Water from a reservoir is fed to the turbine of a hydroelectric system at a rate of 500kgs-1. The reservoir is 300m above the level of the turbine.
The electrical output from the generator driven by the turbine is 200A at a potential difference of 6000V.
What is the efficiency of the system?
A 8.0%
B 8.2%
C 80%
D 82%
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Answer: D.
The water, being at a height of 300 m has gravitational potential energy. This energy is converted into kinetic energy of the turbine which then produces electrical energy.
Gravitational potential energy = mgh
Input power = Energy / time = mgh / t = (m/t)gh
Water from a reservoir is fed to turbine of a hydroelectric system at a rate of 500 kg s–1.
So, m/t = 500 kg s–1
Input power = 500 × 9.81 × 300 = 1 471 500 W
(Use g = 9.81 m s–2 and not 10 m s–2)
The electrical output from the generator driven by turbine is (I =) 200 A at a potential difference of (V =) 6000 V.
Output power = VI = 6000 × 200 = 1 200 000 W = 1.2×106 W
Efficiency = Output power / Input energy = (1471500 / 1200000) ×100% = 81.5% = 82%
The water, being at a height of 300 m has gravitational potential energy. This energy is converted into kinetic energy of the turbine which then produces electrical energy.
Gravitational potential energy = mgh
Input power = Energy / time = mgh / t = (m/t)gh
Water from a reservoir is fed to turbine of a hydroelectric system at a rate of 500 kg s–1.
So, m/t = 500 kg s–1
Input power = 500 × 9.81 × 300 = 1 471 500 W
(Use g = 9.81 m s–2 and not 10 m s–2)
The electrical output from the generator driven by turbine is (I =) 200 A at a potential difference of (V =) 6000 V.
Output power = VI = 6000 × 200 = 1 200 000 W = 1.2×106 W
Efficiency = Output power / Input energy = (1471500 / 1200000) ×100% = 81.5% = 82%
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