How is the installed capacity found from the given discharge and head?

P = 9.81 x eta x Q x H = 9.81 x 0.85 x 25 x 150 = 31.27 MW

P = 9.81 x Q x H without efficiency = 36.8 MW

How is the annual load factor calculated?

Average load (energy/8760 = 25.11 MW) divided by peak load (30 MW) = 0.837

Average load divided by installed capacity

Peak load divided by installed capacity

Energy divided by installed capacity only

Which expression gives the capacity (plant) factor?

Annual energy / (installed capacity x 8760) = 0.803

Installed capacity / annual energy

What is the utilization factor and how does it relate to the other two?

Peak load / installed capacity = 0.959; CF = LF x UF

Average load / installed capacity = 0.803; CF = LF + UF

Installed capacity / peak load = 1.04; CF = LF / UF

Peak load / average load = 1.19; CF = UF / LF

2072 Kartik · 2+2+2 marks

Load, capacity and utilization factors

Short answer

Installed capacity = 9.81×0.85×25×150 = 31.27 MW, average load = 220 GWh/8760 h = 25.11 MW. Annual load factor = 25.11/30 = 0.837; capacity factor = 220×10⁶/(31.27×8760) = 0.803; utilization factor = 30/31.27 = 0.959.

A hydropower plant receives design discharge of 25 m³/s from 150 m height. The annual output of the plant is 220 GWh. If the peak load demand is 30 MW, determine (i) annual load factor, (ii) capacity factor and (iii) utilization factor. Assume the overall efficiency of the plant equals 85% and neglect head loss in the penstock.

Installed capacity

P_{inst} = 9.81\,\eta\,Q\,H = 9.81\times0.85\times25\times150 = 31,269\ \text{kW} = 31.27\ \text{MW}

Average load

Annual energy $=220\ \text{GWh} = 220\times10^6$ kWh.

P_{avg} = \frac{220\times10^6}{8760} = 25,114\ \text{kW} = 25.114\ \text{MW}

(i) Annual load factor

\text{LF} = \frac{P_{avg}}{P_{peak}} = \frac{25.114}{30} = 0.837\ (\approx 83.7\%)

(ii) Capacity (plant) factor

\text{CF} = \frac{E}{P_{inst}\times8760} = \frac{220\times10^6}{31,269\times8760} = 0.803\ (\approx 80.3\%)

(iii) Utilization factor

\text{UF} = \frac{P_{peak}}{P_{inst}} = \frac{30}{31.27} = 0.959\ (\approx 95.9\%)

Check: CF = LF × UF = 0.837 × 0.959 = 0.803.

Answer: load factor = 0.837; capacity factor = 0.803; utilization factor = 0.959.

✓ Answer checked by Claude Opus 5.5 (high effort)

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