Volumetric estimation of hydrocarbons in place

Quick refresher

Oil in place: $N = \mathrm{GRV}\cdot \mathrm{NTG}\cdot\phi\cdot(1 - S_w)/B_{oi}$ . It uses the rock volume, the net-to-gross ratio, the porosity, the oil saturation and the conversion to standard conditions.
Each factor sifts out part of the volume: hundreds of millions of cubic metres of rock give a few million or tens of millions of cubic metres of oil.
We average porosity and water saturation over the same interval for which the net-to-gross ratio was found, that is, over the reservoir and not over the whole section. We weight water saturation by pore volume.
Oil at the surface occupies a smaller volume than in the reservoir ( $B_o > 1$ ), and gas occupies a volume many times larger ( $B_g \ll 1$ ). In both cases we divide the reservoir volume by the formation volume factor.
In oilfield units, 1 acre-foot = 43,560 ft³ ≈ 7758 barrels, and 1 m³ ≈ 35.31 ft³ ≈ 6.29 barrels.
Recoverable volume = hydrocarbons in place × recovery factor (RF). RF depends not only on the reservoir but also on the drive mechanism (the natural energy that pushes the oil to the wells) and the development system.

Key relationships

N = \dfrac{\mathrm{GRV}\cdot \mathrm{NTG}\cdot\phi\cdot(1 - S_w)}{B_{oi}}

oil; with GRV in m³ the result is in m³ at standard conditions

G = \dfrac{\mathrm{GRV}\cdot \mathrm{NTG}\cdot\phi\cdot(1 - S_w)}{B_{gi}}

free gas

N\,[\text{STB}] = 7758\cdot\dfrac{\mathrm{GRV}\,[\text{acre·ft}]\cdot \mathrm{NTG}\cdot\phi\cdot(1 - S_w)}{B_{oi}\,[\text{rb/STB}]}

oilfield units

Q_{\text{rec}} = N\cdot\text{RF}

recoverable volume

Common pitfalls

Multiplying the reservoir volume by $B_o$ instead of dividing: the volume in place is overestimated by a factor of $B_o^2$ , almost 1.7 times at $B_o = 1.3$ .
Taking a porosity averaged over the whole section, shales included, together with a net-to-gross ratio: the non-reservoir rock is counted twice.
Not stating the standard conditions (20 °C, 15 °C or 60 °F): for gas the difference reaches 1.5–2%.
Calling the product of all the minimum values a P90 estimate.
Treating hydrocarbons in place as recoverable: in international terminology reserves means recoverable volumes only.

After this lesson you can

write the formulas for oil and gas in place in metric and oilfield units (barrels, feet and acres), and explain the constants 7758 and 43,560
distinguish gross rock volume, reservoir volume, pore volume and hydrocarbon pore volume, and explain how cut-offs (threshold values) set the net-to-gross ratio
convert reservoir volumes to surface volumes with the formation volume factors of oil and gas (how much a volume changes between the reservoir and the surface), and keep the standard conditions in mind
average porosity and water saturation correctly, and explain why water saturation is weighted by pore volume
distinguish hydrocarbons in place from recoverable volumes (the part that can be produced), and name the factors on which the recovery factor depends
estimate the range of uncertainty, and explain why the product of all the pessimistic values is not a “P90 estimate”

The idea of the method: from rock volume to oil at the surface

Oil and gas underground are not in “lakes” but in the pores of the rock. So we find their amount by moving from a large volume to a small one. First comes the volume of rock inside the accumulation. Then comes the share of this rock that can hold and release fluid. Then the pore space. Then the part of the pores that hydrocarbons occupy. And finally the conversion of the reservoir volume to surface conditions.

Gross rock volume (GRV) is the volume of rock between the top and the base of the productive layer, inside the outline of the accumulation, that is, above the oil–water contact (OWC) or gas–water contact (GWC) (the depth where the hydrocarbons end and the water begins). We find it from structure maps and thickness maps (Gross rock volume from maps).
Net rock volume (reservoir volume) is the part of the GRV made of rocks that can hold and release fluid. Its share is called the net-to-gross ratio (NTG).
Pore volume is the reservoir volume multiplied by the porosity $\phi$ .
Hydrocarbon pore volume (HCPV) is the pore volume multiplied by the share of the pores that water does not occupy, $1 - S_w$ . Here $S_w$ is the water saturation, the share of the pores filled with water.
Hydrocarbons in place are the HCPV divided by the formation volume factor of oil $B_o$ or of gas $B_g$ , that is, brought to standard conditions at the surface.

Gross and net thickness: cut-offs

A productive layer is not uniform from top to bottom. Sandstones alternate with shaly layers, with tight (very low permeability) cemented intervals, and with intervals where the pores are filled with water. We want to count only the part that contains moveable hydrocarbons (hydrocarbons that can flow). So we divide the section, using the data of well logs and core, with the help of cut-offs: threshold values of shale volume, porosity and saturation.

