Determine the size of wire of the solar power system

14/3/2015

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:

\(Resistance\ \alpha(\frac{Length}{Area})\)
\(Resistance=\rho *(\frac{Length}{Area})\)

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,

\(\Delta V=I*R\)

Where \(\Delta V\) is the voltage drop across the wire, the voltage drop up to 5% of the nominal voltage of the panel is permissible.
Putting the value of the R in the above equation, we get

\(\Delta V= I*\rho*(\frac{Length}{Area})\)

\(Area=\frac{\rho * Current * Length}{\Delta V}\)

(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:

Electrical appliance	Watts	Usage in hours	Qty	Total watts	Watt-hours*
Fan	75	8	2	150	1200
CFL	25	4	2	50	200
Bulb	100	2	1	100	200
Total				350	1600
* Watt-hours = Watts x Usage in hours x Qty

Therefore, I want 1600 watt-hours 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 watt-hours
These 2540 watt-hours will reduce to 1600 watt-hours when it reaches to the load, the remaining, 2540 minus 1600 that is 940 watt-hours, 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 watt-hours 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

(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,

\(Area=\frac{\rho * Current * Length}{\Delta V}\)
Area of the wire = (14.28 x 0.0172 x 6) / 0.6volts
Area of the wire = \(2.46\ mm^2\)
Or \(\Pi r^2\) = 2.46

Where \(\Pi\) = 3.14
We get, \(r^2\) = 2.46/3.14 = 0.78 \(mm^2\)
Or r = square root of \(\sqrt 0.78\) = 0.88 mm

Diameter of the wire = 2 x r = 2 x 0.88 mm = 1.77 mm

This is the minimum diameter of the wire which is required to connect the components of the solar power system.

(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.

American wire gauge(AWG)	Diameter (mm)	Area(mm2)
0000	11.68	107.1
000	10.4	84.9
00	9.27	67.5
0	8.25	53.4
1	7.35	42.4
2	6.54	33.6
3	5.83	26.7
4	5.19	21.1
5	4.62	16.8
6	4.11	13.3
7	3.67	10.6
8	3.26	8.3
9	2.91	6.6
10	2.59	5.3
11	2.3	4.2
12	2.05	3.3
13	1.83	2.6
14	1.63	2.1
15	1.45	1.7
16	1.29	1.3
17	1.15	1
18	1.02	0.8
19	0.91	0.7
20	0.81	0.5
21	0.72	0.4
22	0.65	0.3
23	0.57	0.3
24	0.51	0.2
25	0.45	0.2
26	0.4	0.1

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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