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Equations: Phase equilibrium and PVT studies

The equations of the “Phase equilibrium and PVT studies” lessons in “Reservoir engineering”, with their conditions and units, and what each symbol means in them.

Equilibrium ratio
Ki=yixiK_i = \frac{y_i}{x_i}

dimensionless

Holds when: The gas and the liquid are in equilibrium at the same pressure and temperature. The name “constant” is conventional: KiK_i changes with pressure and temperature, and at high pressures also with the compositions of the phases.

Vapour–liquid equilibrium →
Equilibrium ratio of an ideal mixture
Ki=ps,i(T)pK_i = \frac{p_{s,i}(T)}{p}

absolute pressures, in the same units

Holds when: An ideal solution (Raoult's law) and an ideal gas (Dalton's law): low pressures, and molecules close in size and structure. The temperature is below the critical temperature of the component (the highest temperature at which it can be a liquid). Above it there is no vapour pressure.

Vapour–liquid equilibrium →
Initial estimate of equilibrium ratios by Wilson's formula
Ki=pc,ip exp⁡[5.373 (1+ωi)(1−Tc,iT)]K_i = \frac{p_{c,i}}{p}\,\exp\left[5.373\,(1 + \omega_i)\left(1 - \frac{T_{c,i}}{T}\right)\right]

pressures in the same units, temperatures in kelvins

Holds when: This is the ideal ratio ps,i/pp_{s,i}/p, with the vapour pressure found from two reference points. The compositions of the phases are not taken into account. For components above their critical temperature (methane and nitrogen at reservoir temperature), the “vapour pressure” is only the straight line continued beyond the critical point. It has no physical meaning. It works as a first guess for iterations, and as a rough estimate at low pressures. The more accurate empirical correlations are in the lesson Equilibrium ratios: correlations and equations of state.

Vapour–liquid equilibrium →
Phase compositions in a flash calculation
xi=zi1+V(Ki−1),yi=Ki xix_i = \frac{z_i}{1 + V(K_i - 1)},\qquad y_i = K_i\,x_i

dimensionless

Holds when: The phases are in equilibrium and the KiK_i are known. There are two hydrocarbon phases; formation water does not take part in the hydrocarbon equilibrium and is accounted for separately. The balances are written per 1 mole of the mixture; the mole fraction of the liquid phase is L=1−VL = 1 - V.

Vapour–liquid equilibrium →
Rachford–Rice equation
g(V)=∑izi (Ki−1)1+V(Ki−1)=0g(V) = \sum_i \frac{z_i\,(K_i - 1)}{1 + V(K_i - 1)} = 0

dimensionless

Holds when: The equilibrium ratios at the given pp and TT are known and do not depend on VV (in an equation-of-state calculation they are refined in an outer loop).

Vapour–liquid equilibrium →
Conditions for the bubble point pressure and the dew point
∑iziKi(pb,T)=1,∑iziKi(pd,T)=1\sum_i z_iK_i(p_b, T) = 1,\qquad \sum_i \frac{z_i}{K_i(p_d, T)} = 1

absolute pressures

Holds when: The KiK_i are calculated for the pair “mixture and new phase”: at the bubble point pressure, for a liquid of composition ziz_i and a bubble of composition KiziK_iz_i. At one temperature a gas condensate can have two dew points: an upper one (retrograde) and a lower one.

Vapour–liquid equilibrium →
Stability test: iterating the trial phase
ln⁡Yi=ln⁡zi+ln⁡φi(z)−ln⁡φi(y),yi=Yi∑jYj\ln Y_i = \ln z_i + \ln\varphi_i(\mathbf{z}) - \ln\varphi_i(\mathbf{y}),\qquad y_i = \frac{Y_i}{\sum_j Y_j}

dimensionless

Holds when: The iterations are repeated until YiY_i stops changing. Both trials are checked: the gas-like one and the liquid-like one. If the trial converges to the composition of the mixture itself (yi=ziy_i = z_i, the trivial solution), it tells us nothing. The mixture is unstable (two-phase) if ∑Yi>1\sum Y_i > 1 for at least one non-trivial trial.

Vapour–liquid equilibrium →
Convergence criterion of the equilibrium calculation
ε=∑i(1−fi Lfi V)2<ε0\varepsilon = \sum_i\left(1 - \frac{f_i^{\,L}}{f_i^{\,V}}\right)^2 < \varepsilon_0

dimensionless

Holds when: The fugacities are calculated with one equation of state for both phases at the same pp and TT.

