Worked Example: Estimating Gas Reservoir OGIP & Reserves using p/Z Material Balance

The general material balance equation (MBE) is one of the most powerful diagnostic and reserves evaluation tools available to the petroleum reservoir engineer. For a dry gas reservoir under volumetric pressure depletion, the relationship between reservoir pressure, gas compressibility, and cumulative production forms a fundamental straight line when plotted as $p/Z$ versus $G_p$.

This worked example walks through the complete reservoir engineering analysis of a volumetric gas reservoir located at a true vertical depth of $9,500\text{ ft}$, evaluating dynamic Original Gas In Place ($G$), recoverable reserves ($EUR$), recovery factor ($RF$), and drive mechanism verification across five static pressure surveys.

1. Reservoir Parameters & Production History

Reservoir & Fluid Properties:

• Initial Static Bottomhole Pressure ($p_i$): $250.0\text{ barg}$ ($3,640.7\text{ psia} = 251.01\text{ bar(a)}$)
• Reservoir Temperature ($T_{res}$): $85.0^\circ\text{C}$ ($185.0^\circ\text{F} = 644.67\text{ R}$)
• Gas Specific Gravity ($\gamma_g$): $0.65$ ($\text{Air} = 1.0$)
• Economic Abandonment Pressure ($p_{ab}$): $25.0\text{ barg}$ ($377.3\text{ psia}$)

Historical Static Bottomhole Pressure (BHP) Surveys:

Production surveillance has recorded five high-accuracy shut-in pressure surveys over a 6-year production history:

2. Step 1: Real Gas Compressibility Factor ($Z$) & $p/Z$ Evaluation

The gas deviation factor $Z(p, T)$ is computed at each survey point using the pseudo-critical properties of Standing and the Hall-Yarborough / Dranchuk-Abu-Kassem equation of state:

For $\gamma_g = 0.65$: * $P_{pc} = 670.0 - 15.0 \cdot (0.65 - 0.55) = 668.5\text{ psia}$ * $T_{pc} = 168.0 + 325.0 \cdot (0.65 - 0.55) = 200.5\text{ K} = 360.9\text{ R}$ * Pseudo-reduced temperature: $T_{pr} = \frac{644.67}{360.9} = 1.786$

Point-by-Point $p/Z$ Computation:

1. Initial Survey ($p_i = 3,640.7\text{ psia}$):
2. $P_{pr,i} = \frac{3,640.7}{668.5} = 5.446$
3. $Z_i = \mathbf{0.8543}$

4. $(p/Z)_i = \frac{3,640.7}{0.8543} = \mathbf{4,261.6\text{ psia}} \quad (\mathbf{293.8\text{ bar}})$

5. Survey 2 ($p = 3,278.1\text{ psia}, G_p = 2,500\text{ MMSCF}$):

6. $P_{pr} = 4.904 \rightarrow Z = \mathbf{0.8581}$

7. $p/Z = \frac{3,278.1}{0.8581} = \mathbf{3,819.5\text{ psia}} \quad (\mathbf{263.3\text{ bar}})$

8. Survey 3 ($p = 2,915.5\text{ psia}, G_p = 5,000\text{ MMSCF}$):

9. $P_{pr} = 4.361 \rightarrow Z = \mathbf{0.8642}$

10. $p/Z = \frac{2,915.5}{0.8642} = \mathbf{3,374.0\text{ psia}} \quad (\mathbf{232.6\text{ bar}})$

11. Survey 4 ($p = 2,552.9\text{ psia}, G_p = 7,500\text{ MMSCF}$):

12. $P_{pr} = 3.819 \rightarrow Z = \mathbf{0.8728}$

13. $p/Z = \frac{2,552.9}{0.8728} = \mathbf{2,924.9\text{ psia}} \quad (\mathbf{201.7\text{ bar}})$

14. Survey 5 ($p = 2,190.3\text{ psia}, G_p = 10,000\text{ MMSCF}$):

15. $P_{pr} = 3.276 \rightarrow Z = \mathbf{0.8842}$
16. $p/Z = \frac{2,190.3}{0.8842} = \mathbf{2,477.1\text{ psia}} \quad (\mathbf{170.8\text{ bar}})$

