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Make electricity using your backyard stream

  1. Apr 6, 2012 #1
    I'm thinking about building a generator for my parents powered by a stream in their backyard. It would be a great way to learn about science. I'm trying to find the relevant equations so that I can estimate how much energy the stream produces.

    Right now I have this equation:

    ε = NBAω

    I don't understand the difference between voltage and EMF, they seem like the same thing to me.

    The stream is about 3 feet wide by 6 inches deep and I don't know how much volume flows through it per second, but using this wiki article

    http://en.wikipedia.org/wiki/Streamflow

    it looks like I can find out. The it seems that I would need another equation to figure out how many revolutions per minutes I could get in the turbine. I would imagine that the energy or work generated by water would be equal to mass times gravity times distance times a friction coefficient. So then I would need some equation to figure out how energy efficient the turbine is unless that's covered by the friction coefficient. I have no idea about that.

    I'm not sure how to figure out fast I could get the turbine flowing. I also don't know what a reasonable amount of B (magnetic field I could produce). Wouldn't I have to buy the magnets and figure out what Tesla's they are?

    As far as the A (area) part of the equation, I would think that something 3 feet by 2 feet would be reasonable, right? As for the N (number of turns) part of the equation, I don't really know how to estimate that.
     
  2. jcsd
  3. Apr 6, 2012 #2

    russ_watters

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    Staff: Mentor

    You should start by calculating the energy available. I think you'll be shocked to find out how small it is.

    Typically, hydro dams utilize the potential energy of water, not the kinetic energy of water, since after going through a turbine, the water still needs to move away from the turbine. Even a typical pelton turbine starts with potential energy and only converts it to kinetic energy at turbine by constricting the flow. But you should do both calculations and compare. Here's how:

    1. For kinetic energy, drop a ping pong ball in the stream and measure the time it takes to move a measured distance. Calculate speed. Now multiply by the cross section and you have your volumetric flow rate. Convert to mass flow rate and apply the kinetic energy equation.

    2. Measure the altitude difference from one side of your property to the other either with GPS or Google Earth's altitude feature or perhaps even a laser level. Use the mass flow rate from method 1 and apply the gravitational potential energy equation.

    I think you'll find you are hard pressed to power even a singe light bulb.
     
    Last edited: Apr 6, 2012
  4. Apr 6, 2012 #3
    Ok, here's my calculations.

    The stream is .74m wide. Depth is a bit harder because it's not a uniform depth. Its max depth is .0889m so I estimated that the average depth is .05m. The length is .92m (I was using a yard stick) This volume took about 2.7s to flow, so the flow rate is (.92/2.7) .34 m/s.

    To measure mass I will use cm since water is 1g/cm^3

    Volume = 74 * 5 * 92 = 34040 cm^3
    Mass = 34 kg per 2.7 s

    to convert mass into kg/s is 34/2.7 = 12.6 kg/s

    To me that seems to be way off but I can't figure out where I went wrong in my calculations.

    So the kinetic energy is

    .5 (12.6)(.34)^2 = .728 j/s

    measured in watts per hour

    .728 * 60 * 60 = 2.6 kw/hr

    My parents use 63 kw per day. 63/24 = 2.62 kw/h

    But still I'm deeply pessimistic that I convert all that energy into electricity with 100% efficiency.

    I saw an article some where on the net that 2.5 kw/hr is reasonable with a backyard stream.

    For the water to flow one meter took about 2.7 s, so 1/2.7 = .37 m/s
     
  5. Apr 6, 2012 #4

    Vanadium 50

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    Bob, I am afraid that pretty much every unit in your equations is wrong.

    Energy is measured in Joules.
    Power is measured in Watts, which is a Joule per Second.
    Energy is also measured in kilowatt-hours, which is also 3,600,000 Joules. (Can you see why?)
     
  6. Apr 6, 2012 #5

    AlephZero

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    Homework Helper

    I think that's right, based on the numbers you posted.

    After that, you went completely wrong. You are mistaking watt-hours (number of watts times number of hours) for "watts per hour" which is a fairly meaningless unit, because watts are already a measure of the rate of using energy.

    0.728 j/s = 0.728 watts. That's about enough poiwer to light 10 small LEDs, or to recharge one AA size battery in 3 hours.

    The units for your usage are also probably wrong. I think you mean 63 kW-h per day, which is an average power of 63/24 = 2.62 kW.

    As russ-watters guessed, this is no use at all to save on your energy bills.

    To get something close to 2.62 kW, you need a flow velocity about 15 times higher than what you have. That would increase the mass flow by 15 times so the total power would be about 15 x 15 x 15 = 3375 times bigger, or about 2.45 kW.

    Just for fun, let's calculate the height of a dam you would need to get 2.62 kW, from your tiny mass flow:

    Potential energy = m g h, so 12.6 x 9.81 x h J/s = 2620 W
    So h = 2620 / (12.6 x 9.81) = 21 meters. Unless you live on the side of a mountain, that's probably not practical either.
     
  7. Apr 6, 2012 #6

    Redbelly98

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    You did well up to (and including) this point. Since a J/s is equivalent to a Watt, you could produce at most 0.728 W -- though there will be efficiency losses to lower that power level.

    Compare your 0.7 W to just running one 60 W light bulb, or even a "60-W equivalent" CFL bulb of 15 W. Or to a microwave that might be 700 to 900 Watts.

    All you could hope to do is power small objects like some LEDs (as AlephZero said) or maybe a small DC motor that powers a fan a few inches in diameter. Might be worth it to learn about the science, but your wouldn't get practical benefits like saving on your electric bill.
     
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