = porosity (fractional)
  Sw = water saturation (fractional)
  TARfrac = tar mass fraction (fractional)
  TARwt% = tar weight percent (percent)
  Vsh = volume of shale (fractional)
  WTrock = total rock weight (tonne/m2)
  WTsh = shale weight (tonne/m2)
  WTsnd = sand weight (tonne/m2)
  WTtar = tar weight (tonne/m2)
  WTwtr = water weight (tonne/m2)

COMMENTS:
All densities are in kg/m3 in these formulae.

Results are in tonnes/m2 except where noted. To obtain tonnes in place, multiply by area in square meters. To obtain reserves, multiply this result by a recovery factor.

CAUTION: Some core analysis results do not include water (dry basis analysis). To compare log analysis results to core, eliminate the water term from WTrock.

NOTE: In much of Eastern Europe and Asia, oil quantities are reported in tonnes and not barrels. The "tar" equations provide the conversions needed. When using this method for oil, replace the word “tar” with “oil” to prevent confusion.

RECOMMENDED PARAMETERS:
None.

NUMERICAL EXAMPLE:
1. Assume data as follows:
  PHIe = 0.30
  Sw = 0.10
  Vsh = 0.10
  AREA = 640 acres or AREA = 2 550 000 m2
  DENSW = 1000 kg/m3
  DENSoil = 1000 kg/m3
  DENSSH = 2300 kg/m3
  DENSMA = 2650 kg/m3
  NetPay = 10 meters

  WTtar - 0.30 * (1 - 0.10) * 800 / 1000 = 0.216 tonnes/m2
  WTsh = 0.10 * 2300 / 100 = 0.230 tonnes/m2
  WTsnd = (1 - 0.10 - 0.30) * 265- / 1000 = 1.590 tonnes/m2
  WTwtr = 0.30 * 0.10 * 1000 / 1000 = 0.030 tonnes/m2
  WTrock = 0.216 + 0.230 + 1.590 + 0.030 = 2.066 tonnes/m2
  TARfrac = 0.216 / 2.066 = 0.1045
  TARwt% = 100 * 0.1045 = 10.45%
  OOIP = 0.001 * 0.1045 * 10 * 1000 * 2 550 000 = 2.665 million tonnes/section

Gas Hydrate IN PLACE
Empirically, the ratio of water to gas necessary to form a hydrate is as follows:

                                               Excess Hydrogen
     1. Methane CH4.6H20            4/12 = 33%
     2. Ethane C2H6.8H20             6/16 = 37%
     3. Propane C3H8.17H20         8/34 = 23%

The volume of hydrocarbon in a gas hydrate js a function of the hydrocarbon type and the porosity only. Water saturation is meaningless. The ratio of gas to water would range from 433 scf/bbl for propane to 1230 scf/bbl for methane.

This is equivalent to 170 cubic feet (for methane) per cubic foot of pore space (or 170 m3 per cubic meter of pore space) at standard temperature and pressure, for methane, and 60 cubic feet of propane per cubic foot of pore space, regardless of depth of burial.

Gas Hydrate Volume In Place
Convert pore volume to gas volume:
      1: PV = SUM (PHIe * THICK)
      2: HPV = PV * KG0

Calculate gas in place.
      3: GHIP = KV3 * HPV * AREA

Where:
  KV3 = 43.56 for English units
  KV3 = 1 for Metric units
  KG0 =170 for methane
  KG0 = 60 for propane

Where:
  AREA = reservoir area (acres or m2)
  KG0 = equivalent gas hydrate volume factor (fractional)
  HPV = hydrocarbon volume (feet or meters)
  PV = pore volume (feet or meters)
  GHIP = gas in place as hydrates (mcf or m3)

NUMERICAL EXAMPLE:
1. Assume the following data:
     PHIe = 0.35
     hydrate is methane
     THICK = 300 feet
     KG0 = 170 scf/scf
     AREA = 640 acres

  HPV = (0.35 * 300) * 170 = 17850 ft
  GHIP = 43.56 * 17850 * 640 / 1 000 000  = 497.634 Bcf/section