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NCERT Class XII Chapter
Alternating Current
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Question : 24 of 26
Marks: +1, -0
At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m3 s−1\mathrm{m}^3\,\mathrm{s}^{-1}. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g = 9.8 m s−2\mathrm{m}\,\mathrm{s}^{-2}).
Solution:  
Workdone by liquid pressure = pressure × volume shifted power of flowing water
Hydro-power = Worktime\frac{\text{Work}}{\text{time}} = pressure × volumetime\frac{\text{volume}}{\text{time}}
Hydro-power = hρg × (V/t) = 300 × 10310^3 × 9.8 × 100 = 29.4 × 10710^7 Watt
Efficiency of turbine η = Electric powerHydro - power\frac{\text{Electric power}}{\text{Hydro - power}} ⇒ 0.6 = Electric power29.4×107\frac{\text{Electric power}}{29.4 \times 10^7}
Electric power = 0.6 × 29.4 × 10710^7 = 176.4 × 10610^6 W = 176.4 MW
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