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Ann: Borba Gas Shows Continue in Key Target Formations, page-164

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  1. 122 Posts.
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    It's simple. This is the formula I found.

    711. Calculate the number of pound-moles in gas by dividing the gas-condensate ratio (gas volume in cubic feet per barrel of condensate) by 379.12. Based on known moles of gas and condensate, determine the fraction of produced gas and condensate in the reservoir fluid stream.13. To obtain produced gas volume per acre-foot of reservoir rock, first multiply the gas mole fraction (from step 12) by total number of moles (from step 8) to get the total number of gas moles in one acre-foot of rock and then multiply by 379 to determine cubic feet of gas.14. Multiply the condensate mole fraction (from step 12) by total number of moles (from step 8) to get the number of condensate moles in 1 acre-foot of rock and then divide it by number of moles per barrel of condensate (from step 10) to obtain volume in barrels.ExampleReservoir pressure 3,000 psiaReservoir temperature 240oFPorosity 30 percentInterstitial water saturation 12 percentTank condensate production 400 barrels per dayTank condensate gravity 50o APITank gas production 200 MCF per dayTank gas gravity 1.25 (air = 1)Primary trap gas production 4,000 MCF per dayPrimary trap gas gravity 0.65 (air = 1)[psia, pounds per square in absolute; MCF, thousand cubic feet; API, American Petroleum Institute]SolutionBasis—One acre-foot of reservoir volumeAverage separator gas gravity = {(4,000 × 0.65) + (200 × 1.25)}/ (4,000+200) = 0.679Condensate-gas ratio = 400 × 1,000/ (4,000+200) = 95.2 barrels per million of standard cubic feet (MMSCF)Using the value of condensate-gas ratio of 95.2, gas gravity of 0.679, and condensate gravity of 50o API, the ratio of reservoir fluid gravity to trap gas gravity is 1.39(fig. 2).Therefore, the reservoir fluid gravity = 1.39 × 0.679 = 0.944.Corresponding to gas gravity of 0.944, the pseudo-critical temperature is 435o R and pseudo-critical pressure 647 psia (fig. 3).Pseudo-reduced temperature = (460+240)/435 = 1.61Pseudo-reduced pressure = 3,000/647 = 4.64For the above calculated pseudo-reduced parameters, z = 0.835 (fig. 4).Reservoir hydrocarbon volume in acre-foot rock = 43,560 × 0.3 × (1-0.12) = 11,500 cubic feet.Using general gas equation, calculate number of moles,N = PV/zRT = (3,000 × 11,500)/ (0.835 × 10.73 × 700) = 5,500 pound moles.If all this hydrocarbon were to be produced as gas, gas volume would be = 5,500 × 379 = 2.08 MMCF (million cubic feet) per acre-foot of reservoir rock.

    If anyone can work this out please, please let me know.

 
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