Question

Miscible displacement process is one of the EOR techniques. The performance of this process depends on fluid physical properties that affect flow behavior in a reservoir. Two of the important properties are density and viscosity. Consider the use of CO\(_2\) for one such process. The density of CO\(_2\) at the reservoir condition is \(\underline{\hspace{2cm}}\) lb/ft\(^3\) (round off to one decimal place). Relevant data for this calculation are given below.
- Reservoir temperature = 300 ºF (422 K)
- Reservoir pressure = 1470 psig (100 atm)
- Compressibility factor (\(z\)) at the reservoir condition = 0.5
- Values of Universal Gas Constant (\(R\)) in different units are listed below:
- \( R = 8.314 \, \text{m}^3 \, \text{Pa.K}^{-1} \, \text{mol}^{-1} \)
- \( R = 10.731 \, \text{psi.ft}^3 \, \text{lb.mol}^{-1} \, \text{ºR}^{-1} \)
- \( R = 0.082 \, \text{L} \, \text{atm.K}^{-1} \, \text{mol}^{-1} \)

Hint:

To calculate the density of a gas at given conditions, use the ideal gas law and make sure to use appropriate units for pressure, volume, and temperature.

Answer 1

Correct Answer: 15.7

To calculate the density of CO\(_2\) at reservoir conditions, we use the ideal gas law in the form: \[ \text{Density} = \frac{P \times M}{R \times T \times z} \] Where:
- \( P \) is the pressure in atm,
- \( M \) is the molar mass of CO\(_2\) (44.01 g/mol),
- \( R \) is the universal gas constant in appropriate units (we use \( R = 10.731 \, \text{psi.ft}^3 \, \text{lb.mol}^{-1} \, \text{ºR}^{-1} \)),
- \( T \) is the temperature in °R,
- \( z \) is the compressibility factor.
First, convert the reservoir temperature to Rankine (°R): \[ T = 300 \, \text{ºF} + 459.67 = 759.67 \, \text{°R} \] Now, calculate the density using the formula: \[ \text{Density} = \frac{1470 \times 44.01}{10.731 \times 759.67 \times 0.5} \] Substituting the values: \[ \text{Density} = \frac{64786.7}{4082.6} \approx 15.87 \, \text{lb/ft}^3 \] Thus, the density of CO\(_2\) at the reservoir condition is \( \boxed{15.9} \, \text{lb/ft}^3 \).