Question

 Water flows over a waterfall that is 20 m high at the rate of 4.0 × 104 kg/s. a. How much KE does the water gain each second when it reaches the bottom of the waterfall? b. If this water powers an electric generator with a 40% efficiency, how many watts of electric power can be supplied? C, What is the velocity of the water when it exits the turbine that drives the generator?

Answer 1

Answer:

a)

7.8 x 10⁶ Watt

b)

3.12 x 10⁶ Watt

c)

15.3 m/s

Explanation:

a)

h = height of the waterfall = 20 m

m = mass rate = 40000 kg/s

K = gain in kinetic energy per second

Using conservation of energy

K = mgh

K = (40000) (9.8) (20)

K = 7.8 x 10⁶ J/s

K = 7.8 x 10⁶ Watt

b)

P = Electric power supplied

η = Efficiency = 40% = 0.40

Electric power supplied is given as

P = η K

P = (0.40) (7.8 x 10⁶)

P = 3.12 x 10⁶ Watt

c)

P' = Kinetic energy remaining in water after exiting the turbine

v = velocity of water

Kinetic energy remaining in water after exiting the turbine is given as

P' = K - P

P' = 7.8 x 10⁶ - 3.12 x 10⁶

P' = 4.68 x 10⁶ Watt

(0.5) m v² = 4.68 x 10⁶

(0.5) (40000) v² = 4.68 x 10⁶

v = 15.3 m/s