Question

In lab, fluids stored were measured 5.5 cc of oil and 2 cc of water. If one sample has 3 cm diameter and 9 cm length and oil formation volume factor was 1.18 bbl/ST, gas formation factor 0.004 cu ft/ SCF. If formation has area of 160 acres and thickness 90 ft.

Find:

1) porosity formation

2) oil , gas and water saturation

3) OOIP in bbl and STB 4) IGIP in cu ftand SCF

Answer 1

SOLUTION:

length=9cm

diameter=3cm

radius r=1.5cm

volume of core=pie*r^2*l=3.14*(1.5)^2*9

=63.62 cm3

pore volume of the core

=5cc+2cc+2cc=9cm3

1)Porosity=9cm3/63.62cm3=0.1414=14.14%

2) gas saturation

=2/9=0.2222

=22.22%

oil saturation

=5/9=0.5555

55.55%

water saturation=100-22.22-55.55=22.23%

4)

thickness=90ft

IGIP = [43560 Ah phi Sgi]/Bgi

IGIP=43560*160*90*0.1414*0.2222/0.004

=4927014449 scf

A - Reservoir Area(acres)
h - Reservoir Thickness (ft)
phi - Reservoir Porosity
Sgi - Initial Gas Saturation
Bgi - Initial formation volume factor of gas(rb/scf)

3) OOIP = [7758 Ah phi So]/Bo

OOIP=7758*160*90*.1414*.5555/1.18

=7436416.962 scf

OOIP - Original Oil in Place (scf)

A - Reservoir Area(acres)

h - Reservoir Thickness (ft)

phi - Reservoir Porosity

So - Initial oil Saturation

Bo - Initial formation volume factor of oil(rb/scf)=1.18