Reservoir Fluids and Drive Mechanisms

Petroleum reservoir fluid properties

Reservoir fluids are described by properties that depend on pressure, temperature and composition, and several of them are tested directly in GATE. Combustion is a rapid exothermic oxidation reaction. The flame reactions of liquid and solid fuels take place in the gas phase, because these fuels first release a vapour at the surface that then burns. Air is not an absolute requirement, since pure oxygen, oxygen-enriched gases, chlorine and nitrous oxide can also sustain combustion.

Two temperatures are often confused. The flash point is the lowest temperature at which a liquid releases enough vapour to flash momentarily when ignited. The fire point is the temperature at which the vapour sustains burning. A fire and an explosion differ in the rate of energy release: a fire releases energy comparatively slowly, whereas an explosion releases it very rapidly.

The liquid shrinkage curve plots the percentage of liquid volume against pressure as gas is liberated below the bubble point. A low-shrinkage oil keeps nearly 100 % of its liquid volume, an ordinary black oil shows a moderate decline, and a high-shrinkage oil shows a steep rise in shrinkage just below the bubble point. A near-critical crude oil has the steepest and lowest curve.

Hydrocarbon fluids are classified by where the reservoir and separator points lie relative to the two-phase envelope and the critical point. A black oil lies inside the envelope to the left of the critical point. A gas condensate lies inside the envelope to the right of the critical point, and a wet gas lies outside the envelope while its separator point falls inside it. A dry gas has both the reservoir and separator points outside the envelope.

API gravity is defined as ^\circ\mathrm{API} = \dfrac{141.5}{\mathrm{SG}} - 131.5, where SG is the specific gravity of the oil at 60 °F relative to water. Water has SG = 1.0 and therefore gives 10 °API, so any oil denser than water has an API gravity below 10 °API. Specific gravity is the ratio of oil density to water density at the same standard conditions. For example, an oil of density 60 lbm/ft^3 against water at 62.4 lbm/ft^3 has SG \approx 0.962 and API \approx 15.7 °API.

The volume of a mixture is not the simple sum of the component volumes. Mass is conserved on mixing, so the volume of the mixture equals the total mass divided by its density, V = m/\rho. For instance, if water and a dissolved solid are mixed and the total mass and brine density are known, the brine volume follows directly.

The pressure dependence of black oil properties is central to the subject. Above the bubble point P_b, the solution gas-oil ratio R_s is constant. Below P_b, R_s falls as pressure drops, so its curve rises with pressure up to P_b and then flattens. The oil formation volume factor B_o reaches its maximum at P_b, because dissolved gas expands the oil as pressure falls toward P_b, and it decreases above P_b as the oil compresses. Oil density reaches a minimum at P_b, since dissolved gas lightens the oil below P_b and compression increases the density above it.

Real gas compressibility factor

The gas deviation factor Z corrects the ideal gas law, pV = ZnRT. Its effect on gas compressibility appears through the isothermal compressibility, which is defined as c_g = -\dfrac{1}{v}\left(\dfrac{\partial v}{\partial p}\right)_T. This is the fractional volume change per unit pressure change at constant temperature. The negative sign makes c_g positive, because volume decreases as pressure increases. For an ideal gas, pv = RT gives c_g = 1/p.

Writing v in terms of Z gives c_g = \dfrac{1}{p} - \dfrac{1}{Z}\left(\dfrac{\partial Z}{\partial p}\right)_T. The difference between ideal and real gas compressibility is therefore \dfrac{1}{Z}\left(\dfrac{\partial Z}{\partial p}\right)_T. If \left(\partial Z/\partial p\right)_T equals Z^2, this difference simplifies to Z.

The density of a real gas is \rho = \dfrac{p M_a}{Z R T}, where M_a = \sum y_i M_i is the apparent molecular weight of the mixture and T is the absolute temperature. For a gas of 90 % methane and 10 % ethane by mole, M_a = 0.9(16) + 0.1(30) = 17.4 lb/lb-mol. The density then follows once Z, p and T are known.

The gas formation volume factor in field units is B_g = 0.02827\,\dfrac{zT}{p} ft^3/scf, with T in °R and p in psia. For z = 0.9, T = 600 °R and p = 1500 psia, B_g \approx 1.02\times10^{-2} ft^3/scf. The apparent molecular weight is not needed for this calculation.

The real gas pseudo-pressure is m(p) = 2\displaystyle\int_0^p \dfrac{p}{\mu z}\,dp. When correlations make the product \mu z proportional to p, the integrand becomes constant and m(p) is linear in p. For example, if \mu z = Kp, then m(p) = 2p/K.

Material balance for oil reservoir

The material balance equation (MBE) states that the fluid volume produced equals the volume released by the expansion of oil, gas and water, plus any water influx. The total pore volume is the sum of the oil-zone and gas-zone pore volumes, each being the reservoir volume of fluid divided by its pore fraction (1 - S_{wi}). With m = GB_{gi}/(NB_{oi}), the total is V_p = \dfrac{NB_{oi}(1+m)}{1 - S_{wi}} rb. The gas cap adds to the pore volume, and the connate water saturation appears in the denominator because hydrocarbons occupy only the non-water pore space.

