RoR at 40% exceedence; storage regulation

Data

$H=100$ m, $\eta=0.85$, so $9.81\,\eta\,H = 833.85$ kW per m³/s. One month $=730$ h $=2.628\times10^6$ s.

i) RoR designed for $Q_{40}$

Flows in descending order: 2500, 2100, 2000, 1800, 900, 500, 320, 300, 300, 140, 120, 100. By the Weibull formula $p=\frac{m}{N+1}\times100$ with $N=12$, $p=40\%$ gives $m=0.4\times13=5.2$, which lies between 900 (m = 5) and 500 (m = 6):

$$Q_{40} = 900 - 0.2\,(900-500) = 820\ \text{m}^3/\text{s}$$ $$P = 833.85\times820 = 683,757\ \text{kW} = 683.8\ \text{MW}$$

• Spill (flow above $Q_{40}$) by month (m³/s): Jun 980, Jul 1180, Aug 1680, Sep 1280, Oct 80, total $=5,200$ m³/s-months.

$$E_{spill} = 833.85\times5,200\times730 = 3,165\ \text{GWh}$$

• Firm energy (minimum flow 100 m³/s available all year):

$$E_{firm} = 833.85\times100\times8760 = 730.5\ \text{GWh}$$

• (Total energy of the RoR plant $= 833.85\times5,880\times730 = 3,579$ GWh, so the non-firm energy is 2,849 GWh.)

ii) Storage with full regulation

The design discharge equals the mean flow:

$$\bar Q = \frac{11,080}{12} = 923.33\ \text{m}^3/\text{s}$$

Deficits ($\bar Q-Q$) in the months with flow below the mean, from Oct to May (m³/s-months): Oct 23.33, Nov 423.33, Dec 623.33, Jan 823.33, Feb 803.33, Mar 783.33, Apr 623.33, May 603.33. Sum $=4,706.67$ (this equals the surplus of Jun-Sep, which refills the reservoir).

$$V_s = 4,706.67\times2.628\times10^6 = 12,369\times10^6\ \text{m}^3 \approx 12.37\ \text{billion m}^3$$

iii) Installed capacity and energy of the storage project

$$P = 833.85\times923.33 = 769,922\ \text{kW} = 769.9\ \text{MW}$$ $$E = P\times8760 = 6,745\ \text{GWh/year}$$

(The head is taken as constant at 100 m.)

Answer: (i) P = 683.8 MW at $Q_{40}=820$ m³/s; spill energy ≈ 3,165 GWh; firm energy = 730.5 GWh. (ii) storage = 12,369 Mm³. (iii) P = 769.9 MW; E = 6,745 GWh.