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30 June, 12:38

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

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  1. 30 June, 12:45
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    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
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