If there is no water influx and the pore volume is constant, the remaining oil saturation can be found from the oil volume balance. The oil saturation is S_o = \dfrac{(N - N_p)B_o}{N B_{oi}}\,(1 - S_{wi}) when there is no initial gas. The gas saturation then follows from S_g = 1 - S_{wc} - S_o, since connate water is unchanged. For a case with no initial gas, 20 % of the oil produced, and B_o falling from 1.5 to 1.3 rb/STB at S_{wi} = 0.2, the gas saturation is about 24.5 %.

For a volumetric depleted reservoir with no water influx, the initial oil pore volume is NB_{oi}/(1-S_{wc}), and the remaining oil occupies (N - N_p)B_o rb. Dividing the remaining oil volume by the initial oil pore volume gives the oil saturation, and subtracting oil and connate water saturations gives the gas saturation.

For an undersaturated oil reservoir with negligible rock and water compressibility, and with no water influx or production, the only expansion mechanism is oil expansion above P_b. The MBE reduces to N_p B_o = N(B_o - B_{oi}), so the recovery factor is RF = \dfrac{N_p}{N} = \dfrac{B_o - B_{oi}}{B_o}.

Water influx models

A water influx model describes the volume of water that flows from an aquifer into the hydrocarbon zone as reservoir pressure falls. In the pot aquifer model the aquifer is treated as a single compressible volume of water with total compressibility c_t = c_f + c_w. The cumulative influx is W_e = c_t W_i \Delta p, where the aquifer pore volume is W_i = \pi(r_e^2 - r_R^2)\,h\,\phi\,f and f is the fractional encroachment angle. When f = 1, the full annular aquifer contributes. For example, an aquifer of thickness 30 ft, porosity 0.20 and total compressibility 10\times10^{-6} psi^{-1}, with a 700 psi pressure drop, gives an influx from the same expression.

In a steady-state water drive, reservoir pressure and production rates are held constant, so the water influx must equal the reservoir volume withdrawn each day. The withdrawal rate is Q_o\left[B_o + (\mathrm{GOR} - R_s)B_g\right] + Q_w B_w, where the bracketed term accounts for the free gas produced in excess of solution gas. With no water production, the influx rate equals the oil withdrawal term alone.

Volumetric oil in place

For a volumetric reservoir with known drainage area, pay thickness and porosity, the original oil in place (OOIP) is N = \dfrac{7758\,A\,h\,\phi\,(1 - S_{wr})}{B_o} STB, where A is in acres, h is in ft, and 7758 is the number of barrels in one acre-foot. The factor (1 - S_{wr}) is the hydrocarbon saturation, with S_{wr} the irreducible water saturation. For example, 40 acres of pay 20 ft thick, with porosity 0.20, S_{wr} = 0.30 and B_o = 1.2 rb/STB, gives about 7.2\times10^5 STB of oil in place. Recovery from a secondary or enhanced process is added to the primary recovery to obtain the total recovery as a fraction of OOIP.

For a volumetric reservoir produced at a constant rate under pseudo-steady state radial flow, the pressure decline is \Delta p = \dfrac{q_t\,t}{c_t V_p}, where V_p = \pi r_e^2 h\phi is the pore volume of the drainage area. The bottomhole flowing pressure falls by the same amount as the average reservoir pressure, so the productivity index is constant. Units must be consistent throughout, with the rate converted to reservoir volume per time.

Solution gas drive mechanism

In a solution gas drive reservoir, the energy that moves oil comes from the expansion of oil and the release of dissolved gas. An undersaturated reservoir, whose pressure remains above the bubble point, is driven first by the compressibility of oil, connate water and formation. The material balance for this stage is N_p B_{ob} = N\left[(B_{ob} - B_{oi}) + B_{oi}\dfrac{c_w S_{wc} + c_f}{1 - S_{wc}}\,\Delta p\right], where \Delta p = p_i - p_b is the pressure drop from initial to bubble point. Because the compressibility terms are small, the recovery during this stage is typically only a few per cent of OOIP.

The driving force in flow is a pressure gradient, and the mathematical type of each field quantity matters. The gradient of a scalar field such as temperature or pressure is a vector, and the divergence of a velocity field is a scalar. The gradient of a vector field, such as velocity, is a second-order tensor rather than a scalar or a vector.

Black oil classification

A black oil is a crude oil whose liquid shrinkage on pressure depletion below the bubble point is gradual. Only moderate amounts of gas come out of solution as pressure falls, so the liquid volume declines smoothly and retains a substantial fraction of its volume at low pressure. A very lean, gas-like fluid shows almost no shrinkage, whereas a volatile oil shows a sharp drop in liquid volume near the bubble point because it liberates a large amount of gas.

Flow comparisons require the mobility \lambda = k/\mu, the ratio of effective permeability to viscosity. For a crude oil, specific gravity follows from API gravity as \mathrm{SG} = \dfrac{141.5}{131.5 + {}^\circ\mathrm{API}}. When viscosity is taken as proportional to specific gravity, the mobility ratio of two sands is \dfrac{\lambda_A}{\lambda_B} = \dfrac{k_A}{k_B}\cdot\dfrac{\mu_B}{\mu_A}. A lighter oil with lower viscosity therefore raises the mobility of its sand relative to a heavier oil in a less permeable sand.