Question 3 of 5: Continuous Gas Lift Design

Question 3: Continuous Gas Lift Design (25 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) Depth of gas injection; (b) required gas injection rate.

(a) Depth of gas injection

Approach. First fix the required flowing bottomhole pressure from the IPR. Then find the depth at which the OPERATING casing gas-column pressure, less the valve differential, just balances the tubing pressure obtained by unloading formation fluid (GLR = 100 SCF/STB only, no lift gas yet) up from $P_{wf}$ at 8000 ft — that intersection is the point of injection.

1. Required $P_{wf}$. $P_{wf}=\bar P_r-q_o/J=3000-1000/1=\boxed{P_{wf}=2000\ \text{psi}}$ at 8000 ft.
2. Casing (operating) static gas column. $P_{csg}(D)=P_{op}\exp\!\left[\dfrac{0.01875\,\gamma_g D}{\bar T(D)\,Z}\right]$, with $\bar T(D)$ the average absolute temperature over 0–$D$ (linear geothermal gradient, 100°F at surface to 200°F at 8000 ft) and $Z\approx0.85$.
3. Tubing pressure below the injection point. Marching the homogeneous no-slip mixture pressure (GLR = 100 SCF/STB, dead-oil SG $=141.5/(131.5+35)=0.850$) upward from $P_{wf}=2000$ psi at 8000 ft to a trial depth $D$ gives $P_{tbg,below}(D)$.
4. Balance condition. The valve opens where $P_{csg}(D)-\Delta P_{valve}=P_{tbg,below}(D)$. Solving numerically (bisection) for $D$: $\boxed{D_{inj}=5742\ \text{ft}}$, at which $P_{csg}=1378$ psi and $P_{tbg,below}=P_{inj}=1278$ psi (check: $1378-100=1278$ ✓).
5. Kick-off check. At the same depth the kick-off line reads $P_{csg,ko}(5742\ \text{ft})=1435$ psi, which clears $P_{inj}+\Delta P_{valve}=1378$ psi by a 57 psi margin — the 1250 psi kick-off pressure is therefore sufficient to unload fluid all the way down to the operating valve before switching to the 1200 psi operating pressure.

(b) Required gas injection rate

Approach. Above the injection point the tubing carries formation gas plus injected lift gas. Find the total GLR whose homogeneous no-slip mixture gradient carries $P_{inj}=1278$ psi at 5742 ft up to $P_{wh}=160$ psi at the surface, then subtract the 100 SCF/STB already supplied by the formation.

1. Solve for total GLR. Marching the mixture pressure upward from $(D_{inj},P_{inj})=(5742\ \text{ft},1278\ \text{psi})$ to the surface and requiring the top pressure equal $P_{wh}=160$ psi (bisection on GLR) gives $\boxed{\text{GLR}_{total}=213\ \text{SCF/STB}}$.
2. Injected GLR and rate. $\text{GLR}_{inj}=\text{GLR}_{total}-\text{GLR}_{formation}=213-100=113$ SCF/STB. At the desired $q_o=1000$ STB/day: $\boxed{Q_{gas}=113\times1000/1000=113\ \text{MSCF/day}}$.