Discharge exceedance and turbine criterion

Discharge exceedance

Discharge exceedance is the percentage of time a given river discharge is equalled or exceeded. It is read from the flow duration curve (FDC). For example, $Q_{45} = 10$ m³/s means the river flow is 10 m³/s or more for 45% of the time (about 164 days a year).

• Low exceedance (e.g. $Q_{25}$) → large flow, available for a short time.
• High exceedance (e.g. $Q_{95}$) → small but dependable (firm) flow.
• In Nepal, ROR projects are usually designed at about $Q_{40}$–$Q_{45}$; firm power is judged from $Q_{90}$–$Q_{95}$.
• Exceedance is found by ranking the recorded flows in descending order and computing $P = \dfrac{m}{n+1}\times 100\%$ for rank $m$ out of $n$ values.
• The choice of design exceedance is a trade-off: a lower exceedance gives more installed capacity and energy but a lower plant factor and higher cost per kW.

 Q
 |\
 | \
 |  \___ Q45
 |      \_____ Q65
 |            \______ Q95
 +---------------------> % exceedance

Criteria for selecting turbine

1. Net head: Pelton for high head (50–1300 m), Francis for medium head (10–350 m), Kaplan/propeller for low head (2–40 m). Crossflow (Michell–Banki) and Turgo are used in small plants.
2. Specific speed: $N_s = N\sqrt{P}/H^{5/4}$; Pelton 10–35, Francis 60–300, Kaplan 300–1000 (metric, kW).
3. Discharge and its variation: if flow varies greatly, choose a turbine with good part-load efficiency (Pelton, Kaplan) or use several units.
4. Part-load operation: Francis efficiency falls below about 40–50% load; Pelton remains efficient down to 20%.
5. Rotational speed: higher speed gives smaller generator; limited by cavitation and runaway speed.
6. Cavitation: Thoma's coefficient fixes the setting below tailwater for reaction turbines.
7. Sediment and abrasion: important for Himalayan rivers.
8. Cost, efficiency, maintenance and local experience.

Example

For a site with $H_{net} = 250$ m and $Q_{45} = 6$ m³/s split into two units of 3 m³/s, unit power $\approx 9.81 \times 3 \times 250 \times 0.9 \approx 6.6$ MW. At 750 rpm, $N_s = 750\sqrt{6622}/250^{5/4} \approx 61$, at the boundary of multi-jet Pelton and slow Francis; the Pelton is preferred if the river is silty or long part-load running is expected.