Answer:
7.22 m³ of water should be stored behind the hydroelectric dam.
Explanation:
First we find the total energy that can be stored in the battery. For that purpose we use the formula:
P = VI
where,
P = Power delivered by battery
V = Voltage of Battery = 12 volts
I = Current in Battery
Multiplying both sides by time interval (t), we get:
Pt = VIt
where,
Pt = (Power)(Time) = Total Energy Stored in Battery = E = ?
It = Current Rating of Battery = (50 A.min)(60 s/min) = 3000 A.s
Therefore,
E = (12 volt)(3000 A.s)
E = 36000 J
Now, we have to find the energy required to fully charge the battery:
Energy Required = ΔE = Total Energy Capacity(E) - Already Stored Energy
ΔE = 36000 J - 600 J
ΔE = 35400 J
Now, this energy must be equal to the potential energy of water stored behind hydroelectric dam to fully charge the battery, provided that the conversion efficiency is 100 %.
Therefore,
ΔE = mgh
where,
m = mass of water behind hydroelectric dam
g = 9.8 m/s²
h = height of dam = 50 cm = 0.5 m
Therefore,
35400 J = m(9.8 m/s²)(0.5 m)
m = (35400 J)/(4.9 m²/s²)
m = 7224.5 kg
Now, to find the volume of stored water, we use:
ρ = m/V
V = m/ρ
where,
ρ = density of water = 1000 kg/m³
V = Volume of water behind dam = ?
V = (7224.5 kg)/(1000 kg/m³)
<u>V = 7.22 m³</u>
