Reserves are defined as the amount of oil or gas that can be produced from a reservoir with current technology at current prices and current costs. Since these change on a daily basis, economic reserves can vary with time, increasing or decreasing with changes in the overall world economic conditions. Decline curve analysis and material balance methods are used to calculate remaining reserves based on actual production and pressure data. Coupled with the volumetric analysis from petrophysical data, a reasonable solution can usually be found, although differences between the results from the three models may lead to a revised reservoir description. Additional data might be needed to resolve discrepancies, such as additional production history, new wells, better PVT data, or a new geological interpretation. Calculate
oil or gas in place
Where:
NOTE:
AREA is in acres for English units
Spacing unit for oil is usually 160 acres (640 000 m2 approx) In
some parts of the world, oil is measured in metric tonnes
instead of barrels or cubic meters. Where:
Where:
Sxo = 0.5 Sw = 0.2 HPV = 1.56 ft GOR = 500 scf/bbl AREA = 640 acres gas gravity = 0.85 TF
= 460 + 160 = 620 degrees R Remember to round your answers to two or three significant digits, which you started with.
The following formulas are for use in areas where reserves are measured in metric tonnes, or as weight fraction (or weight percent).
1: WTtar = PHIe * (1 - Sw) * DENShy / 1000 Shale weight. 2: WTsh = Vsh * DENSSH / 1000 Sand weight. 3: WTsnd = (1 - Vsh - PHIe) * DENSMA / 1000 Water weight. 4: WTwtr = PHIe * Sw * DENSW / 1000 Total rock weight. 5: WTrock = WTtar + WTsh + WTsnd + WTwtr Tar mass fraction. 6: TARfrac = WTtar / WTrock Tar weight percent. 7: TARwt% = 100 * WTtar / WTrock Bitumen in place. (OOIP) 8: OOIP = 0.001 * TARfrac * NetPay * DENSoil * AREA (tonnes, AREA in m2) Or 8A: OOIP = HPV * NetPay * AREA / Bo (m3, AREA in m2) Or 8B: OOIP = 7758 * HPV * NetPay * AREA / Bo (bbl, AREA in acres) Where:
Results are in tonnes/m2 except where noted. To obtain tonnes in place, multiply by area in square meters. To obtain reserves, multiply this result by a recovery factor. CAUTION: Some core analysis results do not include water (dry basis analysis). To compare log analysis results to core, eliminate the water term from WTrock. NOTE: In much of Eastern Europe and Asia, oil quantities are reported in tonnes and not barrels. The "tar" equations provide the conversions needed. When using this method for oil, replace the word “tar” with “oil” to prevent confusion.
PHIe = 0.30 Sw = 0.10 Vsh = 0.10 AREA = 640 acres or AREA = 2 550 000 m2 DENSW = 1000 kg/m3 DENSoil = 1000 kg/m3 DENSSH = 2300 kg/m3 DENSMA = 2650 kg/m3 NetPay = 10 meters WTtar
- 0.30 * (1 - 0.10) * 800 / 1000 = 0.216 tonnes/m2
Excess Hydrogen 1. Methane CH4.6H20 4/12 = 33% 2. Ethane C2H6.8H20 6/16 = 37% 3. Propane C3H8.17H20 8/34 = 23% The volume of hydrocarbon in a gas hydrate js a function of the hydrocarbon type and the porosity only. Water saturation is meaningless. The ratio of gas to water would range from 433 scf/bbl for propane to 1230 scf/bbl for methane. This is equivalent to 170 cubic feet (for methane) per cubic foot of pore space (or 170 m3 per cubic meter of pore space) at standard temperature and pressure, for methane, and 60 cubic feet of propane per cubic foot of pore space, regardless of depth of burial.
1: PV = SUM (PHIe * THICK) 2: HPV = PV * KG0 Calculate gas in place. 3: GHIP = KV3 * HPV * AREA Where: Where:
PHIe = 0.35 hydrate is methane THICK = 300 feet KG0 = 170 scf/scf AREA = 640 acres HPV = (0.35 * 300) * 170 = 17850 ft GHIP = 43.56 * 17850 * 640 / 1 000 000 = 497.634 Bcf/section |
||||

Page Views ---- Since 01 Jan 2015
Copyright 1978 - 2022 E. R. Crain, P.Eng. All Rights Reserved |