The General Material-Balance Equation

Material balance treats the reservoir as a tank - cumulative underground withdrawal equals the total expansion of what remains, plus any water influx - and for a volumetric gas reservoir it collapses to the famous p/z straight line.

Petroleum EngineeringReservoir EngineeringFree preview
⏱️ About 20 min
The General Material-Balance Equation — illustration
Illustrative image (AI-generated).

Treat the reservoir as a tank: every barrel removed underground must be matched by expansion of what remains, plus any water that flows in.

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The big idea: Material balance is conservation of mass on reservoir volumes - cumulative underground withdrawal equals total expansion (oil + dissolved gas, gas cap, connate water and rock) plus water influx - and for a volumetric dry-gas reservoir it reduces to the p/z straight line whose intercept at p/z = 0 is the original gas in place.
🎯 By the end, you'll be able to
  • State material balance as conservation of mass: cumulative underground withdrawal equals total expansion of what remains plus water influx
  • Name the material-balance terms qualitatively (oil + dissolved-gas expansion, gas-cap expansion, connate-water and rock expansion, water influx We, production)
  • Derive and use the volumetric dry-gas p/z straight line: p/z = (p_i/z_i)(1 - G_p/G)
  • Extrapolate the p/z line to p/z = 0 to find OGIP, and recognize the water-drive signature that bends the line upward
📎 Helpful to know first

The Reservoir as a Tank

Material balance is the other great reservoir-engineering accounting tool, and it treats the reservoir as a single tank. Its statement is conservation of mass: the cumulative underground withdrawal - the reservoir-barrel equivalent of everything produced so far - must equal the total expansion of everything still in the reservoir, plus any water influx $W_e$ that has entered from a connected aquifer. As fluid is removed, what remains expands to fill the space, and the pressure drops as a direct result. Material balance ties production and pressure together through the physics of expansion.

The Terms of the Equation

The general material-balance equation gathers several expansion terms, each named here qualitatively. Oil and its dissolved gas expand as pressure falls below the bubble point and gas comes out of solution. A gas cap, if one exists, expands as its gas swells. The connate water and the rock pore volume themselves expand slightly as pressure drops (compaction). And water influx $W_e$ from an aquifer adds fluid from outside the original reservoir boundary. Setting cumulative production equal to the sum of these terms is what closes the balance - and once every term but one is known, you solve for that remaining unknown.

✨ The p/z Straight Line (Volumetric Gas)

For a volumetric dry-gas reservoir - one with no water drive and no gas cap - the general equation collapses to a famous straight line. Because the gas in place is fixed and only pressure falls as gas is produced, the ratio $p/z$ declines linearly with cumulative gas produced $G_p$: $\dfrac{p}{z} = \dfrac{p_i}{z_i}\left(1 - \dfrac{G_p}{G}\right)$. Plot $p/z$ versus $G_p$ and the data should fall on a straight line whose intercept at $p/z = 0$ is the original gas in place $G$. A water-drive reservoir behaves differently: as the aquifer pushes in to support pressure, the $p/z$ line flattens (bends upward) instead of tracking the straight line - the diagnostic shape explored in this module's interactive $p/z$ material-balance plot.

📝 Worked example: A volumetric (no water drive, no gas cap) dry-gas reservoir has initial pressure $p_i = 4{,}000$ psia with gas deviation factor $z_i = 0.80$ (so $p_i/z_i = 5{,}000$ psia) and original gas in place $G = 100$ Bscf. After producing $G_p = 30$ Bscf, find $p/z$ from $\dfrac{p}{z} = \dfrac{p_i}{z_i}\left(1 - \dfrac{G_p}{G}\right)$, and if the gas deviation factor at that condition is $z \approx 0.85$, find the pressure $p$.
  1. Initial $p/z$: $p_i/z_i = 4000/0.80 = 5000$ psia.
  2. Production fraction: $G_p/G = 30/100 = 0.30$.
  3. Current $p/z = 5000 \times (1 - 0.30) = 5000 \times 0.70 = 3500$ psia.
  4. If $z \approx 0.85$ at that condition: $p = (p/z) \times z = 3500 \times 0.85 = 2975$ psia.
✓ After producing 30 Bscf, $p/z = 3{,}500$ psia; with $z \approx 0.85$ that corresponds to a pressure $p \approx 2{,}975$ psia.
🎮 p/z Material-Balance Plot LIVE
Predict first: With aquifer support at 0 (volumetric), where does the straight line cross p/z = 0? Now raise the aquifer support and watch what the naive straight-line estimate of OGIP does.
Illustrative gas material-balance model. At the default inputs (p_i/z_i = 5000 psia, G = 100 Bscf, volumetric) the line passes through p/z = 3500 psia at G_p = 30 Bscf, matching the worked example. Turning up aquifer support flattens the curve so the early data extrapolate to an apparent OGIP far above the true value.

Check your understanding

1. A volumetric gas reservoir has p_i/z_i = 5,000 psia and OGIP G = 100 Bscf. After producing G_p = 30 Bscf, what is p/z = (p_i/z_i)(1 - G_p/G)?
$p/z = 5000 \times (1 - 30/100) = 5000 \times 0.70 = 3500$ psia.
2. On a p/z-versus-G_p plot for a volumetric gas reservoir, what does the straight-line extrapolation to p/z = 0 give?
For a volumetric dry-gas reservoir, $p/z$ plots as a straight line versus cumulative gas produced $G_p$; extrapolating to $p/z = 0$ (where $G_p = G$) gives the original gas in place.
🔎 Related lessons
Drive Mechanisms: Solution-Gas, Gas-Cap, Water-Drive & Compaction · soon
✅ Key takeaways
  • Material balance treats the reservoir as a tank: cumulative underground withdrawal equals the total expansion of what remains plus any water influx - conservation of mass on reservoir volumes.
  • The general equation gathers oil + dissolved-gas expansion, gas-cap expansion, connate-water and rock (compaction) expansion, water influx We, and production; once all but one term are known, you solve for the rest.
  • For a volumetric dry-gas reservoir the equation collapses to the p/z straight line, p/z = (p_i/z_i)(1 - G_p/G); extrapolating to p/z = 0 gives OGIP, while a water drive bends the line upward.
➡️ Knowing how much is there and how pressure tracks production, the next question is what is actually pushing the oil to the well - the drive mechanisms.
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