A hydropower plant has a head of 100 m and water volumetric flowrate of 1000 m^3 /s. What is the power (in MW) that the plant can produce if the efficiency is 0.75 (assume ?w = 1000 kg/m^3, g=10m/s^2)? Answer:
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Substitute the values into the formula: \(P_g = 1000 \, \text{kg/m}^3 \times 1000 \, \text{m}^3/s \times 10 \, \text{m/s}^2 \times 100 \, \text{m}\) \(P_g = 100 \times 10^6 \, \text{W} = 100 \, \text{MW}\) Show moreā¦
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