How cut-offs divide the section

Interval: Gross thickness (gross interval)
What it includes: the whole layer from top to base
Cut-offs applied: none

Interval: Reservoir by lithology (gross sand)
What it includes: everything except the shaly layers
Cut-offs applied: on shale volume $V_{sh}$

Interval: Net thickness (net sand)
What it includes: the reservoir that can hold and release fluid
Cut-offs applied: on shale volume and porosity

Interval: Net pay thickness (net pay)
What it includes: the reservoir that contains hydrocarbons
Cut-offs applied: on shale volume, porosity and water saturation

Interval	What it includes	Cut-offs applied
Gross thickness (gross interval)	the whole layer from top to base	none
Reservoir by lithology (gross sand)	everything except the shaly layers	on shale volume $V_{sh}$
Net thickness (net sand)	the reservoir that can hold and release fluid	on shale volume and porosity
Net pay thickness (net pay)	the reservoir that contains hydrocarbons	on shale volume, porosity and water saturation

We choose the cut-offs for each layer from core and logs. We often start from the relation of porosity to permeability. A rock whose permeability is below a chosen limit is taken as non-reservoir (for oil this is often of the order of 1 mD, and for gas it is lower). At real pressure differences, such a rock gives up practically no fluid.

The net-to-gross ratio is the ratio of the net thickness to the gross thickness, $\mathrm{NTG} = h_{\text{eff}}/h_{\text{t}}$ (or the ratio of the corresponding volumes). Here $h_{\text{eff}}$ is the net (effective) thickness and $h_{\text{t}}$ is the gross (total) thickness. The difficulty is that “net” thickness is understood in different ways: most often it means net sand, but sometimes net pay. So when we get the NTG from a petrophysicist (a specialist who interprets rock properties from logs and core), we always ask which cut-offs went into it.

The averages must refer to the same interval

If the NTG was found for the reservoir (net sand), we average porosity and water saturation over the reservoir only. A porosity averaged over the whole section together with the shales is too low: the non-reservoir has been counted twice, through the NTG and again through the porosity.
The opposite mistake: the NTG was found for net pay (the water-filled intervals are already excluded), but the water saturation was averaged over the net sand together with those intervals. Water is counted twice again, and the volume in place is underestimated.
Net pay can be identified only when a well has been drilled into the accumulation and the saturation measured. Before drilling we use net sand and the expected water saturation of the reservoir above the contact.

The formula for oil in place

Oil in the reservoir is under high pressure and at a high temperature, and gas is dissolved in it. As the oil rises to the surface, the gas comes out of solution, and the oil cools and “shrinks”. The oil formation volume factor $B_o$ (formation volume factor, Black oil model: solution gas–oil ratio and formation volume factors) compares two volumes. One is the volume of the oil in the reservoir, with its dissolved gas. The other is the volume of the same oil after the gas has been separated off at standard conditions (a fixed reference pressure and temperature). $B_o$ is the ratio of the first to the second. It is larger than one: usually 1.1–1.5, and for light oils with a lot of dissolved gas up to 2 and more.

Oil initially in place (volumetric method)

N = \frac{\mathrm{GRV}\cdot \mathrm{NTG}\cdot\phi\cdot(1 - S_w)}{B_{oi}}

where:

$N$ oil initially in place at standard conditions (STOIIP), m³
$\mathrm{GRV}$ gross rock volume inside the oil outline: between the top and the base of the layer, above the OWC, m³
$\mathrm{NTG}$ net-to-gross ratio: the share of reservoir in the gross rock volume, fraction
$\phi$ mean porosity of the reservoir, fraction
$S_w$ mean initial water saturation of the reservoir; $1 - S_w$ is the oil saturation $S_o$ , fraction
$B_{oi}$ oil formation volume factor at the initial reservoir pressure and temperature, m³/m³ (reservoir per standard)

Units: metric: volumes in m³, the other quantities are dimensionless

Holds when

The pores of the reservoir contain only oil and connate water (the water that has stayed in the pores since the rock was formed). If there is a gas cap (a layer of free gas on top of the oil), the volume above the gas–oil contact is calculated separately, with the formula for gas.
The mean $\phi$ and $S_w$ refer to the same volume of reservoir as the NTG.
$B_{oi}$ corresponds to the initial reservoir conditions and to the adopted scheme of separating the oil at the surface.
We use mean values for the whole accumulation. If the parameters change noticeably across the area, we calculate from maps or from the cells of the geological model and add up the results.

Each factor is a “sieve”: the share of reservoir, the share of pores, the share of oil in the pores and, finally, the conversion from reservoir volume to surface volume. A relative error in GRV, NTG, $\phi$ or $1 - S_w$ passes into the volume in place one to one, and the volume in place is inversely proportional to $B_{oi}$ .

In other unit systems (2)

Metric units, mass of oil

Q\,[\text{t}] = \mathrm{GRV}\,[\text{m}^3]\cdot \mathrm{NTG}\cdot\phi\cdot(1 - S_w)\cdot\frac{\rho_o\,[\text{t/m}^3]}{B_{oi}}

$\rho_o$ is the oil density at standard conditions (0.85 t/m³ = 850 kg/m³); the other quantities are as in the main formula.

There is no constant. In Russian-language practice the factor $1/B_{oi}$ is called the shrinkage factor, and the product $\mathrm{GRV}\cdot \mathrm{NTG}$ is written as the oil-bearing area multiplied by the mean net oil-saturated thickness.

Oilfield units

N\,[\text{STB}] = 7758\cdot\frac{\mathrm{GRV}\,[\text{acre·ft}]\cdot \mathrm{NTG}\cdot\phi\cdot(1 - S_w)}{B_{oi}\,[\text{rb/STB}]}

GRV is in acre-feet (the area in acres multiplied by the thickness in feet); $B_{oi}$ is in reservoir barrels per barrel at standard conditions (rb/STB); $N$ is in barrels of oil at standard conditions (stock-tank barrels, STB).

