Engineering lesson
Build a volumetric gas in place calculation from a rock volume
Build volumetric gas in place by calculating bulk rock volume, connected pore volume and gas-filled pore volume, then dividing by Bg on a compatible reservoir-to-standard volume basis.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Keep a physical volume ledger
Area times net thickness gives bulk rock volume. Multiply by porosity to obtain pore volume, then by 1 − Sw to obtain gas-filled pore volume in this gas–water example.
Use net thickness once. If gross thickness and net-to-gross have already been combined, applying net-to-gross again removes the same nonpay twice.
Convert at the last physical boundary
Bg maps a specified standard gas volume to reservoir gas volume. If Bg is ft³/scf, divide reservoir cubic feet by Bg to obtain scf. A Bg in reservoir bbl/scf requires a different numerator unit.
The worked example assumes 160 acres, 50 ft net pay, porosity 0.15, Sw 0.25 and Bg 0.005 ft³/scf. Every assumption appears in the data.
Interpret in-place volume correctly
The result is a volumetric estimate at the specified property basis. It is not automatically recoverable gas, sales gas or reserves.
Heterogeneity, contacts and correlated input uncertainty can matter more than displayed decimal places. Preserve the measurement source and distribution behind each average.
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.
Porosity varies while bulk rock volume, Sw and Bg are fixed. The result should change linearly.
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.
Test your understanding
Work through the question before revealing its answer. Use the stated assumptions and include units in every numerical step.
1. What pore-space fraction is gas filled when Sw = 0.25 in a gas–water system?
Show the worked answer
0.75 of pore space. Relative to bulk rock, the gas-filled fraction is φ × 0.75.
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.
Before using the result
- Use net thickness once.
- Match reservoir volume to Bg units.
- Keep in-place volume separate from recovery.
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.