Preg 3 Y 8.... Original.docx
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Demuestre matemΓ‘ticamente la π 47% y π 26% ademΓ‘s seΓ±ale las π y K queud conoce nombrΓ‘ndolas. PARA π 47% d3 β d3 Οβ6 VT β VS Ο= Γ 100 β Ο = Γ 100 VT d3 d3 (1 β Οβ6) Ο Ο= Γ 100 β Ο = (1 β ) Γ 100 β Ο = 47.64% 3 d 6 PARA π 26% Vb = π2 sen β. d = π3 1ππ3 Vg = 6 d3 π ππ β βd3 Οβ6) Ο= Γ 100 d3 π ππ β π ππ45Β° β Οβ6) Ο=[ ] Γ 100 = 25.95% = 26% π ππ45Β°
8
Calcular las reservas de gas inicial del yacimiento Jibaro, que tiene los siguientes datos: A= 160 Acres, π = 22 %, Sw=23%, Gas residual despuΓ©s de la depletacion es 34%. Bgi = 0.00533 cu. Ft/scf a P1= 3250 psia Bg = 0.00667 cu. Ft/scf a P2= 2500 psia Bg = 0.03623 cu. Ft/scf a P3= 500 psia Net. Productive Thickness = 40 feet. SOLUCION: ππππ’πππ πππππ = 43560 β 0.22 β 160 β 40 = 61.33 β 106 ππ’ β ππ‘ 1 β 0.23 πΊππ πΌππππππ ππ ππ ππ’πππ: πΊ1 = 61.33 β 106 β = 8860 ππππΆπΉ 0.0053 πΊππ ππ ππ ππ’πππ πππ ππ’ππ ππ ππ πππππππππ π£πππ’πππ‘ππππ π 2500 ππ ππ: 1 β 0.23 πΊ2 = 61.33 β 106 β = 7080 ππππΆπΉ 0.00667 πΊππ ππ ππ ππ’πππ πππ ππ’ππ ππ ππ πππππππππ π£πππ’πππ‘ππππ π 500 ππ ππ: 1 β 0.23 πΊ3 = 61.33 β 106 β = 1303 ππππΆπΉ 0.03623 πΊππ ππ ππ ππ’πππ πππ ππ’ππ ππ ππ πππ£πππππ ππ πππ’π π 3250 ππ ππ: 0.34 πΊ4 = 61.33 β 106 β = 3912 ππππΆπΉ 0.00533 πΊππ ππ ππ ππ’πππ πππ ππ’ππ ππ ππ πππ£πππππ ππ πππ’π π 2500 ππ ππ: 0.34 πΊ5 = 61.33 β 106 β = 3126 ππππΆπΉ 0.00533 π ππ πππ£π πππππππ πππ πππππππππ π 500 ππ ππ: 0.34 πΊ1 β πΊ3 = (8860 β 1303) β 106 β = 7557 ππππΆπΉ 0.00533 π ππ πππ£π πππππππ πππ πππππ’πππππ ππ πππ’π π 3250 ππ ππ: πΊ1 β πΊ4 = (8860 β 3912) β 106 = 4948 ππππΆπΉ π ππ πππ£π πππππππ πππ πππππ’πππππ ππ πππ’π π 2500 ππ ππ: πΊ1 β πΊ5 = (8860 β 3126) β 106 = 5734 ππππΆπΉ ππ βππ¦ π’π πππ§π ππ ππππππ πππ, ππ πππ πππ£π πππππππ πππ πππππ’ππππ ππ πππ’π π 3250 ππ ππ, ππ : 1 1 (πΊ1 β πΊ4 ) = (8860 β 3912) β 106 = 2474 ππππΆπΉ 2 2
Demuestre matemΓ‘ticamente la π 47% y π 26% ademΓ‘s seΓ±ale las π y K queud conoce nombrΓ‘ndolas. PARA π 47% d3 β d3 Οβ6 VT β VS Ο= Γ 100 β Ο = Γ 100 VT d3 d3 (1 β Οβ6) Ο Ο= Γ 100 β Ο = (1 β ) Γ 100 β Ο = 47.64% 3 d 6 PARA π 26% Vb = π2 sen β. d = π3 1ππ3 Vg = 6 d3 π ππ β βd3 Οβ6) Ο= Γ 100 d3 π ππ β π ππ45Β° β Οβ6) Ο=[ ] Γ 100 = 25.95% = 26% π ππ45Β°
8
Calcular las reservas de gas inicial del yacimiento Jibaro, que tiene los siguientes datos: A= 160 Acres, π = 22 %, Sw=23%, Gas residual despuΓ©s de la depletacion es 34%. Bgi = 0.00533 cu. Ft/scf a P1= 3250 psia Bg = 0.00667 cu. Ft/scf a P2= 2500 psia Bg = 0.03623 cu. Ft/scf a P3= 500 psia Net. Productive Thickness = 40 feet. SOLUCION: ππππ’πππ πππππ = 43560 β 0.22 β 160 β 40 = 61.33 β 106 ππ’ β ππ‘ 1 β 0.23 πΊππ πΌππππππ ππ ππ ππ’πππ: πΊ1 = 61.33 β 106 β = 8860 ππππΆπΉ 0.0053 πΊππ ππ ππ ππ’πππ πππ ππ’ππ ππ ππ πππππππππ π£πππ’πππ‘ππππ π 2500 ππ ππ: 1 β 0.23 πΊ2 = 61.33 β 106 β = 7080 ππππΆπΉ 0.00667 πΊππ ππ ππ ππ’πππ πππ ππ’ππ ππ ππ πππππππππ π£πππ’πππ‘ππππ π 500 ππ ππ: 1 β 0.23 πΊ3 = 61.33 β 106 β = 1303 ππππΆπΉ 0.03623 πΊππ ππ ππ ππ’πππ πππ ππ’ππ ππ ππ πππ£πππππ ππ πππ’π π 3250 ππ ππ: 0.34 πΊ4 = 61.33 β 106 β = 3912 ππππΆπΉ 0.00533 πΊππ ππ ππ ππ’πππ πππ ππ’ππ ππ ππ πππ£πππππ ππ πππ’π π 2500 ππ ππ: 0.34 πΊ5 = 61.33 β 106 β = 3126 ππππΆπΉ 0.00533 π ππ πππ£π πππππππ πππ πππππππππ π 500 ππ ππ: 0.34 πΊ1 β πΊ3 = (8860 β 1303) β 106 β = 7557 ππππΆπΉ 0.00533 π ππ πππ£π πππππππ πππ πππππ’πππππ ππ πππ’π π 3250 ππ ππ: πΊ1 β πΊ4 = (8860 β 3912) β 106 = 4948 ππππΆπΉ π ππ πππ£π πππππππ πππ πππππ’πππππ ππ πππ’π π 2500 ππ ππ: πΊ1 β πΊ5 = (8860 β 3126) β 106 = 5734 ππππΆπΉ ππ βππ¦ π’π πππ§π ππ ππππππ πππ, ππ πππ πππ£π πππππππ πππ πππππ’ππππ ππ πππ’π π 3250 ππ ππ, ππ : 1 1 (πΊ1 β πΊ4 ) = (8860 β 3912) β 106 = 2474 ππππΆπΉ 2 2
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