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3 years ago

6

A hydraulic turbine-generator unit placed at the bottom of a 75-m-high dam accepts water at a rate of 1020 L/s and produces 630

kWof electricity. Determine A) the overall efficiency of the turbine-generator unit and B) the turbine efficiency if the generator efficiency is 96 percent, and C) the power losses due to inefficiencies in the turbine and generator.

1 answer:

Bond [772]3 years ago

5 0

Answer:

A. Overall efficiency =83.95%

B. 87.45%

C.120.465kw

Explanation:

H=75m

Flow rate= 120L/s=1.02m^2/s

Po=Output power= 630kw

As power input Pi= Total power available

=gQH/1000

Where

g=9.81

Q=120L/s=1.02m^2/s

H=75m

= (9.81m/s^2 × 1.02m^2/s×75m)/1000

750.465kw

A. Overall efficiency of generator turbine unit (Po/Pi)

= Output power/ power input

= 630/750.465= 0.8395

Overall efficiency of the unit= 0.8395 × 100

=83.95%

B. Turbine efficiency if generator efficiency is 96%

no × nt= 0.96

nT= 87.45%

C. Power losses= Power input - Power output

Pi - Po

=750.465- 630

=120.465kw

Power losses due to inefficiencies in turbine of the generator is 120.465kw

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