Novice with a project

DerkDerk Registered Users Posts: 1
I'm very much a novice, but I'm here to learn and grow. I am designing a system using solar cells and lights. I would store the power in a battery off the grid and it would be outdoors at all times. I live in Wisconsin, so the shortest day of the year yields likely 9 hours of light (best to measure from the worst scenario). I'm looking at lights rated at 50W. Can I assume that is a watt used over an hour? On the shortest day of the year, there would be 15 hours of darkness. With a 50 watt bulb, that would be (15x50) 750 watts used over the course of the night (then scaled linearly for as many lights as I'd want). Correct? A solar panel would need to generate a minimum of 750 watts during the day to power this since no lights run during the 9 hours for this example (theoretically). This does not account for the decreased power generated during cloudy days. The max load I'm estimating for my project is about 2300 watts in a 24 hour period. Any thoughts on sizing batteries? I'm looking at 120v batteries due to the lighting requiring it. Another concern I have is insulating the battery against very cold and very warm temperatures. Any thoughts there? Thanks for any help you guys give.

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  • BB.BB. Super Moderators, Administrators Posts: 29,518 admin
    Welcome to the forum Derk!
    Derk wrote: »
    I'm very much a novice, but I'm here to learn and grow. I am designing a system using solar cells and lights. I would store the power in a battery off the grid and it would be outdoors at all times.

    That is why we all volunteer our time here--To help others after others have helped us.
    I live in Wisconsin, so the shortest day of the year yields likely 9 hours of light (best to measure from the worst scenario).

    Unfortunately will solar, the power (current) from a solar panel is based on how much direct sunlight is hitting the panel... Roughly from 9 am to 3pm is the peak times of the day to capture the energy. And clouds, humidity, dust, pollution, etc. all serve to reduce the amount of sunlight hitting those panels.

    One basic unit we use here is "Hours of Sun per day"... That is actually "hours of noon-time equivalent sun per day"... Full noon-time sun is around 1,000 Watts per square meter... A hazy day can reduce that by 1/2. A really dark/rainy day, you may get 5% of the the nominal sunny day power collection.

    We use programs like PV Watts to estimate the average daily sun... Here is what a 1,500 Watt fixed array tilted to 45 degrees for Eau Claire Wsconsin looks like.
    "Station Identification"
    "City:","Eau_Claire"
    "State:","Wisconsin"
    "Lat (deg N):", 44.87
    "Long (deg W):", 91.48
    "Elev (m): ", 273
    "PV System Specifications"
    "DC Rating:"," 1.5 kW"
    "DC to AC Derate Factor:"," 0.520"
    "AC Rating:"," 0.8 kW"
    "Array Type: Fixed Tilt"
    "Array Tilt:"," 44.9"
    "Array Azimuth:","180.0"

    "Energy Specifications"
    "Cost of Electricity:"," 9.1 cents/kWh"

    "Results"
    "Month", "Solar Radiation (kWh/m^2/day)", "AC Energy (kWh)", "Energy Value ($)"
    1, 3.29, 85, 7.74
    2, 4.63, 106, 9.65
    3, 5.24, 129, 11.74
    4, 4.84, 108, 9.83
    5, 5.42, 119, 10.83
    6, 5.61, 115, 10.46
    7, 5.66, 121, 11.01
    8, 5.31, 112, 10.19
    9, 4.68, 98, 8.92
    10, 3.82, 87, 7.92
    11, 2.46, 55, 5.01
    12, 2.33, 57, 5.19
    "Year", 4.44, 1190, 108.29

    So, if we toss the bottom three months (you use a genset to make up for loss power/use utility power) and use October as your "break even" month (may or may not need a genset):
    • 1,500 Watts * 0.52 AC off grid system efficiency * 3.82 hours of sun per day (average October day) = 2,979 Watt*Hours per average October day
    I'm looking at lights rated at 50W. Can I assume that is a watt used over an hour?

    Yes, Watts are a "rate" (like gallons per hour pumped).

    Watts*Hours is an amount (like total gallons pumped).

    50 Watts * 8 hours = 400 Watt*Hours = 0.4 kWatt*Hours

    Note, kWatt*Hours is what our electric meters bill... However, we usually use Watt*Hours in our calculations.
    On the shortest day of the year, there would be 15 hours of darkness.

    As you can see from above--The hours that the sun is above the horizon is not enough... It has to be "bright sun" with the panel pointing towards the sun (if you are 60 degrees of angle, you will only collect 50% of the energy from the sun, etc.). In Eau Clair, 2.33 hours of "average sun" per day in December...

    If you have tracking arrays, you can collect more sun (a good thing), but trackers are not cheap and do require some maintenance.
    With a 50 watt bulb, that would be (15x50) 750 watts used over the course of the night (then scaled linearly for as many lights as I'd want). Correct?
    • 50 Watt bulb * 15 Hours = 750 Watt*Hours of energy per day
    Solar power is expensive--So if you used a 50 Watt filament bulb--You could replace it with a 9 Watt LED bulb--And use 80% less electrical energy--And need a 1/5 the size solar array+battery bank.

    Especially in solar, measuring your loads and conservation are your friends. Get a Kill-a-Watt type meter to measure 120 VAC plug in load.
    A solar panel would need to generate a minimum of 750 watts during the day to power this since no lights run during the 9 hours for this example (theoretically). This does not account for the decreased power generated during cloudy days.

    From above, solar panels generate much less power than you estimated... And losses (panel derating, charge controller losses, battery losses, AC inverter losses) all add up--So that you get almost 1/2 the panel's rating at 120 VAC power at the outlet (times hours of sun per day).

    Even the most sunniest places in the world are hard pressed to have more than 7-8 hours of "noon time equivalent sun" per day.
    The max load I'm estimating for my project is about 2300 watts in a 24 hour period. Any thoughts on sizing batteries?

    Note that 2,300 Watt*Hours of energy per day... Battery sizing is usually around 2 days of backup power and 50% maximum discharge... For an AC power system on a 24 volt battery bank, you would sse:
    • 2,300 Watt*Hours per day * 1/0.85 AC inverter efficiency * 1/24 volt battery bank * 2 days of storage * 1/0.50 maximum discharge (for long battery life) = 451 Amp*Hour @ 24 volt battery bank
    I'm looking at 120v batteries due to the lighting requiring it. Another concern I have is insulating the battery against very cold and very warm temperatures. Any thoughts there? Thanks for any help you guys give.

    Slow down on the 120 Volt Battery bank... First, there are few (if any) standard "high voltage capable" solar charge controllers that can charge a 120 VAC battery bank. If there are a MPPT type controller available, they can be quite expensive.

    Also, 120 VDC loads--While you can run some items from DC--It may limit your options (LED and CFL's may not work, some computer power supplies may not work, etc.).

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • BB.BB. Super Moderators, Administrators Posts: 29,518 admin
    PS: I moved your thread to the Beginner's corner.

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
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