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Tidal Energy.

If we model 2 blades as thin uniform steel bars 15.0m long and .25m in diameter, at what rate (radians/second and revolutions per minute) must they spin for a turbine to store 10^6 J of energy? Keep in mind that the density of steel is 7800kg/m^3 and density is equal to mass/volume).

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7 answers

  1. This sounds like a flywheel energy problem, not a tidal energy problem.

    The kinetic energy of each blade is
    (1/2) I w^2 . The moment of inertia of each blade (assuming that it is attached at the axis of rotation) is
    I = (1/3) ML^2
    L is the length of the blade (15 m)and M is the mass (volume x density)

    Set the total rotational energy equal to 10^6 J and solve for the angular velocity w. Remember that there are two blades.

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  2. How do I determine the volume of the blade so I can find the mass? I'm not quite sure what shape the figure actually is...

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  3. Very good question. Since they specify the diameter of the "bars", which I will call D, the blades are cylinders and the volume of each is
    (1/4) pi D^2 * L

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  4. M=7800*[(1/4)*PI*(.25)^2*15]

    KE=(1/2[(1/3)M(15)2]w^2=10^6
    KE=2.15... x2 (for the two blades)
    KE=4.31

    would that be in radians or revolutions?

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  5. Your KE calculation would be in Joules, but you forgot the w^2 in the formula.

    You are supposed to be solving for w, which would be in radians per second

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  6. I can't seem to find where I've gone wrong...

    Volume=(1/4)PI(.25)^2*15=.736...
    Mass= V*7800= 5743.2...
    I=(1/3)M*15^2=430741.8...
    10^6=(1/2)Iw^2
    w=sqrt(4.64...)
    w=2.15...
    w*2[for the 2 blades]=4.309...

    I tried the above value and was incorrect.

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  7. I can't seem to find where I've gone wrong...

    Volume=(1/4)PI(.25)^2*15=.736...
    Mass= V*7800= 5743.2...
    I=(1/3)M*15^2=430741.8...
    10^6=(1/2)Iw^2
    w=sqrt(4.64...)
    w=2.15...
    w*2[for the 2 blades]=4.309...

    I tried the above value and was incorrect.

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