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