An acre is 43,560 square feet, so an acre-foot is 43,560 ft³. A barrel is 42 US gallons, or 5.6146 ft³. So $43{,}560/5.6146 \approx 7758$ barrels per acre-foot.

Worked exampleOil in place in metric and oilfield units

Given (assumed data)

Gross rock volume above the OWC, GRV: $1.8\cdot10^{8}$ m³ (180 million m³)
Net-to-gross ratio NTG: 0.70
Mean porosity $\phi$: 0.19
Mean water saturation $S_w$: 0.28
Oil formation volume factor $B_{oi}$: 1.25 m³/m³
Oil density at standard conditions: 850 kg/m³

Solution

Reservoir volume: $180\cdot0.70 = 126$ million m³.
Pore volume: $126\cdot0.19 = 23.94$ million m³.
Oil-filled pore volume: $23.94\cdot(1 - 0.28) = 17.24$ million m³ at reservoir conditions.
Oil in place: $N = 17.24/1.25 = 13.79$ million m³ at standard conditions. By mass, $13.79\cdot0.85 = 11.72$ million t.
In barrels: $13.79\cdot6.2898 = 86.7$ million STB.
Check with the oilfield formula: $\mathrm{GRV} = 1.8\cdot10^{8}/1233.5 = 145{,}928$ acre-feet, and $N = 7758\cdot145{,}928\cdot0.70\cdot0.19\cdot0.72/1.25 = 86.7$ million STB, the same number.

Answer: Oil in place is about 13.79 million m³ of oil at standard conditions, or 11.72 million t, or 86.7 million barrels.

What it means. Oil at standard conditions is only 7.7% of the gross rock volume. This shows why errors in the structure and the contacts are so expensive: an error in GRV is multiplied by all the other factors. The recoverable volume will be smaller still (see the section on the recovery factor). The data in this example are assumed for illustration.

CalculatedThe chain of the volumetric method for the assumed accumulation of the example; the recoverable volume is for a recovery factor of 0.30.

Gas: the formula and the gas formation volume factor

For gas the formula is the same, and only the conversion to the surface changes. As the gas rises, the pressure falls hundreds of times, and the gas expands. The gas formation volume factor $B_g$ is the ratio of the volume of the gas in the reservoir to the volume of the same gas at standard conditions, so it is much smaller than one. The quantity $1/B_g$ shows how many times the gas expands: usually tens to hundreds of times.

Free gas initially in place

G = \frac{\mathrm{GRV}\cdot \mathrm{NTG}\cdot\phi\cdot(1 - S_w)}{B_{gi}}

where:

$G$ free gas initially in place at standard conditions (GIIP), m³
$B_{gi}$ gas formation volume factor at the initial reservoir pressure and temperature, m³/m³
$\mathrm{GRV},\ \mathrm{NTG},\ \phi,\ S_w$ as in the formula for oil; GRV is the volume of rock above the gas–water (or gas–oil) contact, m³ and fraction

Units: metric: volumes in m³

Holds when

The pores of the reservoir are filled with gas and connate water (a gas accumulation or a gas cap).
$B_{gi}$ is calculated for the same standard conditions to which the volumes in place are referred.
Gas dissolved in the oil is not included here. We count it separately as $N\cdot R_{si}$ , where $R_{si}$ is the initial solution gas–oil ratio of the oil (how much gas is dissolved in each cubic metre of oil).

The formula differs from the oil formula only in the denominator. Since $B_{gi} \ll 1$ , dividing by it turns millions of reservoir cubic metres into billions of standard cubic metres.

In other unit systems (1)

Oilfield units

G\,[\text{scf}] = 43{,}560\cdot\frac{\mathrm{GRV}\,[\text{acre·ft}]\cdot \mathrm{NTG}\cdot\phi\cdot(1 - S_w)}{B_{gi}\,[\text{ft}^3/\text{scf}]}

GRV is in acre-feet; $B_{gi}$ is in reservoir cubic feet per standard cubic foot (ft³/scf); $G$ is in standard cubic feet (scf), and in practice in billions of scf (Bcf).

43,560 is the number of cubic feet in an acre-foot. If $B_{gi}$ is expressed in reservoir barrels per scf, we put 7758 in place of 43,560.

Gas formation volume factor

B_g = \frac{p_{sc}}{T_{sc}}\cdot\frac{Z\,T}{p}

where:

$p_{sc},\ T_{sc}$ absolute pressure and temperature at standard conditions, MPa, K
$p$ reservoir pressure, absolute, MPa
$T$ reservoir temperature, absolute ( $T[\text{K}] = t[{}^{\circ}\text{C}] + 273.15$ ), K
$Z$ gas deviation factor at the reservoir $p$ and $T$ , —

Units: any consistent units: pressures in the same units, temperatures absolute (K or °R)

Holds when

The real-gas equation of state $pV = ZnRT$ (it links the pressure, volume and temperature of the gas); at standard conditions the gas is taken as ideal ( $Z = 1$ ).
The pressures are absolute (measured from zero), not gauge (measured above atmospheric pressure). The temperatures are in kelvins or in degrees Rankine (the absolute scale for degrees Fahrenheit).

$B_g$ is inversely proportional to the pressure: the higher the reservoir pressure, the more gas fits into the same pore volume. For more about $Z$ , see the lesson Ideal and real gas: the gas deviation factor (Z).

In other unit systems (2)

Metric, standard conditions 20 °C and 0.101325 MPa

B_g = 3.456\cdot10^{-4}\,\frac{Z\,T\,[\text{K}]}{p\,[\text{MPa}]}

$B_g$ is in m³/m³; $T$ is in K; $p$ is in MPa, absolute.

