Chapter 10. The Displacement of Oil and Gas

The tendency for a solid to prefer one fluid over another is called wettability. Wettability is a function of the chemical composition of both the fluids and the rock. Surfaces can be either oil wet or water wet, depending on the chemical composition of the fluids. The degree to which a rock is either oil wet or water wet is strongly affected by the absorption or desorption of constituents in the oil phase. Large, polar compounds in the oil phase can absorb onto the solid surface, leaving an oil film that may alter the wettability of the surface.

The concept of wettability leads to another significant factor in the recovery of oil. This factor is capillary pressure. To illustrate capillary pressure, consider a capillary tube that contains both oil and brine, the oil having a lower density than the brine. The pressure in the oil phase immediately above the oil-brine interface in the capillary tube will be slightly greater than the pressure in the water phase just below the interface. This difference in pressure is called the capillary pressure, P_c, of the system. The greater pressure will always occur in the nonwetting phase. An expression relating the contact angle, θ; the radius, r_c, of the capillary in feet; the oil-brine interfacial tension, σ_wo, in dynes/cm; and the capillary pressure in psi is given by

It has been shown that the pressure required to force a nonwetting phase through a small capillary can be very large. For instance, the pressure drop required to force an oil drop through a tapering constriction that has a forward radius of 0.00002 ft, a rearward radius of 0.00005 ft, a contact angle of 0°, and an interfacial tension of 25 dynes/cm is 0.71 psi. If the oil drop were 0.00035-ft long, a pressure gradient of 2029 psi/ft would be required to move the drop through the constriction. Pressure gradients of this magnitude are not realizable in reservoirs. Typical pressure gradients obtained in reservoir systems are of the order of 1 psi/ft to 2 psi/ft.

If the water is replaced by an oil of 3.0-cp viscosity and 1.2-bbl/STB formation volume factor, then, under the same pressure differential, the flow rate will be 0.0834 STB/day, and again the absolute permeability is

If the same core is maintained at 70% water saturation (S_w = 70%) and 30% oil saturation (S_o = 30%), and at these and only these saturations and under the same pressure drop, it flows 0.18 STB/day of the brine and 0.01 STB/day of the oil, then the effective permeability to water is

The effective permeability, then, is the permeability of a rock to a particular fluid when that fluid has a pore saturation of less than 100%. As noted in the foregoing example, the sum of the effective permeabilities (i.e., 298 md) is always less than the absolute permeability, 413 md.

Figure 10.1 shows a typical plot of oil and water relative permeability curves for a particular rock as a function of water saturation. Starting at 100% water saturation, the curves show that a decrease in water saturation to 85% (a 15% increase in oil saturation) sharply reduces the relative permeability to water from 100% down to 60%, and at 15% oil saturation, the relative permeability to oil is essentially zero. This value of oil saturation, 15% in this case, is called the critical saturation, the saturation at which oil first begins to flow as the oil saturation increases. It is also called the residual saturation, the value below which the oil saturation cannot be reduced in an oil-water system. This explains why oil recovery by water drive is not 100% efficient. If the initial connate water saturation is 20% for this particular rock, then the maximum recovery from the portion of the reservoir invaded by high-pressure water influx is

Returning to Fig. 10.1, as the water saturation decreases further, the relative permeability to water continues to decrease and the relative permeability to oil increases. At 20% water saturation, the (connate) water is immobile, and the relative permeability to oil is quite high. This explains why some rocks may contain as much as 50% connate water and yet produce water-free oil. Most reservoir rocks are preferentially water wet—that is, the water phase and not the oil phase is next to the walls of the pore spaces. Because of this, at 20% water saturation, the water occupies the least favorable portions of the pore spaces—that is, as thin layers about the sand grains, as thin layers on the walls of the pore cavities, and in the smaller crevices and capillaries. The oil, which occupies 80% of the pore space, is in the most favorable portions of the pore spaces, which is indicated by a relative permeability of 93%. The curves further indicate that about 10% of the pore spaces contribute nothing to the permeability, for at 10% water saturation, the relative permeability to oil is essentially 100%. Conversely, on the other end of the curves, 15% of the pore spaces contribute 40% of the permeability, for an increase in oil saturation from zero to 15% reduces the relative permeability to water from 100% to 60%.