Vapour–liquid equilibrium →
Derivative of the Rachford–Rice function and the Newton step
g′(V)=−∑izi (Ki−1)2[1+V(Ki−1)]2<0,Vn+1=Vn−g(Vn)g′(Vn)g'(V) = -\sum_i \frac{z_i\,(K_i - 1)^2}{[1 + V(K_i - 1)]^2} < 0,\qquad V_{n+1} = V_n - \frac{g(V_n)}{g'(V_n)}

dimensionless

Holds when: At least one equilibrium ratio is greater than one and at least one is less than one. Otherwise there are no poles on both sides, and the mixture is single-phase for any VV. If a Newton step goes outside the current interval in which gg changes sign, take a bisection step instead.

Vapour–liquid equilibrium →
Tangent plane distance in the form for a trial phase
tm(Y)=1+∑iYi[ln⁡Yi+ln⁡φi(y)−ln⁡zi−ln⁡φi(z)−1]tm(\mathbf{Y}) = 1 + \sum_i Y_i\left[\ln Y_i + \ln\varphi_i(\mathbf{y}) - \ln z_i - \ln\varphi_i(\mathbf{z}) - 1\right]

dimensionless

Holds when: The fugacity coefficients come from the equation of state. For the original mixture, take the root with the lower Gibbs energy.

Vapour–liquid equilibrium →
Moles of gas and stock-tank oil in multistage separation
ng,j=Vj∏k=1j−1(1−Vk),nod=∏k=1N(1−Vk)n_{g,j} = V_j\prod_{k=1}^{j-1}(1 - V_k),\qquad n_{\text{od}} = \prod_{k=1}^{N}(1 - V_k)

kmol per 1 kmol of reservoir oil

Holds when: At each stage the gas and the liquid are in equilibrium and all the gas is removed. The feed of the first stage is single-phase reservoir oil (for BobB_{ob}, at the bubble point pressure); water is not included in the calculation. The ratios Ki,jK_{i,j} of each stage are calculated for its pressure, temperature and feed composition (here from an equation of state, Equilibrium ratios: correlations and equations of state).

Separator calculations and optimum separator pressure →
Stage gas–oil ratio and solution gas–oil ratio from the separator calculation
Rj=ng,j Vm,scVod,Vod=nod Modρod,Rsb=∑j=1NRjR_j = \frac{n_{g,j}\,V_{m,\text{sc}}}{V_{\text{od}}},\qquad V_{\text{od}} = \frac{n_{\text{od}}\,M_{\text{od}}}{\rho_{\text{od}}},\qquad R_{sb} = \sum_{j=1}^{N} R_j

m³/m³: standard cubic metres of gas per cubic metre of stock-tank oil

Holds when: We treat gas at standard conditions as ideal. Its ZZ (the compressibility factor, which is 1 for an ideal gas) is 0.99–1.00, and about 0.99 for the heavy stock-tank gas. The assumption overestimates the solution gas–oil ratio by less than one per cent: by 0.5% in the lesson example. The density of the stock-tank oil is known. It is measured, or calculated from an equation of state with volume translation (a correction that improves the density of liquids), or calculated from the densities of the components, assuming that volumes add up. The stock-tank gas is included in the sum: it too was dissolved in the reservoir oil.

Separator calculations and optimum separator pressure →
Oil formation volume factor at the bubble point pressure from the separator calculation
Bob=Mres/ρobVodB_{ob} = \frac{M_{\text{res}}/\rho_{ob}}{V_{\text{od}}}

m³/m³ (reservoir cubic metres per cubic metre of stock-tank oil)

Holds when: The feed is single-phase oil at the bubble point pressure; at a higher pressure the numerator is its molar volume at that pressure. The density of the reservoir oil is measured or calculated from an equation of state with volume translation; such an estimate of liquid densities is off by a few per cent, and BobB_{ob} by as much.

Separator calculations and optimum separator pressure →
Stage pressures with equal ratios (first estimate)
r=(p1pN)1/(N−1),pj+1=pjrr = \left(\frac{p_1}{p_N}\right)^{1/(N-1)},\qquad p_{j+1} = \frac{p_j}{r}

absolute pressures in the same units

Holds when: It is an empirical rule for the first choice, not a result of theory. The separator calculation and the economics refine the final pressures. The temperatures of the stages are close. If the stages differ greatly in temperature, the rule is less reliable.

Separator calculations and optimum separator pressure →
Hoffman characterisation factor
Fi=bi(1Tb,i−1T),bi=log⁡10(pc,i/psc)1/Tb,i−1/Tc,iF_i = b_i\left(\frac{1}{T_{b,i}} - \frac{1}{T}\right),\qquad b_i = \frac{\log_{10}\left(p_{c,i}/p_{\text{sc}}\right)}{1/T_{b,i} - 1/T_{c,i}}

absolute temperatures in the same units (in Standing's correlation, °R); pressures in the same units

Holds when: A vapour pressure line through two points is an approximation: the real curve is slightly bent in these coordinates. For components above their critical temperature (nitrogen and methane at separator temperature; CO₂ and ethane too, above 31–32 °C), FiF_i is only the line continued beyond the critical point.

