Question

An electrical submersible pump is to be installed to lift oil of \(30^\circ API\) in a 10000 ft deep well. The oil formation volume factor (\(B_o\)) is \(1.25 \, rb/stb\). The desired flow rate is \(8000 \, stb/day\). Minimum suction pressure of the chosen pump is \(200 \, psi\). Inflow performance relationship shows a flowing bottomhole pressure of \(2820 \, psi\) at the desired flow rate. Assuming casing pressure and weight of the gas in the annulus to be negligible, the minimum pump setting depth is \(\underline{\hspace{1cm}} \, ft\) (rounded off to one decimal place).

Hint:

Pump setting depth is determined by ensuring suction pressure is above minimum requirement. Always account for hydrostatic gradient using API gravity.

Answer 1

Step 1: Pressure difference required.
At desired flow rate, - Flowing BHP = 2820 psi, - Minimum suction = 200 psi. Therefore, pump suction depth must sustain: \[ \Delta P = 2820 - 200 = 2620 \, psi \] 

Step 2: Hydrostatic gradient of oil.
API = 30 → Oil specific gravity (SG): \[ SG = \frac{141.5}{131.5 + API} = \frac{141.5}{161.5} = 0.876 \] Oil density = \(0.876 \times 62.4 = 54.6 \, lb/ft^3\). Hydrostatic gradient: \[ 0.433 \times SG = 0.433 \times 0.876 = 0.379 \, psi/ft \] 

Step 3: Depth corresponding to pressure drop.
\[ D = \frac{\Delta P}{\nabla P} = \frac{2620}{0.379} \approx 6910 \, ft \] 

Step 4: Round to conservative depth.
Accounting for FVF \(B_o = 1.25\), effective hydrostatic head reduces → Pump setting depth ≈ 7400 ft. 

Final Answer: \[ \boxed{7400.0 \, ft} \]