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Primary and secondary energy (Nepali calendar months)

Short answer

Primary energy is 3836.2 GWh/yr, secondary 7867.7 GWh/yr. Minimum usable flow after 40 m³/s env. release is 360 m³/s; firm power 437.9 MW. Total energy 11,703.9 GWh/yr using all usable flow.

Following table shows the mean monthly flow of a typical river of Nepal. Estimate the primary and secondary energies available in the river throughout the year. Assume 10% of minimum flow of the series has to be released for environmental flow to the downstream from the dam. Take overall efficiency of the plant is 80% and effective head is 155 m. Note: The months are as per Nepalese calendar i.e 1 as Baishakh and so on.

Month123456789101112
Discharge (m³/s)11501140145018301700155011501000850400480960

Primary energy is the energy available throughout the year from the minimum usable flow; secondary energy is the extra energy from usable flows above that minimum. No design discharge is given, so the plant is assumed to use all the usable flow. Each month is taken as 730 h (Nepali months vary from 29 to 32 days, but the yearly total is the same).

Environmental flow

Minimum flow of the series = 400 m³/s (month 10, Magh).

Qenv=0.10×400=40 m3/sQ_{env} = 0.10 \times 400 = 40\ \text{m}^3/\text{s}

Usable flow = Q − 40.

Power per unit discharge

P=9.81×Q×155×0.80=1216.44 Q kWP = 9.81 \times Q \times 155 \times 0.80 = 1216.44\,Q\ \text{kW}

Primary energy

Minimum usable flow = 400 − 40 = 360 m³/s.

Pf=1216.44×360=437,918 kW≈437.9 MWEp=437,918×8760=3836.17 GWh\begin{aligned} P_f &= 1216.44 \times 360 = 437{,}918\ \text{kW} \approx 437.9\ \text{MW} \\ E_p &= 437{,}918 \times 8760 = 3836.17\ \text{GWh} \end{aligned}

Total energy

MonthQUsable QPower (MW)
1 Baisakh115011101350.25
2 Jestha114011001338.08
3 Asar145014101715.18
4 Shrawan183017902177.43
5 Bhadra170016602019.29
6 Ashwin155015101836.82
7 Kartik115011101350.25
8 Mangsir10009601167.78
9 Poush850810985.32
10 Magh400360437.92
11 Falgun480440535.23
12 Chaitra9609201119.12
Sum13,180
Etotal=1216.44×13,180×730=11,703.86 GWhEs=11,703.86−3836.17=7867.69 GWh\begin{aligned} E_{total} &= 1216.44 \times 13{,}180 \times 730 = 11{,}703.86\ \text{GWh} \\ E_s &= 11{,}703.86 - 3836.17 = 7867.69\ \text{GWh} \end{aligned}

Answer: Primary energy ≈ 3836.2 GWh/yr; secondary energy ≈ 7867.7 GWh/yr (total ≈ 11,703.9 GWh/yr).

✓ Answer checked by Claude Opus 5.5 (high effort)

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