In the previous article, How to size the panels of your solar power system?, we learned the sizing of the solar array. Now, what should be the size of the wires that will connect the all components (solar panels, charge controller, battery and the load) of the solar power system?
The size of the wire is very important, while an undervalued wire may result in excessive heating and ultimately it will burn out on the other side a overvalued wire may cost your more. (1) The concept of resistance
Whenever energy flows through wire, it offers some kind of resistance. The resistance offered by the wire is directly proportional to its length and inversely proportional to its area , or we can write in a mathematical equation like this:
Where \(\rho\) is the resistivity of the wire, generally copper wires are used; therefore we will take the resistivity of copper, which is \(0.0172\ \Omega.m\) Ohms law says,
Putting the value of the R in the above equation, we get
(2) Calculating the load and the energy required
Let us understand all this with one simple example:
I want to run two ceiling fans of 75 watts each, 2 CFLs of 25 watts each and a 100 watt bulb. I do not require the battery backup. Summarizing my requirements in the load chart, this is as follows:
Therefore, I want 1600 watthours as an output in a day from the solar panels. Assuming 70% conversion efficiency (conversion from solar energy into electrical energy) of the solar panels and 90% efficiency of the inverter (The energy will be lost when inverter will convert DC electrical energy to AC electrical energy), we get, 1600/ (0.7 x 0.9) = 2540 watthours
These 2540 watthours will reduce to 1600 watthours when it reaches to the load, the remaining, 2540 minus 1600 that is 940 watthours, will be lost during energy conversion process and heat losses while transferring through wires. (3) Sizing the solar panels
The average solar insolation value of Delhi/NCR is around 5.5, therefore I am dividing the above value of 2540 watthours by 5.5 to get the size of my solar panels, and this comes out as:
2540/5.5 = 462 watts, this is the size of the solar panels which will run the above mentioned electrical appliances. I am taking two 250 watt panels to make it 500 watts. The nominal voltage of each panel is 12 volts and the current at the maximum power, Imp, is 7.14 amps. (4) Calculating the current flow through the wire
I have connected the solar panels in parallel (In parallel combination, the voltage remains the same and the current adds up, whereas in series combination, the voltage adds up and the current remains the same). Therefore, the value of the current (I) is 7.14 x 2 = 14.28 amperes. This is the amount of the current that will flow through the wires.
(5) Calculating the diameter of wire
I have connected the solar panels in parallel (In parallel combination, the voltage remains the same and the current adds up, whereas in series combination, the voltage adds up and the current remains the same). Therefore, the value of the current (I) is 7.14 x 2 = 14.28 amperes. This is the amount of the current that will flow through the wires.
I want not more than 5% voltage drop across the wires. The nominal voltage of the panels is 12 volts; therefore the voltage drop \((\Delta V)\) is 5% of 12 volts i.e. 0.6 volts. The length between the my solar panels and the charge controller is, say 3 meters (multiply it by 2 as two wires will run simultaneously to connect the positive and negative terminals of the panels to the terminals of the charge controller), we get L as 6 meters. Putting all the above values in the equation below, we get,
We get, \(r^2\) = 2.46/3.14 = 0.78 \(mm^2\) Or r = square root of \(\sqrt 0.78\) = 0.88 mm
(6) Know the gauge of the wire
In the market the wires are differentiated by the gauge, therefore we have to find the relation between the wire diameter and the wire gauge.
The nearest gauge comes out to be 13. Therefore, I need to purchase wire of gauge 13 to connect my components of the solar power system.
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