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Annual firm and secondary energy (900 m³/s, Nov 100)

Short answer

Annual firm energy 773.4 GWh from minimum flow 100 m³/s. Secondary energy 3016.3 GWh, total 3789.8 GWh. Installed capacity 794.6 MW at design discharge 900 m³/s. Power per unit discharge 882.9 kW per m³/s. Monthly energy summed using flows capped at 900 m³/s over 730 hours per month.

Following are the mean monthly flow of the river. Where head (H) = 100 m, overall efficiency = 90% and design discharge selected is 900 m³/s. Calculate the annual firm and secondary energy.

MonthJanFebMarAprMayJunJulyAugSepOctNovDec
Flow (m³/s)1001201403003201800200025102000900100300

Firm energy is the energy from the minimum flow (available all year). Secondary energy is the extra energy from flows above the minimum, limited by the design discharge of 900 m³/s.

Power per unit discharge

P=9.81×Q×100×0.90=882.9 Q kWP = 9.81 \times Q \times 100 \times 0.90 = 882.9\,Q\ \text{kW}

Installed capacity =882.9×900=794.61= 882.9 \times 900 = 794.61 MW.

Firm energy

Minimum flow = 100 m³/s (January and November).

Pf=882.9×100=88,290 kWEf=88,290×8760=773.42 GWh\begin{aligned} P_f &= 882.9 \times 100 = 88{,}290\ \text{kW} \\ E_f &= 88{,}290 \times 8760 = 773.42\ \text{GWh} \end{aligned}

Total energy (730 h per month)

MonthQQ usedPower (MW)
Jan10010088.29
Feb120120105.95
Mar140140123.61
Apr300300264.87
May320320282.53
Jun1800900794.61
Jul2000900794.61
Aug2510900794.61
Sep2000900794.61
Oct900900794.61
Nov10010088.29
Dec300300264.87
Sum5880
Etotal=882.9×5880×730=3789.76 GWhEs=3789.76−773.42=3016.34 GWh\begin{aligned} E_{total} &= 882.9 \times 5880 \times 730 = 3789.76\ \text{GWh} \\ E_s &= 3789.76 - 773.42 = 3016.34\ \text{GWh} \end{aligned}

Answer: Annual firm energy ≈ 773.4 GWh; annual secondary energy ≈ 3016.3 GWh (total ≈ 3789.8 GWh).

✓ Answer checked by Claude Opus 5.5 (high effort)

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