Primary and secondary energy at 45% exceedance

Primary (firm) energy is the energy that can be produced all year from the minimum (dependable) flow. Secondary energy is the extra energy produced when flow is above the minimum, up to the plant capacity.

Step 1: Flow duration (Weibull plotting position $P = \frac{m}{N+1}\times100$, N = 12)

Step 2: Flow at 45% time (linear interpolation)

$$Q_{45} = 42 - \frac{45-38.46}{46.15-38.46}(42-39) = 39.45\ \text{m}^3/\text{s}$$

Step 3: Plant capacity

$P = \gamma Q H \eta = 9.81 \times Q \times 170 \times 0.85 = 1417.55\,Q$ kW

$$P_{inst} = 1417.55 \times 39.45 = 55{,}922\ \text{kW} \approx 55.9\ \text{MW}$$

Step 4: Primary energy (minimum flow 18 m³/s, May)

$$\begin{aligned} P_f &= 1417.55 \times 18 = 25{,}516\ \text{kW} \\ E_p &= 25{,}516 \times 8760 = 223.52\ \text{GWh} \end{aligned}$$

Step 5: Total energy (flow used = smaller of Q and 39.45; 730 h per month)

$\sum Q_{used} = 402.25$ m³/s-months.

$$\begin{aligned} E_{total} &= 1417.55 \times 402.25 \times 730 = 416.25\ \text{GWh} \\ E_s &= 416.25 - 223.52 = 192.73\ \text{GWh} \end{aligned}$$

Answer: Plant capacity ≈ 55.9 MW; primary energy ≈ 223.5 GWh/yr; secondary energy ≈ 192.7 GWh/yr (total ≈ 416.3 GWh/yr).

(If $P = m/N$ is used for the duration curve, $Q_{45} \approx 40.8$ m³/s and the secondary energy rises slightly; the method is the same.)