Question

Natural gas is produced at a flow rate of 2 MMscf/day at the wellhead having temperature and pressure of \(560^\circ R\) and 200 psi, respectively. The apparent molecular weight and compressibility factor (\(z\)) of the gas are estimated to be 20 g/g-mole and 0.8, respectively, at wellhead conditions. The gas formation volume factor (\(B_g\)) at the wellhead condition is ............ \(\times 10^{-2} \, \text{ft}^3/\text{scf}\) (rounded off to one decimal place).

Hint:

Remember the shortcut formula for gas formation volume factor: \[ B_g = \frac{0.02827 \, zT}{p} \] with \(T\) in \(^\circ R\), \(p\) in psia, and result in \(\text{ft}^3/\text{scf}\).

Answer 1

Step 1: General formula.
The gas formation volume factor is defined as: \[ B_g = \frac{V_g \text{ at reservoir conditions}}{V_g \text{ at standard conditions}}. \] In petroleum engineering units: \[ B_g = 0.02827 \, \frac{zT}{p} \] where, - \(T\) = absolute temperature in Rankine, - \(p\) = pressure in psia, - \(z\) = gas compressibility factor. Step 2: Substitution of given values.
\[ T = 560^\circ R, \quad p = 200 \, \text{psia}, \quad z = 0.8 \] \[ B_g = 0.02827 \times \frac{0.8 \times 560}{200} \] Step 3: Simplification.
\[ 0.8 \times 560 = 448 \] \[ \frac{448}{200} = 2.24 \] \[ B_g = 0.02827 \times 2.24 = 0.0633 \, \text{ft}^3/\text{scf} \] Step 4: Express in \(10^{-2}\).
\[ 0.0633 = 6.33 \times 10^{-2} \, \text{ft}^3/\text{scf} \] Wait – let's check carefully with units again: Step 5: Verification.
Formula often used: \[ B_g = \frac{0.02827 \, z T}{p} \] Substitute: \[ = \frac{0.02827 \times 0.8 \times 560}{200} = \frac{12.68}{200} = 0.0634 \, \text{ft}^3/\text{scf}. \] \[ \Rightarrow B_g = 6.3 \times 10^{-2} \, \text{ft}^3/\text{scf}. \] Final Answer: \[ \boxed{6.3 \times 10^{-2} \, \text{ft}^3/\text{scf}} \]