Simple algorithm for solar power generation

Solar gel battery, backup power lead-acid battery, lithium battery.

The solar AC power generation system is composed of solar panels, charge controllers, inverters and batteries; the solar DC power generation system does not include inverters. In order to enable the solar power generation system to provide sufficient power for the load, it is necessary to select various components reasonably according to the power of the electrical appliances. The design of solar power system needs to consider the following factors:
Q1. Where is the solar power system used? What is the solar radiation situation in this place?
Q2. What is the load power of the system?
Q3. What is the output voltage of the system, DC or AC?
Q4. How many hours does the system need to work every day?
Q5. In case of rainy weather without sunlight, how many days does the system need to supply power continuously?

Let’s take (load) 100W output power and use it for 6 hours a day as an example to introduce the calculation method:

  1. First, calculate the number of watt-hours consumed per day (including the loss of the inverter):
    If the conversion efficiency of the inverter is 90%, when the output power is 100W, the actual required output power should be 100W/90%=111W; if it is used for 6 hours per day, the power consumption is 111W*6 hours= 666Wh, or 0.666 kilowatt-hours of electricity.
  2. Calculate solar panels:
    Calculated according to the effective daily sunshine time of 5 hours, and taking into account the charging efficiency and the loss during the charging process, the output power of the solar panel should be 666Wh÷5h÷70% =190W. Among them, 70% is the actual power used by the solar panels during the charging process.
  3. Daily power generation of 180 watt modules
    180×0.7×5=567WH=0.63 degrees
    1MW daily power generation = 1000000×0.7×5=3500,000=3500 degrees

Example 2: Installing a 10w lamp, lighting for 6 hours a day, 3 consecutive rainy days, how to calculate the solar panel wp? and 12V battery ah?
Daily power consumption: 10W X 6H = 60WH,
Calculate solar panels:
Assume that the average peak sunshine hours at your installation site is 4 hours.
Then: 60WH/4 hours, = 15WP solar panels.
Then calculate the charge and discharge loss, and the daily supplement of the solar panel:
15WP/0.6= 25WP,
That is, a 25W solar panel is enough.

Then calculate the battery.
Use 12V5AH of electricity every day.
It is 12V15AH for three days.
The battery configuration needs to be designed such that the daily power consumption does not exceed 20%, or the power consumption does not exceed 50% during continuous rainy days. In order to achieve the longest battery life requirement.
In this way, we conclude that the battery of this system is sufficient for 26AH-30AH.

Example 3: How many watts of solar panels are needed to fill a 12V45A battery in 6 hours?
The 12V45A battery is 648 watt-hours (?) If it is fully charged in 6 hours, the solar panel theoretically only needs to be 108 watts, but the actual solar panel is affected by factors such as sunshine intensity, temperature, and overall efficiency of the photovoltaic controller. The overall efficiency of the battery is calculated by 0.8. You need to choose a 135 watt solar cell module. By the way, the best charging current of a lead-acid battery is 1/10 of the battery capacity current, which is 4.5A. Excessive charging current will speed up the battery plate. Sulfuration affects battery life.
The simplest calculation method:
Battery: 12V×45A=540WH
Solar panel power = 540/6/0.8 (loss) = 112.5W

Example 4: How many hours does it take for two 20-watt (36 pieces) solar panels to charge a 12-volt 17-amp battery? How many hours does it take to charge an ordinary 12v4AH battery with those two solar panels?
The working voltage of 1.20W solar panels is generally 17.2V, and the current is 1.15A. If the board is of good quality, the measured current is generally 1.1A (I tested it).

  1. Assuming that the 6 hours of light you said is the period from noon to afternoon, then 4 hours of full power generation can be calculated, which means that 2 20W boards can generate 21.14=8.8A per day
  2. In this way, the 17AH battery can be fully charged in 2 days; the 4AH battery is almost the same in 2 hours.
    Or the total w of solar panels is 20+5=25W
    The total number of w of the battery is 12v*17A=204w
    Full time is 204/25=8 hours

4A battery:
4A 12=48w 48w /25w = 1.92 hours Or because of the inaccurate relationship between sunlight intensity and battery capacity, actuarial calculations are unnecessary and cumbersome. Estimate, Solar cell current: 20/12=1.7A Charging time 1: 17/1.71.5 charging constant = 15 hours,
Charging time 2: 4/1.71.5 charging constant=3.5 hours, In fact, you can charge two batteries and two solar panels in parallel, the same is true. Charging time 3: (17AH+4AH)/(1.72 blocks)*1.5 charging constant=9 hours,
If the sunlight is good in your place, it will last for almost two days.
There is nothing to pay attention to when charging. If you have a multimeter, always measure the voltage at both ends of the battery during charging, and it does not exceed 14V. Remember not to be less than 10.5V when discharging. Both overcharge and overdischarge affect battery life.

