A wildcat is drilled into an oil reservoir. It encounters a Gas-Oil contact at 4500 ft;...

  

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A wildcat is drilled into an oil reservoir. It encounters a Gas-Oil contact at 4500 ft; the pressure at the Gas-Oil contact is 2185 psia. Several samples of water collected from surrounding aquifers indicated an average water density of 64.8 lb ft. A pressure measurement in a nearby aquifer at a depth of 4300 ft was 2045 psia. Hydrocarbon samples from the reservoir were analyzed; oil density was determined to be 50.4 lbm/ lb/ft, while gas density was estimated at 10.08 lbm/ lb/ft³. Answer the following questions: (a) (10 points) What is the water pressure equation? (b) (10 points) Estimate the depth to the Oil-Water contact. (c) (5 points) If the well had been drilled into the gas cap at 4350 ft (i.c., no Gas-Oil contact was encountered). approximately what value would you expect for the pressure of gas at 4350 ft? Explain your answer. A wildcat is drilled into an oil reservoir. It encounters a Gas-Oil contact at 4500 ft; the pressure at the Gas-Oil contact is 2185 psia. Several samples of water collected from surrounding aquifers indicated an average water density of 64.8 lb ft. A pressure measurement in a nearby aquifer at a depth of 4300 ft was 2045 psia. Hydrocarbon samples from the reservoir were analyzed; oil density was determined to be 50.4 lbm/ lb/ft, while gas density was estimated at 10.08 lbm/ lb/ft³. Answer the following questions: (a) (10 points) What is the water pressure equation? (b) (10 points) Estimate the depth to the Oil-Water contact. (c) (5 points) If the well had been drilled into the gas cap at 4350 ft (i.c., no Gas-Oil contact was encountered). approximately what value would you expect for the pressure of gas at 4350 ft? Explain your answer.

Answer rating: 100% (QA)

SOLUTION a The water pressure equation is given by the hydrostatic pressure formula Pwater Psurface water g h where Pwater is the water pressure at a certain depth Psurface is the pressure at the surf... View the full answer