Practice and answer pack
Reservoir volumetrics and material balance problem set
Practice volumetric inventories, fluid-volume conversions and a simplified material balance. The answer pack requires units and assumptions as well as numerical results.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Build the rock-to-fluid sequence
Start with gross or net geometry, state which is supplied and calculate bulk rock volume. Apply net-to-gross only when starting from gross thickness, then porosity and phase occupancy.
Convert the reservoir fluid volume with the matching formation volume factor. Keep standard conditions and the reservoir property state visible.
Compare different evidence paths
The material-balance questions use a simplified withdrawal–expansion–influx relation. They ask which quantities are known and which remain assumptions.
A numerical answer from one pressure point should therefore be accompanied by an identifiability explanation. The full linked calculator has additional terms that must be evaluated for a real application.
Interpret the result category
An in-place number is not automatically recoverable volume or reserves. The exercises keep those classifications distinct even when an assumed recovery factor is introduced.
Print the worksheet or download its synthetic case table. Answers show the physical ledger so a volume-basis error can be distinguished from simple arithmetic.
Explore the example
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
Select one of 5 recorded scenarios. Intermediate states are not interpolated.
Porosity (fraction) = 0.15. Gas in place (scf): 7840800000; Gas-filled pore volume (ft³): 39204000.
- Gas in place (scf)
Worked example
Rock volume to standard gas inventory
160 acres, 50 ft net pay, Sw = 0.25, Bg = 0.005 reservoir ft³/scf. Porosity varies.
Relation and unit basis
G = 43,560 A h φ(1 − Sw)/Bg
A in acres, h in ft, φ and Sw as fractions; 43,560 converts acre-ft to ft³.
1. Define the baseline
Porosity (fraction) = 0.15
160 acres, 50 ft net pay, Sw = 0.25, Bg = 0.005 reservoir ft³/scf. Porosity varies.
2. Select the relation
G = 43,560 A h φ(1 − Sw)/Bg
A in acres, h in ft, φ and Sw as fractions; 43,560 converts acre-ft to ft³.
3. Substitute the values
43,560 × 160 × 50 = 348,480,000 ft³; × 0.15 × 0.75 = 39,204,000 reservoir ft³; /0.005 = 7,840,800,000 scf
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
Gas in place (scf) = 7840800000; Gas-filled pore volume (ft³) = 39204000
This is initial gas in place, without a recovery or reserves classification.
Result and interpretation
Gas in place (scf) = 7840800000; Gas-filled pore volume (ft³) = 39204000
This is initial gas in place, without a recovery or reserves classification.
Work the volumetric baseline first and use the subsequent questions to test units, sensitivity and classification.
Exact scenario data
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
| Porosity (fraction) | Gas in place (scf) | Gas-filled pore volume (ft³) |
|---|---|---|
| 0.1 | 5.2272e+9 | 2.6136e+7 |
| 0.125 | 6.5340e+9 | 3.2670e+7 |
| 0.15 | 7.8408e+9 | 3.9204e+7 |
| 0.175 | 9.1476e+9 | 4.5738e+7 |
| 0.2 | 1.0454e+10 | 5.2272e+7 |
Scroll the table to see all columns. The highlighted row is the worked baseline. Downloads retain full numerical precision.
Practice problems and answer key
Work through the question before revealing its answer. Use the stated assumptions and include units in every numerical step.
1. Why is a computed initial gas volume not automatically reserves?
Show the worked answer
Reserves require a recoverable, commercial project classification and supporting evidence; volumetrics alone estimates in-place quantity.
2. Reproduce the baseline Gas in place (scf) without rounding intermediate values.
Show the worked answer
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. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
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. What would be needed to apply this example to a different physical system?
Show the worked answer
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. At Bg = 0.005 reservoir ft³/scf, convert 1,000 reservoir ft³ to standard volume.
Show the worked answer
1,000/0.005 = 200,000 scf. The complementary expansion factor is 200 scf/reservoir ft³.
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.
Show the worked answer
N = (1,000,000 − 200,000)/0.1 = 8,000,000 STB. Other expansion terms are explicitly excluded.
Before using the result
- Apply net-to-gross once.
- Match all fluid-volume bases.
- State the interpretation and remaining unknowns.
This is initial gas in place, without a recovery or reserves classification.
Continue learning
Start with the broader workflow guide · Related method guide
Sources and method
The explanation and synthetic examples are original PetroCalcHub teaching material. The references below support the definitions and model context. They do not supply field validation for the synthetic scenarios.