Understanding Insolation (W/m^2)

My residence uses around 1500 kWhr / month averaged over the year. In researching PV, I am trying to calculate the amount of kW*hr that can be stored in a month for various sized systems. Most panels use 1000 W/m^2 for their ratings, and also run tests at 800 W/m^2. These appear to be standard test conditions.

When I look up the amount of sunlight energy reaching the earth I usually get numbers of around 300 W/m^2 for Tucson, or in one case 1000 W/m^2 during peak sunlight hours (value for 5 hours/day). Some sites give W*h/M^2/day as the units. Its a bit confusing.

I assume that the 300 W/m^2 is averaged over daylight hours, or an entire 24 hour day.

Specifically, on an average sunny August day in Tucson -- at noon -- what would be the expected amount of sunlight energy that hits a panel pointed towards the sun. Not averaged out over 24 hours, or seasonally -- but peak instantaneous energy.


Any help would be appreciated.

Thanks!

-- edit. And I do understand that in sizing a system it would need to be averaged out over off peak conditions. I'm just trying to clarify the definitions of the numbers I am pulling up...

Comments

  • BB.
    BB. Super Moderators, Administrators Posts: 33,431 admin
    Re: Understanding Insolation (W/m^2)

    The US government a few decades ago ran ~20 year solar isolation measurements around the US. They then put the data into a very simple to use program to calculate how much power (per month, or even per hour) you can get from XX kW array and yy% efficiency mounted on a 0/1/2 axis tracker.

    Here is a link to the PV Watts program. Enter, for example Tucson, 10 kW of solar panels, 0.77 for Grid Tied, and defaults for everything else:
    "Station Identification"
    "City:","Tucson"
    "State:","Arizona"
    "Lat (deg N):", 32.12
    "Long (deg W):", 110.93
    "Elev (m): ", 779
    "PV System Specifications"
    "DC Rating:"," 10.0 kW"
    "DC to AC Derate Factor:"," 0.770"
    "AC Rating:"," 7.7 kW"
    "Array Type: Fixed Tilt"
    "Array Tilt:"," 32.1"
    "Array Azimuth:","180.0"

    "Energy Specifications"
    "Cost of Electricity:"," 8.5 cents/kWh"

    "Results"
    "Month", "Solar Radiation (kWh/m^2/day)", "AC Energy (kWh)", "Energy Value ($)"
    1, 5.70, 1282, 108.97
    2, 6.11, 1212, 103.02
    3, 7.03, 1550, 131.75
    4, 7.50, 1561, 132.69
    5, 7.29, 1513, 128.60
    6, 7.15, 1418, 120.53
    7, 6.44, 1316, 111.86
    8, 6.85, 1411, 119.94
    9, 7.06, 1434, 121.89
    10, 6.72, 1444, 122.74
    11, 5.99, 1287, 109.39
    12, 5.27, 1199, 101.92
    "Year", 6.59, 16627, 1413.30

    You will get 1,199 to 1,561 kWHrs per month for Grid Tied solar.

    Now, assuming that you want off grid system (batteries and AC inverter), change the derating factor to 0.52 (flooded cell batteries at 80% efficiency and 85% inverter efficiency * the 77% derating for a grid tied system):
    "Station Identification"
    "City:","Tucson"
    "State:","Arizona"
    "Lat (deg N):", 32.12
    "Long (deg W):", 110.93
    "Elev (m): ", 779
    "PV System Specifications"
    "DC Rating:"," 10.0 kW"
    "DC to AC Derate Factor:"," 0.520"
    "AC Rating:"," 5.2 kW"
    "Array Type: Fixed Tilt"
    "Array Tilt:"," 32.1"
    "Array Azimuth:","180.0"

    "Energy Specifications"
    "Cost of Electricity:"," 8.5 cents/kWh"

    "Results"
    "Month", "Solar Radiation (kWh/m^2/day)", "AC Energy (kWh)", "Energy Value ($)"
    1, 5.70, 853, 72.50
    2, 6.11, 805, 68.42
    3, 7.03, 1034, 87.89
    4, 7.50, 1041, 88.48
    5, 7.29, 1005, 85.42
    6, 7.15, 941, 79.98
    7, 6.44, 868, 73.78
    8, 6.85, 935, 79.48
    9, 7.06, 953, 81.00
    10, 6.72, 963, 81.86
    11, 5.99, 856, 72.76
    12, 5.27, 797, 67.75
    "Year", 6.59, 11049, 939.16

    Now you are down to 856-1,041 kWHrs per month.

    Your question was a bit confusing for me... You asked about battery bank sizing but talked about the amount of sun.

    So, the above is an example of PV array sizing.

    For Battery Bank Sizing, a good rule of thumb is to plan for 3 days of no-sun and a 50% maximum discharge. Or 6x your daily DC power needs (remember, the inverter is only 85% efficient, so you have to add more power to the battery bank). So, for 1,500 kWhrs per month:
    • 1,500,000 Watt*Hour * 3 days * 1/0.50 * 1/0.85 inverter * 1/48 VDC bank = 220,588 Amp*Hours at 48 volts
    That is one heck of a battery bank.

    At this point in time, for home sized systems, lead acid battery banks tend to be the most cost efficient storage devices available. However, your energy usage is about 15x that what a "typical" off-grid home would aim for (try for 100 kWHrs or less).

    Grid Tied is typically the best way to use Solar PV array for home type systems. There are variations that can include a Hybrid GT/Off-Grid inverter (Grid Tied for normal operations, battery bank + generator backup for emergency off-grid). Sort of depends on how much you want to spend and what it is you want to protect against. Also, not all utilities will allow a Grid Tied connection with 1 year net metering...

