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How many watts soler ???

Discussion in 'Home Power and Microgeneration' started by no one that you know, Sep 3, 2006.

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  1. To get 2 kwh per day north of [email protected] degrees stored in batts.
    How many watts of 12 volt panels?


    also has anyone or does anyone know of what type of panel that can
    produce some power from the moon or a street light out front of there
    house. (This is a night light experiment.)
    I upgraded my camper with led... about 21 fixtures in total. My guess is
    about 300 12v led's all together......I couldn't believe that a crappy
    tire five watt panel could make them all light up no problem. Was just
    thinking that a bigger panel whould make a simpler low wattage
    night/accent lighting project for the home if a panel could react to low
    light. I dont expect full power from the panel but I would size the
    panels to get the wattage I need for night time use and aim them in the
    daytime for max to store power.
     
  2. Eeyore

    Eeyore Guest

    The first thing you need to know is the insolation for your area ( how much
    light it gets ).

    You'll get a lot less in winter btw, so size with that in mind.

    Graham
     
  3. Thats why I gave the cordinates....I live in sunny south western
    Alberta.
    I am allready taking into account for twice my needs to make up for
    losses and cloudy days.
    I believe my area is about average 6 usable hours per day. I cant
    remember where I got that from.
     
  4. Malc

    Malc Guest

    <Googles for definition of kwh>

    Do you seriously want 2000 watts per hour? If so you'll need assuming 100%
    efficient panels and inverters etc 2000 W of solar panels. But as you want
    it for the whole day then you'll need probably at least 3 times as much and
    then some because of inefficiencies.
     
  5. Guest

    What do you mean by "watts per hour"?

    Nick
     
  6. Huh? 2000 watts a day 2kwh a day what am I missing or did you miss it?
     
  7. Eeyore

    Eeyore Guest

    In winter you may only get a few hours of useful sunlight ( say just a few hours )
    so you'll need one heck of a lot of panels.

    Graham
     
  8. Eeyore

    Eeyore Guest

    No it isn't. There is no such thing as a 'kW per hour' at all !
    It says............ " equivalent to one kilowatt (1 kW) of power expended for one
    hour (1 h) of time "

    So it's power *times* time not power divided by time.
    No he doesn't. He's talking about *energy*.

    Graham
     
  9. Guest

    Here's a clue: this is the alt.ENERGY.homePOWER group.

    Nick
     
  10. daestrom

    daestrom Guest

    No, you've misread the definition (and that page's wording isn't all that
    great).

    Energy is power *times* time. A kilowatt-hour is a measure of energy. A
    kilowatt is a measure of power.

    When you say, "...a kwh is 1 kw per hour", you are using the word 'per'.
    And mathematically, that means 'divided'. So you are essentially saying
    "...a kwh is 1 kw divided over an hour". And that is wrong, it makes no
    sense. It would be like saying an amp-hour (AH) is an amp divided over an
    hour. Doesn't make sense.

    An amp-hour is much like a kilowatt-hour. It is a measurement of a flow
    *multiplied* by a time period. An ampere is a measure of the flow of
    current, a kilowatt is a measure of power (which is the flow of energy).
    When you multiply either by time, you get a measure of the total amount of
    what has 'flowed'. In the case of amp-hour, since amperes is a measure of
    electric-charges flowing in a given amount of time, when you multiply it by
    the length of time it has been flowing, you get a measure of how many
    electric charges have gone past a point in the wire (such as into or out of
    a battery). In the case of kilowatts, since a kilowatt is a measure of
    energy flowing in a given amount of time, when you *multiply* it by the
    length of time it has been flowing, you get a measure of how much energy has
    gone past a point in a wire.

    A usage of 2kwh in one day means the OP uses just 2kwh of energy over a 24
    hour period. The fact that he's specified his energy usage over a 24 hour
    period, using a unit of measure that has a 'time unit' built into the name
    *does* add a lot to the confusion. One could use energy at the rate of 2
    kilowatts for one hour. That would be a total energy of 2 kwh. Or one
    could use energy at the rate of 100 watts. After twenty hours, they would
    have used 2 kwh.

