Question

An oil well (wellbore radius = 0.5 inch) in a heavy oil reservoir (drainage radius = 745 ft, oil viscosity = 500 cP) is being operated at 200 rb/day and 150 psi under the radial steady state flow regime. A huff and puff steam injection is planned to reduce the oil viscosity to 35 cP. The steam soaks into the reservoir up to a distance of 65 ft from the centre of the wellbore. The new production rate at the downhole condition after the steam stimulation is ____________ rb/day (rounded off to two decimal places).

Hint:

The production rate after viscosity reduction depends on the change in oil viscosity and the extent of steam injection.

Answer 1

Correct Answer: 661

The production rate after steam injection is affected by the change in viscosity and the distance the steam has soaked into the reservoir. The new production rate can be calculated using the following formula for radial flow under steady state conditions:
\[ q = \frac{2\pi k h (P_1 - P_2)}{\mu L} \times \ln\left(\frac{r_d}{r_w}\right) \] Where:
- \( k \) is the permeability,
- \( h \) is the reservoir thickness,
- \( P_1 - P_2 \) is the pressure difference,
- \( \mu \) is the viscosity, \ - \( L \) is the length,
- \( r_d \) is the drainage radius,
- \( r_w \) is the wellbore radius.
For oil viscosity reduction, the new viscosity is 35 cP. Using the given values and the fact that viscosity inversely affects production, we can adjust the initial production rate by the ratio of viscosities. The ratio of viscosities is: \[ \text{Viscosity ratio} = \frac{500}{35} = 14.29 \] Thus, the new production rate is: \[ q_{\text{new}} = 200 \times 14.29 = 2858 \, \text{rb/day}. \] Final Answer: 2858.0