In describing two-phase flow mathematically, it is typically the relative permeability ratio that enters the equations. Figure 10.2 is a plot of the relative permeability ratio versus water saturation for the same data of Fig. 10.1. Because of the wide range of k_ro/k_rw values, the relative permeability ratio is usually plotted on the log scale of a semilog graph. The central or main portion of the curve is quite linear on the semilog plot and in this portion of the curve, the relative permeability ratio may be expressed as a function of the water saturation by

Solving simultaneously, the intercept a = 1220 and the slope b = 13.0. Equation (10.3) indicates that the relative permeability ratio for a rock is a function of only the relative saturations of the fluids present. Although it is true that the viscosities, the interfacial tensions, and other factors have some effect on the relative permeability ratio, for a given rock, it is mainly a function of the fluid saturations.

In many rocks, there is a transition zone between the water and the oil zones. In the true water zone, the water saturation is essentially 100%, although in some reservoirs, a small oil saturation may be found a considerable distance vertically below the oil-water contact. In the oil zone, there is usually connate water present, which is essentially immobile. For the present example, the connate water saturation is 20% and the oil saturation is 80%. Only water will be produced from a well completed in the true water zone, and only oil will be produced from the true oil zone. In the transition zone (Fig. 10.3), both oil and water will be produced, and the fraction that is water will depend on the oil and water saturations at the point of completion. If the well in Fig. 10.3 is completed in a uniform sand at a point where S_o = 60% and S_w = 40%, the fraction of water in reservoir flow rate units or reservoir watercut may be calculated using Eq. (8.19):

Either Eq. (10.5) or (10.6) can be used with the data of Fig. 10.1 and with viscosity data to calculate the watercut. From Fig. 10.1, at S_w = 0.40, k_rw = 0.045, and k_ro = 0.36. If μ_w = 1.0 cp and μ_o = 3.0 cp, then the reservoir watercut is

Mobility is a relative measure of how easily a fluid moves through porous media. The apparent mobility, as defined in Chapter 8, is the ratio of effective permeability to fluid viscosity. Since the effective permeability is a function of fluid saturations, several apparent mobilities can be defined. When a fluid is being injected into a porous medium containing both the injected fluid and a second fluid, the apparent mobility of the displacing phase is usually measured at the average displacing phase saturation when the displacing phase just begins to break through at the production site. The apparent mobility of the nondisplacing phase is measured at the displacing phase saturation that occurs just before the beginning of the injection of the displacing phase.

Areal sweep efficiencies are a strong function of the mobility ratio. The mobility ratio M, as defined in Chapter 8, is a measure of the relative apparent mobilities in a displacement process and is given by

A phenomenon called viscous fingering can take place if the mobility of the displacing phase is much greater than the mobility of the displaced phase. Viscous fingering simply refers to the penetration of the much more mobile displacing phase into the phase that is being displaced.

Figure 10.6(b) shows the effect of mobility ratio on areal sweep efficiency at initial breakthrough for a five-spot network (shown in Fig. 10.6(a)) obtained using the X-ray shadowgraph. The pattern at breakthrough for a mobility ratio of 1 obtained with an electrolytic model is included for comparison.

The arrangement of injection and production wells depends primarily on the geology of the formation and the size (areal extent) of the reservoir. For a given reservoir, an operator has the option of using the existing well arrangement or drilling new wells in other locations. If the operator opts to use the existing well arrangement, there may be a need to consider converting production wells to injection wells or vice versa. An operator should also recognize that, when a production well is converted to an injection well, the production capacity of the reservoir will have been reduced. This decision can often lead to major cost items in an overall project and should be given a great deal of consideration. Knowledge of any directional permeability effects and other heterogeneities can aid in the consideration of well arrangements. The presence of faults, fractures, and high-permeability streaks can dictate the shutting in of a well near one of these heterogeneities. Directional permeability trends can lead to a poor sweep efficiency in a developed pattern and can suggest that the pattern be altered in one direction or that a different pattern be used.

We now apply Eq. (10.15) to a reservoir under active water drive where the walls are located in uniform rows along the strike on 40-ac spacing, as shown in Fig. 10.8. This gives rise to approximate linear flow, and if the daily production of each of the three wells located along the dip is 200 STB of oil per day, then for an active water drive and an oil volume factor of 1.50 bbl/STB, the total reservoir throughput,

will be 900 bbl/day.