Equilibrium ratios: correlations and equations of state →
Standing's correlation for equilibrium ratios at low pressures
log⁡10(Kip)=A0+A1Fi,A0=1.2+4.5⋅10−4 p+15⋅10−8 p2,A1=0.89−1.7⋅10−4 p−3.5⋅10−8 p2\log_{10}(K_ip) = A_0 + A_1F_i,\qquad A_0 = 1.2 + 4.5\cdot10^{-4}\,p + 15\cdot10^{-8}\,p^2,\qquad A_1 = 0.89 - 1.7\cdot10^{-4}\,p - 3.5\cdot10^{-8}\,p^2

oilfield units only: pp in psia, temperatures in °R (p [psia]=145.04 p [MPa]p\,[\text{psia}] = 145.04\,p\,[\text{MPa}], T [∘R]=1.8 T [K]T\,[{}^{\circ}\text{R}] = 1.8\,T\,[\text{K}])

Holds when: The pressure is up to 6.9 MPa (1000 psia) and the temperature up to 93 °C (200 °F): the range of the data to which the correlation was fitted. For nitrogen, CO₂, hydrogen sulfide, methane and ethane, bb and TbT_b were fitted to measurements and differ from those calculated by Hoffman's formula. The composition of the mixture does not enter the formula. The correlation describes an “average” oil, and for unusual mixtures the error grows.

Equilibrium ratios: correlations and equations of state →
The C7+ fraction in Standing's correlation
n=7.30+0.0075 TF+0.0016 p,b=1013+324n−4.256n2,Tb=301+59.85n−0.971n2n = 7.30 + 0.0075\,T_F + 0.0016\,p,\qquad b = 1013 + 324n - 4.256n^2,\qquad T_b = 301 + 59.85n - 0.971n^2

TFT_F in °F (TF=1.8 t+32T_F = 1.8\,t + 32, tt in °C), pp in psia, bb and TbT_b in °R

Holds when: The C7+ fraction is represented by an imaginary normal alkane (a straight-chain hydrocarbon) with nn carbon atoms. The molar mass and density of the fraction do not enter the formula. The estimate is rough: in the lesson example the ratio of C7+ at atmospheric pressure comes out several times greater than from the equation of state (see the table below).

Equilibrium ratios: correlations and equations of state →
Linear estimate of the convergence pressure of an oil from the molar mass of the C7+ fraction
pk≈60 M7+−4200p_k \approx 60\,M_{7+} - 4200

pkp_k in psia; pk [MPa]=pk [psia]/145.04p_k\,[\text{MPa}] = p_k\,[\text{psia}]/145.04

Holds when: An empirical estimate for oils that gives only the order of magnitude; temperature does not enter it.

Equilibrium ratios: correlations and equations of state →
Equilibrium ratios with the convergence pressure (Whitson–Torp formula)
Ki=(pc,ipk)A−1pc,ipexp⁡[5.373 A (1+ωi)(1−Tc,iT)],A=1−(p−pscpk−psc)0.6K_i = \left(\frac{p_{c,i}}{p_k}\right)^{A-1}\frac{p_{c,i}}{p}\exp\left[5.373\,A\,(1 + \omega_i)\left(1 - \frac{T_{c,i}}{T}\right)\right],\qquad A = 1 - \left(\frac{p - p_{\text{sc}}}{p_k - p_{\text{sc}}}\right)^{0.6}

absolute pressures, in the same units (their differences are gauge pressures); temperatures in kelvins

Holds when: At p=pscp = p_{\text{sc}} the formula coincides with Wilson's formula, and at p=pkp = p_k it gives Ki=1K_i = 1. In between it is an interpolation, not a physical model. Composition enters only through pkp_k. The exponent 0.6 is empirical. In the original, as with Wilson, the factor is 5.37. Here it is 5.373, as in the lesson on phase equilibrium.

Equilibrium ratios: correlations and equations of state →
Acentric factor from the boiling point (Edmister's formula)
ω=37 log⁡10(pc/psc)Tc/Tb−1−1\omega = \frac{3}{7}\,\frac{\log_{10}(p_c/p_{\text{sc}})}{T_c/T_b - 1} - 1

pressures in the same units, absolute temperatures

Holds when: The vapour pressure follows a straight line in the coordinates log⁡10ps\log_{10} p_s versus 1/T1/T between TbT_b and TcT_c.

