Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Hydro-power output question(s)

  1. Dec 27, 2011 #1

    D-C

    User Avatar

    Hi all. I'm feeling a little new here. Here's the deal. Question involving my micro-hydro installation. Pretty basic in this crowd I would guess but a chore for me, and am not sure the numbers I come up with end up making practical sense. I've been tinkering with it for years, but still seem to manage to get a very paltry amount of power out of quite a bit of water, falling a long ways. All home-made or scrounged materials with the exception of the 5-1/4 in. split cup Pelton impeller.

    If the weight of water works out to .43333 lbs/sq.in/foot, and my static pressure at the nozzle/valve is 34psi that would put my head just under 80' (78.5).

    The 600 or so feet of pipe is sized well enough that with my largest nozzle, (Rain Bird) 17/64, or about .28 in, dynamic pressure is only 2 lbs less, 32psi. A little tough running that into a container there in the gully, but a gallon in 5 secs = 12/Min..... 17,280/Day. (if I cut some brush and devise a better way to catch the water in a larger container I could be more accurate!)
    (I do have nozzles down to .22)

    So I'm dropping about 12 gals a minute 80 feet and measuring under 3 Amps of 12 volt. (2.6-2.8), or around 30 watts. I've been off the grid for over 30 years, this is my main power when the sun isn't shining.
    Any thoughts on if/how it might be improved would be very welcome!

    And...am I even in the ballpark?
     
    Last edited: Dec 27, 2011
  2. jcsd
  3. Dec 27, 2011 #2

    russ_watters

    User Avatar

    Staff: Mentor

    Welcome to PF!

    Unfortunately, there isn't as much energy in falling water as one might think. The energy is easily calculated as height * weight, using the appropriate conversions to get it into watts...

    For your water source, 12 gpm is 1.668 lb/sec. Multiplied by your height gives 130.9 ft-lb/sec or 0.24 hp or 177.6 Watts.

    If you do a spectacular job of recovering this energy, you might achieve an efficiency of 50%, or 88 Watts. So you'll want to get at least double what you're getting now to feel good about your effort.

    In optimizing the performance, you'll want to make sure you're operating at the right rpm and your jet is at the right velocity. Do you have more info about your turbine with which we can figure this out? This might get you started: http://en.wikipedia.org/wiki/Specific_speed
     
  4. Dec 27, 2011 #3

    D-C

    User Avatar

    Great, thank you Russ! That link is a mouthful. I'll read that through, and also put together better info for the "hardware". Meantime:

    I have a couple different permanent magnet DC motors. (one can be fixed up while the other is wearing down). I just make the jet as big, and fast as I can.

    **just now I put in a smaller nozzle, (.20) and am getting a solid 2A, 1/2 the water, 6 Gal./Min.** (the creek is low, driest Dec. here since records)***

    Again my thanks, I'll get back, Don

    Sooo, at the lower flow, litll' slower, my #'s get allot closer to the parameters you mentioned.
     
    Last edited: Dec 28, 2011
  5. Dec 29, 2011 #4

    D-C

    User Avatar

    Still getten my facts straight. I'm going to need jet velocity with a 1/4" nozzle, @ 32psi, (dynamic).

    The impeller should recede at 1/2 Jet V , leaving the, "tail-water" with no momentum.

    This is close to my layout, (without the moss and frogs!)
    http://www.youtube.com/watch?feature=endscreen&NR=1&v=M3OMha0Qv_M

    I need to find the ideal RPM for two different DC perm-mag motor/generators. That will need to match my wheel. Might be hard to find anything smaller than my 5 1/4" though.

    OK, I did find V= 138.9 x Sq.Rt. of Height in inches. (did not mention water though!)
    That there would give me 4,303.66 feet/min. About 4/5 ths. of 60mph...reasonable.

    OK, just found "similar threads" below.-----pressure times volumetric flow rate (or potential energy flow rate, which is essentially the same equation)

    I'm simply going to essentially find the length of a (long narrow), cylinder of known volume, then its "travel time"through the nozzle gives me Velocity. Pie R Sq x H yada yada...

    4,708 @ 1.604 cu/ft/min ?
     

    Attached Files:

    Last edited: Dec 29, 2011
  6. Dec 30, 2011 #5

    D-C

    User Avatar

    4708 feet is the length of a cylinder ¼ inch in diameter “encompassing” 12 gallons.
    “Transiting” the nozzle in one minute gives me a jet velocity of 4708 ft./min.

    A 5 ¼ in. impulse wheel would rotate at 1,714.5 RPM to receive the jet at ½ the jet velocity.