The cross-sectional area is the product of the width, 1320 ft, and the true formation thickness, 20 ft, so that for a porosity of 25%, Eq. (10.15) becomes

If we let x = 0 at the bottom of the transition zone, as indicated in Fig. 10.8, then the distances the various constant water saturation planes will travel in, say, 60, 120, and 240 days are given by

The derivative (∂f_w/∂S_w) may also be obtained mathematically using Eq. (10.3) to represent the relationship between the relative permeability ratio and the water saturation. Differentiating Eq. (10.16), the following is obtained:

Figure 10.9 shows the fractional watercut, f_W, and also the derivative (∂f_w/∂S_w) plotted against water saturation from the data of Table 10.1. Equation (10.17) was used to determine the values of the derivative. Since Eq. (10.3) does not hold for the very high and for the quite low water saturation ranges, some error is introduced below 30% and above 80% water saturation. Since these are in the regions of the lower values of the derivatives, the overall effect on the calculation is small.

The lowermost curve of Fig. 10.10 represents the initial distribution of water and oil in the linear sand body of Fig. 10.8. Above the transition zone, the connate water saturation is constant at 20%. Equation (10.16) may be used with the values of the derivatives, calculated in Table 10.1 and plotted in Fig. 10.9, to construct the frontal advance curves shown in Fig. 10.10 at 60, 120, and 240 days. For example, at 50% water saturation, the value of the derivative is 2.86; so by Eq. (10.16), at 60 days, the 50% water saturation plane, or front, will advance a distance of

This distance is plotted as shown in Fig. 10.10 along with the other distances that have been calculated in Table 10.1 for the other time values and other water saturations. These curves are characteristically double valued or triple valued. For example, Fig. 10.10 indicates that the water saturation after 240 days at 400 ft is 20%, 36%, and 60%. The saturation can be only one value at any place and time. The difficulty is resolved by dropping perpendiculars so that the areas to the right (A) equal the areas to the left (B), as shown in Fig. 10.10.

Figure 10.11 represents the initial water and oil distributions in the reservoir unit and also the distributions after 240 days, provided the flood front has not reached the lowermost well. The area to the right of the flood front in Fig. 10.11 is commonly called the oil bank and the area to the left is sometimes called the drag zone. The area above the 240-day curve and below the 90% water saturation curve represents oil that may yet be recovered or dragged out of the high-water saturation portion of the reservoir by flowing large volumes of water through it. The area above the 90% water saturation curve represents unrecoverable oil, since the critical oil saturation is 10%.

Figure 10.11 also indicates that a well in this reservoir unit will produce water-free oil until the flood front approaches the well. Thereafter, in a relatively short period, the watercut will rise sharply and be followed by a relatively long period of production at high, and increasingly higher, watercuts. For example, just behind the flood front at 240 days, the water saturation rises from 20% to about 60%—that is, the watercut rises from zero to 78.4% (see Table 10.1). When a producing formation consists of two or more rather definite strata, or stringers, of different permeabilities, the rates of advance in the separate strata will be proportional to their permeabilities, and the overall effect will be a combination of several separate displacements, such as described for a single homogeneous stratum.

Due to the high oil-gas viscosity ratios and the high gas-oil relative permeability ratios at low gas saturations, the displacement efficiency by gas is generally much lower than that by water, unless the gas displacement is accompanied by substantial gravitational segregation. This is basically the same reason for the low recoveries from reservoirs produced under the dissolved gas drive mechanism. The effect of gravitational segregation in water-drive oil reservoirs is usually of much less concern because of the higher displacement efficiencies and the lower oil-water density differences, whereas the converse is generally true for gas-oil systems. Welge showed that capillary forces may generally be neglected in both, and he introduced a gravitational term in Eq. (10.5), as will be shown in the following equations.¹⁰ As with water displacement, a linear system is assumed, and a constant gas pressure throughout the system is also assumed so that a constant throughput rate may be used. These assumptions also allow us to eliminate changes caused by gas density, oil density, oil volume factor, and the like. Equation (8.1) may be applied to both the oil and gas flow, assuming the connate water is essentially immobile, so that the fraction of the flowing reservoir fluid volume, which is gas, is