Equilibrium ratios: correlations and equations of state →
Greatest conditioning rate before sampling
qmax⁡=J (pˉ−psat−δ)q_{\max} = J\,(\bar p - p_{\text{sat}} - \delta)

metric oilfield units

Holds when: The inflow is single-phase and proportional to the drawdown: q=J(pˉ−pwf)q = J(\bar p - p_{wf}) above the bubble point pressure (Well inflow: the well performance diagram). JJ was measured on this well after cleaning. While the skin factor is changing, JJ changes with it. (The skin factor is a number that shows how much the zone around the well helps or hinders flow.)

Reservoir fluid sampling →
Pressure change of single-phase liquid in a rigid chamber
(∂p∂T)V=βoco\left(\frac{\partial p}{\partial T}\right)_V = \frac{\beta_o}{c_o}

MPa and °C

Holds when: The chamber is rigid. The thermal expansion of the steel, a few per cent of that of the oil, slightly softens the pressure drop. The oil is single-phase. When gas comes out, the pressure falls more slowly, because the gas phase supports it.

Reservoir fluid sampling →
Contamination from optical density and extrapolation to clean fluid
η=1−ODOD∞,OD(V)=OD∞−a V−5/12\eta = 1 - \frac{\mathrm{OD}}{\mathrm{OD}_\infty},\qquad \mathrm{OD}(V) = \mathrm{OD}_\infty - a\,V^{-5/12}

optical density is dimensionless; volume in any units

Holds when: The filtrate is transparent in the chosen channel; if not, η=(OD∞−OD)/(OD∞−ODf)\eta = (\mathrm{OD}_\infty - \mathrm{OD})/(\mathrm{OD}_\infty - \mathrm{OD}_{\text{f}}), where ODf\mathrm{OD}_{\text{f}} is the optical density of the filtrate. Mixing is linear by volume. The exponent 5/12 is empirical and depends on the geometry of the inflow and on how uneven the reservoir is. At the start of pumping the law does not hold. For gas condensates and light oils the colour contrast is small, so the gas–oil ratio, density and methane bands are used instead.

Reservoir fluid sampling →
Contamination as a fraction of stock-tank oil
xsc=ηη+(1−η)/Box_{\text{sc}} = \frac{\eta}{\eta + (1 - \eta)/B_o}

dimensionless

Holds when: The volume of the filtrate hardly changes on going to standard conditions, and there is no gas in it.

Reservoir fluid sampling →
Correction of the gas–oil ratio for gas density and gas deviation factor
Rsep=Rmeas γset Zsetγg ZR_{\text{sep}} = R_{\text{meas}}\,\sqrt{\frac{\gamma_{\text{set}}\,Z_{\text{set}}}{\gamma_g\,Z}}

gas at standard conditions (here 0.101325 MPa and 20 °C)

Holds when: The gas is measured by an orifice (differential-pressure) flow meter, whose reading is inversely proportional to γgZ\sqrt{\gamma_g Z}. For other types of meter the correction is different. Both volumes are referred to the same standard conditions.

Reservoir fluid sampling →
Recombination proportion and total gas–oil ratio
Rrec=RsepBsep,R=Rsep+RtankR_{\text{rec}} = \frac{R_{\text{sep}}}{B_{\text{sep}}},\qquad R = R_{\text{sep}} + R_{\text{tank}}

gas at standard conditions

Holds when: The stock-tank gas is already dissolved in the separator oil and is not added separately in recombination. The gas and oil samples were taken at the same time, in a steady flow, at a bottomhole pressure above the bubble point pressure.

Reservoir fluid sampling →
Pressure of a sample with a nitrogen cushion after cooling
p1=p0 Vg0Vg1 T1T0,Vg1=Vg0+Vo[βo (T0−T1)−co (p0−p1)]p_1 = p_0\,\frac{V_{\text{g}0}}{V_{\text{g}1}}\,\frac{T_1}{T_0},\qquad V_{\text{g}1} = V_{\text{g}0} + V_{\text{o}}\left[\beta_o\,(T_0 - T_1) - c_o\,(p_0 - p_1)\right]

MPa, cm³, K (the temperature difference in °C or K)

Holds when: Nitrogen is an ideal gas, the piston is frictionless, and the chamber is rigid. The changes in the volume of the oil are small and linear in temperature and pressure. The equations are solved for p1p_1 by trial and error.

Reservoir fluid sampling →
Oil compressibility above the bubble point pressure from two readings
co=−1V(∂V∂p)T≈V2−V1Vˉ (p1−p2)c_o = -\frac{1}{V}\left(\frac{\partial V}{\partial p}\right)_T \approx \frac{V_2 - V_1}{\bar V\,(p_1 - p_2)}

MPa and 1/MPa; the volumes can be in any units, as long as both are in the same unit (relative volumes work too)

Holds when: The temperature is constant (reservoir temperature) and the oil is single-phase. Between the two readings the compressibility changes little. Reports sometimes divide the change in volume not by the mean volume but by the volume at the higher pressure. The difference is about 1%, but you need to know it when you compare numbers.

