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Annual firm and secondary energy (900 m³/s, Nov 650)

Short answer

Annual firm energy 696.1 GWh from minimum flow 90 m³/s. Secondary energy 3467.5 GWh, total 4163.6 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 table shows the mean monthly flow of a typical river in Nepal. Calculate annual firm and secondary energy if design discharge selected is 900 m³/s. Take overall efficiency = 90% and head 100 m.

MonthJanFebMarAprMayJunJulAugSepOctNovDec
Discharge (m³/s)120100901503501800200025002000900650500

Firm energy comes from the minimum flow, which is available throughout the year. Secondary energy is the extra energy from flows above the minimum, up to the design discharge (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=794,610= 882.9 \times 900 = 794{,}610 kW ≈ 794.6 MW.

Firm energy

Minimum flow = 90 m³/s (March).

Pf=882.9×90=79,461 kWEf=79,461×8760=696.08 GWh\begin{aligned} P_f &= 882.9 \times 90 = 79{,}461\ \text{kW} \\ E_f &= 79{,}461 \times 8760 = 696.08\ \text{GWh} \end{aligned}

Total energy (flow used = smaller of Q and 900; 730 h per month)

MonthQ (m³/s)Q usedPower (MW)
Jan120120105.95
Feb10010088.29
Mar909079.46
Apr150150132.44
May350350309.02
Jun1800900794.61
Jul2000900794.61
Aug2500900794.61
Sep2000900794.61
Oct900900794.61
Nov650650573.89
Dec500500441.45
Sum6460
Etotal=882.9×6460×730=4163.58 GWhEs=4163.58−696.08=3467.50 GWh\begin{aligned} E_{total} &= 882.9 \times 6460 \times 730 = 4163.58\ \text{GWh} \\ E_s &= 4163.58 - 696.08 = 3467.50\ \text{GWh} \end{aligned}

Answer: Annual firm energy ≈ 696.1 GWh; annual secondary energy ≈ 3467.5 GWh (total ≈ 4163.6 GWh).

✓ Answer checked by Claude Opus 5.5 (high effort)

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