Calculate the pressure of the following mixture.

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AI Thread Summary
The discussion focuses on calculating the pressure at various depths in an oil reservoir, specifically at the water/oil contact (WOC) and depths of 12500, 11700, and 11000 feet. The pressure is calculated using the formula P = rho g h + Patm, with densities for water, oil, and gas provided. The initial calculation for pressure at WOC is approximately 64.7 MPa, but there is confusion regarding the correct height to use for subsequent calculations. Various attempts to calculate pressures at different depths yield different results, indicating a need for clarity on which densities and heights to apply. The conversation emphasizes the importance of correctly applying the principles of hydrostatic pressure in multi-phase systems.
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Homework Statement



An oil reservoir contains at water/oil contact (WOC) of 12000ft contains a water column at 800ft, oil at 500ft and gas at 600ft height

Calculate the pressure at WOC, 12500, 11700 and 11000 ft?

Homework Equations



P = rho g h + Patm

1 ft = 0.305m
Rho water = 1000
Rho Oil = 800
Rho gas = 120

The Attempt at a Solution



P WOC = 1000*9.81*(12000*0.305) + 800*9.81*(12000*0.305)+patm = 64.7Mpa

For the water column for example would the pressure be:

Pwoc + (1000*9.81*(800*0.305) = 67.1Mpa ?
or
Pwoc + (1000*9.81*((12000-800)*0.305)) = 98.2Mpa
or
Pwoc+(1000*9.81*((12000+800)*0.305)) = 102.3Mpa

Basically what height do I have to use?

And what densitys do I need to use for the depths at 12000, 11700 and 11000ft please?
 
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New approach does this look better?

Pwoc = (rhoWgh ) + (rhoOgh)
= (1000*9.8*(12000*0.305)) + (800*9.8*(12000*0.305)
= 64.6 Mpa

Pgoc = (rhoG gh) + (rhoO gh)
= ((120*9.8*((12000-800)*0.305))) + (800*9.8*((12000-800)*0.305)))
= 30.8 Mpa

P12500 = (rhoW gh)
= 1000*9.8*(12500*0.305)
= 34.7 Mpa
 
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