Laboratory PVT tests →
Oil formation volume factor above the bubble point pressure
Bo(p)=Bob V(p)Vb,p≥pbB_o(p) = B_{ob}\,\frac{V(p)}{V_b},\qquad p \ge p_b

m³ of reservoir oil per m³ of stock-tank oil

Holds when: Above the bubble point pressure the composition of the oil does not change, so the ratio of the volumes at pp and at pbp_b is the same in the cell and in the reservoir. BobB_{ob} must be the value for the separation scheme to which the result should refer.

Laboratory PVT tests →
Y-function of two-phase readings
Y=pb−pp (V/Vb−1)Y = \frac{p_b - p}{p\,\left(V/V_b - 1\right)}

absolute pressures in any identical units; YY is dimensionless

Holds when: Only for readings below the bubble point pressure. For black oils YY is almost linear in pressure. This is an empirical observation, not a law. For volatile oils, and close to the critical point, the line is curved.

Laboratory PVT tests →
Solution gas–oil ratio and oil volume in differential liberation
RsD(pk)=1Vr∑j>kVg,j,BoD(pk)=Vo(pk)VrR_{sD}(p_k) = \frac{1}{V_{\text{r}}}\sum_{j > k} V_{g,j},\qquad B_{oD}(p_k) = \frac{V_o(p_k)}{V_{\text{r}}}

gas at 0.101325 MPa and 20 °C; volume ratios in m³/m³

Holds when: The gas is let out completely and at the constant pressure of the step. At the bubble point pressure RsD=RsDbR_{sD} = R_{sDb}: all the gas that came out during the test. At the last step RsD=0R_{sD} = 0 by definition. The results refer to the residual oil, not to the stock-tank oil of separation.

Laboratory PVT tests →
Formation volume factor and gas deviation factor of the gas let out at a step
Bg=Vg(p,T)Vg,sc,Z=p Vg(p,T) Tscpsc Vg,sc TB_g = \frac{V_g(p, T)}{V_{g,\text{sc}}},\qquad Z = \frac{p\,V_g(p, T)\,T_{\text{sc}}}{p_{\text{sc}}\,V_{g,\text{sc}}\,T}

absolute pressures and temperatures; volumes in identical units

Holds when: At standard conditions we take the gas as ideal (Zsc=1Z_{\text{sc}} = 1). Only gas leaves the cell, with no drops of oil.

Laboratory PVT tests →
Solution gas–oil ratio and formation volume factor from the separator test
Rsb=∑iVg,iVod,Bob=VbVodR_{sb} = \frac{\sum_i V_{g,i}}{V_{\text{od}}},\qquad B_{ob} = \frac{V_b}{V_{\text{od}}}

gas at 0.101325 MPa and 20 °C; m³/m³

Holds when: While it is pushed out, the oil in the cell stays single-phase: the pressure is kept at or above the bubble point pressure. The pressures and temperatures of the stages are the same as in the field. Otherwise the result refers to another scheme.

Laboratory PVT tests →
Conversion of differential data to separator conditions
Bo=BoD BobBoDb,Rs=Rsb−(RsDb−RsD)BobBoDb,p<pbB_o = B_{oD}\,\frac{B_{ob}}{B_{oDb}},\qquad R_s = R_{sb} - \left(R_{sDb} - R_{sD}\right)\frac{B_{ob}}{B_{oDb}},\qquad p < p_b

all quantities in m³/m³; in oilfield units (rb/STB, scf/STB) the formulas are the same

Holds when: We take the ratio Bob/BoDbB_{ob}/B_{oDb} (the cubic metres of residual oil per cubic metre of stock-tank oil obtained from the same reservoir oil) to be the same at all pressures. This is an approximation, and close to atmospheric pressure it breaks down (advanced section). We convert the gas that came out in the reservoir (RsDb−RsDR_{sDb} - R_{sD} per 1 m³ of residual oil) to stock-tank oil with the same factor. Above the bubble point pressure Rs=RsbR_s = R_{sb}, and Bo=Bob V/VbB_o = B_{ob}\,V/V_b from the constant-composition test.

Laboratory PVT tests →
Laminar flow in a capillary (Hagen–Poiseuille equation)
μ=π d4 Δp128 L q\mu = \frac{\pi\,d^4\,\Delta p}{128\,L\,q}

SI

Holds when: The flow is laminar (smooth, in layers) and steady: the Reynolds number Re=4ρq/(πdμ)\mathrm{Re} = 4\rho q/(\pi d\mu), which shows whether a flow is smooth or turbulent, is much less than 2000. The liquid is single-phase and Newtonian (its viscosity does not depend on the rate of shear). The effects of the inlet and the outlet of the capillary are small or are included in the calibration. The pressure in the capillary is above the bubble point pressure of the liquid, otherwise gas comes out.

