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Describe the energy flow in a hydroelectric power plant and calculate its efficiency given specific parameters.

Physics

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Here are the solutions for the questions from Section III, Option 1:

1. (a) (i)
The completed diagram showing the energy flow sequence in a hydroelectric power plant is:
Gravitational potential energy of water in the reservoir →\rightarrow Kinetic energy of water in the turbine →\rightarrow Mechanical energy of the turbine →\rightarrow Electrical energy from the generator.

1. (a) (ii)

  1. Convert the volume of water discharged per second to mass per second.

    Given volume V=150 cm3/sV = 150 \text{ cm}^3/\text{s}.
    Convert cm3\text{cm}^3 to m3\text{m}^3:

    150 cm3=150×(10−2 m)3=150×10−6 m3=1.5×10−4 m3150 \text{ cm}^3 = 150 \times (10^{-2} \text{ m})^3 = 150 \times 10^{-6} \text{ m}^3 = 1.5 \times 10^{-4} \text{ m}^3

    Density of water ρ=1000 kg/m3\rho = 1000 \text{ kg/m}^3.
    Mass of water discharged per second mt=ρ×Vt\frac{m}{t} = \rho \times \frac{V}{t}:

    mt=1000 kg/m3×1.5×10−4 m3/s=0.15 kg/s\frac{m}{t} = 1000 \text{ kg/m}^3 \times 1.5 \times 10^{-4} \text{ m}^3/\text{s} = 0.15 \text{ kg/s}
  2. Calculate the input power (gravitational potential energy per second).

    Given height h=38 mh = 38 \text{ m} and acceleration due to gravity g=9.81 m/s2g = 9.81 \text{ m/s}^2.

    Pinput=mght=(mt)ghP_{input} = \frac{mgh}{t} = \left(\frac{m}{t}\right)gh Pinput=0.15 kg/s×9.81 m/s2×38 m=55.803 WP_{input} = 0.15 \text{ kg/s} \times 9.81 \text{ m/s}^2 \times 38 \text{ m} = 55.803 \text{ W}
  3. Calculate the power output from the turbine.

    Given efficiency η=72%=0.72\eta = 72\% = 0.72.

    Poutput=Pinput×ηP_{output} = P_{input} \times \eta Poutput=55.803 W×0.72=40.17816 WP_{output} = 55.803 \text{ W} \times 0.72 = 40.17816 \text{ W}

    Rounding to three significant figures:
    The power output from the turbine is 40.2 W\boxed{\text{40.2 W}}.

    1. (a) (iii)
    Fossil fuels are non-renewable energy sources formed over millions of years from the decomposition of ancient organic matter (e.g., coal, oil, natural gas). Bio-fuels are renewable energy sources derived from recently living organisms or their metabolic byproducts (e.g., ethanol from corn, biodiesel from vegetable oils).

    1. (b) (i)
    One advantage nuclear fusion has over nuclear fission is that it produces less radioactive waste with shorter half-lives, and the fuel sources (isotopes of hydrogen) are abundant.

    1. (b) (ii)
    The fusion reaction is: 12H+13H→24He+01n+ΔE^2_1\text{H} + ^3_1\text{H} \rightarrow ^4_2\text{He} + ^1_0\text{n} + \Delta E
    Given masses:
    Mass of 12H=2.015 u^2_1\text{H} = 2.015 \text{ u}
    Mass of 13H=3.017 u^3_1\text{H} = 3.017 \text{ u}
    Mass of 24He=4.004 u^4_2\text{He} = 4.004 \text{ u}
    Mass of 01n=1.009 u^1_0\text{n} = 1.009 \text{ u}

  4. Calculate the total mass of the reactants.

    Mreactants=mass(12H)+mass(13H)=2.015 u+3.017 u=5.032 uM_{reactants} = \text{mass}(^2_1\text{H}) + \text{mass}(^3_1\text{H}) = 2.015 \text{ u} + 3.017 \text{ u} = 5.032 \text{ u}
  5. Calculate the total mass of the products.

    Mproducts=mass(24He)+mass(01n)=4.004 u+1.009 u=5.013 uM_{products} = \text{mass}(^4_2\text{He}) + \text{mass}(^1_0\text{n}) = 4.004 \text{ u} + 1.009 \text{ u} = 5.013 \text{ u}
  6. Calculate the mass defect (Δm\Delta m).

    Δm=Mreactants−Mproducts=5.032 u−5.013 u=0.019 u\Delta m = M_{reactants} - M_{products} = 5.032 \text{ u} - 5.013 \text{ u} = 0.019 \text{ u}
  7. Convert the mass defect to energy (ΔE\Delta E) using the conversion factor 1 u=931.5 MeV1 \text{ u} = 931.5 \text{ MeV}.

    ΔE=Δm×931.5 MeV/u\Delta E = \Delta m \times 931.5 \text{ MeV/u} ΔE=0.019 u×931.5 MeV/u=17.7005 MeV\Delta E = 0.019 \text{ u} \times 931.5 \text{ MeV/u} = 17.7005 \text{ MeV}

    Rounding to three significant figures:
    The energy released is 17.7 MeV\boxed{\text{17.7 MeV}}.

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