The relative permeability ratio (k_ro/k_rg) may be used for the effective permeability ratio in the denominator of Eq. (10.22); however, the permeability to oil, k_o, in the numerator is the effective permeability and cannot be replaced by the relative permeability. It may, however, be replaced with (k_rok), where k is the absolute permeability. The total velocity, ν_t, is the total throughput rate,

divided by the cross-sectional area, A_c. Inserting these equivalents, the fractional gas flow equation with gravitational segregation becomes

Although Eq. (10.24) is not rate sensitive (i.e., it does not depend on the throughput rate), Eq. (10.23) includes the throughput velocity

and is therefore rate sensitive. Since the total throughput rate,

, is in the denominator of the gravitational term of Eq. (10.23), rapid displacement (i.e., large

) reduces the size of the gravitational term, and so causes an increase in the fraction of gas flowing, f_g. A large value of f_g implies low displacement efficiency. If the gravitational term is sufficiently large, f_g becomes zero, or even negative, which indicates countercurrent flow of gas updip and oil downdip, resulting in maximum displacement efficiency. In the case of a gas cap that overlies most of an oil zone, the drainage is vertical, and cos α = 1.00; in addition, the cross-sectional area is large. If the vertical effective permeability k_o is not reduced to a very low level by low permeability strata, gravitational drainage will substantially improve recovery.

The use of Eq. (10.23) is illustrated using the data given by Welge for the Mile Six Pool, Peru, where advantage was taken of good gravitational segregation characteristics to improve recovery.¹⁰ Pressure maintenance by gas injection has been practiced since 1933 by returning produced gas and other gas to the gas cap so that reservoir pressure has been maintained within 200 psi of its initial value. Figure 10.12 shows the average relative permeability characteristics of the Mile Six Pool reservoir rock. As is common in gas-oil systems, the saturations are expressed in percentages of the hydrocarbon porosity, and the connate water, being immobile, is considered as part of the rock. The other pertinent reservoir rock and fluid data are given in Table 10.2. Substituting these data in Eq. (10.25),

The values of f_g have been calculated in Table 10.2 for three conditions: (1) assuming negligible gravitational segregation by using Eq. (10.24); (2) using the gravitational term equal to 2.50k_ro for the Mile Six Pool, Eq. (10.25); and (3) assuming the gravitational term equals 1.25k_ro, or half the value at Mile Six Pool. The values of f_g for these three conditions are shown plotted in Fig. 10.13. The negative values of f_g for the conditions that existed in the Mile Six Pool indicate countercurrent gas flow (i.e., gas updip and oil downdip) in the range of gas saturations between an assumed critical gas saturation of 5% and about 17%.

The values of the derivatives (∂f_g/∂S_g) given in Table 10.2 have been determined graphically from Fig. 10.13. Figure 10.14 shows the plots of Eq. (10.26) to obtain the gas-oil distributions and the positions of the gas front after 100 days. The shape of the curves will not be altered for any other time. The distribution and fronts at 1000 days, for example, may be obtained by simply changing the scale on the distance axis by a factor of 10.

Welge showed that the position of the front may be obtained by drawing a secant from the origin as shown in Fig. 10.13.¹⁰ For example, the secant is tangent to the lower curve at 40% gas saturation. Then, in Fig. 10.14, the front may be found by dropping a perpendicular from the 40% gas saturation as indicated. This will balance the areas of the S-shaped curve, which was done by trial and error in Fig. 10.10 for water displacement. In the case of water displacement, the secant should be drawn, not from the origin, but from the connate water saturation, as indicated by the dashed line in Fig. 10.9. This is tangent at a water saturation of 60%. Referring to Fig. 10.10, the 240-day front is seen to occur at 60% water saturation. Owing to the presence of an initial transition zone, the fronts at 60 days and 120 days occur at slightly lower values of water saturation.