$3.456\cdot10^{-4} = 0.101325/293.15$ . For 15 °C the constant is $0.101325/288.15 = 3.516\cdot10^{-4}$ .

Oilfield units, 60 °F and 14.696 psia

B_g = 0.02828\,\frac{Z\,T\,[{}^{\circ}\text{R}]}{p\,[\text{psia}]}

$B_g$ is in ft³/scf; $T$ is in degrees Rankine, $T[{}^{\circ}\text{R}] = t[{}^{\circ}\text{F}] + 459.67$ ; $p$ is the absolute pressure in psia (pounds per square inch).

$0.02828 = 14.696/519.67$ (the pressure and the absolute temperature of standard conditions).

Worked exampleGas in place and the effect of standard conditions

Given (assumed data)

Gross rock volume above the GWC, GRV: $2.0\cdot10^{8}$ m³
NTG, $\phi$ , $S_w$: 0.80, 0.17, 0.25
Reservoir pressure (absolute): 26 MPa
Reservoir temperature: 85 °C
Gas deviation factor $Z$: 0.90

Solution

Gas-filled pore volume: $2.0\cdot10^{8}\cdot0.80\cdot0.17\cdot0.75 = 20.4$ million m³ at reservoir conditions.
Formation volume factor at 20 °C: $B_{gi} = 3.456\cdot10^{-4}\cdot\dfrac{0.90\cdot358.15}{26} = 0.00429$ m³/m³. The gas expands by a factor of $1/B_{gi} = 233$ .
Gas in place: $G = 20.4\cdot10^{6}/0.00429 = 4.76$ billion m³ at 20 °C.
Oilfield units: $26$ MPa $= 3771$ psia, $85$ °C $= 644.7$ °R, $B_{gi} = 0.02828\cdot0.90\cdot644.7/3771 = 0.004351$ ft³/scf; $\mathrm{GRV} = 162{,}143$ acre-feet; $G = 43{,}560\cdot162{,}143\cdot0.102/0.004351 = 165.6$ billion scf (here $0.102 = 0.80\cdot0.17\cdot0.75$ ).
Comparison: $165.6/35.315 = 4.69$ billion m³, which is 1.5% less than at 20 °C. The gas is the same. The difference is only in the standard temperature (60 °F against 20 °C).

Answer: About 4.76 billion m³ of gas at 20 °C, or about 165.6 billion scf at 60 °F.

What it means. Twenty million cubic metres of pore space hold almost five billion cubic metres of gas. A compression of about 230 times explains why gas accumulations with a fairly small pore volume contain so much gas. The second conclusion is practical: when we compare volumes in place from different reports, we first bring them to the same standard conditions. The data in this example are assumed for illustration.

Averaging porosity and saturation

Logs give porosity and water saturation for every thin interval of the section, but the formula needs mean values. The way of averaging matters: it must preserve what is most important, the pore volume and the volume of hydrocarbons.

Mean porosity over the net thickness

\bar\phi = \frac{\sum_k \phi_k\,h_k}{\sum_k h_k}

where:

$\phi_k$ porosity of the $k$ th interval that passed the cut-offs, fraction
$h_k$ thickness of this interval, m

Units: any units, the same for all intervals

Holds when

We add up only the intervals of net thickness.
All the intervals belong to one well (one point on the map).

This is weighting by thickness: a thick interval weighs more than a thin one, and the product $\bar\phi\sum h_k$ equals the total pore volume per unit area.

Mean water saturation weighted by pore volume

\bar S_w = \frac{\sum_k S_{w,k}\,\phi_k\,h_k}{\sum_k \phi_k\,h_k}

where:

$S_{w,k}$ water saturation of the $k$ th interval, fraction
$\phi_k\,h_k$ pore volume of the interval per unit area, m

Units: any units, the same for all intervals

Holds when

We add up the same intervals as for the mean porosity.
Saturation is a share of the pore volume, so the weight of an interval is its pore volume, not its thickness.

With such weights, the product $h_{\text{eff}}\,\bar\phi\,(1 - \bar S_w)$ is exactly equal to the sum of the hydrocarbons over the intervals. A simple average, or a thickness-weighted average, of the saturations does not have this property.

Hydrocarbon pore thickness

H = \sum_k h_k\,\phi_k\,(1 - S_{w,k}) = h_{\text{eff}}\,\bar\phi\,(1 - \bar S_w)

where:

$H$ the thickness of the layer that the hydrocarbons would form if we gathered them out of the pores into a solid layer; the volume of hydrocarbons at reservoir conditions per unit area, m
$h_{\text{eff}}$ net thickness, $\sum_k h_k$ , m

Units: m (or ft)

Holds when

$\bar\phi$ and $\bar S_w$ are averaged by the formulas above.

We calculate $H$ in each well and build a map of it. Adding up (integrating) the map over the area of the accumulation gives the HCPV directly. It also takes into account that thickness, porosity and saturation change across the area together.

