Calculate the pressure (psig) at the bottom of a 7500 foot column of fluid for each of the following fluids:
a) water
b) drilling mud (sg = 1.3)
c) gas lift column (s. g. = 0.85)
d) natural gas column (sg = 0.6*) ‘*sg for natural gas is compared against air (density = 0.0765 lb/(cu ft))

Pressure at the botton of 7500 ft column is:

a) P (water) = 3256.508 psig

b) P (drilling mud) = 4229.035 psig

c) P (gas lift) = 18.092 psig

d) P (Natural gas) = 17.092 psig

Explanation:

hidrostatic pressure:

∴ density of fluid ( d.f ) ≡ Kg/m³

∴ aceleration o gravity ( g ) = 9.807 m/s²

∴ column height ( h ) = 7500 ft * ( m / 3.28084 ft ) = 2285.99 m

∴ atmospheric pressure (Po) = 101300 Pa ( Kg /m*s²)

specific gravity:

∴ d H2O = 997 Kg/m³

a) water:

⇒ P = ( 997 * 9.807 * 2285.99 ) + 101300

⇒ P = 22452844,88 Pa * ( psig / 6894.76 Pa ) = 3256.508 psig

b) Drilling Mud

⇒ SG = 1.3 = d.f / 997

⇒ d.f = 1296.1 Kg/m³

⇒ P = ( 1296.1 * 9.807 * 2285.99 ) + 101300

⇒ P = 29158182.16 Pa = 4229.035 psig

c) Gas lift column

⇒ SG = 0.85 = d.f / d.air (ref)

∴ d.air = 0.0765 Lb/ft³ = 1.23 Kg/m³

⇒ d.f = 1.23 * 0.85 = 1.0455 Kg/m³

⇒ P = ( 1.0455 * 9.807 * 2285.99 ) + 101300

⇒ P = 124738.755 Pa = 18.092 psig

d) Natural Gas column:

⇒ SG = 0.6 = d.f / 1.23

⇒ d.f = 0.738 Kg/m³

⇒ P = ( 0.738 * 9.807 * 2285.99 ) + 101300

⇒ P = 117845.003 Pa = 17.092 psig

the product of the pressure and volume is also constant

i think it is true since you can test the answer and can use observations to complete it.