    Can I “reverse engineer” (for lack of a better term), the characteristics of the motor/gen. by metering it's current draw under load and somehow measuring it's RPM? Sheez..A known load. String winding on a drum could provide both load and RPM.

    Right now matching my work to that wheel and that jet is my issue. If the above motors aren't reasonably close enough I might find another, or a larger wheel may be affordable.

    (I did put in a .30 nozzle today, up to 3.2 Amps (40 watts). Inefficient, but it's raining buckets!
    The .25 will remain the ideal. Also I should have made clear, it's a 12 volt system with 440 A-Hr of battery capacity.
     
    Last edited: Dec 30, 2011
  7. Dec 30, 2011 #6

    D-C

    User Avatar

    Posted above;
    “””For your water source, 12 gpm is 1.668 lb/sec. Multiplied by your height gives 130.9 ft-lb/sec or 0.24 hp or 177.6 Watts.

    If you do a spectacular job of recovering this energy, you might achieve an efficiency of 50%, or 88 Watts. So you'll want to get at least double what you're getting now to feel good about your effort.”””

    So then, if I set a goal of 35% of the (GPE), comes out about 60 Watts. 5 amps is handy as it's about the max I might go unregulated. Also being about double what I'm seeing is fine!

    Below are some specs I found on the Ametek motor/Gen, I have here in the house. Once while it was running, I left the circuit open a second, (makes a siren sound down there in the gully when it freewheels). I then hooked it back to the house circuit forgetting that the batteries where not connected. I discovered that the 60+ volts will fry my radio, even when it's off! (memory ect.)

    Rpm........Open V.........Shorted Amps
    100........3.48..............1.18
    400........13.92.............4.51
    500........17.4............5.5
    600........22.88.............6.30
    Then I found that it will go 60+ open at 2-3K RPM

    I wonder how "shorted", correlates with the resistance of my battery system. Could this motor/gen. be near enough to optimum for my system with only impeller size as the variable?
     
    Last edited: Dec 31, 2011
  8. Jan 1, 2012 #7

    D-C

    User Avatar

    Power output of Ametek 38Volt Motor/Generator.

    OK, I found a guy's PDF of a graph for the Power output of Ametek 38Volt Motor/Generator. Pretty much linear. Watts up to 100, RPM over to 3000. He shows what looks like a “sweet spot”, (slightly steeper line), with 520 ohm's resistance and posts 3.3 amps. (but over a range of output?)

    10Watts @ 750 RPM
    50Watts @ 1800 RPM
    75Watts @ 2300 RPM
    (these numbers are my approximation from multiple plots on his graph)

    Tough search! (for me anyway). Found myself here even! Happy New Year!

    Since I can't always EDIT!, I'll post later if I find these figures way off.
     
    Last edited: Jan 1, 2012
  9. Jan 1, 2012 #8

    D-C

    User Avatar

    Power output of Ametek 38Volt Motor/Generator
    I've been trying to figure how to present the graph, maybe this PDF.....
    OK! Ignore the red square.
    I can in no way verify it's accuracy. He does seem competent though.
     

    Attached Files:

    Last edited: Jan 2, 2012
  10. Jan 2, 2012 #9

    D-C

    User Avatar

    So if it's a given that 60 whats is attainable with the Ametek 18 Volt generator with that ,^ jet, and that ,^ graph, I need to calculate wheel size to accommodate, lets say 2200 RPM.

    So with 2,354 Ft/Min, (Wheel velocity at the circumference), what size wheel will rotate at 2,200 rpm?

    OK..C isn't really a second variable..knowable..since 2,354 fpm (1/2 nozzle velocity, is known)

    Well duh,..about 1 foot. Without an equation I "found" .953.

    .29729 ft. diameter, just under 4 inches. On the one hand it doesn't feel right. On the other hand it's not an "unreasonable" number. (the splash pattern/tail water on the current unit, the 95 volt treadmill motor/gen. is angling out 45* PAST perpendicular, suggesting it too would like a smaller wheel than 5 1/4.)
     
    Last edited: Jan 2, 2012
  11. Jan 2, 2012 #10
    Water can generate power in more than one way. Ideally, to get the most power out of the water you would utilize as many methods as possible.

    Kinetically - water falling is used to create energy.
    Temperature - differences in temperature, costly to get it to work, but it could be controlled theoretically to have temperature fluctuate at various areas with the use of albedo effect for heating and heat sinks for cooling. And then add in some peltier plates, or something similar.
    And the other method I know about is through ion charges - kelvin's thunderstorm method of collecting high voltage static from drops of water.

    Realistically it would require a lot of work, and money, to utilize all of the potential energy that water would have to offer though.
     
    Last edited: Jan 2, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook