1a. Assuming 100% efficient energy conversion, how much water stored behind a 50cm high hydroelectric dam would be required to charge a 50 ampere-minute 12 volt battery with 600j of energy stored in it

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)

V = 7.22 m³

amplitude

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