Worked exampleMean parameters for a well

Given (assumed data)

Interval 1: 4 m; $\phi = 0.22$ ; $S_w = 0.20$
Interval 2: 3 m; $\phi = 0.16$ ; $S_w = 0.35$
Interval 3: 2 m; $\phi = 0.12$ ; $S_w = 0.55$
Shaly layers: 3 m (did not pass the cut-offs)

Solution

The net thickness is $h_{\text{eff}} = 4 + 3 + 2 = 9$ m, with a gross thickness of 12 m, so $\mathrm{NTG} = 0.75$ .
Mean porosity: $\bar\phi = \dfrac{4\cdot0.22 + 3\cdot0.16 + 2\cdot0.12}{9} = \dfrac{1.60}{9} = 0.178$ .
Mean water saturation by pore volume: $\bar S_w = \dfrac{0.20\cdot0.88 + 0.35\cdot0.48 + 0.55\cdot0.24}{1.60} = 0.2975$ .
Hydrocarbon pore thickness: $H = 9\cdot0.178\cdot(1 - 0.2975) = 1.12$ m. The sum of $h_k\phi_k(1 - S_{w,k})$ over the intervals gives exactly the same.
For comparison, the simple average of the water saturations, 0.367, would give $H = 1.01$ m, which is 10% less.

Answer:

\bar\phi = 0.178

\bar S_w = 0.2975

; under every square metre of area at the well there are 1.12 m³ of oil at reservoir conditions.

What it means. The simple average gave the same weight to a thin, low-porosity, water-bearing interval and to a thick oil-bearing one, and in this way it “added” water. When we weight by pore volume, the contribution of each interval is proportional to how much fluid it really contains. The data in this example are assumed for illustration.

Across the area we do the same. We build maps of the net thickness, the porosity and the saturation, or straight away a map of the hydrocarbon pore thickness $H$ , and integrate it inside the outline of the accumulation. On the maps, the thickness is given perpendicular to the bedding or vertically. In dipping layers these two differ, and we must not mix them. In a digital geological model, we do the same by cells: the volume in place is the sum of $V_i\,\mathrm{NTG}_i\,\phi_i\,(1 - S_{w,i})/B_{oi}$ over all the cells above the contact (What a reservoir simulation model is and why it is needed).

Where the input data come from

Parameters of the calculation and their sources

Parameter: GRV
Main sources: seismic (structure maps of the top and base), the depths of the layers in the wells, the positions of the contacts and the faults
What most often introduces errors: converting the arrival times of reflections to depths, an unknown contact, an unclear role of the faults

Parameter: NTG
Main sources: well logs (shale volume from the gamma ray log), core, a model of the depositional environment between the wells
What most often introduces errors: the choice of cut-offs; thin interbedding that the log cannot resolve

Parameter: $\phi$
Main sources: density, neutron and sonic logs, calibrated with core
What most often introduces errors: shale, gas in the pores, converting core values to reservoir conditions

Parameter: $S_w$
Main sources: resistivity logs (Archie's equation), capillary pressure from core, height above the free water level
What most often introduces errors: the parameters of Archie's equation, the resistivity of the formation water, the transition zone

Parameter: $B_{oi}$ , $B_{gi}$
Main sources: laboratory tests of reservoir fluid samples: the dependence of volume on pressure and temperature (PVT); correlations
What most often introduces errors: a sample that does not represent the reservoir fluid; separation conditions

Parameter: RF
Main sources: analogues, analytical calculations, reservoir simulation
What most often introduces errors: the drive mechanism and the heterogeneity, which are poorly known at an early stage

Parameter	Main sources	What most often introduces errors
GRV	seismic (structure maps of the top and base), the depths of the layers in the wells, the positions of the contacts and the faults	converting the arrival times of reflections to depths, an unknown contact, an unclear role of the faults
NTG	well logs (shale volume from the gamma ray log), core, a model of the depositional environment between the wells	the choice of cut-offs; thin interbedding that the log cannot resolve
$\phi$	density, neutron and sonic logs, calibrated with core	shale, gas in the pores, converting core values to reservoir conditions
$S_w$	resistivity logs (Archie's equation), capillary pressure from core, height above the free water level	the parameters of Archie's equation, the resistivity of the formation water, the transition zone
$B_{oi}$ , $B_{gi}$	laboratory tests of reservoir fluid samples: the dependence of volume on pressure and temperature (PVT); correlations	a sample that does not represent the reservoir fluid; separation conditions
RF	analogues, analytical calculations, reservoir simulation	the drive mechanism and the heterogeneity, which are poorly known at an early stage

Contacts. We find the oil–water (OWC), gas–oil (GOC) and gas–water (GWC) contacts from logs, from samples, and from the pressure measurements of a formation tester (a tool that measures pressure at points in the rock behind the wall of the well). The contact is where the pressure gradients of oil and water cross (a pressure gradient shows how fast pressure grows with depth; see Pressure gradients, fluid contacts and capillary equilibrium). If a well has not reached the contact, then for a cautious estimate we take the lowest depth at which oil was found (oil down to, ODT) as the lower boundary of the accumulation. For an optimistic estimate we take the highest depth at which water was found (water up to, WUT), or the spill point of the trap (the level down to which the trap can hold hydrocarbons).

Water saturation and height above the contact. Water saturation depends not only on the rock but also on the height above the free water level (the level where the capillary pressure is zero). At the contact the pores are almost entirely filled with water, and higher up the share of water falls to its irreducible value (the lowest value it can reach). It falls faster where the pores are larger. In thick accumulations with a wide transition zone (where the share of water changes gradually with height), we calculate the mean $S_w$ from the dependence of saturation on height, built from the capillary pressure of core.

Formation volume factors. Across the area they change much less than the properties of the rocks. The exception is thick accumulations in which the composition of the oil changes with depth (heavier below and lighter above): there we average $B_o$ over depth instead of taking it from one sample.

