Answer:
1) 364,564 houses
2) Approximately 2.7 × 10⁸ litres of water annually
Explanation:
The power consumption of the average household = 10,972 kWh/year
The power consumption, [tex]P_{(consumed/house)}[/tex] of the average household = 914 kWh/month
The amount of power produced, [tex]P_{produced}[/tex], by the Hoover Dam = 4×10⁹ kW/year
1) The number of houses, n, the Hoover Dam can power is given by the relation;
[tex]n = \dfrac{P_{produced}}{P_{(consumed/house)}} = \dfrac{4 \times 10^9}{10972} = 364,564.35 \ houses[/tex]
Which is approximately 364,564 houses
2) Given the height, h = 15 m
We have at 100% efficiency,
The potential energy of the water per year = 10,972 kWh
The potential energy of the water = m×g×h
Where:
g = Acceleration due to gravity = 9.81 m/s²
∴ The potential energy of the water = m×9.81×15
Therefore, we have;
m×9.81×15 = 10,972 kWh = 10972×60 min/hour ×60 seconds/minute Joules
m×9.81×15 = 39,499,200 kJ = 39,499,200,000 J
m = 39,499,200,000/(9.81 × 15) = 268,428,134.6 kg
The volume, V, of water that would have to drop out of the dam to power a house for a year is given by the relation;
Volume = (Mass of water)/(Density of water)
V = (268,428,134.6 kg)/(1000 kg/m³)
V = 268,428.135 m³ ≈ 2.7 × 10⁸ litres of water annually.