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Micro hydro net head, power and energy

Short answer

Net head = 60 - 0.08×60 = 55.2 m. Hydraulic power P_h = 1000×9.81×0.15×55.2 = 81.2 kW. Electrical output P_e = 0.80×0.92×81.2 = 59.8 kW. Annual energy = 59.8×0.6×8760 = 3.14×10⁵ kWh (314 MWh).

A micro hydro site has a gross head of 60 m and a design flow of 0.15 m³/s. Penstock and other losses are 8 % of gross head. The turbine efficiency is 80 % and the generator efficiency is 92 %. Calculate (a) the net head, (b) the hydraulic power, (c) the electrical output, and (d) the annual energy if the plant factor is 0.6.

(a) Net head

Hn=Hg−hL=60−0.08(60)=55.2 mH_n = H_g - h_L = 60 - 0.08(60) = 55.2\ \text{m}

(b) Hydraulic power at turbine inlet

Ph=ρgQHn=1000×9.81×0.15×55.2=81 227 W=81.2 kWP_h = \rho g Q H_n = 1000 \times 9.81 \times 0.15 \times 55.2 = 81\,227\ \text{W} = 81.2\ \text{kW}

(c) Electrical output

ηo=0.80×0.92=0.736\eta_o = 0.80 \times 0.92 = 0.736 Pe=ηoPh=0.736×81.2=59.8 kWP_e = \eta_o P_h = 0.736 \times 81.2 = 59.8\ \text{kW}

The scheme falls in the micro hydro range (5–100 kW). Gross hydraulic power was ρgQHg=88.3\rho g Q H_g = 88.3 kW, so overall efficiency from the gross head is 59.8/88.3=67.7 %59.8/88.3 = 67.7\ \%.

(d) Annual energy

E=Pe×8760×PF=59.8×8760×0.6=3.14×105 kWhE = P_e \times 8760 \times PF = 59.8 \times 8760 \times 0.6 = 3.14\times10^{5}\ \text{kWh}

Answer: Hn=55.2H_n = 55.2 m; Ph=81.2P_h = 81.2 kW; Pe=59.8P_e = 59.8 kW; E≈3.14×105E \approx 3.14 \times 10^5 kWh (314 MWh) per year.

✓ Answer checked by Claude Opus 5.5 (high effort)

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