Lake 0 mHead 0 mIn 0 m³/sTurbines 0Spill 0Power 0 MWLevel okday 1 00:00

Hydroelectric Dam Cutaway — 2D Section, Phase Plane and Turbine Hill

The 2D companion: the same plant model as the 3D lab, drawn as a slice through the dam so you can see pressures, flows and the stability of the lake level.

Switch to 3D

Educational model: one lumped reservoir, steady-flow turbines, model efficiency curves. Not a design, dam-safety or flood-forecast tool.

Scenarios

River and reservoir

Water balance: the lake rises when inflow is bigger than what leaves through turbines and gates.

Turbines

Governor on: the wicket gates open just enough for the demand. Off: you pick the opening and the flow follows the head.

Spillway

Each gate lifted by a lets water pass under it (orifice) and, when lifted clear of the water, over the sill (weir).

Results

Level and power over time

Level: green = normal full level (65 m), orange = flood limit (72 m). Power in MW.

Experiments: measured vs predicted

Each test runs the plant model offline and compares it with a prediction made with a different formula.

Press a test.

Independent numerical checks

Controls help

Tabs switch between the cross-section, the phase plane (level versus rate of change), time series and the turbine hill.
With the governor on the units hold the demanded power; switch it off to set the wicket-gate opening yourself.
Clock speed is hours of plant time per real second; Pause stops it.

The science

Water balance. A(z)·dz/dt = Q_in − Q_turbines − Q_spill, with surface area A(z) = A₀ + k(z − z₀) (10 km² at 60 m). It is integrated with a 4th-order Runge-Kutta step. Evaporation, seepage and sediment are ignored. The phase plane plots dz/dt against z: where the curve crosses zero the lake is in equilibrium, and it is stable only if the curve crosses downward (a higher level then means more outflow than inflow).

Hydrostatic pressure. Water of depth d presses on the upstream face with p = ρ·g·d, a triangle. Per metre of dam width the resultant is F = ½ρg d² acting d/3 above the base, so doubling the depth quadruples the force and the overturning moment M = F·d/3 grows like d³. Here d is measured from the reservoir bed at the dam (20 m).

Power. P = η·ρ·g·Q·H_n, net head H_n = gross head − penstock friction h_f = K·Q². With fixed efficiency the penstock power peaks at H_n = 2H/3 (a check below).

Turbine. A Francis unit passes Q = x·Q_r√(H_n/H_r) for wicket-gate fraction x. The efficiency is a model parabola, best near 82 % of full gate and falling away from the design head. The numbers are illustrative, not from a manufacturer.

Spillway gates. Orifice Q = C_d·b·a·√(2g(h − a/2)); once lifted clear, a free weir Q = (2/3)C_d·1.06·b√(2g)·h^3/2 with h the depth over the 55 m sill. The two meet at a = h. The thickness of the sheet on the chute in the drawing is q / v with q = Q/b per metre and v about √(2gh).

Simplifications. Steady flow in the penstocks (no water hammer), one tailwater level, no ice, silt or fish passage, no grid-frequency control. The cross-section is a schematic, not to scale in its details. Not affiliated with any operator or manufacturer. Typical-value constants should be checked against a hydraulic-machinery textbook.