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# Predicting behavior of in-situ combustion

Many useful and reasonably accurate calculations can be made on in-situ combustion to predict the behavior of a proposed project. This page discusses the calculation process involved with behavior prediction.

## Combining material- and heat-balance calculations

In-situ combustion prediction calculations will be explained in the following diagrams and example calculations. They start with a very simple heat balance and are then extended to more closely represent what happens in the reservoir.

### First assumptions

Start by assuming that no combustion data are available to get an initial idea of the feasibility of a project. This preliminary work gives the engineer a sound basis to decide whether further work has economic promise.

Assume a sandstone formation with:

• Porosity of 22%
• Temperature of 100°F
• 24°(API) oil at a saturation of 65%
• Injection pressure of 300 psia

Also assume the CO2/CO atomic ratio m to be ....................(1)

This is a reasonable ratio to assume, based on both laboratory and field experience.

Because there are no tube run data, generalized correlation curves (Figs. 1 and 2) will be used to calculate expected results. From Fig. 1, the fuel availability, W, for 24°API crude is ....................(2)

The apparent H/C atomic ratio (n) of the fuel is also needed. This is a function of the combustion-front temperature, as shown in Fig. 2. Selected data from the graph are listed in Table 1. These data for 21.8°API crude are close enough to 24°API crude for initial estimates.

### Calculate initial heat balances and temperatures

Start by assuming that all heat generated is used to heat the rock formation through which the combustion front has moved. This assumption is not accurate, but it simplifies the understanding of the mathematics and concepts involved in heat-balance calculations. A sketch of the temperature profile generated is shown in Fig. 3. Corrections to this heat balance calculation will be discussed later.

Assuming 1.0 ft3 of rock formation burned and a front temperature of 1,000°F, from Eq. 2 and Table 1, ....................(3)

Quartz weighs approximately 164 lbm/ft3. The amount of fuel for a cubic foot of formation equals ....................(4)

Using the heat of combustion, Eq. 5, with the appropriate parameters, becomes ....................(5)

where Hc = heating value, Btu/lbm fuel. To convert to J/kg, multiply by 2,326. ....................(6)

Thus, the total heat generated is ....................(7)

Next, calculate the temperature rise of the formation behind the front to see if it matches the temperature assumed by performing a heat balance behind the front. Because, for practical purposes, the only fluid in the formation behind the front is air (which has a very small volumetric heat capacity), we only need to calculate a heat balance on the sandstone itself. A good equation for average sandstone heat capacity is ....................(8a) ....................(8b)

where

• T1 = initial reservoir temperature, °F
• T2 = final reservoir temperature, °F

From a heat-balance calculation, the reservoir sand temperature is as follows: ....................(9)

The result from Eq. 9 does not agree with the assumed temperature of 1,000°F. Calculations with other assumed temperatures result in the calculated temperature values shown in Table 2.

The tabular data are graphed as circles in Fig. 3. The two temperatures match at 801°F. This is the calculated combustion-front temperature if all the heat generated is used to heat the formation behind the combustion front.

### Correction for water of combustion

These results do not include all the processes occurring in the reservoir. First, the water formed by combustion will condense beyond the combustion front, absorb some heat of combustion, and reduce the heat of the formation behind the combustion front. This effect can be calculated as follows.

From Eq. 3, 0.0317 lbm of H2 are burned per 100 lbm of rock at 1,000°F. Assuming that a pound of steam will release 1,000 Btu when cooling from combustion temperature and condensing (this number is only approximately correct but is adequate for estimation purposes), the amount of heat carried forward by the steam is calculated with concepts similar to Eqs. 4 and 7: ....................(10)

In this equation, 18/2 is the ratio of molecular weight of water and hydrogen. The heat given up by the steam is 1,000 Btu/lbm, and the other numbers are similar to those in Eq. 4. Thus, the calculated temperature is lower than it was in Eq. 9, as shown below: ....................(11)

Other temperatures were calculated in a similar way, and the results, graphed as triangles in Fig. 3, show a corrected combustion temperature of 788°F. At this temperature, the H/C ratio is 0.85, as indicated in Fig. 4.

### Calculating the volume and temperature of the steam plateau

No calculations on the steam plateau were necessary in the above calculations. The steam-plateau temperature and volume directly affect the volume of oil moved as a result of the combustion process. To calculate these terms, use the H/C ratio of 0.85 and calculate the partial pressure of the water as follows.

The stoichiometric equation is then ....................(12)

where

• n = hydrogen/carbon atomic ratio of fuel
• m = CO2/CO concentration ratio produced

The other symbols indicate the various components in the chemical-balance equation.

The fuel composition is CH 0.85 . From Eq. 12, the moles of oxygen used per mole of fuel are ....................(13)

Combustion products are calculated in a similar way: ....................(14) ....................(15) ....................(16) ....................(17)

The operating pressure is 300 psia. The partial pressure of H2O in the combustion gas is

From steam tables, the saturation temperature for 21.6 psia is 232°F. This is the temperature of the steam plateau.

The volume of the steam plateau is a function of the amount of H2O formed. Knowing that there are 0.95 lbm C/100 lbm rock burned, and knowing from Fig. 2 that the H/C ratio is 0.85, an equation similar to Eq. 4 yields the amount of water formed per cubic foot of rock burned. ....................(18)

Thus, the total heat carried forward by the water formed is ....................(19)

Using Eq. 8a, the heat capacity of the formation is ....................(8c)

The amount of heat required to raise the temperature of a cubic foot of sand from 100 to 305°F, from a heat balance, is ....................(20)

Thus, the volume of rock heated by condensing steam is Eq. 19 divided by Eq. 20: ....................(21)

A sketch of the resulting temperature profile is shown in Fig. 5.

