Material Balance in Saturated Reservoirs

The general Schilthuis material balance equation was developed in and is as follows:

Equation (3.7) can be rearranged and solved for N, the initial oil in place:

If the expansion term due to the compressibilities of the formation and connate water can be neglected, as they usually are in a saturated reservoir, then Eq. (7.1) becomes

Example 7.1 shows the application of Eq. (7.2) to the calculation of initial oil in place for a water-drive reservoir with an initial gas cap. The calculations are done once by converting all barrel units to cubic feet units and then a second time by converting all cubic feet units to barrel units. It does not matter which set of units is used, only that each term in the equation is consistent. Problems sometimes arise because gas formation volume factors are reported either in ft³/SCF or in bbl/SCF. Typically, when applying the material balance equation for a liquid reservoir, gas formation volume factors are reported in bbl/SCF. Use care to ensure that the units are correct.

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Example 7.1 Calculating the Stock-Tank Barrels of Oil Initially-in-Place in a Combination Drive Reservoir

Given

Volume of bulk oil zone = 112,000 ac-ft

Volume of bulk gas zone = 19,600 ac-ft

Initial reservoir pressure = 2710 psia

Initial formation volume factor = 1.340 bbl/STB

Initial gas volume factor = 0.006266 ft³/SCF

Initial dissolved GOR = 562 SCF/STB

Oil produced during the interval = 20 MM STB

Reservoir pressure at the end of the interval = 2000 psia

Average produced GOR = 700 SCF/STB

Two-phase formation volume factor at 2000 psia = 1.4954 bbl/STB

Volume of water encroached = 11.58 MM bbl

Volume of water produced = 1.05 MM STB

Formation volume factor of the water = 1.028 bbl/STB

Gas volume factor at 2000 psia = 0.008479 ft³/SCF

Solution

In the use of Eq. (7.2),

B_ti = 1.3400 × 5.615 = 7.5241 ft³/STB

B_t = 1.4954 × 5.615 = 8.3967 ft³/STB

W_e = 11.58 × 5.615 = 65.02 MM ft³

B_w = 1.028 × 5.615 = 5.772 ft³/STB

B_wW_p = 1.028 × 5.615 × 1.05 = 6.06 MM res ft³

Assuming the same porosity and connate water for the oil and gas zones,

Substituting in Eq. (7.2) with all barrel units converted to cubic feet units,

The calculation will be repeated using Eq. (7.2), with B_t in barrels per stock-tank barrel, B_g in barrels per standard cubic foot, and W_e and W_p in barrels.