Published on *PNG 520: Phase Behavior of Natural Gas and Condensate Fluids* (https://www.e-education.psu.edu/png520)

Due to the dramatically different conditions prevailing at the reservoir when compared to the conditions at the surface, we do not expect that 1 barrel of fluid at reservoir conditions could contain the same amount of matter as 1 barrel of fluid at surface conditions. Volumetric factors were introduced in petroleum and natural gas calculations in order to readily relate the *volume* of fluids that are obtained at the surface (stock tank) to the volume that the fluid actually occupied when it was compressed in the reservoir.

For example, the volume that a *live oil* occupies at the reservoir is *more* than the volume of oil that leaves the stock tank at the surface. This may be counter-intuitive. However, this is a result of the evolution of gas from oil as pressure decreases from reservoir pressure to surface pressure. If an oil had no gas in solution (i.e., a *dead oil*), the volume that it would occupy at reservoir conditions is less than the volume that it occupies at the surface. In this case, only liquid compressibility plays a role in the change of volume.

**The formation volume factor of a natural gas** (B_{g}) relates the volume of 1 lbmol of gas at reservoir conditions to the volume of the same lbmol of gas at standard conditions, as follows:

$${B}_{g}=\frac{\text{Volumeof1lbmolofgasatreservoirconditions,RCF}}{\text{Volumeof1lbmolgasatstandardconditions,SCF}}$$

Those volumes are, evidently, the specific molar volumes of the gas at the given conditions. The reciprocal of the specific molar volume is the molar density, and thus, Equation (18.5) could be written:

$${B}_{g}=\frac{{\tilde{v}}_{g}/{}_{res}}{{\tilde{v}}_{g}/{}_{sc}}=\frac{{\overline{\rho}}_{g}/{}_{sc}}{{\overline{\rho}}_{g}/{}_{res}}=\frac{({\rho}_{g}/M{W}_{g}){|}_{sc}}{({\rho}_{g}/M{W}_{g}){|}_{res}}$$

Introducing the definition for densities in terms of compressibility factor,

$${B}_{g}=\frac{\frac{{P}_{sc}}{R{T}_{sc}{Z}_{sc}}}{\frac{P}{RTZ}}$$

Therefore, recalling that ${Z}_{sc}\approx 1$ ,

$${B}_{g}=\frac{{P}_{sc}}{{T}_{sc}}\frac{ZT}{P}=0.005035{\frac{ZT}{P}}_{\left[RCF/SCF\right]}$$

Gas formation volume factors can be also expressed in terms of [RB/SCF]. In such a case, 1 RB = 5.615 RCF and we write:

$${B}_{g}=0.005035{\frac{ZT}{P}}_{\left[RCF/SCF\right]}$$

The formation volume factor of an **oil or condensate** (B_{o}) relates the volume of 1 lbmol of liquid at reservoir conditions to the volume of that liquid once it has gone through the surface separation facility.

The total volume occupied by 1 lbmol of liquid at reservoir conditions (V_{o})_{res} can be calculated through the compressibility factor of that liquid, as follows:

Upon separation, some gas is going to be taken out of the liquid stream feeding the surface facility. Let us call “n_{st}” the moles of liquid leaving the stock tank per mole of feed entering the separation facility. The volume that 1 lbmol of reservoir liquid is going to occupy after going through the separation facility is given by:

$${({V}_{o})}_{res}={\left(\frac{{n}_{st}{Z}_{o}RT}{P}\right)}_{sc}\text{wheren=1lbmol,}$$

Here we assume that the last stage of separation, the stock tank, operates at standard conditions. Introducing Equations (18.12) and (18.11) into (18.10), we end up with:

$${B}_{o}=\frac{{\left(\frac{n{Z}_{o}RT}{P}\right)}_{res}}{{\left(\frac{{n}_{st}{Z}_{0}RT}{P}\right)}_{sc}}$$

or,

$${B}_{o}=\frac{1}{{n}_{st}}\frac{{\left({Z}_{o}\right)}_{res}}{{\left({Z}_{o}\right)}_{sc}}\frac{T}{P}{\frac{{P}_{sc}}{{T}_{sc}}}_{\left[RB/STB\right]}$$

Please notice that (Z_{o})_{sc} — unlike Z_{sc} for a gas — is never equal to one. Oil formation volume factor can be also seen as the volume of reservoir fluid required to produce one barrel of oil in the stock tank.