The much greater displacement efficiency with gravity segregation than without is apparent from Fig. 10.14. Since the permeability to oil is essentially zero at 60% gas saturation, the maximum recovery by gas displacement and gravity drainage is 60% of the initial oil in place. Actually, some small permeability to oil exists at even very low oil saturations, which explains why some fields may continue to produce at low rates for quite long periods after the pressure has been depleted. The displacement efficiency may be calculated from Fig. 10.14 by the measurement of areas. For example, the displacement efficiency at Mile Six Pool with full gravity segregation is in excess of

One interesting application of gravity segregation is to the recovery of updip or “attic” oil in active water-drive reservoirs possessing good gravity segregation characteristics. When the structurally highest well(s) has gone to water production, high-pressure gas is injected for a period. This gas migrates updip and displaces the oil downdip, where it may be produced from the same well in which the gas was injected. The injected gas is, of course, unrecoverable.

It appears from the previous discussions and examples that water is generally more efficient than gas in displacing oil from reservoir rocks, mainly because (1) the water viscosity is of the order of 50 times the gas viscosity and (2) the water occupies the less conductive portions of the pore spaces, whereas the gas occupies the more conductive portions. Thus, in water displacement, the oil is left to the central and more conductive portions of the pore channels, whereas in gas displacement, the gas invades and occupies the more conductive portions first, leaving the oil and water to the less conductive portions. What has been said of water displacement is true for preferentially water wet (hydrophilic) rock, which is the case for most reservoir rocks. When the rock is preferentially oil wet (hydrophobic), the displacing water will invade the more conductive portions first, just as gas does, resulting in lower displacement efficiencies. In this case, the efficiency by water still exceeds that by gas because of the viscosity advantage that water has over gas.

10.3.3 Oil Recovery by Internal Gas Drive

Oil is produced from volumetric, undersaturated reservoirs by expansion of the reservoir fluids. Down to the bubble-point pressure, the production is caused by liquid (oil and connate water) expansion and rock compressibility (see Chapter 6, section 6.6). Below the bubble point, the expansion of the connate water and the rock compressibility are negligible, and as the oil phase contracts owing to release of gas from solution, production is a result of expansion of the gas phase. When the gas saturation reaches the critical value, free gas begins to flow. At fairly low gas saturations, the gas mobility, k_g/μ_g, becomes large, and the oil mobility, k_o/μ_o, becomes small, resulting in high gas-oil ratios and in low oil recoveries, usually in the range of 5% to 25%.

The reader will recall that the material balance is successful in predicting the performance of volumetric reservoirs down to pressures at which free gas begins to flow. In the study of the Kelly-Snyder Field, Canyon Reef Reservoir, for example, in Chapter 6, section 6.4, the produced gas-oil ratio was assumed to be equal to the dissolved gas-oil ratio, down to the pressure at which the gas saturation reached 10%, the critical gas saturation assumed for that reservoir. Below this pressure (i.e., at higher gas saturations), both gas and oil flow to the wellbores, their relative rates being controlled by their viscosities, which change with pressure, and by their relative permeabilities, which change with their saturations. It is not surprising, then, that the material balance principle (static) is combined with the producing gas-oil ratio equation (dynamic) to predict the performance at pressures at which the gas saturation exceeds the critical value.

Equation (10.32) is simply an expression of the material balance for volumetric, undersaturated reservoirs in differential form. The producing gas-oil ratio may also be written as

Equation (10.33) applies both to the flowing free gas and to the solution gas that flows to the wellbore in the oil. These two types of gas make up the total surface producing gas-oil ratio, R, in SCF/STB. Equation (10.33) may be equated to Eq. (10.32) and solved for dS_o/dp to give

To simplify the handling of Eq. (10.34), the terms in the numerator that are functions of pressure may only be grouped together and given the group symbols X(p), Y(p), and Z(p) as follows:

Equation (10.36) gives the change in oil saturation that accompanies a pressure drop, Δp. The functions X(p), Y(p), and Z(p) are obtained from the reservoir fluid properties using Eq. (10.35). The values of the derivatives dR_so/dp, dB_o/dp, and dB_g/dp are found graphically from the plots of R_so, B_o, and B_g versus pressure. It has been found that when determining dB_g/dp, the numbers are more accurately obtained by plotting 1/B_g versus pressure. When this is done, the following substitution is used:

2. Solve Eq. (10.36) for ΔS_o/Δp using the oil saturation that corresponds to the initial pressure of the given Δp.

5. Determine an average value for ΔS_o/Δp from the two values calculated in steps 2 and 4.

6. Using (ΔSo/Δp)_ave, solve for S_oj using Eq. (10.38). This value of S_oj becomes S_o₍_j_–1) for the next pressure drop interval.

