1,705
edits
Changes
Jump to navigation
Jump to search
Predicting hydrocarbon recovery (view source)
Revision as of 20:57, 31 January 2014
, 20:57, 31 January 2014Initial import
{{publication
| image = exploring-for-oil-and-gas-traps.png
| width = 120px
| series = Treatise in Petroleum Geology
| title = Exploring for Oil and Gas Traps
| part = Predicting the occurrence of oil and gas traps
| chapter = Predicting reservoir system quality and performance
| frompg = 9-1
| topg = 9-156
| author = Dan J. Hartmann, Edward A. Beaumont
| link = http://archives.datapages.com/data/specpubs/beaumont/ch09/ch09.htm
| pdf =
| store = http://store.aapg.org/detail.aspx?id=545
| isbn = 0-89181-602-X
}}
The volume of hydrocarbon contained in a reservoir is a function of pore volume and water saturation (S<sub>w</sub>). Reservoir size and [[porosity]] determine pore volume. Pore throatsize distribution, pore geometry, and hydrocarbon column height determine S<sub>w</sub>. Estimating hydrocarbon volume in place before drilling a well is a matter of predicting pore volume and S<sub>w</sub>. Recovery of hydrocarbons depends on the efficiency of the reservoir drive mechanism. Predicting recovery depends on predicting [[reservoir quality]] and reservoir drive.
==Calculating oil volume in place==
To calculate volume of original oil in place (OOIP), use the following formula:
:<math>\mbox{OOIP} = \frac{7758 \cdot \mbox{A} \cdot \mbox{h} \cdot \Phi \cdot (1 - \mbox{S}_{\rm w})}{\mbox{B}_{\rm oi}}</math>
where:
* 7758 = conversion factor from acre-ft to bbl
* A = area of reservoir, acres from map data
* h = thickness of reservoir pay, ft
* Φ = porosity (decimal, not percent)
* S<sub>w</sub> = water saturation (decimal, not percent)
* B<sub>oi</sub> = formation volume factor = 1.05 + (N × 0.05), where N = number of ft<sup>3</sup> of gas produced per bbl of oil (GOR). For example, if a well has a GOR of 1,000, then B<sub>oi</sub> = 1.05 + (10 × 0.05) = 1.1.
==Calculating gas volume in place==
To calculate volume of original gas in place (OGIP), use the following formula:
:<math>\mbox{OGIP} = 43{,}560 \cdot \mbox{A} \cdot \mbox{h} \cdot \Phi \cdot (1 - \mbox{S}_{\rm w}) \cdot \left(\mbox{depth} \cdot \frac{0.43}{15}\right)</math>
where:
* 43,560 = conversion factor from acre-ft to ft<sup>3</sup>
==Estimating recoverable volume of oil or gas==
Estimating recoverable oil or gas depends on predicting reservoir quality and reservoir drive. Reservoir analogs help narrow the range of values for variables that determine recovery factor (R.F.). Use the equation below to estimate the recoverable oil or gas in a reservoir:
{{quotation|Recoverable oil or gas = OHIP. R.F.
}}
where:
* OHIP = original hydrocarbons in place
==Estimating recovery factor==
Drive mechanism has the greatest geological impact on recovery factor. Narrowing the range in recovery factor is a matter of estimating how much difference pore type and reservoir heterogeneity impact the efficiency of the drive mechanism. To estimate the recovery factor, use the procedure below.
{| class = "wikitable"
|-
! Step
! Action
|-
| 1
| Decide which drive mechanism is most likely from the geology of the prospective reservoir system and by comparing it with reservoir systems of nearby analog fields or analog fields in other basins.
|-
| 2
| Multiply OOIP or OGIP by the recovery factor for the expected drive.
|-
| 3
| Narrow the recovery factor range by predicting the thickness of the reservoir by port type. Port type affects recovery rate. For example, in a reservoir with strong water drive and macroporosity, recovery will be up to 60%, mesoporosity recovery will be up to 20%, and microporosity recovery will be 0%.
|}
==Recovery factors for different drive types==
The table below shows recovery factor percentages for different drive mechanisms for oil vs. gas reservoirs.