Laboratory PVT tests →
Two-phase gas deviation factor in a constant volume test
Z2ph=p Vd(n0−np) R T,n0=pd VdZd R TZ_{2\text{ph}} = \frac{p\,V_d}{(n_0 - n_p)\,R\,T},\qquad n_0 = \frac{p_d\,V_d}{Z_d\,R\,T}

MPa·cm³ = J, so pressure in MPa and volume in cm³ give the amount of substance in moles

Holds when: After the gas is let out, the volume of the cell is again equal to VdV_d, like the pore volume of a reservoir without water influx. n0−npn_0 - n_p includes both the gas and the liquid that remain in the cell.

Laboratory PVT tests →
Mass balance of differential liberation
ρob BoDb Vr=ρr Vr+ρair∑jVg,j γj\rho_{ob}\,B_{oDb}\,V_{\text{r}} = \rho_{\text{r}}\,V_{\text{r}} + \rho_{\text{air}}\sum_{j} V_{g,j}\,\gamma_j

grams on both sides: g/cm³·cm³ and g/L·L

Holds when: Only gas leaves the cell, and all the gas of each stage is collected and measured. The density of the gas at standard conditions is ρair γj\rho_{\text{air}}\,\gamma_j, which means that we take the gas as ideal at 0.1 MPa. For the heavy gas of the last stages this slightly underestimates the mass.

How to read a PVT report →
The condition of positive apparent compressibility between the rows of the table
coapp≈Bˉg (Rs,i−Rs,i+1)−(Bo,i−Bo,i+1)Bˉo (pi−pi+1)>0c_o^{\text{app}} \approx \frac{\bar B_g\,(R_{s,i} - R_{s,i+1}) - (B_{o,i} - B_{o,i+1})}{\bar B_o\,(p_i - p_{i+1})} > 0

pressures in MPa, formation volume factors and solution gas–oil ratio in m³/m³

Holds when: This is the apparent compressibility −1Bo∂Bo∂p+BgBo∂Rs∂p-\frac{1}{B_o}\frac{\partial B_o}{\partial p} + \frac{B_g}{B_o}\frac{\partial R_s}{\partial p} written for a finite interval: the derivatives are replaced by differences between two rows (apparent compressibility). We take RsR_s, BoB_o and BgB_g from one table, on one base.

How to read a PVT report →
Ratio of relative permeabilities in the zone near the well
krgkro=μgμo(1Vro−1)\frac{k_{rg}}{k_{ro}} = \frac{\mu_g}{\mu_o}\left(\frac{1}{V_{ro}} - 1\right)

dimensionless ratios; viscosities in the same units

Holds when: Steady flow of a mixture of constant composition: as much liquid flows out into the well as drops out. Capillary and gravity forces are neglected. Gas and condensate are in equilibrium.

Retrograde condensation, wax, asphaltenes and high-pressure fluids →
Share of the rate and radius of the condensate zone (two-zone estimate)
qq0=(pe−pd)+m (pd−pwf)pe−pwf,ln⁡rbrw=m (pd−pwf) ln⁡(re/rw)(pe−pd)+m (pd−pwf)\frac{q}{q_0} = \frac{(p_e - p_d) + m\,(p_d - p_{wf})}{p_e - p_{wf}},\qquad \ln\frac{r_b}{r_w} = \frac{m\,(p_d - p_{wf})\,\ln(r_e/r_w)}{(p_e - p_d) + m\,(p_d - p_{wf})}

pressures in any common units; radii in common units

Holds when: Steady-state radial inflow, with the gas flow proportional to the pressure difference (high pressures). In a wet (rich) gas, p/(μZ)p/(\mu Z) rises slightly towards low pressures. So the drawdown below the dew point weighs a little more in the real inflow than in the estimate, and for this reason the estimate is slightly optimistic. The mobility in the zone is constant, which makes the estimate conservative, because the saturation is smaller at the boundary of the zone. The gas viscosity is taken to be the same in the zone and outside it. But the gas in the zone is leaner and less viscous, so this assumption too slightly overstates the losses. Inertial losses and the improvement of gas permeability at high velocities near the wellbore are not taken into account.