    To use 2 kwh in one day, you could use energy at the rate of 4000 watts for
    just 1/2 an hour, then turn everything off for the remaining 23.5 hours. Or
    you could use energy at the rate of 200 watts for 10 hours and then turn
    everything off. Or you could use energy at the rate of 83.3333 watts for a
    full 24 hours. All of these will total the same amount of *energy*, 2
    kilowatt-hours. But of course the equipment to support those different
    power levels, and the losses change with the different power levels.

    If the OP gets just 5 hours of good, strong sunlight, then if he can collect
    solar energy at the rate of 400 watts, he would collect 2 kwh of energy in
    those five hours of sunshine. Of course if he's storing it and using an
    inverter, there will be losses in each step of the process. So he may want
    to collect solar energy at a higher rate, say 500 watts. Then over the same
    5 hours of good strong sunlight, he will have collected 500*5/1000 = 2.5 kwh
    of energy. Some of that energy will be 'lost' in inefficiencies in storage
    battery so he may get just 2kwh 'delivered' to his loads. If he wants to
    collect enough energy in one day to support two days of operation (prepare
    for a cloudy day tomorrow), he might have to collect energy at the rate of
    1000 watts. Then over the same 5 hours of good strong sunlight, he will
    collect 5*1000 / 1009 = 5 kwh. He can then use 2.5 kwh of that energy today
    (2 kwh in loads and 0.5 kwh in losses), and the remaining 2.5 kwh of energy
    tomorrow.

    Hope this clears up some of the confusion...

    daestrom
     
  11. I need 400 kwh per year minimum. 400/365=1.096 round up= 2 kwh per day
    or 730 kwh per year. This is where I get lost ...if I have minimum 5
    usable hours a day (lots of days with more) is it 2000w/5hr= 400w ?
     
  12. Thanks Graham.
    I think I have it figured out? I would need minimum 400 watts PV with 5
    hours a day minimum usable sun. If I spend a total of 4000 buks for
    panels and batts and stuff etc... at todays prices for power that is
    about 40 years of power I am paying for in advance.
    This a hobby I want to start. I have been researching it for years.
    It would be the 2nd hobby that I had that has some payback so no big
    deal if I ever see payback. I am going to make my fridge green powered.
    It is a small apartment size fridge and uses very little power. I really
    dont care if it reduces my power bill becasue my power bill is minor
    anyways.
    I would get two things that I need more.... like a hobby and back up
    power when the grid goes down. I have wasted far more on stupid shit
    like the big dish and computers....bet all told since 1980 when the big
    dish came out I spent 20k and now it's all in the garbage.....and when
    win95 came out lol I remember running out and buying 8 meg of ram for
    400 buks. I got caught up in the upgrade upgrade upgrade upgrade craze
    just of the top of my head thats another 20k. So somehow me thinks PV is
    a better hobby lol!
    I just happen to have the ideal location for panels on my bungalow. I
    have easy acsess to the roof the angle and direction is perfect for all
    year round use. Even when it snows the drift side is on the other side
    that I cant use. The locals call that the "east wind" are these ones any
    good
    http://cgi.ebay.ca/Sun-Elec-SUN-180...2QQihZ009QQcategoryZ41981QQrdZ1QQcmdZViewItem
    ? Or these ones?
    http://cgi.ebay.ca/Solar-Panels-160...5QQihZ014QQcategoryZ41981QQrdZ1QQcmdZViewItem
     
  13. oops I just screwed up ignore my last post. The math is all wrong.
    I need 730 kwh of 120vac per year. @ 12 volts how many watts of PV is
    that per 5hr a day.
    I want to make 2kwh of 120 vac to use over 24 hours from 12 volt PV.
    I can figure out the dollars. I cant figure out how many watts of dc to
    get 2 kwh of ac at 5 hours a day. Dont worry about the cloudy days I can
    figure that out. We are talking a 60 kwh per month home/cabin system.
     
  14. Technically, yes, this is how it works.

    Unfortunately, PV panels are sold based on a STC rating (Standard
    Test Conditions) which do not exactly match real world conditions.
    The main difference is that solar insolation isn't usually quite
    as much as they use in the test and panel temperature is usually
    higher. Both of these factors conspire to make a PV panel sold as
    400W produce somewhat less than 400W. A safe derating would be to
    assume the panel produces only 80% of it's STC rating.

    Then you'll need to account for losses in your system design.
    Batteries only return around 80% (more or less) of the energy
    that is put into them and if you use an inverter you may only
    expect 90% (or less) of the energy out of it as AC that you
    put in as DC.

