clarification on pannel efficiency and unitery conversion/meaning

JNSMAR043JNSMAR043 Registered Users Posts: 4
Hi,

So when calculating the size of a PV system at which stage in the calculations do you take into account the solar panels inefficiency? Also if say as an example that in my location the average solar insolation is 4 Kwh/m^2/day (hours/day) how do you tell how much energy you will yield from the array assuming 250W panels? And finally I cant make sense of how Kwh/m^2/day equates to equivalent sun hours or peak sun hours.
Thanks and praise
JNSMAR043

Comments

  • BB.BB. Super Moderators, Administrators Posts: 29,235 admin
    Re: clarification on pannel efficiency and unitery conversion/meaning

    I am not quite sure how I can answer your question without making for more confusion--In general. But I will give it a try. If you have an example of what you would like to do (such as power 3,300 Watt*Hours per day in Johannesburg SA), it will probably make more sense.

    First, there are conversion losses every time you go from one form of energy to another (sunlight to electricity, through a charge controller, to stored energy in a battery bank, back to 12 VDC from the battery, changing from 12 VDC to 230 VAC, down the wiring to the AC load, etc.). We use rules of thumbs to describe the losses at various points in the system.

    When we say there are 4 hours of sun equivalent (average) in February... We are saying that if you add up all of the (useful for solar) energy for the sun during the day, we may get something like 4 kWatt*Hours when measured on a fixed solar cell at a specific angle. The full noon time average sea level sun is defined as about 1,000 Watts per square meter. We take the 4,000 WH per sqmeter and divide by 1,000 W/sqmtr average noon time sun, we get "4 hours" of energy from the sun per day at that location, with a specific "flat plate" set at a specific angle facing south (or north for folks below the equator), etc.

    You would get a different hours of sun per day if the panel was at a different angle, was not facing true south (or north), or you are using a "tracker" that follows the sun across the sky.

    We also try to use actual ~20 year average data to account for weather too. When we use the ratings for the panels (i.e., 250 watts at Vmp=25 volts and Imp=10 amps) this takes care of the per sqrmtr conversion. Roughly a crystalline solar panel will converter around 15% of the solar energy from the sun into electricity.

    And there are other losses too:

    ~81% * marketing rating for solar panels (solar panels do not generate as much voltage when they are hot--Marketing numbers overestimate panel output)
    ~95% efficiency through MPPT type charge controller (or GT inverter)
    ~80% efficiency for charging/discharging flooded cell battery (AGM, LiFePO4, etc. batteries are more efficient; NiCad, NiMH are less efficient)
    ~85% efficiency for DC to AC inverter

    0.81*0.95*0.80*0.85=0.52 end to end "typical" off grid system efficiency (from Marketing solar panel rating to useful AC power out the inverter).

    There are other losses too (wire can drop upwards of 3% of power, PWM charge controllers+solar panels have "different" matching losses, but the overall derating for other reasons is close enough to the 0.81*0.95=0.77 panel+MPPT controller deratings for our needs).

    We can talk about specific types of losses for each device (AC inverter have both a "tare loss", i.e., use 10 Watts of power just "turned on" and no loads due to running the electronics and capacitive switching losses of the FETs that "chop" the DC current to send through a transformer for the high voltage AC power; and a second loss due to wiring resistance, etc.--The efficiency will be different between running 1,000 Watt inverter at 10 watt AC output--perhaps 50% eff; vs running at 500 Watts and 95% eff, and 1,00 Watts at 80% eff., etc.).

    So--We use rules of thumbs and what people actually "see" from their systems... For example, a "typical" cost effective well designed off grid power system with 4*250 Watt panels, in late fall getting 4 hours of sun per day, charging the batteries during the day and discharging at night will produce an "average" useful AC power of:

    4* 250 Watt panels * 0.52 end to end eff * 4.0 hours of sun = 2,080 Watt*Hours of "useful" 230 VAC 50 Hz power

    Of course in deep winter you may get 2 hours of sun per day, and in summer you may get 6 hours per day. And on dark/cloudy/rainy days, you may get very close to zero hours of sun per day.

    We try to give "conservative" answers--But we have to be careful. You can see we are already taking 1/2 of the energy away as losses (from the Marketing numbers). It is possible to do better (cold/clear weather, new AGM batteries, using power during the day and skipping battery storage, using DC rather than AC inverter power), and it is also possible to do worse (shaded panels, dirty panels, very hot climate, old fork lift batteries, etc.).

    And with solar power, it is use it or lose it. The battery bank can store only ~2 days of useful power. If you are not there, the power is lost forever. If you pull more power than the sun put in the battery bank for a day or two, you need to use a generator or turn off your loads to bring the battery bank back up to full...

    Every system/need/end user is different. And between us here, we even have a few differences in how we do our calculations (and specific numbers/conditions). In the end, if you get within 10% of predicted output with your system--The calculations were dead on.

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • JNSMAR043JNSMAR043 Registered Users Posts: 4
    Re: clarification on pannel efficiency and unitery conversion/meaning

    Wow great answer just one further question.

    "Roughly a crystalline solar panel will converter around 15% of the solar energy from the sun into electricity"

    Does 15% efficiency of the panels converting sunlight into electricity ever feature in the efficiency calculations of the system? Or do we ignore it because we know the panels will output approximately there rated watt output in full sunlight.
  • CariboocootCariboocoot Banned Posts: 17,615 ✭✭
    Re: clarification on pannel efficiency and unitery conversion/meaning
    JNSMAR043 wrote: »
    Wow great answer just one further question.

    "Roughly a crystalline solar panel will converter around 15% of the solar energy from the sun into electricity"

    Does 15% efficiency of the panels converting sunlight into electricity ever feature in the efficiency calculations of the system? Or do we ignore it because we know the panels will output approximately there rated watt output in full sunlight.

    We ignore it because the panels' Wattage ratings are done under Standard Test Conditions. As such a 200 Watt panel from one company will put out about the same power as a same rated panel from a different company.

    For example polycrystaline panels are about 16% efficient, monocrystaline panels about 18% efficient. Some more, some less for either. But a 200 Watt poly panel and a 200 Watt mono panel are both "200 Watts".

    The application derating is also the same, that of typical output averaged over the hours of good sun, because the two types behave very similar. They will not, however, average their rated Wattage. Instead it's about 77% of that.

    Additional considerations are made for the particular installation if conditions are not expected to be average. For example temperature extremes: hot climate, panel runs hotter and Voltage is reduced. Cold climate, panel runs cooler and Voltage is increased. Another case would be atmospheric conditions which limit (marine layer) or enhance (high elevation) insolation.
  • BB.BB. Super Moderators, Administrators Posts: 29,235 admin
    Re: clarification on pannel efficiency and unitery conversion/meaning

    The only time panel efficiency comes into consideration is that a more efficient panel takes up less area.

    So, if you have a small roof (home, travel trailer), you may want to get the most efficient panel you can.

    There are another class of solar panels called thin film panels. They are less expensive and about 1/2 the cost (at least a few years ago) vs crystalline panels. They take more area (square meters) and also take more mounting racks (again, more costs) than the more expensive crystalline panels.

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