    In the end, we usually try to convince people to spend their money first on conservation (insulation, double pane windows, energy star appliances, CFL lighting, downsizing loads such as a laptop instead of a desktop computer farm, turning things off, etc.)--because it is almost always more cost effective to conserve energy than to generate it.

    Do you have more questions or more information on what you are trying to achieve?

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • System2
    System2 Posts: 6,290 admin
    Re: Understanding Insolation (W/m^2)

    Bill -

    Thanks for the very fast reply to my thread.

    I am trying to understand panel output more so than battery sizing. My question was poorly worded / misstated.

    No doubt, grid tie makes the most sense, especially at the consumption rate I currently have. The battery requirements are an eye opener, to say the least!

    With the incentives from the state, even partially covering the kWhr usage would be cost effective. But, first -- there is clearly much I could/should do to reduce consumption...

    Still trying to understand the 300 W/m^2 vs 1000 W/m^2 numbers I keep seeing. Does a 205 Watt panel put out 205 Watts at midday in Tucson, or would it be expected to put out 30% of that amount given the 300 W/m^2 numbers I see for isolation?

    I have an engineering background, so I dont mind crunching my own numbers, but the PV nomenclature is completely new to me. Just trying to get my arms around it..


    One additional question: Is it possible to have a grid tie system, but have a limited number of batteries to store energy for brief periods of power outages (at MUCH reduced levels of consumption than the 1500 kWhr/mo numbers I currently have? (ie: charge up the batteries, and then when they are at capacity, send the rest into the grid.)

    Thanks again!
  • niel
    niel Solar Expert Posts: 10,300 ✭✭✭✭
    Re: Understanding Insolation (W/m^2)

    i'll throw a few cents worth into here. pvs usually don't reach their 1000w/m^2 rating and is more closely about 90% of that stc for ptc or even better you can go by the california ratings known as cec. there are other losses that could bring the figures lower too like wire resistance and inverter efficiency.
    to answer your last concern, the answer is yes you can, but the efficiency will drop even further and up the costs some. if you need the backup power then it is worth it, but be sure you have to use no more than about 50% of the battery capacity to help keep battery life.
  • System2
    System2 Posts: 6,290 admin
    Re: Understanding Insolation (W/m^2)

    Thanks for the quick responses guys!

    The more I think about it the more an emergency gas generator probably makes the most sense, long with a grid tie system. (Noticed Bill's tag line) Plus it could be portable...

    I'm sure I'll have more questions as I figure out my path ahead.
  • BB.
    BB. Super Moderators, Administrators Posts: 33,431 admin
    Re: Understanding Insolation (W/m^2)

    Here is a nice I*V curve data sheet for a crystalline silicon panel.

    The solar PV cell is basically a current source whose output current is proportional to the amount of solar energy falling on the surface. Once the panel is exposed to weak sunlight (or strong back scatter from the sky), Voc/Vmp is attained and the current above that point is pretty much proportional to the amount of sunlight. In fact, you can make a pretty decent sensor by just placing a silicon cell directly on an amp meter or low resistance shunt. Here is an example that can be home made--website also has some automated loggers available too (kit/turnkey).

    When people type about "hours of sun"--they basically are assuming 1,000 Watts/sq.meter of "noon day" sun spread out over the entire day.

    The output of the panel is defined on a 1,000 watt/sq.meter day, Standard Temperature and Wind. While the results are accurate, they are not really representative of real world average installations where there is not as much wind and the solar cells tend to run hotter, which depresses the Vmp (voltage maximum power). Imp does go up a little bit with temperature, but not near as much as Vmp goes down as temperature rises.

    How much power can a panel output--on a sub-zero windy day with lots of white snow in the foreground / edge of cloud event, you can probably get upwards of 130% rated power. The NEC (current National Electric Code) has requiements on how to calculate the Isc and Voc with safety factors if you need those.

    But assuming that the average V*I output + controller losses is in the range of 77% is not a bad place to start.

    The PV Watts program has a list of deratings and the standard values they use here.

    Because there is both an absolute value of solar radiation, plus mounting/tracking options and Cosine Error, you have a multitude of variables to address and integrate over.

    The PV Watts Program (and other software/data) is open source (as I recall) if you hunt round in the website if you want to code something up yourself.

    I would not count on much better than 10% accuracy given all of the variables out there (including weather, temperature, and specific equipment performance).

    Regarding a GT/Off-Grid system... There are Hybrid systems out there that can do both. The Xantrex XW system is one very nice one. Up to 6kW 120/240 VAC of GT/Off-Grid power and a minimum of 600 Amp*Hour 48 Volt Battery Bank. With an Aux AC input for a backup Generator. Note that there are multiple components to the XW system (naming convention is a bit confusing). There is a Hybrid Inverter, a 60 amp MPPT solar charge controller, and a pre-wired panel to connect everything together. Plus other options (generator controller and such). You can price the major components from our host's website" Northern Arizona Wind and Sun.

    A hybrid system is a great way to go... In general, it is best when you have lots or long power outages and generator fuel is difficult to store. Otherwise, a GT system + regular backup genset is hard to beat.

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • a0128958
    a0128958 Solar Expert Posts: 316 ✭✭✭
    Re: Understanding Insolation (W/m^2)
    BB. wrote: »
    ...Here is an example that can be home made--website also has some automated loggers available too (kit/turnkey). ...

    -Bill

    This reference has a great tutorial on Solar Power Basics, particularly a discussion on measuring solar radiation, and what insolation is.

    Many thanks!

    Best regards,

    Bill
  • bobdog
    bobdog Solar Expert Posts: 192 ✭✭
    Re: Understanding Insolation (W/m^2)

    Another basic tutorial on insolation....stuff I didn't know....I'm kinda slow though...

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