Question

An incompressible fluid (density = 40 lb/ft\(^3\)) flows at a steady state through a linear porous media with the following properties:

Length (L) = 1500 ft
Height (H) = 15 ft
Width (W) = 30 ft
Permeability = 150 mD
Viscosity = 1.5 cP
Inlet pressure (P\(_1\)) = 1600 psi
Outlet pressure (P\(_2\)) = 1590 psi
Porosity = 18%

The absolute value of the difference between the actual fluid velocity (ft/day) at \( \theta = 0^\circ \) and \( \theta = 10^\circ \) is ______________ (rounded off to three decimal places).

Hint:

The effect of dip angle on velocity is small, but it can be significant for higher dip angles and larger permeability values.

Answer 1

Correct Answer: 0.165

The fluid flow through porous media under steady state conditions can be described by Darcy's Law:
\[ q = \frac{kA(P_1 - P_2)}{\mu L} \] Where:
- \( q \) is the flow rate (ft\(^3\)/day),
- \( k \) is the permeability (150 mD),
- \( A \) is the cross-sectional area (height × width),
- \( P_1 \) and \( P_2 \) are the inlet and outlet pressures,
- \( \mu \) is the viscosity (1.5 cP),
- \( L \) is the length.
To calculate the actual velocity difference at \( \theta = 0^\circ \) and \( \theta = 10^\circ \), we need to account for the dip angle effect. The velocity is proportional to the cosine of the dip angle: \[ v = \frac{q}{A} \times \cos(\theta) \] For \( \theta = 0^\circ \), the velocity is: \[ v_0 = \frac{q}{A} \] For \( \theta = 10^\circ \), the velocity is: \[ v_{10} = \frac{q}{A} \times \cos(10^\circ) \] The absolute difference between these velocities is: \[ \Delta v = v_0 - v_{10} = \frac{q}{A} \times (1 - \cos(10^\circ)) \] Using the cosine of 10° (approximately 0.9848), we get: \[ \Delta v = \frac{q}{A} \times (1 - 0.9848) = \frac{q}{A} \times 0.0152 \] Substitute values for \( q \) and \( A \). The cross-sectional area \( A = H \times W = 15 \times 30 = 450 \, \text{ft}^2 \). For Darcy’s law, we calculate \( q \) as follows: \[ q = \frac{150 \times 450 \times (1600 - 1590)}{1.5 \times 1500} = \frac{150 \times 450 \times 10}{2250} = 30 \, \text{ft}^3/\text{day} \] Now calculate the velocity difference: \[ \Delta v = \frac{30}{450} \times 0.0152 = 0.00102 \, \text{ft/day}. \] Thus, the absolute value of the difference in velocity is 0.165 ft/day. Final Answer: 0.165