3. Step 2: Linear Regression Analysis ($p/Z$ vs $G_p$)

Under the classical Craft & Hawkins material balance relation:

$$\frac{p}{Z} = \frac{p_i}{Z_i} - m \cdot G_p$$

Applying least-squares linear regression to the five coordinates $(G_p, p/Z)$:

• Slope ($m$): $$m = \frac{\sum (G_p - \bar{G}_p)(p/Z - \overline{p/Z})}{\sum (G_p - \bar{G}_p)^2} = \mathbf{-0.17845\text{ psia/MMSCF}}$$ Or in oilfield units: $\mathbf{-178.45\text{ psia/BCF}}$

• Intercept ($p_i / Z_i$): $$\text{Intercept} = \mathbf{4,261.6\text{ psia}}$$

• Coefficient of Determination ($R^2$): $$R^2 = \mathbf{0.9998} \quad (\text{Near-perfect linearity})$$

4. Step 3: Original Gas In Place (OGIP $G$) Determination

In a volumetric gas reservoir, extrapolating the linear regression line to $p/Z = 0$ yields the Original Gas In Place ($G$):

$$G = \frac{(p_i / Z_i)}{|m|}$$

$$G = \frac{4,261.6\text{ psia}}{0.17845\text{ psia/MMSCF}} = \mathbf{23,881.2\text{ MMSCF}} = \mathbf{23.88\text{ BCF}} \quad (\mathbf{0.676\text{ BSm}^3})$$

5. Step 4: Abandonment Pressure & Recoverable Reserves ($EUR$)

At field abandonment pressure $p_{ab} = 25.0\text{ barg}$ ($377.3\text{ psia}$): * $P_{pr,ab} = \frac{377.3}{668.5} = 0.564 \rightarrow Z_{ab} = \mathbf{0.9702}$ * $(p/Z)_{ab} = \frac{377.3}{0.9702} = \mathbf{388.9\text{ psia}} \quad (\mathbf{26.8\text{ bar}})$

Estimated Ultimate Recovery ($EUR$):

$$G_{p,ab} = G \left[ 1 - \frac{(p/Z)_{ab}}{(p_i / Z_i)} \right]$$

$$G_{p,ab} = 23,881.2 \times \left[ 1 - \frac{388.9}{4,261.6} \right] = 23,881.2 \times [1 - 0.09126] = \mathbf{21,701.9\text{ MMSCF}} = \mathbf{21.70\text{ BCF}}$$

Field Recovery Factor ($RF$):

$$RF = \frac{G_{p,ab}}{G} \times 100\% = \frac{21,701.9}{23,881.2} \times 100\% = \mathbf{90.87\%}$$

Remaining Recoverable Reserves:

Given current cumulative production of $G_p = 10,000.0\text{ MMSCF}$ ($10.00\text{ BCF}$): $$G_{rem} = G_{p,ab} - G_{p,current} = 21,701.9 - 10,000.0 = \mathbf{11,701.9\text{ MMSCF}} = \mathbf{11.70\text{ BCF}}$$

6. Step 5: Drive Mechanism Diagnostics (Cole & Havlena-Odeh Plots)

To confirm whether active water drive or aquifer influx is present, the reservoir expansion terms are calculated:

$$B_g = 0.02827 \frac{Z \cdot T (R)}{p (\text{psia})} \quad \left[\frac{\text{rcf}}{\text{scf}}\right]$$ $$B_{gi} = 0.02827 \frac{0.8543 \times 644.67}{3,640.7} = 0.004273\text{ rcf/scf}$$

Evaluating the Cole diagnostic grouping $F / E_g$ across historical points: * Underground voidage $F = G_p B_g$ * Gas expansion $E_g = B_g - B_{gi}$

Because $R^2 = 0.9998$ and the residuals show zero upward curvature ($residuals[-1] \approx 0$), the Havlena-Odeh plot of $F$ vs $E_g$ forms a straight line passing through the origin with slope $G = 23.88\text{ BCF}$. This positively confirms a 100% Volumetric Depletion Drive with zero active aquifer water support.

7. Engineering Summary & Results Dashboard