The Schilthuis method begins with the general material balance equation, which reduces to the following for a volumetric, undersaturated reservoir, using the single-phase formation volume factor:

The first term on the right-hand side of Eq. (10.33) accounts for the production of solution gas, and the second term accounts for the production of free gas in the reservoir. The second term is a ratio of the gas to oil flow equations discussed in Chapter 8. To calculate R with Eq. (10.33), information about the permeabilities to gas and oil is required. This information is usually known from laboratory measurements as a function of fluid saturations and is often available in graphic form (see Fig. 10.15). The fluid saturation equation is also needed:

All the parameters in Eq. (10.41) are known as functions of pressure from laboratory studies except N_p/N and R_p. When the correct values of these two variables are used in Eq. (10.41) at a given pressure, then the left-hand side of the equation equals zero. The trial-and-error procedure follows this sequence of steps:

1. Guess a value for an incremental oil production (ΔN_p/N) that occurs during a small drop in the average reservoir pressure (Δp).

2. Determine the cumulative oil production to pressure p_j = p_j_–1 – Δp by adding all the previous incremental oil productions to the guess during the current pressure drop. The subscript, j – 1, refers to the conditions at the beginning of the pressure drop and j to the conditions at the end of the pressure drop.

4. Knowing S_L, determine a value for k_g/k_o from permeability ratio versus saturation information, and then solve Eq. (10.33) for R_j at the current pressure.

5. Calculate the incremental gas production using an average value of the gas-oil ratio over the current pressure drop:

Any one of a number of iteration techniques can be used to assist in the trial-and-error procedure. One that has been used is the secant method,¹⁷ which has the following iteration formula:

To apply the secant method to the foregoing procedure, the left-hand side of Eq. (10.41) becomes the function, f, and the cumulative oil recovery becomes x. The secant method provides the new guess for oil recovery, and the sequence of steps is repeated until the function, f, is zero or within a specified tolerance (e.g., ±10^–4). The solution procedure described earlier is fairly easy to program on a computer. The authors are keenly aware that programs like Excel can be used to solve this problem without writing a separate program to include a solution procedure like the secant method. However, an understanding of the secant method may help the reader to visualize how these solvers work, and for that reason, it is presented here.

The Tarner method for predicting reservoir performance by internal gas drive is presented in a form proposed by Tracy.¹⁸ Neglecting the formation and water compressibility terms, the general material balance in terms of the single-phase oil formation volume factor may be written as follows:

5. Determine k_g/k_o by calculating the total liquid saturation, S_L, from Eq. (10.40) and using k_g/k_o versus saturation information.

6. Calculate a value of R_j by using Eq. (10.33), and compare it with the assumed value in step 2. If these two values agree within some tolerance, then the ΔN_p calculated in step 3 is correct for this pressure drop interval. If the value of R_j does not agree with the assumed value in step 2, then the calculated value should be used as the new guess and steps 2 through 6 repeated.

As a further check, the value of R_ave can be recalculated and Eq. (10.56) solved for ΔN_p. Again, if the new value agrees with what was previously calculated in step 3 within some tolerance, it can be assumed that the oil recovery is correct. The three methods are illustrated in Example 10.1.

Example 10.1 Calculating Oil Recovery as a Function of Pressure

This example uses the (1) Muskat, (2) Schilthuis, and (3) Tarner methods for an undersaturated, volumetric reservoir. The recovery is calculated for the first 200-psi pressure increment from the initial pressure down to a pressure of 2300 psia.

Given

Initial reservoir pressure = 2500 psia

Initial reservoir temperature = 180°F

Initial oil in place = 56 × 10⁶ STB

Initial water saturation = 0.20

Fluid property data are given in Table 10.4.

Permeability ratio data are plotted in Fig. 10.16.

Table 10.4 Fluid Property Data for Example 10.1

Figure 10.16 Permeability ratio relationship for Example 10.1.

