P = η·ρ·g·Q·H (W). ρ = 1000 kg/m³, g = 9.81 m/s², Q in m³/s, H in metres. Net head H_n = H_gross − h_f. The generator adds a loss of 1.5 % (η_gen = 0.985). Example: 4 m³/s falling 300 m at η = 0.898 gives 10.57 MW on the shaft, 10.41 MW to the town. Energy chain: height energy → speed → spinning wheel → electricity, with losses at each step. One m³ at 80 m with η = 0.9 makes 0.196 kWh.

Darcy–Weisbach: h_f = (f·L/D + K)·v²/(2g), f = 0.015, L = 1200 m, D = 2.8 m, K = 1. At 14 m³/s: v = 2.27 m/s, h_f = 1.96 m (about 2.5 % of the head). Left out: f really depends on the Reynolds number and the roughness; water hammer is not modelled.

Nq = n·√Q / H^0.75 (n in rpm, Q in m³/s, H in m). Sites: A 300 m, 4 m³/s, 600 rpm gives Nq 16.6 (Pelton); B 80 m, 30 m³/s, 250 rpm gives 51.2 (Francis); C 10 m, 60 m³/s, 100 rpm gives 137.7 (Kaplan). Model efficiency η = η_pk − 0.18·ln²(Nq/Nq*) with (η_pk, Nq*) = Pelton (0.90, 15), Francis (0.93, 60), Kaplan (0.94, 200). The curves cross at Nq ≈ 28 and ≈ 107 (textbooks say about 30 and 100–110). Typical heads: Pelton above ~200–300 m, Francis ~20–300 m, Kaplan ~2–40 m. Pelton jet v = 0.97·√(2gH) = 74.4 m/s at 300 m; best wheel speed ≈ 0.46·v. Part load: η = η_site − k·(load − 0.85)²; Pelton and Kaplan can adjust needle or blades, so they keep their efficiency better; Francis does not (so one unit works better at low power).

The lake is V-shaped: S(h) = 60 Mm³·(h/70 m)³, area A = dS/dh = 2.57 km² at full pool. Balance: dS/dt = Q_in − Q_turbine − Q_river − Q_spill. Catchment 600 km², rain 1.1 m/year, runoff 45 % gives 297 Mm³/year, 9.41 m³/s on average; 54 % arrives in the four wet months (Jun–Sep). The town needs more in winter. Smoothing a year takes about 61 Mm³ of storage (sequent peak), so the lake is just big enough. Rationing: when the lake is low, make less power: f = clamp((S/S_full − 0.05)/0.25, 0, 1). For 3.5 MW with 30 % river flow the rationing year covers about 91 % of the need; making power greedily empties the lake and covers only about 52 % (our calculation; an earlier estimate said 64 %).

Spillway: Q = C·L·H^1.5, C = 2.2, L = gate width (three gates × 10 m, Model: a half-open gate really acts like an opening, not a shorter crest), H = height over the crest. 30 m at 2 m gives 186.7 m³/s. Storm: Q(t) = Q̄ + (Q_p − Q̄)(x e^(1−x))², x = t/6 h. Routed from a full lake with turbines off (conservative): a 400 m³/s storm rises 2.48 m over the crest (ok, freeboard is 3 m); 600 m³/s rises 3.38 m (too much!). Three low sluices (Model: orifices, Cd 0.62) let you lower the lake before the storm — that is the lesson: open early. A wider spillway (45 m) would give 2.79 m, but this dam only has three gates.

A fish ladder is a staircase of pools. Slot speed v = √(2g·Δh) = 2.43 m/s for 0.3 m; discharge Q = 0.65·b·h·v = 0.47 m³/s; power per pool p = ρ g Q Δh / V = 155 W/m³ in a 9 m³ pool (fish like ≲ 150: Typical). A 6 m weir needs 6/0.3 = 20 pools; a 70 m dam would need 233, so tall dams use fish lifts. Success (Model) = step score × roughness score × attraction score (the ladder should carry ≥ 5 % of the flow below the weir so fish find the door). Between the dam and the powerhouse the river only gets the river-flow setting; habitat (Tennant style, Typical): 10 % of the average flow is "poor", 30 % "good", 60 % "best". Each 10 % costs about 0.57 MW.

Water power makes electricity with no smoke, stores water for dry months and can reduce floods. A dam also changes the river: fish need a way past, the stretch below needs enough water, and the lake covers land. Good engineering looks for a balance. This game has no villain: every knob has a price.

Hill-chart efficiency surfaces, cavitation, water hammer, ice, sediment filling the lake, year-to-year weather, dam stability, fish passing through turbines. Real fish criteria depend on the species and the river. Nothing here is a forecast for a real dam.

Each row compares the game's code with a hand calculation written separately.