Current Method for Photovoltaic Calculations

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Introduction: Current Method for Photovoltaic Calculations

Hi everyone!

This is my firs attempt to make an Instructable, being a huge fan of this site, I am pretty excited.

I am from Argentina, so it is probable that some misspellings and grammatical errors pop up here an there, I apologize in advance for them.

Current Method for Stand Alone Photovoltaic Systems Sizing
This particular method is known as "Current method" since it uses the currents (in Amps, Amps Hour / Day) to select the components.

The components:
1) The modules
In Argentina we use this scale: cell → Module → PanelThe PANEL is an array of several modules, plugged in series or parallels to achieve the right configuration. In this example we will have everything in 12 V systems, so 32-cells modules are going to be OK.
The MODULE is composed of several cells in series and parallel to achieve the right current and voltage.The CELL is the smaller component. Is made of silicon, with the electric contacts in front and rear to pick up some electrons, running around the silicon when some photons hits them.
2) The charge regulator
This is a critic component, since it keeps your batteries fresh and healthier. A bad chosed charge regulator will end up in an early death of the batteries.
3) The battery bank
Deep cycle solar batteries aren't cheap, nor light weight. So choosing the right size of that battery bank, suitable for your budget and energy needs is going to be criticall.

Hope someone finds this text helpful!
Lets begin!

Step 1: Basic Knowledge

Ok, let's start with the basic knowledge.

You will need to know here that some skills on the electric field are required. Not much, but some really helps.
One thing you will found very often is:
Power = Current x Voltage
Power in Wats [W]
Current in Amps [A]
Voltage in Volts [V]

We are going to need an “Isolation” or “Irradiance” chart of the area where we are located, to calculate the average monthly energy that reaches our module every day.

In Argentina we have some charts, and tables with measured Irradiance at several angles. This is really helpful. But in fact, we use the “Peak Sun Hours” wich means the equivalent in hours at 1000 W m^2 / day (a constant known as 1 SUN).

You probably found your Irradiance charts or tables in Kw (Kilowatts = 1000 Wats) or MJ (Mega Joules).
If you have them in Kilowats, you are lucky, so there is no need to make any change or calculation. But if you have them in Megajoules, you will have to divide that number by 3,6 (we use comma as decimal separator. Take notice of this) to convert that unit to Kw.

So lets recap:
W = A x V
MJ / 3,6 = Peak Hours (HSP) in further formulas referred as "H"

The full formula looks like this:
1 HSP = [(1000 W x 1 h ) / m2] x [3600 s / 1 h] x [1 J/s / 1 W] = 3,6 MJ/m2

Step 2: Energy Needs

Here we will found our energy needs, or “demand”.

This is nothing else that our hours of use on any appliance on a day bassis. This means that you have to go and figure out how many hours do you use some appliances during a day.

I have to say, this method is really good for small scale installations, but if you have to make a whole house installation... well you better talk to a local spetialist.

To make this easier, we will use a table.
You will have to note like this:
--------------------------------------------------------------------------------------------------------
Room or ambient | Appliance | Power [W] | Quantity | Hours of Use [h] | [W . h / day]
--------------------------------------------------------------------------------------------------------
Bedroom                | Light          | 9 W             | 2             | 3 h                         | 54 Wh/day
--------------------------------------------------------------------------------------------------------
Kitchen                   | Light          | 18 W           | 1             | 5 h                         | 90 Wh/day
--------------------------------------------------------------------------------------------------------
Living-room           | Light          | 18 W           | 1             | 3 h                         | 54 Wh/day
--------------------------------------------------------------------------------------------------------

So we have here some subtotals, on how much energy in watts are we going to need on each room for light.

Notice that I am using small appliances and a few hours for each, just to make things easy to follow. Once you have some practice with this, you could add as many appliances as you want to.

So here we are going to take this numbers and make 2 more calculations on this subject.

1) The “maximum simultaneous demand” or MSD (in spanish is DPMS).
2) The total daily energy demand, which we will cal ED.

For the MSD you have to take the “Power [W]” column and multiply this by the “Quantity” column, and make a sum on each result.In this example:
Bedroom        | Light | 9 W x 2 = 18 W
-----------------------------------------------------
Kitchen           | Light | 18 W x 1 = 18 W
-----------------------------------------------------
Living-room   | Light | 18 W x 1 = 18 W
-----------------------------------------------------
                                               SUM = 54 W

So here you have the maximum energy you will have to manage on you cables, and with your regulator.