Hydrocarbons in place and recoverable volumes: the recovery factor

The oil recovery factor (RF) is the share of the hydrocarbons in place that we will manage to produce over the whole life of development. RF is not a property of the rock. It depends on the source of reservoir energy (Reservoir drive mechanisms), on the viscosity of the oil (how thick it is), on the heterogeneity of the reservoir, and on the number and placing of wells. It also depends on the technology used: waterflooding (injecting water to push the oil out), gas injection, or enhanced oil recovery methods. Finally, it depends on the flow rate or water cut (the share of water in the produced liquid) at which production is stopped for economic reasons.

Recoverable volume

Q_{\text{rec}} = N\cdot\text{RF}

where:

$Q_{\text{rec}}$ recoverable oil, m³ or t (as for N)
$N$ oil initially in place, m³ or t
$\text{RF}$ oil recovery factor, fraction

Units: the same as for $N$

Holds when

The RF refers to a particular development system and particular economic conditions. When these change, the recoverable volume changes too.
For gas we use the gas recovery factor in the same way.

The volume in place is a property of the accumulation. The recoverable volume is the result of combining the accumulation with the project for developing it. So the same accumulation has different recoverable volumes under different technologies.

It is convenient to think of RF as the product of two factors. The displacement efficiency shows what share of the oil the water or gas pushes out of the part of the pores that it has reached. We find it from experiments on core. The sweep efficiency shows what share of the volume of the accumulation the displacing agent has touched at all. It depends on the heterogeneity of the reservoir, the viscosity ratio and the placing of wells.

Approximate values of the recovery factor

Drive mechanism: Solution gas drive (oil without pressure maintenance)
RF, fraction: 0.05–0.30
Why: the only energy is the expansion of the oil and of the gas that comes out of solution; the gas quickly breaks through to the wells

Drive mechanism: Gas cap drive
RF, fraction: 0.20–0.40
Why: the expanding cap pushes the oil from the top downwards

Drive mechanism: Water drive or waterflooding
RF, fraction: 0.30–0.60
Why: water maintains the pressure and displaces the oil; the result depends strongly on the viscosity of the oil and on the heterogeneity

Drive mechanism: Gas accumulation, volumetric depletion drive
RF, fraction: 0.70–0.90
Why: the gas is mobile and compressible; production is limited by the pressure needed at the wellhead

Drive mechanism: Gas accumulation with an active aquifer (a large body of water-bearing rock beside the accumulation)
RF, fraction: 0.50–0.70
Why: water traps part of the gas and floods the wells

Drive mechanism	RF, fraction	Why
Solution gas drive (oil without pressure maintenance)	0.05–0.30	the only energy is the expansion of the oil and of the gas that comes out of solution; the gas quickly breaks through to the wells
Gas cap drive	0.20–0.40	the expanding cap pushes the oil from the top downwards
Water drive or waterflooding	0.30–0.60	water maintains the pressure and displaces the oil; the result depends strongly on the viscosity of the oil and on the heterogeneity
Gas accumulation, volumetric depletion drive	0.70–0.90	the gas is mobile and compressible; production is limited by the pressure needed at the wellhead
Gas accumulation with an active aquifer (a large body of water-bearing rock beside the accumulation)	0.50–0.70	water traps part of the gas and floods the wells

The values are orders of magnitude for guidance. The actual values depend on the reservoir, the fluids and the development system.

At an early stage we take RF from similar fields, from statistical correlations, or from simple analytical calculations. The correlations are built from the data of many accumulations and relate RF to porosity, saturation, formation volume factor, oil mobility $k/\mu$ (permeability divided by viscosity) and the ratio of the initial pressure to the pressure when production stops. The scatter of the actual values around them is large, so they are good only for first estimates. Later we refine RF from the reservoir simulation model (What a reservoir simulation model is and why it is needed) and from the actual production history. The uncertainty of RF at the start of a project is often no smaller than the uncertainty of the volume in place.

Uncertainty and the development of the estimate

Every parameter is known with some error, and usually the largest error is in GRV: before drilling it comes from the uncertainty of the structure and the contacts, and later from the faults and the edges of the reservoir. As we study the accumulation, the estimate changes in a typical way.

How the estimate is refined

1
Before drilling
The volume of the trap is known only from seismic, and the contact is a guess. We use such an estimate for the decision to drill an exploration well. These are prospective resources (oil and gas that have not yet been found), not reserves.
2
Discovery
The first well gives the section, core, fluid samples and sometimes the contact. The zone around the well is studied most reliably, and the far parts of the accumulation are estimated less confidently.
3
Appraisal wells
They refine the boundaries and find faults, separate blocks with their own contacts, and the gas cap. The estimate may either grow or shrink.
4
Development
Material balance and the reservoir simulation model are added to the volumetric method. They check the volume in place against the behaviour of pressure and production.

The simplest way to show the uncertainty is to calculate three deterministic cases (each one uses a fixed value for every parameter): a pessimistic one, a base one and an optimistic one. This method has a pitfall.

