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

h

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current