{| class = "wikitable"
|-
! Reservoir drive mechanism
! Percent ultimate recovery
! Gas
! Oil
|-
| Strong water
| 30–40
| 45–60
|-
| Partial water
| 40–50
| 30–45
|-
| Gas expansion
| 50–70
| 20–30
|-
| Solution gas
| N/A
| 15–25
|-
| Rock
| 60–80
| 10–60
|-
| Gravity drainage
| N/A
| 50–70
|}
==See also==
* [[Interpreting water saturation]]
* [[Interpreting Sw distribution in a reservoir]]
* [[Interpreting hydrocarbon shows]]
==External links==
{{search}}
* [http://archives.datapages.com/data/specpubs/beaumont/ch09/ch09.htm Original content in Datapages]
* [http://store.aapg.org/detail.aspx?id=545 Find the book in the AAPG Store]
[[Category:Predicting the occurrence of oil and gas traps]]
[[Category:Predicting reservoir system quality and performance]]
| image = exploring-for-oil-and-gas-traps.png
| width = 120px
| series = Treatise in Petroleum Geology
| title = Exploring for Oil and Gas Traps
| part = Predicting the occurrence of oil and gas traps
| chapter = Predicting reservoir system quality and performance
| frompg = 9-1
| topg = 9-156
| author = Dan J. Hartmann, Edward A. Beaumont
| link = http://archives.datapages.com/data/specpubs/beaumont/ch09/ch09.htm
| pdf =
| store = http://store.aapg.org/detail.aspx?id=545
| isbn = 0-89181-602-X
}}
The volume of hydrocarbon contained in a reservoir is a function of pore volume and water saturation (S<sub>w</sub>). Reservoir size and [[porosity]] determine pore volume. Pore throatsize distribution, pore geometry, and hydrocarbon column height determine S<sub>w</sub>. Estimating hydrocarbon volume in place before drilling a well is a matter of predicting pore volume and S<sub>w</sub>. Recovery of hydrocarbons depends on the efficiency of the reservoir drive mechanism. Predicting recovery depends on predicting [[reservoir quality]] and reservoir drive.
==Calculating oil volume in place==
To calculate volume of original oil in place (OOIP), use the following formula:
:<math>\mbox{OOIP} = \frac{7758 \cdot \mbox{A} \cdot \mbox{h} \cdot \Phi \cdot (1 - \mbox{S}_{\rm w})}{\mbox{B}_{\rm oi}}</math>
where:
* 7758 = conversion factor from acre-ft to bbl
* A = area of reservoir, acres from map data
* h = thickness of reservoir pay, ft
* Φ = porosity (decimal, not percent)
* S<sub>w</sub> = water saturation (decimal, not percent)
* B<sub>oi</sub> = formation volume factor = 1.05 + (N × 0.05), where N = number of ft<sup>3</sup> of gas produced per bbl of oil (GOR). For example, if a well has a GOR of 1,000, then B<sub>oi</sub> = 1.05 + (10 × 0.05) = 1.1.
==Calculating gas volume in place==
To calculate volume of original gas in place (OGIP), use the following formula:
:<math>\mbox{OGIP} = 43{,}560 \cdot \mbox{A} \cdot \mbox{h} \cdot \Phi \cdot (1 - \mbox{S}_{\rm w}) \cdot \left(\mbox{depth} \cdot \frac{0.43}{15}\right)</math>
where:
* 43,560 = conversion factor from acre-ft to ft<sup>3</sup>
==Estimating recoverable volume of oil or gas==
Estimating recoverable oil or gas depends on predicting reservoir quality and reservoir drive. Reservoir analogs help narrow the range of values for variables that determine recovery factor (R.F.). Use the equation below to estimate the recoverable oil or gas in a reservoir:
{{quotation|Recoverable oil or gas = OHIP. R.F.
}}
where:
* OHIP = original hydrocarbons in place
==Estimating recovery factor==
Drive mechanism has the greatest geological impact on recovery factor. Narrowing the range in recovery factor is a matter of estimating how much difference pore type and reservoir heterogeneity impact the efficiency of the drive mechanism. To estimate the recovery factor, use the procedure below.
{| class = "wikitable"
|-
! Step
! Action
|-
| 1
| Decide which drive mechanism is most likely from the geology of the prospective reservoir system and by comparing it with reservoir systems of nearby analog fields or analog fields in other basins.
|-
| 2
| Multiply OOIP or OGIP by the recovery factor for the expected drive.
|-
| 3
| Narrow the recovery factor range by predicting the thickness of the reservoir by port type. Port type affects recovery rate. For example, in a reservoir with strong water drive and macroporosity, recovery will be up to 60%, mesoporosity recovery will be up to 20%, and microporosity recovery will be 0%.
|}
==Recovery factors for different drive types==
The table below shows recovery factor percentages for different drive mechanisms for oil vs. gas reservoirs.
{| class = "wikitable"
|-
! Reservoir drive mechanism
! Percent ultimate recovery
! Gas
! Oil
|-
| Strong water
| 30–40
| 45–60
|-
| Partial water
| 40–50
| 30–45
|-
| Gas expansion
| 50–70
| 20–30
|-
| Solution gas
| N/A
| 15–25
|-
| Rock
| 60–80
| 10–60
|-
| Gravity drainage
| N/A
| 50–70
|}
==See also==
* [[Interpreting water saturation]]
* [[Interpreting Sw distribution in a reservoir]]
* [[Interpreting hydrocarbon shows]]
==External links==
{{search}}
* [http://archives.datapages.com/data/specpubs/beaumont/ch09/ch09.htm Original content in Datapages]
* [http://store.aapg.org/detail.aspx?id=545 Find the book in the AAPG Store]
[[Category:Predicting the occurrence of oil and gas traps]]
[[Category:Predicting reservoir system quality and performance]]