Question 5 of 6: Reservoir Oil & Gas Density and Flow Rates

Question 5: Reservoir Oil & Gas Density and Flow Rates (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) ρ_o,res, ρ_g,res; (b) q_o,res, q_g,res; (c) phase state and saturation.

Approach. Build a mass balance per stock-tank barrel of oil (oil mass + dissolved-gas mass, divided by the reservoir volume that mass occupies) for ρ_o,res; use ρ_g,sc/B_g (converted to consistent volume units) for ρ_g,res; convert surface rates to reservoir rates using B_o and B_g; then compare the produced GOR against R_s to test for free gas.

(a) Reservoir densities

1. Gas density at standard conditions. $\rho_{g,sc}=\gamma_g\times\rho_{air,sc}=0.67\times0.076=0.0509\text{ lb/ft}^3$
2. Mass balance per STB of oil. 1 STB = 5.615 ft³, so the oil mass is $53\times5.615=297.6$ lb; the dissolved-gas mass carried with it is $R_s\times\rho_{g,sc}=347\times0.0509=17.67$ lb. Total mass $=297.6+17.67=315.3$ lb.
3. Reservoir oil density. That same STB of oil occupies $B_o\times5.615=1.19\times5.615=6.682\text{ ft}^3$ at reservoir conditions, so $$\rho_{o,res}=\frac{315.3}{6.682}=\boxed{47.2\text{ lb/ft}^3}$$
4. Reservoir gas density. 1 SCF of gas (mass $\rho_{g,sc}=0.0509$ lb) occupies $B_g\times5.615=0.0012\times5.615=0.00674\text{ ft}^3$ at reservoir conditions: $$\rho_{g,res}=\frac{0.0509}{0.00674}=\boxed{7.56\text{ lb/ft}^3}$$

(b) Reservoir flow rates

1. Oil rate in the reservoir converts directly through B_o: $$q_{o,res}=q_{o,sc}\times B_o=2000\times1.19=\boxed{2380\text{ rb/day}}$$
2. Split the produced gas into solution gas and free gas. Solution gas travelling with the oil $=q_{o,sc}\times R_s=2000\times347=694{,}000$ SCF/day. The remainder is free gas: $2{,}000{,}000-694{,}000=1{,}306{,}000$ SCF/day.
3. Free-gas rate in the reservoir converts through B_g: $$q_{g,res}=1{,}306{,}000\times0.0012=\boxed{1567\text{ rb/day}}$$

(c) Phase state and saturation

The produced (total) GOR is $q_{g,sc}/q_{o,sc}=2{,}000{,}000/2000=1000$ SCF/STB, which is greater than the solution GOR at current reservoir pressure, R_s = 347 SCF/STB. Because more gas is being produced than the oil can hold in solution at this pressure, free gas must already exist and be flowing in the reservoir near the wellbore. The reservoir is therefore currently in a two-phase (oil + gas) state, and because free gas is present at the current pressure, the remaining oil is at its bubble point – a saturated reservoir.