Question

In a 770 kW hydroelectric plant, 300 m^3 of water passes through the turbine each minute. Assuming complete conversion of the water's initial gravitational potential energy to electrical energy, what distance does the water fall? Assume two significant figures.

Answer 1

Answer:

16 m

Explanation:

given,

power of hydraulic plant = 770 kW

volume of water pass through the turbine = 300 m³

density of water = 1000 kg/m³

m  =ρ × V

mass of water pass each minute  = 300 × 1000 = 3 × 10⁵

assume height of the fall be h

potential head of the water = mgh

$\dfrac{mgh}{60}= 770 \times 10^3$

$3\times 10^5 \times 9.81\times h= 770 \times 10^3\times 60$

$h = \dfrac{770 \times 10^3\times 60}{3\times 10^5 \times 9.81}$

h = 15.69 m ≈ 16 m

the distance of the water fall is equal to 16 m.