Reservoir volumetrics and material balance problem set PetroCalcHub · Original synthetic teaching exercise · 2026-10-24 Practice volumetric inventories, fluid-volume conversions and a simplified material balance. The answer pack requires units and assumptions as well as numerical results. 160 acres, 50 ft net pay, Sw = 0.25, Bg = 0.005 reservoir ft³/scf. Porosity varies. 1. Why is a computed initial gas volume not automatically reserves? Working: ________________________________________________ 2. Reproduce the baseline Gas in place (scf) without rounding intermediate values. Working: ________________________________________________ 3. Compare the first and last recorded scenarios. Which input changes? Working: ________________________________________________ 4. What would be needed to apply this example to a different physical system? Working: ________________________________________________ 5. At Bg = 0.005 reservoir ft³/scf, convert 1,000 reservoir ft³ to standard volume. Working: ________________________________________________ 6. For F = 1,000,000 RB, Eo = 0.1 RB/STB and We = 200,000 RB, solve the simplified F = N Eo + We balance. Working: ________________________________________________ ANSWER KEY 1. Reserves require a recoverable, commercial project classification and supporting evidence; volumetrics alone estimates in-place quantity. 2. 43,560 × 160 × 50 = 348,480,000 ft³; × 0.15 × 0.75 = 39,204,000 reservoir ft³; /0.005 = 7,840,800,000 scf. Gas in place (scf) = 7840800000. This is initial gas in place, without a recovery or reserves classification. 3. Porosity (fraction) changes from 0.1 to 0.2. Other controls remain as stated in the premise. Gas in place (scf) changes from 5227200000 to 10454400000. 4. Replace the synthetic controls with compatible documented inputs and check the relation's assumptions. A in acres, h in ft, φ and Sw as fractions; 43,560 converts acre-ft to ft³. This is initial gas in place, without a recovery or reserves classification. 5. 1,000/0.005 = 200,000 scf. The complementary expansion factor is 200 scf/reservoir ft³. 6. N = (1,000,000 − 200,000)/0.1 = 8,000,000 STB. Other expansion terms are explicitly excluded. Method: Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements. Sources: Texas A&M: reservoir material-balance concepts: https://blasingame.engr.tamu.edu/z_zCourse_Archive/P620_09C/P620_09C_Lectures/P620_09C_Lec_15_Mod3_FunFld_03_MatBal.pdf SLB: gas formation volume factor: https://glossary.slb.com/en/terms/g/gas_formation_volume_factor