Question

One dimensional oil flow occurs in a horizontal porous medium of length 2000 ft, with a constant cross sectional area of 6000 ft\(^2\) and absolute permeability of 50 mD. The oil has a viscosity of 2 cP and flows along the length of the medium. The pressure at the inlet is 2500 psig and at the outlet is 2200 psig. Using Darcy's equation for single phase, linear flow, the flow rate of oil (in bbl/day, rounded to one decimal place) is _______. [1 bbl = 5.61 ft\(^3\)]

Hint:

Use the field unit form of Darcy's law, q(bbl/day) = 1.127x10^-3 x k x A x deltaP / (mu x L), and plug in the given values directly.

Answer 1

Correct Answer: 25.4

Step 1: Write Darcy's equation for linear, horizontal, single phase flow in field units:
For steady state, horizontal, incompressible linear flow, Darcy's law in commonly used petroleum field units is: \[ q \, (\text{bbl/day}) = \frac{1.127 \times 10^{-3} \, k \, A \, \Delta p}{\mu \, L} \] where k is permeability in mD, A is cross sectional area in ft\(^2\), \(\Delta p\) is the pressure drop in psi, \(\mu\) is viscosity in cP, and L is the length in ft.
Step 2: List the given values:
k = 50 mD, A = 6000 ft\(^2\), L = 2000 ft, \(\mu\) = 2 cP. The pressure drop is \[ \Delta p = 2500 - 2200 = 300 \text{ psi} \]
Step 3: Substitute the values into Darcy's equation and compute step by step:
First compute the numerator piece by piece: \[ 1.127 \times 10^{-3} \times 50 = 0.05635 \] \[ 0.05635 \times 6000 = 338.1 \] \[ 338.1 \times 300 = 101430 \] Next compute the denominator: \[ 2 \times 2000 = 4000 \] Now divide the numerator by the denominator: \[ q = \frac{101430}{4000} = 25.3575 \text{ bbl/day} \]
Step 4: Round the result to one decimal place:
\[ q = 25.3575 \approx 25.4 \text{ bbl/day} \]
Final Answer:
\[ \boxed{25.4 \text{ bbl/day}} \]