Question

. Oil is flowing in a 150 ft long rock with a cross-sectional area of 35 ft2, and permeability of 100 millidarcies

at a rate of 6 B/D.  The pressures at outlet end is 20 psia.  Calculate the pressure at the midpoint of the rock.  

Oil viscosity is 5 cp

Answer 1

FROM DARCY'S EQUATION WE KNOW THAT

Q=(A k p) / (n x)

where Q=oil flow rate = 6B/D = 1.104x10^-5 m³/s [ 1B/D =1/150.96 m³/hour =1/(150.96x3600)  m³/s ]

k= permeability =100 millidarcies =9.87x10^-14 m² [1 md =9.87x10^-16 m² ]

n=viscosity of oil =5cp =0.005 kg/ms

p=pressure difference between outlet end and mid point of the rock

x=distance between outlet and mid point = 150/2 ft =75 ft =22.86m

A= cross sectional area =35 ft² = 3.25161 m²

Now from the equation Q=(A k p) / (n x)

p=(Q n x)/Ak

putting the values we get p= (1.104x10^-5x0.005x22.86) / (3.25161x9.87x10^-14 )

=3.932x10⁶ pascals = 570 psia [ 1psia =6894.75 pascals ]

so the pressure difference is 570 psia

pressure in outlet end is 20 psia

so pressure in the mid point is 570+20=590 psia