### Calculating the effects of injected air and water

Further corrections are needed to increase the accuracy of the temperature profile in Fig. 5. Injected air will partially cool the burned zone, raise the temperature as it approaches the combustion front, and carry heat forward. This will have little effect on the combustion kinetics or the amount of heat generated by combustion, so, in essence, this amount of energy is merely carried forward to extend the size of the steam plateau.

A sketch of this idea is shown in Fig. 6. In this sketch, the area marked "1" is the temperature profile behind the burning front; Area 2 is the steam plateau, which is now larger than calculated before because of the heat carried forward by the combustion gases.

This temperature profile can be approximated as indicated in Fig. 7, where the profiles of the burned zone and steam plateau are treated as square waves that have been adjusted so that the total heat in Areas 1 and 2 is the same as in Fig. 6.

There are several reasons for using this square-wave concept. One is that it makes it easier to calculate heat losses to be expected from either a laboratory or field combustion operation using superposition calculations similar to those discussed by Ramey as seen in Prats. The reference also indicate that the heat losses calculated with Fig. 7 are quite adequate.

When wet combustion is used, the temperature behind the front tends to change abruptly, as shown in Fig. 7. As a result, heat and material balances of the sort discussed next can be used to calculate the movement of the resulting cooling front, burn front, and steam plateau.

The amount of air injected per cubic foot of rock burned and the heat capacity of air are needed to calculate this heat-transfer process for dry combustion. The volume of combustion gas should also theoretically be calculated, but normally this isn’t necessary because its volume is nearly identical to the air volume. Further, its heat capacity is nearly the same—remember that most of the combustion gas is nitrogen (assuming that air is injected).

The moles of air injected are calculated by adding the O2 from Eq. 13 to the N2 from Eq. 17: ....................(22)

The heat capacity for air is 7.00 Btu/lbm mol-°F. With previously determined factors of 0.95 lbm of carbon burned for 100 lbm rock, 164(1 – 0.22) pounds of rock per cubic foot of rock, 12 lbm of carbon per mole, a combustion-zone temperature of 788°F (Fig. 1, and the results of Eq. 22, we obtain the amount of heat carried forward by the injected air as follows. ....................(23)

This heat extracted behind the burned zone is deposited into the steam plateau. The resulting size of the steam plateau can be calculated in a way similar to Eq. 21 by adding the heat carried by the combustion gas to that carried by the water as follows: ....................(24)

This calculation shows that the condensing steam front is far enough ahead of the combustion front to displace oil efficiently, and it is unnecessary to have the combustion front cover the entire reservoir to obtain good recovery. Recovery can be estimated by knowing the amount of fuel burned, by estimating the residual oil saturation in the steam plateau, and by estimating the volumetric sweep efficiency of the process.

### Heat losses

An estimate of heat losses using the superposition concepts seen in Prats, based on Ramey’s work, will make the calculations just presented more accurate. These estimates are particularly important if a laboratory heat balance indicates significant heat losses. The temperature profiles just calculated assuming no heat losses can be used to make a first estimate of heat losses and a recalculated steam-plateau size. This is a reasonable way to handle the heat balance. Because all heat transferred was assumed to be in the steam plateau, any reduction in that transferred heat because of losses will reduce the amount of heat in the plateau.

Because the size of the steam zone and the size of the calculated heat losses are interdependent, iterative calculations are necessary until the assumed and calculated heat balances match. This will usually require only two to three iterations to achieve this match.

Data used in the previous calculations were based on generalized predictions of combustion behavior (i.e., the amount of fuel per cubic feet of formation) and the H/C ratio of the fuel. If combustion tube runs are made in the laboratory, those parameters will be known and should be used in the calculations. In addition, accurate temperature and saturation profiles vs. time will allow reasonably accurate heat-balance calculations to determine the heat losses from the experiment. As an alternative, reasonable assumptions about the heat losses can be used to check the heat-balance calculations and indicate if there is significant experimental error.

Computer assisted tomography (CAT) scanner measurements produce the most accurate saturation histories. Accurate measurements of temperature profile and accurate oil, water, and gas production data also make it possible to estimate reasonable saturation histories. These are the major sources of error in the overall heat-balance calculations, but they are fairly small compared to the amount of heat stored in the hot matrix.

## Nomenclature

 A = Arrhenius rate constant cp/cv = heat capacity ratio Cf = concentration of fuel, kg/m3 E = activation energy, kJ/mol Hc = heating value, Btu/lbm fuel K = reaction rate m = CO2/CO atomic ratio m = CO2/CO concentration ratio produced n = hydrogen/carbon atomic ratio of fuel Np = oil produced, m3 (O2)prod = oxygen concentration produced = oxygen partial pressure, Pa R = 0.00831 universal gas constant, kJ/mol °K RC = reaction rate of crude, kg/m3s Sf = oil saturation burned, fraction So = oil saturation, fraction Soi = initial oil saturation, fraction Swf = water saturation resulting from the combustion process, fraction T = absolute temperature, °K T1 = initial reservoir temperature, °F T2 = final reservoir temperature, °F Vb = volume burned, m3 Vp = volume of the pattern, m3 W = fuel availability Wp = water produced, m3 Ф = porosity, fraction μ = oil viscosity