Retrograde condensation, wax, asphaltenes and high-pressure fluids →
Radius of the zone cooled by water injection
rT=Vinj (ρc)wπ h (ρc)rock,(ρc)rock=(1−ϕ)(ρc)ma+ϕ [Sw(ρc)w+So(ρc)o]r_T = \sqrt{\frac{V_{\text{inj}}\,(\rho c)_w}{\pi\,h\,(\rho c)_{\text{rock}}}},\qquad (\rho c)_{\text{rock}} = (1 - \phi)(\rho c)_{\text{ma}} + \phi\,\big[S_w(\rho c)_w + S_o(\rho c)_o\big]

SI; heat capacities in the same units

Holds when: The injected water gives up all its “cold” to the rock: behind the front the temperature equals the injection temperature, and ahead of it the reservoir temperature. Heat exchange with the cap rock and the base rock (the rocks above and below the reservoir) is not taken into account. For thin reservoirs and long injection the radius comes out too large. The flow is radial and the reservoir is homogeneous.

Retrograde condensation, wax, asphaltenes and high-pressure fluids →
Colloidal instability index
CII=wsat+wasphwarom+wresin\text{CII} = \frac{w_{\text{sat}} + w_{\text{asph}}}{w_{\text{arom}} + w_{\text{resin}}}

dimensionless; fractions in the same units

Holds when: It is an empirical screening of stock-tank oil: pressure and dissolved gas are not taken into account. The accepted limits: below 0.7 the asphaltenes are stable, 0.7–0.9 is uncertain, above 0.9 they are unstable.

