The solar panels are the most important component of any solar power system and are the first interphase between the sun and your energy needs.
They convert sunlight into direct current.
Their proper sizing can solve your energy needs and you enjoy smooth electricity for decades.
Online Solar Course: System Design and Financial Feasibility
The sizing of the panels depends on your solar energy needs that we learned in the previous post and the amount of sunlight falling on a place where you live.
Let us see how these variables are related to the sizing of the panels?
My daily energy need is 22 units.
I further analyzed and found that I consume 16 units in the day time when there is sunlight and the remaining 8 units in the night.
Now, 2-panel sizing conditions arise, one when I want to limit my solar energy needs to 16 units (daytime only), and another when I need complete 22 units from Solar Power.
First Condition: It is the simplest configuration when you restrict your solar energy needs to daytime only (16 units). I don’t need any storage system as my energy needs are fulfilled by the solar panels only.
Second Condition: The panels can produce current in the daytime only whereas I need these 8 units in the night when there is no sun and solar panels cannot work.
Therefore, I need an add-on system (solar batteries) that can store some energy for night use.
How I should do it?
Here is the answer,
I will size my panels so that they produce whole 22 units in the daytime.
I’m going to consume 16 units in the daytime as it is my daytime energy needs.
The remaining 8 units from the panels will go inside the batteries and they will store them.
In the night, these batteries will produce these 8 units and fulfill my night energy needs.
My night time energy needs are indirectly fulfilled by Solar Power through batteries.
Therefore, the batteries are required if you want to fulfill your night-time energy needs.
It is understood that if my energy needs are more than my panel size will also be more.
Now, coming to the second variable, the sunlight.
If your region or area receives good amount of sunlight then you need fewer solar panels than the area with not so good sunlight.
In short, good sunlight less panel size.
Let me share Sun Peak Hours/Solar Insolation (amount of sunlight per square meter area in a day) of some of the major cities:
Sunlight is inversely proportional to the panel size
The complete formula is:
If I say that my daily energy needs is 22 units and I live in Delhi where the average sun peak hours are 5.35.
In that case, my panel size would be = 22 kWhr/5.35 hr
= 4.11 kW
But when you live in Arizona with the same energy needs then your panel size would be:
= 22/6.44 = 3.42 kW
Now, you can see that increase in the Peak Sun Hours reduces the panel size.
Here is the panel size chart of the 4 major cities which we have discussed right now:
What you think you divide your energy needs by sun peak hours, you get the right size of the panels that will fulfil your energy needs.
That would be an ideal system with no losses.
Let me share the diagram of a solar power system with battery backup:
You can see that sunlight falls on the solar panels and they convert it into direct current.
This direct current is get stored inside the batteries in the form of chemical energy.
Also, when D.C. current passes into the inverter gets converted into A.C. current that finally run our electrical appliances.
You know that when energy undergoes conversions while traveling from one component to another, it suffers losses at each conversion.
In short, what is produce at the source will not reach the same at the destination.
You will receive a diminished or decreased value.
If your requirement at the destination is 22 units then the source (Solar Panels) must produce more than 22 units so that you get the desired value.
Therefore, we need to slightly over-size the panel to compensate for the energy losses and you get the required units to run your electrical appliances.
Let us see the different types of losses:
I have taken the following de-rating factors (when the loss is subtracted from the value, it is called the de-rating factor) to know how much my solar panels should produce so that I get 22 units daily to run my appliances.
= 22 units/ (Temp. de-rating factor x Dirt de-rating factor x Transmission efficiency x Shading de-rating factor x Power tolerance de-rating factor x Inverter efficiency)
= 22 units/ (0.95 x 0.95 x 0.97 x 0.95 x 0.95 x 0.95)
= 22/ 0.75
= 29.3 units
My solar panels must produce at least 29 units so that I get 22 units for running electrical appliances.
Right Panel Size:
= Units Produced by the Panel/ Sun Peak Hours
= 29/5.35 = 5.4 kW
Therefore, I should install 5.4 kW solar panels to meet the daily energy needs of 22 units.