Question

Calculate the pressure (psig) at the bottom of a 7500 foot column of fluid for each of the following fluids:

Water

Drilling Mud (SG = 1.3)

Gas Lift column (S.G. = 0.85)

Natural Gas Column (SG = 0.6*)

‘*SG for Natural Gas is compared against air (density = 0.0765 lb/(cu ft))

Answer 1

pressure at the bottom

P = density x g x height

Height = 7500 ft x 12in/ft = 90000 in

g = 32.174 ft/s2 x 12in/ft = 386.088 in/s2

P = density (lbm/in3) x 386.088 (in/s2) x 90000 (in)

P (psi) = 34747920 x density

Part a

Water density = 62.428 lbm/ft3 x (1ft/12in)³

= 0.03613 lbm/in3

P = 0.03613 x 34747920 = 1255442.35 psi

Part b

Density of drilling mud = specific gravity x density of water

= 1.3 x 0.03613 = 0.046969 lbm/in3

P = 0.046969 x 34747920 = 1632075.05 psi

Part c

Density of gas lift column = specific gravity x density of water

= 0.85 x 0.03613 = 0.0307105 lbm/in3

P = 0.0307105 x 34747920 = 1067125.99 psi

Part d

Density of air = 0.0765 lbm/ft3 x (1ft/12in)³

= 4.427 x 10^-5 lbm/in3

Density of natural gas column = specific gravity x density of air

= 0.6 x 4.427 x 10^-5 = 0.0000265625 lbm/in3

P = 0.0000265625 x 34747920 = 922.99 psi