Given. The same well as Question 3 (depth, $J$, $\bar P_R$, API, gas gravity, formation GLR, temperature profile), now equipped with an ESP set at 7000 ft, a higher $P_{wh}=160$ psi, and an explicitly given casing flowing-gradient formula.
Total liquid rate, $q_L$
1200 STBL/day
ESP setting depth
7000 ft
Casing ID below pump
6 in.
Free gas separated at pump
one half
Wellhead pressure, $P_{wh}$
160 psi
Given casing flowing gradient
$dP/dL=0.0001\,q$ psi/ft
Find. The required ESP (hydraulic) horsepower.
Check: as in Question 3, the produced gas is modelled as fully "free" (no PVT/$R_s$ data survives extraction for this well), so "one half of the free gas separated at the pump" is taken as halving the full formation GLR (100 → 50 SCF/STB) in the tubing above the pump; the casing ID below the pump is given but unused, since the flowing gradient there is supplied directly by the problem's own formula.
Approach. Find $P_{wf}$ from the straight-line IPR, step the given casing gradient up to the pump intake, march the homogeneous no-slip mixture traverse (with the reduced above-pump GLR) from the known $P_{wh}$ down to the pump to get the required discharge pressure, then convert the pump's differential pressure to head and hydraulic horsepower.
Pump intake pressure. Casing gradient $dP/dL=0.0001\times1200=0.12$ psi/ft over the 1000 ft from bottom (8000 ft) to the pump (7000 ft): $P_{intake}=P_{wf}-0.12\times1000$. $\boxed{P_{intake}=1680\ \text{psi}}$.
Required discharge pressure. Above the pump, GLR $=100/2=50$ SCF/STB. Marching the homogeneous mixture traverse from $(D,P)=(0,\,160\ \text{psi})$ down to 7000 ft (temperature $T(7000)=187.5^\circ$F) gives $\boxed{P_{disc}\approx2426\ \text{psi}}$ — the discharge pressure needed for that (partly gas-lightened) column to reach $P_{wh}=160$ psi.
Pump differential pressure and total dynamic head. $\Delta P_{pump}=P_{disc}-P_{intake}=2426-1680=746$ psi. Mixture density at average pump conditions ($\bar P\approx2053$ psi, $T=187.5^\circ$F, GLR$=50$) gives $SG_m\approx0.803$. $TDH=\Delta P_{pump}\times2.31/SG_m=746\times2.31/0.803$. $\boxed{TDH\approx2145\ \text{ft}}$.
Check: 15.2 HP is the hydraulic horsepower delivered to the fluid only; an installed nameplate motor must be sized larger to cover pump efficiency (typically 50–65% at this rate/head combination for a centrifugal ESP stage stack), i.e. a nameplate rating on the order of 25–30 HP, though no efficiency curve was given to compute that figure exactly.
Fig. 4 — Pressure traverse below the pump (casing, purple, using the given flowing gradient) and above the pump (tubing, blue, homogeneous no-slip mixture at GLR=50 SCF/STB); the pump boosts the fluid from the 1680-psi intake to the 2426-psi discharge required to reach $P_{wh}=160$ psi.