Example 5 Assuming 2 consecutive rainy days, the load power is 40W and the lighting time is 8 hours per day. To achieve the above lighting time, how many watts of solar panels and how many watts of batteries are needed?
The simplest algorithm is four times.
That is, the load power * 4 times, and 160W solar panels are required.

If you want to be more precise, it’s as follows:
The load power is 40W.
40W * 8 hours / ceiling * = 320WH / 12V (battery voltage) == 27AH.
Use 12V27AH of electricity every day,
It’s best to keep the battery within 30% of the discharge capacity every day. So we need a battery that can easily be 90AH12V. In this case, we can only choose 100AH, because 90AH batteries are difficult to buy, solar cells. 40W*8 Hour=320WH.
320WH removes 20% of the loss in the circuit and the power storage process, and the actual daily demand is 400WH.
If the time is 4 hours per day according to the standard sunshine time, the calculation is as follows:
400WH/4 hours = 100W.

Example 6 Load 2 50w load input voltage 24v 3 consecutive rainy days, working 8 hours a day
Request the required system solar panels and battery calculations

  1. Solar panel 250W8H/0.6/4H=340W (total power consumption/system utilization factor/effective sunshine time)
  2. Battery 250/248*(3+1)/0.7=200AH (total current * self-holding time / margin factor)
    (Solar panel power=load power*working time/loss 0.6/average effective light)
    (Battery capacity = load power * working time * continuous rainy weather / battery voltage / charge and discharge coefficient)

Calculated by the amount of solar radiation
Annual power generation (EP) = PAS * HA * K * 365 (days)
PAS: solar battery string capacity
HA: Cumulative solar radiation of installation location and installation conditions (kWh/m2 *day)
K: Sum design coefficient (0.65~0.8≒0.7 degree)
Calculated by system utilization
Annual power generation = power generation of solar cell array template * system utilization rate * 8760 (hours)
System utilization ratio = 0.1~0.15≒0.12 degree

Total hours in a year = 24 (hours) * 365 (days) = 8760 hours.

Household electricity can be replaced by solar power generation, which will also become a fashion when environmental protection is popular today. We can recommend the best solution for you based on the amount of electricity your home uses, your geographic location and other information.
Although the solar power system has the advantages of safety, environmental protection and pollution-free, its cost is quite high, so it is generally recommended to be used only for lighting.
About the approximate cost calculation, you can calculate according to the following simple method to see how to arrange the scale of solar power generation.

  1. Calculate the total daily power consumption, the average household electricity consumption should be between 5 degrees and 10 degrees per day. You can divide the total monthly electricity bill by the unit price and then the number of days.
  2. You can simply apply the formula 5000W (assuming 5 kilowatt-hours of electricity per day)/5 hours (average effective light time per day, different in different regions)/0.7 (actual efficiency of solar panels)/0.9 (various losses) = 1600W, then Adding a 5% margin, it is almost 1700W.
  3. The above number is the power of the system. Even if the average unit price of the current system is 60 yuan/W (including all materials and installations), then the total investment is 1700X60=102,000, which is more than 100,000. At present, the electricity price in most areas is calculated at 0.6 yuan, 102000/0.6=170,000 kWh, 5 kWh per day, which can be used for 90 years.
  4. From the above point of view, it is basically unrealistic for domestic households to rely solely on solar power for electricity. Foreign countries are developing very well because of state subsidies. We must also have subsidies, and the cost must be greatly reduced, so that solar power can truly enter the homes of the people.

The AC power generation system can be composed of solar panels, batteries, controllers and inverters. When there is sunshine during the day, you can use the battery board with a controller to charge the battery, and use the battery to power the electrical appliances at night.
In this case, it is recommended to use an 80W battery board, a 12V20AH battery (purchased locally), a 12V5A controller and a 300W inverter. When fully charged, it can be used for four 20W lamps for more than 5 hours, which is enough for most people. If it is not enough, you can add one or more panels.
This kind of small system is very suitable for power shortage or low power areas, such as forest areas, mountainous areas, or field work (beekeeping). The cost is not high, and it is convenient to carry. The system can be adjusted according to needs, which can fully meet the daily electricity consumption

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