    Then there is the question of 'usable hours a day'. If you mount
    the panels in a fixed position, which is most common, then the
    angle of the sun at any time other than high noon will mean the
    panels are receiving less than full sunlight. Typically solar
    insolation maps and measurements would account for this by saying
    something like 5 kWh/m^2/day on average. This means that even
    though the day was 12 hours long, the amount of sunlight that
    falls on a flat panel is only equivalent to 5 hours of direct
    high noon sunlight. Some people call this 'sun-hours' as opposed
    to 'hours of sunlight' because it doesn't really measure how
    long the day is but how much sunlight energy there is.

    So, let's assume you live in an area that gets 5 kWh/m^2/day
    on average over the year, with a minimum of 3 kWh/m^2/day in
    the winter. If you need 2kWh/day even in the depths of winter
    and don't use a tracking mount (which adds some 30% by facing
    the sun all day) and you are using an inverter to produce AC
    then it figures out to something like the following...
    2,000Wh/dayAC / .90(inverter) / .80(batt) / .80 (PV) / 3 sun-hours
    1,157 STC Watts of solar PV.

    Add a tracker and then sun-hours becomes around 4 instead of 3
    and you would only need 868 STC Watts of PV.

    Drop the inverter and only use DC with a tracker and it becomes
    781 STC Watts of PV.

    Add a small generator that you use a few hours every couple of
    days in the dead of winter and not at all in the summer months
    and you can drop the PV requirement some more. Adding a small
    wind or water turbine, if you can, would also help.

    Anthony
     
  15. Guest

    Almost. It's 2000 Wh divided by 5 h. See how the h's cancel out?

    Nick
     
  16. Guest

    We can also do this with numbers :)

    2
    ---
    3
    2 2 3
    --- = --- X ---
    3 1 1
    2 2 3
    --- = --- X --- = 2 x 3.
    3 1 1

    Nick
     
  17. Guest

    Very creative...

    Nick
     
  18. Eeyore

    Eeyore Guest

    And very wrong !

    Graham
     
  19. daestrom

    daestrom Guest

    Your trouble is right here....

    You have
    1KW
    ---
    1 hr

    You want to invert and multiply, but what is the inverse of '1 hr'?? It is
    *NOT* ...

    1 hr
    --
    1

    The inverse of '1 hr' is....

    1
    --
    1 hr


    You're doing two inversions, you're taking the denominator, '1hr' and saying
    it is really
    1
    --
    1hr

    when it is not, so when you 'invert' it you're getting

    1hr
    --
    1


    But that is *WRONG*.

    The denominator is '1hr', period.

    As nick tried to show you...

    2 2 1
    -- = -- X --
    3 1 3

    The denominator on the left is '3'. The inverse of that is 1/3.
    Multiplying, the results is simply

    2 2
    -- = --
    3 3

    Which is a valid equality. The fact that nick came up with....

    2 2 3
    -- = -- X -- = 2X3 = 6
    3 1 1

    Since clearly,

    2
    -- is not equal to 6
    3

    means that the method of 'invert and multiply' you used is flawed.



    What you *may* be thinking of is something like....

    2 2 5 10
    -------- = --- X --- = ---
    3 1 3 3
    --
    5
    This is the more common use of 'invert and multiply'. It is used to get rid
    of the fraction in the denominator and simplify.


    Your 'trick' gives you the 'right answer for the wrong reason'. If you try
    to use it to solve the problem in reverse, you may be in for trouble....

    2 kw 6kwh
    ----- = -------
    x hr 1

    Solving for 'x', we might just 'cross-multiply' and set the products equal
    to each other....

    2 kw X 1 = 6 kwh X 'x' hr

    'x' hr 2kw 1
    = ---- = ---
    6 kwh 3 hr

    But the answer *should* be 3 hr, not 1/3 hr. Yes, if you try hard enough
    you can make 2 / (1/3) equal to six again, but think about this. Drawing 2
    kw for *less* than an hour can't possibly be more than 2 kilowatt-hours of
    energy.

    daestrom

    P.S. Hope all the 'spacing' and such comes out, I'm not good with 'ASCII
    art'.
     
  20. Eeyore

    Eeyore Guest

    kWh is kW times hours not divided by !

    Graham
     
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