Check: no pump/motor mechanical efficiency is given, so the boxed figure below is the hydraulic horsepower the fluid column itself demands ($Q\cdot\text{TDH}\cdot\text{SG}/3960$); an installed motor nameplate would be sized higher to cover a typical 50–65% pump efficiency (roughly 25–30 HP). Friction losses in the tubing above the pump are folded into the same homogeneous no-slip gradient used in Question 3 (no separate friction correlation or viscosity was supplied in the source for the tubing leg, only the casing $dP/dL$ formula below the pump).
Approach. Use the IPR to fix $P_{wf}$ at total depth, subtract the given casing flowing-gradient loss to reach the pump intake, compute the required pump discharge pressure by carrying the surviving (half) free gas up the tubing to $P_{wh}$, and convert the resulting differential pressure to head and hydraulic horsepower.
Flowing bottomhole pressure. $P_{wf}=\bar P_r-q_L/J=3000-1200/1=1800$ psi at 8000 ft.
Pump intake pressure. $dP/dL=0.0001\times1200=0.12$ psi/ft over the 1000 ft casing interval (8000 to 7000 ft): $\Delta P=120$ psi, so $\boxed{P_{intake}=1800-120=1680\ \text{psi}}$.
Gas split at the pump. Formation GLR = 100 SCF/STB; half is separated and vented up the casing-tubing annulus, half (50 SCF/STB) continues up the tubing with the liquid.
Required pump discharge pressure. Marching the homogeneous no-slip mixture (GLR = 50 SCF/STB, oil SG = 0.850) downward from $P_{wh}=160$ psi at surface to 7000 ft gives $\boxed{P_{discharge}=2422\ \text{psi}}$.
Pump differential and TDH. $\Delta P_{pump}=2422-1680=742$ psi. Converting to head using the mixture's own density at average pump conditions ($\rho\approx50.1$ lbm/ft$^3$, gradient $=0.348$ psi/ft): $\text{TDH}=742/0.348=\boxed{\text{TDH}=2133\ \text{ft}}$.