Solution

The Muskat method involves the following sequence of steps:

1. R_so, B_o, and 1/B_g are plotted versus pressure to determine the slopes. Although the plots are not shown, the following values can be determined:

The values of X(p), Y(p), and Z(p) are tabulated as follows, as a function of pressure:

2. Calculate ΔS_o/Δp using X(p), Y(p), and Z(p) at 2500 psia:

3. Estimate S_oj:

S_oj = 0.80 – 200(0.000146) = 0.7709

4. Calculate ΔS_o/Δp using the S_oj from step 3 and X(p), Y(p), and Z(p) at 2300 psia:

5. Calculate the average ΔS_o/Δp:

6. Calculate S_oj using (ΔS_oΔp)_ave from step 5:

S_oj = 0.8 – 0.000159(200) = 0.7682

This value of S_o can now be used to calculate the oil recovery that has occurred down to a pressure of 2300 psia:

Because the Schilthuis method involves an interactive procedure, an iterative solver, like Excel’s solver, is used to assist in the solution. For this problem, N_p/N is the guess value that the solver will use to solve Eq. (10.41). The solver requires one guess value of N_p/N to begin the iteration process.

1. Assume incremental oil recovery:

2. Calculate S_L:

3. Determine k_g/k_o from Fig. 10.16 and calculate R:

4. Calculate incremental gas recovery:

5. Calculate R_p:

6. Use Excel’s solver function to solve Eq. (10.41) iteratively by changing N_p/N until the left-hand side of Eq. (10.41) is equal to zero:

Using this method, the correct value of fractional oil recovery down to 2300 psia is 0.0165. To compare with the Muskat method, the recovery ratio must be multiplied by the initial oil in place, 56 M STB, to yield the total cumulative recovery:

N_p = 56(10)⁶ (0.0167) = 935,200 STB

The Tarner method requires the following steps:

1. Calculate Ф_n and Ф_g at 2300 psia:

2. Assume R_j = 670 SCF/STB, which is just slightly larger than R_so, suggesting that only a very small amount of gas is flowing to the wellbore and is being produced:

3. Calculate ΔN_p:

4. Calculate N_p:

N_p = ΔN_p = 939,300 STB

5. Determine k_g/k_o:

From this value of S_L, the permeability ratio, k_g/k_o, can be obtained from Fig. 10.16. Since the curve is off the plot, a very small value of k_g/k_o = 0.00001 is estimated.

6. Calculate R_j and compare it with the assumed value in step 2:

This value agrees very well with the value of 670 that was assumed in step 2, satisfying the constraints of the Tarner method. For the data given in this example, the three methods of calculation yielded values of N_p that are within 0.5%. This suggests that any one of the three methods may be used to predict oil and gas recovery, especially considering that many of the parameters used in the equations could be in error more than 0.5%.

(d) If μ_o = 3.4 cp, μ_w = 0.68 cp, B_o = 1.50 bbl/STB, and B_w = 1.05 bbl/STB, what is the surface watercut of a well completed in the transition zone where the water saturation is 50%?

(f) What percentage of recovery will be realized from this sandstone under high-pressure water drive from that portion of the reservoir above the transition zone invaded by water? The initial water saturation above the transition zone is 30%.

(g) If water drive occurs at a pressure below saturation pressure such that the average gas saturation is 15% in the invaded portion, what percentage of recovery will be realized? The average oil volume factor at the lower pressure is 1.35 bbl/STB and the initial oil volume factor is 1.50 bbl/STB.

(h) What fraction of the absolute permeability of this sandstone is due to the least permeable pore channels that make up 20% of the pore volume? What fraction is due to the most permeable pore channels that make up 25% of the pore volume?

(b) Graphically determine ∂f_w/∂S_w at a number of points and plot versus S_w.

(c) Calculate ∂f_w/∂S_w at several values of S_w using Eq. (10.17), and compare with the graphical values of part (b).

(d) Calculate the distances of advance of the constant saturation fronts at 100, 200, and 400 days. Plot on Cartesian coordinate paper versus S_w. Equalize the areas within and without the flood front lines to locate the position of the flood fronts.

(e) Draw a secant line from S_w = 0.20 tangent to the f_w versus S_w curve in part (b), and show that the value of S_w at the point of tangency is also the point at which the flood front lines are drawn.

(f) Calculate the fractional recovery when the flood front first intercepts a well, using the areas of the graph of part (d). Express the recovery in terms of both the initial oil in place and the recoverable oil in place (i.e., recoverable after infinite throughput).

(g) To what surface watercut will a well rather suddenly rise when it is just enveloped by the flood fronts? Use B_o = 1.50 bbl/STB and B_w = 1.05 bbl/STB.