Retrograde condensation, wax, asphaltenes and high-pressure fluids →

Symbols

SymbolMeaning and context
pˉ\bar p
Vˉ\bar V
Δp\Delta p
(ρc)rock(\rho c)_{\text{rock}}
(ρc)w, (ρc)o, (ρc)ma(\rho c)_w,\ (\rho c)_o,\ (\rho c)_{\text{ma}}
(∂p/∂T)V(\partial p/\partial T)_V
aa
AA
A0, A1A_0,\ A_1
bib_i
BoB_o
Bo, RsB_o,\ R_s
  • formation volume factor and solution gas–oil ratio per 1 m³ of stock-tank oil of the chosen separation scheme, m³/m³Laboratory PVT tests
Bo(p)B_o(p)
BobB_{ob}
  • formation volume factor at the bubble point pressure from the separator test (section below), m³/m³Laboratory PVT tests
Bob, RsbB_{ob},\ R_{sb}
BoD, RsDB_{oD},\ R_{sD}
  • the same quantities from differential liberation at pressure pp, per 1 m³ of residual oil, m³/m³Laboratory PVT tests
BoD(pk)B_{oD}(p_k)
BoDb, RsDbB_{oDb},\ R_{sDb}
BoDb VrB_{oDb}\,V_{\text{r}}
BsepB_{\text{sep}}
b, Tbb,\ T_b
coc_o
coappc_o^{\text{app}}
dd
FiF_i
fi L, fi Vf_i^{\,L},\ f_i^{\,V}
Bˉg\bar B_g
g′(V)g'(V)
g(V)g(V)
  • Rachford–Rice function: the difference between the sums of the mole fractions in the gas and in the liquid, —Vapour–liquid equilibrium
h, ϕ, Sw, Soh,\ \phi,\ S_w,\ S_o
JJ
KiK_i
Ki(p,T)K_i(p, T)
krg, krok_{rg},\ k_{ro}
LL
mm
M7+M_{7+}
Mod, ρodM_{\text{od}},\ \rho_{\text{od}}
  • molar mass of the stock-tank oil ∑xi,odMi\sum x_{i,\text{od}}M_i and its density at standard conditions (in the lesson on the black oil model, ρo,sc\rho_{o,\text{sc}}), kg/kmol, kg/m³Separator calculations and optimum separator pressure
MresM_{\text{res}}
nn
NN
n0, npn_0,\ n_p
  • amount of substance in the cell at the dew point, and the accumulated amount of gas let out, molLaboratory PVT tests
ng,jn_{g,j}
nodn_{\text{od}}
Bˉo\bar B_o
OD\mathrm{OD}
OD∞\mathrm{OD}_\infty
pp
p0, p1p_0,\ p_1
  • pressure of the sample after pressurisation by nitrogen at reservoir temperature and after cooling, MPaReservoir fluid sampling
p1p_1
p1, p2p_1,\ p_2
pbp_b
pc, Tcp_c,\ T_c
pc,i, Tc,ip_{c,i},\ T_{c,i}
pc,i, Tc,i, ωip_{c,i},\ T_{c,i},\ \omega_i
pdp_d
pd, Zdp_d,\ Z_d
pe, pd, pwfp_e,\ p_d,\ p_{wf}
pi>pi+1p_i > p_{i+1}
pkp_k
pNp_N
ps,i(T)p_{s,i}(T)
  • vapour pressure of the pure component (the pressure at which it boils) at the temperature of the mixture, MPaVapour–liquid equilibrium
psatp_{\text{sat}}
pscp_{\text{sc}}
psc, Tscp_{\text{sc}},\ T_{\text{sc}}
p, Tp,\ T
qq
qmax⁡q_{\max}
  • greatest rate at which the bottomhole pressure is not below psat+δp_{\text{sat}} + \delta, m³/d (oil at standard conditions)Reservoir fluid sampling
q, q0q,\ q_0
rr
RR
rb, re, rwr_b,\ r_e,\ r_w
RjR_j
RmeasR_{\text{meas}}
  • separator gas–oil ratio from the field measurement: separator gas per 1 m³ of stock-tank oil, m³/m³Reservoir fluid sampling
RrecR_{\text{rec}}
  • separator gas (at standard conditions) per 1 m³ of separator oil at separator pressure and temperature (the recombination proportion), m³/m³Reservoir fluid sampling
Rs,i, Bo,iR_{s,i},\ B_{o,i}
RsbR_{sb}
Rsb, BobR_{sb},\ B_{ob}
  • solution gas–oil ratio and formation volume factor of the oil at the bubble point pressure for this separation scheme, m³/m³Laboratory PVT tests
RsD(pk)R_{sD}(p_k)
  • gas that is still dissolved in the oil after step kk, at standard conditions, per 1 m³ of residual oil, m³/m³Laboratory PVT tests
RsepR_{\text{sep}}
rTr_T
RtankR_{\text{tank}}
TT
T0, T1T_0,\ T_1
TbT_b
Tb,iT_{b,i}
Tc,i, pc,iT_{c,i},\ p_{c,i}
TFT_F
tmtm
VV
V1, V2V_1,\ V_2
  • volumes of the sample at pressures p1>p2p_1 > p_2 (both above the bubble point pressure), cm³Laboratory PVT tests
VbV_b
  • oil pushed out of the cell, at the bubble point pressure and reservoir temperature, cm³Laboratory PVT tests
VdV_d
Vg0, Vg1V_{\text{g}0},\ V_{\text{g}1}
Vg,scV_{g,\text{sc}}
Vg,iV_{g,i}
Vg,jV_{g,j}
Vg,j, γjV_{g,j},\ \gamma_j
  • volume of the gas of stage jj at standard conditions and its specific gravity relative to air, L, —How to read a PVT report
Vg(p,T)V_g(p, T)
  • volume of the gas let out at the step, in the cell, at the pressure of the step and at reservoir temperature, cm³Laboratory PVT tests
VinjV_{\text{inj}}
VjV_j
Vm,scV_{m,\text{sc}}
VnV_n
VoV_{\text{o}}
Vo(pk)V_o(p_k)
  • volume of the oil in the cell after the gas is let out at step kk, at the pressure of the step and at reservoir temperature, cm³Laboratory PVT tests
VodV_{\text{od}}
VrV_{\text{r}}
VroV_{ro}
V(p)/VbV(p)/V_b
V/VbV/V_b
wresin, wasphw_{\text{resin}},\ w_{\text{asph}}
wsat, waromw_{\text{sat}},\ w_{\text{arom}}
xi, yix_i,\ y_i
xscx_{\text{sc}}
  • fraction of filtrate in the stock-tank liquid of the sample by volume at standard conditions, fractionReservoir fluid sampling
YY
yiy_i
YiY_i
  • “amounts” of the components of the trial phase, not yet scaled to add up to one; the initial values are ziKiz_iK_i (gas-like trial) or zi/Kiz_i/K_i (liquid-like) with KiK_i from Wilson's formula, —Vapour–liquid equilibrium
yi, xiy_i,\ x_i
Yi, yiY_i,\ y_i
Z2phZ_{2\text{ph}}
ziz_i
zi, Kiz_i,\ K_i
βo\beta_o
  • thermal expansion coefficient of reservoir oil; usually (0.7–1.2)⋅10−3(0.7\text{–}1.2)\cdot 10^{-3}, 1/°CReservoir fluid sampling
βo, co\beta_o,\ c_o
γg, Z\gamma_g,\ Z
γset, Zset\gamma_{\text{set}},\ Z_{\text{set}}
δ\delta
ε0\varepsilon_0
η\eta
μ\mu
μg, μo\mu_g,\ \mu_o
ρair\rho_{\text{air}}
ρob\rho_{ob}
ρr, Vr\rho_{\text{r}},\ V_{\text{r}}
φi(z), φi(y)\varphi_i(\mathbf{z}),\ \varphi_i(\mathbf{y})
  • fugacity coefficients of the component in the original mixture and in the trial phase (from the equation of state), —Vapour–liquid equilibrium
ω\omega
ωi\omega_i