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Primary and secondary energy (102 m head)

Short answer

Primary energy 119.9 GWh/yr from minimum flow 18 m³/s. Secondary energy 139.3 GWh/yr, total 259.3 GWh/yr. Installed capacity 41.8 MW at design discharge 55 m³/s. Calculated using power per unit discharge (760.47 kW per m³/s), monthly flows capped at 55 m³/s, 730 hours per month.

Estimate primary and secondary energy from the data given below. Net head available = 102 m, Discharge capacity = 55 m³/s, Overall efficiency = 76%.

MonthJanFebMarAprMayJunJulAugSepOctNovDec
Flow (m³/s)302824291837659074564239

Primary energy is generated by the flow available 100% of the time (minimum flow). Secondary energy is the additional energy from flows above the minimum, limited by the plant (design) discharge of 55 m³/s.

Power per unit discharge

P=9.81 Q H η=9.81×Q×102×0.76=760.47 Q kWP = 9.81\,Q\,H\,\eta = 9.81 \times Q \times 102 \times 0.76 = 760.47\,Q\ \text{kW}

Installed capacity =760.47×55=41,826= 760.47 \times 55 = 41{,}826 kW ≈ 41.8 MW.

Primary energy

Minimum flow = 18 m³/s (May).

Pf=760.47×18=13,688 kWEp=13,688×8760=119.91×106 kWh=119.91 GWh\begin{aligned} P_f &= 760.47 \times 18 = 13{,}688\ \text{kW} \\ E_p &= 13{,}688 \times 8760 = 119.91 \times 10^6\ \text{kWh} = 119.91\ \text{GWh} \end{aligned}

Monthly energy (flow used ≤ 55 m³/s, 730 h per month)

MonthQ (m³/s)Q usedPower (kW)Energy (GWh)
Jan303022,81416.65
Feb282821,29315.54
Mar242418,25113.32
Apr292922,05416.10
May181813,6889.99
Jun373728,13720.54
Jul655541,82630.53
Aug905541,82630.53
Sep745541,82630.53
Oct565541,82630.53
Nov424231,94023.32
Dec393929,65821.65
Total467259.25
Etotal=760.47×467×730=259.25 GWhEs=259.25−119.91=139.34 GWh\begin{aligned} E_{total} &= 760.47 \times 467 \times 730 = 259.25\ \text{GWh} \\ E_s &= 259.25 - 119.91 = 139.34\ \text{GWh} \end{aligned}

Answer: Primary energy ≈ 119.9 GWh/yr; secondary energy ≈ 139.3 GWh/yr; total ≈ 259.3 GWh/yr.

✓ Answer checked by Claude Opus 5.5 (high effort)

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