Built on Facts

Storing Energy

Posted by Matt Springer on October 15, 2008
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This week I’m teaching rotational motion to my students. Here’s an easy problem from their textbook, which comes from the idea of using a flywheel to store energy. I’m modifying it from problem 9.41 in Young and Geller:

Suppose we want to built a flywheel in the shape of a solid cylinder or radius 1.00m using concrete of density 2.2 x 103 kilograms per cubic meter. What must its length be to store 2.5 MJ of energy in its rotational motion, if for stability its speed is limited to 1.75 seconds for each revolution?

That seems a little slow – I bet some decent engineering could get it a lot faster, but that’s the setup. First let’s talk about what’s happening in this situation. A moving object has kinetic energy. Now this spinning cylinder is not going anywhere, but any given part is in motion. The parts of the wheel toward the outside will be moving faster and the parts near the center will be moving slower. There’s a convenient way of taking this into account when writing the total rotational kinetic energy of a spinning object:

i-d4bb9fe09d6b78eb832985821d8a307f-1.png

Where I is the moment of inertia of the object, which is tabulated in many places for differently shaped objects. It describes the way the mass is distributed and how much total energy per angular velocity the object has. That lower case Greek letter omega is the angular velocity of the object. In needs to be in radians per second, but we’re given a value in seconds per revolution. We’ll invert that to get revolutions per second, and then multiply by 2π to get radians per second. I get 3.59 radians per second, though formally radians don’t actually count as a unit.

For a solid cylinder, the moment of inertia is:

i-88516ca505d84f969b7ab15a965c3fb3-2.png

M is the mass, R is the radius. Plugging in, we have:

i-d18063683dcc0d42b9be45451a84d1e3-3.png

So, we can find the total mass we need in order to store the requisite amount of energy. From there, we can find the required length. Solve for M, you scalawags!

i-8d7070ce28f0af28047c7a5fb8246d53-4.png

Plug in our numbers for this problem and I get 776,000 kg. Heavy, but concrete is cheap. Given the density of the concrete, we have a length of 112 meters. Disaster, that’s preposterously large. That’s pretty hideously impractical, considering 2.5 MJ is not all that much energy – about what a good hairdryer uses in half an hour. This is why actual flywheel storage systems spin at a lot more than our pathetic 150ish rpm. Something like 20,000 rpm is more usual, and since energy is proportional to the rotational speed squared, all other things being equal a flywheel spinning at 20,000 rpm will hold something like 18,000 times more energy. That’s much more practical, and would allow the flywheel to be much more compact.

Indeed these kinds of systems have been in use for a long time. Today with the ever increasing interest in ways to generate and store energy, flywheels are not only still around, they’re finding even more new applications. Neat stuff, and a cool way to introduce rotation to a 201 class.

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Comments

  1. #1 Uncle Al
    October 15, 2008

    A broached battery’s energy release is limited by chemical kinetics. A broached flywheel detonates as local imbalance forces global disintegration. Air resistance, hence hard vacuum and moving contact wear issues. Maximum spin rate is defined by binding energy – and flaws.

    usable energy storage/mass
    ——————————————
    matter antimatter annihilation
    nuclear fusion
    nuclear fission
    Radioisotope thermoelectric generator
    Diesel internal combustion
    kerosene internal combustion
    gasoline internal combustion
    lithium hydride battery
    nickel metal hydride battery
    lead acid battery
    flywheel
    capacitance
    inductance

  2. #2 DA
    October 15, 2008

    I suppose the good thing is that the height scales like R^(-4), so if you double the radius to 2 meters, then you only need a 7 meter high cylinder.

  3. #3 Paul Murray
    October 16, 2008

    I’m surprised that the moment of inertia is proportional to R^2. The speed with which a bit of concrete travels will be proportional to it’s distance from the center, and so it’s KE will be proportional to R^2, and so the KE of a hollow cylinder of negligible thickness will be proportional to R^2.

    But to get the KE of the whole cylinder, you’d have to take an integral, so I would have thought that for a *solid* cylinder (which is what the problem specifies), the ke would be proportional to R^3.

  4. #4 Paul Murray
    October 16, 2008

    Actually, it’s worse than that, because the volume (and hence mass) of a cylinder of negligibe thickness also increases with R, so that stacked on top of the velocity means that the KE of a solid cyinder should be proportional to R^4.

  5. #5 james Dempsey
    October 16, 2008

    Speaking of angular momentum… Ever wonder if the iomega corporation, a company which made its money selling little spinning disks, chose their name because angular momentum can be expressed as the moment of inertia times angular velocity? (L = iw)

  6. #6 Bob
    October 16, 2008

    Looks like gasoline/diesel is still the energy storage champ. We’ll be driving cars with gasoline/diesel piston engines for many more decades. Maybe the engine blocks will be ceramic, but they’ll have cast iron cylinder sleeves.

  7. #7 eddie
    October 16, 2008

    Looks like gasoline/diesel is still the energy storage champ.

    A bit nitpicky I know but just who or what is doing the storing in this case. As long as petroleum is a fossil fuel there’s no way it will be a method of storing energy.

    See this wiki link for info on flywheel energy storage and it’s pros and cons. Where they will be most useful is in providing storage for renewable-generated electricity, to flatten out the good-time/lean-time cycle of wind and solar generation. One particular system is marketed in this first link that came up on a web search. Using them in vehicles is limited by the gyroscopic effects mentioned in the wiki.

  8. #8 CCPhysicist
    October 16, 2008

    Paul at #4:
    You formulated your analysis in terms of density, but the moment of inertia is given in terms of mass. The mass is rho*pi*r^2*L, which removes your extra 2 powers of r. The 1/2 is there because the mass close to the axis has less rotational inertia. Remember, the “moment of inertia” is the 2nd moment (integral of dm*r^2 over the mass distribution), written in terms of the total mass M and the outside radius R.

    Uncle Al at #1 alludes, intentionally or unintentionally, to one reason for limiting the rotational speed of a concrete cylinder. Concrete is very weak in tension, so fragments would split off very quickly compared to a steel flywheel. (Steel is strong in tension but weak in compression, so it is great for this sort of application.) Just think about how you break up a sidewalk: undermine it, lift the slab a bit, and hit it so the top is in tension. Note that making it bigger in radius makes this problem worse because R increases the centripetal force needed.

    The other problem with concrete is the low density. It would be much better to use depleted uranium.

    But I don’t quite buy the problem with stability at 34 rpm unless the concrete was not very homogeneous.