Head, nozzle, penstock and load against shaft power.
The science
Water falling through a head H reaches the nozzle with speed v = C_v sqrt(2 g H_net) (Torricelli). The penstock wastes part of the head as friction, h_f = f (L / D) v_p^2 / (2 g) (Darcy-Weisbach), so H_net = H - h_f, and the flow is Q = A_nozzle v. The page solves this by iteration.
A Pelton wheel is an impulse turbine. A bucket moving at speed u turns the relative jet around by almost 180 degrees, so the force is F = rho Q (v - u)(1 + k cos beta) and the efficiency is eta = 2 (u/v)(1 - u/v)(1 + k cos beta). It peaks at u/v = 1/2 with eta = (1 + k cos beta)/2, about 93 percent here.
Energy balance: rho g Q H = rho g Q h_f + (nozzle loss) + (1/2) rho Q v^2, and the shaft power is the bucket efficiency times the jet power. As the lake drains at dh/dt = -Q / A the head falls, and so does the power.
Honest simplifications: one jet, constant friction factor 0.02, constant bucket friction k = 0.9, no wheel windage or generator loss, no water hammer and a flat-sided reservoir.