Two wells were drilled into the same formation. One was completed at a depth of 5800 ft. When it was shut-in, it was filled with brine with a surface pressure of 70 psia. The second well was completed at 4800ft. Upon shut-in, it was filled with oil with a surface pressure of 695 psia. Assuming the oil and water densities are 45 and 63 lb/ft^3 respectively.
a) At what depth is the Water-Oil Contact ( WOC ) ?
b) What is the pressure at that depth ?
a ) PA = 70 + ( 5800 - h1 ) ( 63 ) g/gc * ( 1 / 144 )
PB = 695 + ( 4800 + h2 ) ( 45 ) g/gc * ( 1 / 144 )
h1 + h2 = 1000 ft
PA = PB
695 - 70 = [ 5800 - ( 1000 - h2 ) ] × ( 63 / 144 ) - [ 4800 + h2 ] × ( 45 / 144 )
h2 = 200 ft
WOC depth = 4800 + 200 = 5000 ft
h woc = 5000 ft
b ) PA = PB = 70 + ( 5000 ) ( 63 / 144 ) = 2257.5 Psi
a ) PA = 70 + ( 5800 - h1 ) ( 63 ) g/gc * ( 1 / 144 )
PB = 695 + ( 4800 + h2 ) ( 45 ) g/gc * ( 1 / 144 )
h1 + h2 = 1000 ft
PA = PB
695 - 70 = [ 5800 - ( 1000 - h2 ) ] × ( 63 / 144 ) - [ 4800 + h2 ] × ( 45 / 144 )
h2 = 200 ft
WOC depth = 4800 + 200 = 5000 ft
h woc = 5000 ft
b ) PA = PB = 70 + ( 5000 ) ( 63 / 144 ) = 2257.5 Psi
Given, Density of Oil = 45lb/ft3
Density of water = 63lb/ft3
Oil depth = 4800ft
Water depth = 5800ft
P = PA + ẞ(D)
Where P = Pressure, PA = Surface Pressure, ẞ = Geothermal Gradient, D= Depth
For Oil
Po = PA + ẞo (D)
ẞo = 45(lb/ft3)*(ft2/144in) = 0.3125Psia/ft
ẞw = 63(lb/ft3)*(ft2/144in) = 0.4375 Psia/ft
So, Po = 695+0.3125D
Pw = 70+0.4375D
At the OWC, Po = Pw
So, 695+0.3125D = 70+0.4375D
695-70 = 0.4375D – 0.3125D
625 = 0.125D, D = 5000ft
For the pressure at WOC, either of the two equations can be used
Therefore pressure at WOC = 695+0.3125(5000) = 2257.5 Psia.
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