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For an oil reservoir, the production zone has been encountered at depth 5220 – 5354 ft. The following data have been calculated from the log analysis: Porosity = 16% Water saturation (Sw) = 27% Oil formation volume factor= 1.38 bbl/STB Area= 84 acres 1- Calculate the stock tank oil initially in place (STOIIP)? 2- Calculate the recoverable oil volume if the recovery factor= 0.58?

For an oil reservoir, the production zone has been encountered at depth 5220 – 5354 ft. The following data have been calculated from the log analysis: Porosity = 16% Water saturation (Sw) = 27% Oil formation volume factor= 1.38 bbl/STB Area= 84 acres 1- Calculate the stock tank oil initially in place (STOIIP)? 2- Calculate the recoverable oil volume if the recovery factor= 0.58?

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
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For an oil reservoir, the production zone has been encountered at depth 5220 – 5354 ft. The following data have been calculated from the log analysis: Porosity = 16% Water saturation (Sw) = 27% Oil formation volume factor= 1.38 bbl/STB Area= 84 acres 1- Calculate the stock tank oil initially in place (STOIIP)? 2- Calculate the recoverable oil volume if the recovery factor= 0.58?
Transcribed Image Text:For an oil reservoir, the production zone has been encountered at depth 5220 – 5354 ft. The following data have been calculated from the log analysis: Porosity = 16% Water saturation (Sw) = 27% Oil formation volume factor= 1.38 bbl/STB Area= 84 acres 1- Calculate the stock tank oil initially in place (STOIIP)? 2- Calculate the recoverable oil volume if the recovery factor= 0.58?
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