How to calculate pressure of a mixture of oil Reservoir fuids?

[marcoskids]

1. An oil reservoir with normal reservoir pressure at water oil contact (WOC of =12000ft) contains water, oil and gas columns of 800, 500 and 600 ft height respectively.

Calculate reservoir pressure at WOC, gas oil contact (G0C) and depths of 12500, 11700 and 11000 ft?

ρw = 1000 kg/m^3
ρo = 800 kg/m^3
ρg = 120 kg/m^3

g = 9.8 m/s^2
ft = 0.305m
Patm = 0.1 Mpa
2. P = ρgh
3.Would Pressure at WOC be : Patm + Pw + Po = 0.1x10^6 + (1000*9.8*3660m) + (800*9.8*3660m) = 64.7 Mpa

Also what would be the height to use in the P = ρgh formula for the pressure calculations at GOC,depths of 12500, 11700 and 11000 ft.
Would for example the water contact give a height of 12000 - 800 = 11200ft water column?

 

[marcoskids]

Any help at all would be great :)

 

[marcoskids]

Please anyone?

 

[marcoskids]

Does my initial solution look about right?

 

[asteriatic]

I would like to first clarify that the units used in the given information are inconsistent. The heights are given in feet while the densities are given in kg/m^3. Therefore, before any calculations can be done, the units must be converted to a consistent system.

To calculate the pressure of a mixture of oil reservoir fluids, we can use the ideal gas law, which states that pressure (P) is equal to the product of the density (ρ), the acceleration due to gravity (g), and the height (h).

1. Convert the given heights from feet to meters:
WOC = 12000 ft * 0.305 m/ft = 3660 m
GOC = 12500 ft * 0.305 m/ft = 3813 m
depths:
12500 ft * 0.305 m/ft = 3813 m
11700 ft * 0.305 m/ft = 3566 m
11000 ft * 0.305 m/ft = 3355 m

2. Convert the densities from kg/m^3 to g/cm^3:
ρw = 1000 kg/m^3 * (1 g/1000 kg) * (1 m/100 cm)^3 = 1 g/cm^3
ρo = 800 kg/m^3 * (1 g/1000 kg) * (1 m/100 cm)^3 = 0.8 g/cm^3
ρg = 120 kg/m^3 * (1 g/1000 kg) * (1 m/100 cm)^3 = 0.12 g/cm^3

3. Calculate the pressure at WOC using the ideal gas law:
Pwoc = ρw * g * h = (1 g/cm^3) * (9.8 m/s^2) * (3660 m) = 35.9 MPa

4. Calculate the pressure at GOC using the ideal gas law:
Pgoc = ρg * g * h = (0.12 g/cm^3) * (9.8 m/s^2) * (3813 m) = 4.5 MPa

5. Calculate the pressure at the given depths using the ideal gas law:
P12500 = ρo * g * h = (0.8 g/cm^3)