Electrical Submersible Pump — Required Horsepower: 24-Pet-A5 Petroleum Production Operations, December 2017, Q5

Question 5: Electrical Submersible Pump — Required Horsepower (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. A single-zone oil well (no water) with a 7000 ft total depth, ESP set at 6000 ft, 400 SCF/STBL producing GLR, half the free gas separated at the pump.

Well depth / pump setting depth	7000 ft / 6000 ft
Productivity index, $J$	1 bbl/day/psi
Reservoir pressure, $\bar P_R$	1800 psi
Desired oil rate, $q_o$ ($f_w=0$)	1000 STBO/day
Oil gravity / gas gravity	35°API / 0.65
$B_o$ / GLR	1.1 bbl/STB / 400 SCF/STBL
Wellhead pressure, $P_{wh}$	160 psi
Casing gradient below pump	$dP/dL=0.0001\,q_L$ psi/ft

Find. The required ESP hydraulic horsepower.

Check: the pump handles a no-slip, homogeneous mixture of liquid (constant $B_o=1.1$) plus the un-separated half of the free gas (200 SCF/STBL), with gas volume from the real-gas law ($z$ via the Papay correlation) at local pressure/temperature; friction is neglected above the pump, matching the given below-pump casing gradient being purely a function of rate (no friction term either). Hydraulic HP $=Q_{gpm}\times\Delta P_{pump}/1714$, equivalent to $Q\times\text{TDH}\times SG/3960$; no pump efficiency is given, so this is hydraulic horsepower — a real nameplate motor needs roughly 50–65% additional capacity to cover pump inefficiency.

Approach. Get $P_{wf}$ from the straight-line IPR, march the given casing gradient up 1000 ft to the pump intake, march the no-slip tubing gradient (with the reduced GLR) down from the wellhead requirement to find the pump discharge pressure, then combine the pump's flow rate and differential pressure into hydraulic horsepower.

Flowing bottomhole pressure (IPR). $P_{wf}=\bar P_R-q_o/J=1800-1000/1$. $\boxed{P_{wf}=800\ \text{psi at 7000 ft}}$.
Pump intake pressure. Casing gradient $=0.0001\times1000=0.1$ psi/ft over the 1000 ft from bottom to the 6000 ft pump setting: $\Delta P=0.1\times1000=100$ psi. $P_{intake}=800-100$. $\boxed{P_{intake}=700\ \text{psi at 6000 ft}}$.
Gas entering the pump. With 50% of the free gas separated at the pump, the tubing above the pump carries GLR $=0.5\times400=200$ SCF/STBL.
Pump discharge pressure. Marching the no-slip mixture gradient (GLR = 200 SCF/STBL, $B_o=1.1$) from the wellhead requirement of 160 psi down to the pump (6000 ft) — equivalently, solving for the bottom pressure whose upward traverse reaches exactly 160 psi at surface — gives $\boxed{P_{discharge}\approx1113\ \text{psi}}$.
Pump differential pressure. $\Delta P_{pump}=P_{discharge}-P_{intake}=1113-700$. $\boxed{\Delta P_{pump}\approx413\ \text{psi}}$.
Volumetric rate at pump suction. Liquid: $1000\times1.1=1100$ rb/day. Free gas (200 SCF/STBL, at $P_{intake}=700$ psi, $T=200$°F): $B_g\approx0.00448$ rb/SCF, giving $\approx896$ rb/day. Total $\approx1996$ rb/day $=58.2$ gpm.
Hydraulic horsepower. $HP=\dfrac{Q_{gpm}\times\Delta P_{pump}}{1714}=\dfrac{58.2\times413}{1714}$. $\boxed{HP\approx14.0\ \text{hydraulic hp}}$.

Fig. 5 — ESP well schematic: 7000 ft total depth, pump set at 6000 ft, $P_{wf}=800$ psi, $P_{intake}=700$ psi, $P_{discharge}\approx1113$ psi.

Quantity	Value
$P_{wf}$ (7000 ft)	800 psi
$P_{intake}$ (6000 ft, pump)	700 psi
$P_{discharge}$ (pump)	≈1113 psi
Pump differential, $\Delta P_{pump}$	≈413 psi
Suction rate, $Q$	≈58.2 gpm
Required hydraulic horsepower	≈14.0 hp

Topic: ESP sizing — intake/discharge pressures and hydraulic horsepower for a gassy well with partial downhole gas separation.

Key relations: $HP_{hydraulic}=\dfrac{Q_{gpm}\,\Delta P_{psi}}{1714}\ \left(\equiv\dfrac{Q_{gpm}\,\text{TDH}_{ft}\,SG}{3960}\right)$; no-slip mixture gradient as in Question 3.

Why this works: the pump does work on whatever volume actually reaches its intake at intake conditions — free gas that bypasses the pump (vented up the casing annulus, the other 50% here) neither costs pump horsepower nor helps lift, so only the un-separated fraction belongs in the suction-volume calculation.

Common pitfall: sizing the pump on surface (stock-tank) volumes instead of in-situ suction volumes, which drastically understates the gas volume fraction the pump must actually handle at the lower intake pressure; and forgetting that hydraulic horsepower is not nameplate horsepower once realistic pump efficiency is applied.

Source: Brown, The Technology of Artificial Lift Methods, Vol. 2b, Ch. 4 (ESP sizing); Golan & Whitson, Well Performance, 2nd ed., Ch. 3.

Question 5 of 5: Electrical Submersible Pump — Required Horsepower

Question 5: Electrical Submersible Pump — Required Horsepower (25 marks)