In a hydroelectric dam, water falls 35.0 m and then spins a turbine to generate electricity. suppose the dam is 80% efficient at converting the water's potential energy to electrical energy. how many kilograms of water must pass through the turbines each second to generate 53.0 mw of electricity? this is a typical value for a small hydroelectric dam.

Height through the water falls h = 33 m

Efficiency of the of the unit

η

=

80

%

Power of the production unit

P

=

45

×

10

6

W

Acceleration due to gravity

g

=

9.8

m

/

s

Potential energy of one kg of water

Δ

U

=

m

∗

g

∗

h

=

1

∗

9.8

∗

33

=

323.4

J

Eighty percentage of the above energy is converted into electrical energy.

So eighty percentage of the potential energy of one kg of water

Δ

U

1

=

258.72

J

So mass of water required to flow per second to produce 45 MW of electricity in kilograms

M

=

P

Δ

U

1

=

45

∗

10

6

258.72

=

173933.2096

k

g

/

s

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Hydroelectric Energy: Definition, Uses, Advantages & Disadvantages

from Earth Science 101: Earth Science

Chapter 23 / Lesson 9

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