(h) Do the answers to parts (f) and (g) depend on how far the front has traveled? Explain.

(b) Plot the gas saturation versus distance after 100 days of gas injection both with and without the gravity segregation term.

(c) Using the areas of part (b), calculate the recoveries behind the flood front with and without gravity segregation in terms of both initial oil and recoverable oil.

10.7 Rework the water displacement calculation of Table 10.1, and include a gravity segregation term. Assume an absolute permeability of 500 md, a dip angle of 45°, and a density difference of 20% between the reservoir oil and water and an oil viscosity of 1.6 cp. Plot water saturation versus distance after 240 days and compare with Fig. 10.11.

2. F. F. Craig, The Reservoir Engineering Aspects of Waterflooding, Society of Petroleum Engineers, 1993.

4. D. W. Green and G. P. Willhite, Enhanced Oil Recovery, Vol. 6, Society of Petroleum Engineers, 1998.

5. E. D. Holstein, ed., Petroleum Engineering Handbook, Vol. 5, Reservoir Engineering and Petrophysics, Society of Petroleum Engineers, 2007.

6. R. D. Wycoff, H. G. Botset, and M. Muskat, “Mechanics of Porous Flow Applied to Water-Flooding Problems,” Trans. AlME (1933), 103, 219.

7. R. L. Slobod and B. H. Candle, “X-ray Shadowgraph Studies of Areal Sweep Efficiencies,” Trans. AlME (1952), 195, 265.

8. S. E. Buckley and M. C. Leverett, “Mechanism of Fluid Displacement in Sands,” Trans. AlME (1942), 146, 107.

9. J. G. Richardson and R. J. Blackwell, “Use of Simple Mathematical Models for Predicting Reservoir Behavior,” Jour. of Petroleum Technology (Sept. 1971), 1145.

10. H. J. Welge, “A Simplified Method for Computing Oil Recoveries by Gas or Water Drive,” Trans. AlME (1952), 195, 91.

11. D. R. Shreve and L. W. Welch Jr., “Gas Drive and Gravity Drainage Analysis for Pressure Maintenance Operations,” Trans. AlME (1956), 207, 136.

12. L. R. Kern, “Displacement Mechanism in Multi-well Systems,” Trans. AlME (1952), 195, 39.

13. R. H. Smith reported by J. A. Klotz, “The Gravity Drainage Mechanism,” Jour. of Petroleum Technology (Apr. 1953), 5, 19.

14. M. Muskat, “The Production Histories of Oil Producing Gas-Drive Reservoirs,” Jour. of Applied Physics (1945), 16, 147.

15. E. T. Guerrero, Practical Reservoir Engineering, Petroleum Publishing Co., 1968.

16. J. Tarner, “How Different Size Gas Caps and Pressure Maintenance Programs Affect Amount of Recoverable Oil,” Oil Weekly (June 12, 1944), 144, 32–34.

17. J. B. Riggs, An Introduction to Numerical Methods for Chemical Engineers, Texas Tech. University Press, 1988.

18. G. W. Tracy, “Simplified Form of the Material Balance Equation,” Trans. AlME (1955), 204, 243.

19. T. Ertekin, J. H. Abou-Kassem, and G. R. King, Basic Applied Reservoir Simulation, Vol. 10, Society of Petroleum Engineers, 2001.

20. J. Fanchi, Principles of Applied Reservoir Simulation, 3rd ed., Elsevier, 2006.

21. C. C. Mattax and R. L. Dalton, Reservoir Simulation, Vol. 13, Society of Petroleum Engineers, 1990.

22. M. Carlson, Practical Reservoir Simulation, PennWell Publishing, 2006.

10.3.1 The Buckley-Leverett Displacement Mechanism

Chapter 10. The Displacement of Oil and Gas

10.1 Introduction

10.2 Recovery Efficiency

10.2.1 Microscopic Displacement Efficiency

10.2.2 Relative Permeability

10.2.3 Macroscopic Displacement Efficiency

10.3 Immiscible Displacement Processes

10.3.2 The Displacement of Oil by Gas, with and without Gravitational Segregation

10.3.3 Oil Recovery by Internal Gas Drive

10.4 Summary

Problems

References