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: OOIP = KV3 * HPV * AREA / Bo
Where:
KV3 = 7758 bbl for English units
KV3 = 1.0 m3 for Metric units
2: Bg = (PS *
(TF + KT2)) / (PF * (TS + KT2)) * ZF
3: 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 noncombustible (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.
8: 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)
REServes  Oil and Gas Reserves
Calculate
reserves
9: RF = (Sxo  Sw) / (1  Sw) or from decline curve
analysis or analogy.
10: Roil = RF * OOIP
11: 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 uninvaded 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.
Tar Assay 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).
WTtar  Tar 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