Worked exampleThe pessimistic and optimistic cases

Given (assumed data)

Pessimistic case: GRV $1.4\cdot10^{8}$ m³; NTG 0.60; $\phi$ 0.17; $S_w$ 0.33; $B_{oi}$ 1.30
Optimistic case: GRV $2.3\cdot10^{8}$ m³; NTG 0.78; $\phi$ 0.21; $S_w$ 0.24; $B_{oi}$ 1.20
Base case: 13.8 million m³ (the example above)

Solution

Pessimistic: $N = 1.4\cdot10^{8}\cdot0.60\cdot0.17\cdot0.67/1.30 = 7.4$ million m³. It has the highest water saturation and the highest $B_{oi}$ , and both reduce the volume in place.
Optimistic: $N = 2.3\cdot10^{8}\cdot0.78\cdot0.21\cdot0.76/1.20 = 23.9$ million m³.
The ratio of the optimistic to the pessimistic is 3.2. Of this spread, GRV alone accounts for a factor of 1.64, more than any other parameter.
Suppose each pessimistic value is such that the parameter will turn out even worse with a probability of 10%. If the parameters are independent, the probability that all five will turn out worse at once is $0.1^{5} = 10^{-5}$ .

Answer: The range is 7.4–23.9 million m³ around the base estimate of 13.8 million m³.

What it means. The product of all the pessimistic values is not a “P90 estimate”. It is an extremely unlikely scenario, and the same is true of the product of all the optimistic values. We get a realistic range by multiplying not numbers but distributions, with the Monte Carlo method (Probabilistic volumetrics by Monte Carlo simulation). We also take into account the links between the parameters: for example, in more porous rocks the water saturation is usually lower. The data in this example are assumed for illustration.

Terms and symbols

GRV, gross rock volume: The volume of rock between the top and the base of the layer, inside the outline of the accumulation (above the contact), including the non-reservoir rock.
NTG, net-to-gross ratio: The share of reservoir in the gross thickness or volume. We always ask which cut-offs it was found with.
Net pay: The net thickness that contains hydrocarbons: the part of the section that passed the cut-offs on shale volume, porosity and saturation.
HCPV: Hydrocarbon pore volume at reservoir conditions, $\mathrm{GRV}\cdot \mathrm{NTG}\cdot\phi\cdot(1 - S_w)$ .
STOIIP, GIIP: Stock-tank oil initially in place and gas initially in place: the oil and the free gas in the reservoir, counted at standard conditions. OOIP and OGIP are also used.
$B_o$ , $B_g$: The formation volume factors of oil and gas: the reservoir volume per unit volume at standard conditions.
Acre-foot: The volume of a layer 1 foot thick over an area of 1 acre: 43,560 ft³ ≈ 1233.5 m³ ≈ 7758 barrels.
RF: The recovery factor: the share of the hydrocarbons in place that we expect to produce.

Common mistakes and misconceptions

Wrong

“ $B_o = 1.3$ , so we multiply the reservoir volume by 1.3 and get the oil at the surface.”

In fact

At the surface the oil occupies a smaller volume, so we divide the reservoir volume by $B_o$ . Multiplying overestimates the volume in place by a factor of $B_o^2$ , almost 1.7 times at $B_o = 1.3$ .

Wrong

“We take the NTG from the petrophysicists and average the porosity over the whole section, which is safer.”

In fact

We average porosity and saturation over the same interval for which the NTG was found. Otherwise the non-reservoir is counted twice, and the volume in place is underestimated.

Wrong

“The pessimistic estimate is the product of the minimum values of all the parameters, and this is the P90.”

In fact

A simultaneous combination of all the worst values is extremely unlikely. We get probabilistic estimates by multiplying distributions with the Monte Carlo method, taking the links between the parameters into account.

Wrong

“Hydrocarbons in place are 100 million t, so we will produce 100 million t.”

In fact

Only a part can be produced. The recoverable volume is the volume in place multiplied by RF, and RF for oil is usually from 0.05 to 0.6, depending on the drive mechanism and the technology.

Wrong

“A cubic metre of gas is a cubic metre, whatever the temperature.”

In fact

Standard conditions differ (20 °C, 15 °C, 60 °F). For gas the difference reaches 1.5–2%, and for large fields this is billions of cubic metres.

In depth: advanced detail

Conversion factors without memorising

All the factors that we need to convert volumes between systems of units follow from two exact definitions: 1 foot = 0.3048 m, and 1 barrel = 42 US gallons = 0.158987 m³. It is useful to be able to recalculate them yourself instead of taking them from tables.

The main conversion factors

Quantity: 1 foot (ft)
Definition: exact definition
Metric equivalent: 0.3048 m

Quantity: 1 cubic foot (ft³)
Definition: $0.3048^3$ m³
Metric equivalent: 0.028317 m³; 1 m³ = 35.315 ft³

Quantity: 1 barrel (bbl)
Definition: 42 US gallons = 5.6146 ft³
Metric equivalent: 0.158987 m³; 1 m³ = 6.2898 bbl

Quantity: 1 acre
Definition: 43,560 ft²
Metric equivalent: 4046.9 m²; 1 km² = 247.1 acres

Quantity: 1 acre-foot
Definition: 43,560 ft³ = 7758.4 bbl
Metric equivalent: 1233.5 m³

Quantity: 1 billion m³ of gas
Definition: —
Metric equivalent: 35.315 billion ft³ (Bcf) at the same standard conditions

Quantity: 1 t of oil
Definition: depends on the density
Metric equivalent: at 860 kg/m³: 1.163 m³ = 7.31 bbl

Quantity	Definition	Metric equivalent
1 foot (ft)	exact definition	0.3048 m
1 cubic foot (ft³)	$0.3048^3$ m³	0.028317 m³; 1 m³ = 35.315 ft³
1 barrel (bbl)	42 US gallons = 5.6146 ft³	0.158987 m³; 1 m³ = 6.2898 bbl
1 acre	43,560 ft²	4046.9 m²; 1 km² = 247.1 acres
1 acre-foot	43,560 ft³ = 7758.4 bbl	1233.5 m³
1 billion m³ of gas	—	35.315 billion ft³ (Bcf) at the same standard conditions
1 t of oil	depends on the density	at 860 kg/m³: 1.163 m³ = 7.31 bbl

The number of barrels in a tonne of oil is not constant: it is inversely proportional to the density of the oil.

