Volumetric OOIP & OGIP

Before a barrel is produced, the first question is how much oil and gas the reservoir holds - and the volumetric estimate counts it: bulk rock volume, times porosity, times hydrocarbon saturation, converted to surface barrels.

Petroleum EngineeringReservoir EngineeringFree preview
⏱️ About 16 min
Volumetric OOIP & OGIP — illustration
Illustrative image (AI-generated).

How much oil is down there? Before production, you count it - bulk volume, porosity, saturation, and a shrink factor back to surface barrels.

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The big idea: The volumetric (static) estimate multiplies bulk rock volume by porosity and hydrocarbon saturation, then divides by the formation volume factor to convert reservoir barrels into surface barrels; original oil in place N and original gas in place G are the size of the tank, not the amount recoverable.
🎯 By the end, you'll be able to
  • Explain the volumetric (static) estimate of hydrocarbons in place before production begins
  • Apply the OOIP formula N = (7758 A h phi (1-Sw)) / B_oi, naming each term and its unit
  • Relate the analogous OGIP form using the gas formation volume factor B_gi
  • Distinguish original oil/gas in place from recoverable reserves, and locate the estimate's uncertainty in A, h, phi, and Sw

The Volumetric Idea

Before a single barrel is produced, the first question is simply: how much oil or gas does this reservoir hold? The volumetric (or static) estimate answers that by counting. You take the bulk rock volume of the pay zone, multiply by the porosity $\phi$ (the fraction of that rock that is pore space), multiply again by the hydrocarbon saturation $(1 - S_w)$ (the fraction of the pore space filled with hydrocarbon rather than water), and finally convert from reservoir conditions down to surface barrels using the formation volume factor. It is a bookkeeping of what is physically in the ground - no production history required.

Original Oil In Place (OOIP)

For oil, the original oil in place is $N = \dfrac{7758\, A\, h\, \phi\, (1-S_w)}{B_{oi}}$ STB, where the constant 7758 is the number of barrels in one acre-foot, $A$ is the area in acres, $h$ is the net pay thickness in feet, $\phi$ is porosity, $S_w$ is water saturation, and $B_{oi}$ is the initial oil formation volume factor in reservoir barrels per stock-tank barrel (rb/STB). The numerator counts the reservoir barrels of oil sitting in the pores; dividing by $B_{oi}$ shrinks each reservoir barrel down to the smaller surface barrel it becomes at the stock tank. Original gas in place uses the same idea, with the gas formation volume factor $B_{gi}$ in place of $B_{oi}$.

✨ In-Place Is Not Recoverable

A common mistake is to read $N$ as the amount the well will actually produce. It is not. Original oil in place ($N$) or original gas in place ($G$) is everything physically present before production begins. What is actually recoverable is only a fraction of that - the in-place volume times a recovery factor - and the recovery factor depends on the reservoir's drive mechanism and on engineering choices, the topics of the later lessons in this module. For now, treat $N$ and $G$ as the size of the tank, not the amount you can drain from it.

Where the Uncertainty Lives

Because the volumetric estimate multiplies several measured quantities, its uncertainty is the product of their individual uncertainties. The area $A$ and net pay $h$ come from mapping and well logs; porosity $\phi$ and water saturation $S_w$ come from petrophysics; and $B_{oi}$ comes from fluid (PVT) analysis. Each carries error, and they compound multiplicatively - so the headline number is really a distribution, not a single sharp value. Knowing where that uncertainty sits is half of using the volumetric estimate wisely.

📝 Worked example: Estimate the original oil in place (OOIP) for a reservoir with: area $A = 640$ acres, net pay $h = 50$ ft, porosity $\phi = 0.20$, water saturation $S_w = 0.30$, and initial oil formation volume factor $B_{oi} = 1.30$ rb/STB. Use $N = \dfrac{7758\, A\, h\, \phi\, (1-S_w)}{B_{oi}}$.
  1. Build up the numerator: $7758 \times A \times h \times \phi \times (1-S_w) = 7758 \times 640 \times 50 \times 0.20 \times 0.70$.
  2. $7758 \times 640 = 4{,}965{,}120$.
  3. $\times 50 = 248{,}256{,}000$.
  4. $\times 0.20 = 49{,}651{,}200$ (applying porosity).
  5. $\times 0.70 = 34{,}755{,}840$ (applying $1 - S_w = 1 - 0.30$).
  6. $\div 1.30 = 26{,}735{,}262$ STB (dividing by $B_{oi}$ to convert to surface barrels).
✓ $N \approx 26{,}735{,}262$ STB, i.e. about 26.7 million STB (26.7 MMSTB) of original oil in place.

Check your understanding

1. A reservoir has A = 640 acres, h = 50 ft, phi = 0.20, Sw = 0.30, and B_oi = 1.30 rb/STB. Using N = (7758 A h phi (1-Sw)) / B_oi, what is the original oil in place?
$7758 \times 640 \times 50 \times 0.20 \times 0.70 / 1.30 \approx 26{,}735{,}262$ STB $\approx 26.7$ MMSTB.
2. Which statement best distinguishes original oil in place (N) from recoverable reserves?
$N$ is the hydrocarbon physically in the ground before production; recoverable reserves are $N \times RF$, fixed later by the drive mechanism and recovery factor. So in-place is not recoverable.
✅ Key takeaways
  • The volumetric (static) estimate counts hydrocarbons in place before production: bulk rock volume times porosity times hydrocarbon saturation, converted to surface barrels by the formation volume factor.
  • Original oil in place is N = (7758 A h phi (1-Sw)) / B_oi, with 7758 the barrels per acre-foot; OGIP uses the analogous form with the gas formation volume factor B_gi.
  • In-place (N, G) is the size of the tank; recoverable reserves are only a fraction of it (N times a recovery factor), and the estimate's uncertainty lives in A, h, phi, and Sw.
➡️ Counting what is in place is the start. The next lesson shows how production history and pressure let you check that count with the material-balance equation.