Why we calculate from maps and not from averages

Thickness, porosity and saturation usually change across the area in step with one another: where the layer is thicker, the reservoir is often better too. Then the product of the simple averages is not equal to the average of the products, and a calculation “from averages” makes a systematic error.

Worked exampleTwo zones of an accumulation

Given (assumed data)

Zone A: area 1 km², net thickness 20 m, $\phi = 0.22$
Zone B: area 1 km², net thickness 5 m, $\phi = 0.12$

Solution

Pore volume by zones: $10^{6}\cdot20\cdot0.22 + 10^{6}\cdot5\cdot0.12 = 5.00$ million m³.
From simple averages: area 2 km², thickness 12.5 m, porosity $(0.22 + 0.12)/2 = 0.17$ : $2\cdot10^{6}\cdot12.5\cdot0.17 = 4.25$ million m³, which is 15% less.
Porosity weighted by thickness: $(20\cdot0.22 + 5\cdot0.12)/25 = 0.20$ . With it, $2\cdot10^{6}\cdot12.5\cdot0.20 = 5.00$ million m³, which is correct.

Answer: The simple averages underestimated the pore volume by 15%; weighting by thickness removed the error.

What it means. Maps and the cells of a geological model take the joint change of the parameters into account automatically. If we calculate from averages, the weights must match the meaning of the quantity: porosity by rock volume, and saturation by pore volume. The data in this example are assumed for illustration.

Connections to other topics

Gross rock volume from maps — how the gross rock volume is found from structure maps and contacts.
Porosity — what porosity is and how it is found.
Fluid saturation — saturation and its link with the height above the free water level.
Black oil model: solution gas–oil ratio and formation volume factors — the formation volume factor and the solution gas–oil ratio of oil.
Probabilistic volumetrics by Monte Carlo simulation — probabilistic estimation of hydrocarbons in place, and the meaning of P90, P50 and P10.
Classification of reserves and resources — the categories of reserves and resources.
General material balance equation for oil reservoirs — an independent check of the volume in place against the history of pressure and production.
Rock permeability — why the same volumes in place are produced at different rates.

Check yourself

1
All else being equal, the oil formation volume factor turned out to be 1.40 instead of the 1.20 that was assumed. How will the oil in place at standard conditions change?
1. A.It will grow by about 17%
2. B.It will fall by about 14%
3. C.It will not change: $B_o$ affects only production
4. D.It will fall by 40%
Show the answer
Correct answer: B. It will fall by about 14%
The volume in place is inversely proportional to $B_{oi}$ : $1.20/1.40 = 0.857$ , that is, 14% less. The same reservoir volume of oil with more dissolved gas gives less oil at the surface (and more gas).
2
Why are there about 7758 barrels in one acre-foot? Derive this number.
Show the answer
An acre is 43,560 square feet, so a layer 1 foot thick over an area of 1 acre is 43,560 ft³. A barrel is 42 US gallons, or 5.6146 ft³. $43{,}560/5.6146 \approx 7758$ .
3
The net-to-gross ratio was defined as the ratio of the reservoir thickness (net sand) to the gross thickness. Which porosity do we put into the formula for the volume in place?
1. A.The mean over the reservoir intervals, weighted by thickness
2. B.The mean over the whole section, including shales
3. C.The maximum from core
4. D.The mean over the reservoir intervals, weighted by water saturation
Show the answer
Correct answer: A. The mean over the reservoir intervals, weighted by thickness
The porosity must refer to the same volume whose share the NTG describes, that is, to the reservoir, and it must be averaged by thickness (rock volume). A porosity for the whole section would count the non-reservoir a second time.
4
A gas accumulation: GRV = 50 million m³, NTG = 0.9, $\phi = 0.20$ , $S_w = 0.20$ , $B_{gi} = 0.005$ m³/m³. Estimate the gas initially in place.
Show the answer
Gas-filled pore volume: $5\cdot10^{7}\cdot0.9\cdot0.20\cdot0.80 = 7.2\cdot10^{6}$ m³. Gas in place: $G = 7.2\cdot10^{6}/0.005 = 1.44\cdot10^{9}$ m³, that is, about 1.4 billion m³ at standard conditions.
5
A pessimistic case of the calculation was found by multiplying the minimum values of all the parameters. Is it right to call it a P90 estimate?
1. A.No: it is a P50 estimate
2. B.No: such a combination is far less likely than 10%
3. C.Yes, if there are no more than five parameters
4. D.Yes, by definition
Show the answer
Correct answer: B. No: such a combination is far less likely than 10%
If each minimum is a value such that the parameter will be worse with a probability of 10%, then for five independent parameters the probability that all of them are worse at once is $10^{-5}$ . We find the P90 of the volume in place with the Monte Carlo method.
6
Two accumulations have the same oil in place. Why can their recoverable volumes differ several times?
Show the answer
RF depends on the drive mechanism (solution gas, gas cap, water), the viscosity of the oil, the heterogeneity of the reservoir, the well pattern, the methods used (waterflooding, gas injection) and the economic limit of production. With solution gas drive, RF can be about 0.1, and with a good water drive, about 0.5.

Updated: 22 September 2026