Since we use Amps on each case to run some calculations, you will have to divide this power (54 W) by the system voltage (12 V).
A = W/V
A = 54 W / 12 V = > A = 4,5 A

NOTE: We are using small appliances, and small energy needs, so 12 V will be more than right. If you have heavier appliances, and a lot of power, you are probably going to need a higher voltage system and an inverter. Those are not mentioned here, and therefor is something you sould probably consult with a professional on your area, with knowledge in the electric code.

The total daily energy demand is the sum of each Wh/day on the last column

--------------------------------------------------------------------------------------------------------
Room or ambient | Appliance | Power [W] | Quantity | Hours of Use [h] | [W . h / day]
--------------------------------------------------------------------------------------------------------
Bedroom                | Light          | 9 W             | 2             | 3 h                         | 54 Wh/day
--------------------------------------------------------------------------------------------------------
Kitchen                    | Light          | 18 W          | 1             | 5 h                         | 90 Wh/day
--------------------------------------------------------------------------------------------------------
Living-room            | Light          | 18 W          | 1             | 3 h                         | 54 Wh/day
--------------------------------------------------------------------------------------------------------
                                  | MSD          |                 54 W           | ED                         | 198 Wh/day

As we stated earlier, we use currents, so you have to take this Wh/day, and transform them in Ah/day. Pice of cake...
198 Wh/day / 12 V = 16,6 Ah/day.

So, you have here some interesting values.

You know that you charge controller is going to handle, on the demand side, at least 4,5 A. And on a standard day, you will consume 16,6 Ah/day. This is called our “Daily current needs” or Iday. “I” stand for current.

Those are our basic numbers to continue to the next step: Calculate your energy resource

Step 3: Resource

 Lets go now with our resource.

Here you will have to do some research, since I can only provide some data from my country and city as an example, but you could found that data on the WWW.

Ask around, and it will reveal to you.

As a first impression I have to say that is quite easy to acquire this data.

One of the things with solar modules is that they work it's best when they are at 90º to the sun rays. So, if you could afford a solar tracker, or you are skilled enough to make one, good for you, this is almost solved.

But there is some things to take into account:
In southern hemisphere, your panel should be facing south.
In northern hemisphere, it should be facing south.
This is you ORIENTATION.

Now, the next thing is INCLINATION.

To choose the best inclination for your modules, if they are going to be stuck in one position is the one according to this table:

Your Latitude | Angle of the panel
                          | In Winter | In Summer
0º to 5º             | 15º             | 15º
15º to 25º        | Your Lat.   | Your Lat.
25º to 30º        | Lat + 5º     | Lat. - 5º
30º to 35º        | Lat + 10º   | Lat. - 10º
35º to 40º        | Lat + 15º   | Lat. - 15º
40º <                | Lat + 20º   | Lat. - 20º

As a practical note, I have to say that here, with a latitude of 34º we use a plain fix, Lat + 10º, which compromises a small amount in the summer, when we need less energy, and gives some advantage in the winter.

So you have two mayor problems solved: where to face them, and in which angle.

Next is our Kilowatt or Megajoule thing.

Here we have (take a look at the picture) a table with the average monthly energy on an angle of Lat+10º in MJ m^2 /day: 11,2

So we took this 11,2 and divide it by 3,6. This is 3,11 “equivalent hours” or “peak sun hours”. We are going to call them just “H” for HOURS.

This means that, if the sun where a full 1000 W lamp facing our modules, it will shine right above them for 3,11 hours straight. So if you take a look at the specifications of the modules, you will notice that they say: 80 W @ 1000 W m^2 – 25º C – 1,5 MA
That means: this particular module will produce 80 W when the sun gives 1000 W on each square meter at 25º Celsius and 1,5 mass of air.

But this is something you could investigate in further readings. For now, let's continue with the next step: Choose our module 

Step 4: Generator

Ok, next thing: chose the right generator.

We have to use here some of the data of our energy needs.

The formula for this method is:(1,2 x Iday) / (H x Im)
Where:1,2 is a factor, which includes aging of the module, dust, and several losses.
Iday is as we have seen, the energy need in Amps.
H is the hours, we have seen this on the “Resource”.
And Im is our new inquiry. Whis is nothing more than the current of our chosen module at standard test conditions (STC = 1000 W m^2 – 25º C, and so on).

So, here we could choose our desired module, and see what happens.
(1,2 x 16,5 Ah/day) / (3,11 H x 3,45 A) = 1,85
This means we are going to need 1,85 modules. Since this is a simplified calculation, we are just taking modules in parallels.

We are going to need 2 KS60T modules in parallel.

TIP: If the result is near 1,3 just keep 1. But if its more than 1,3 it is recommended to chose 2.

Pretty easy, right? 

Step 5: Battery Bank

The battery bank is related to our cloudy days.

So you are going to need more research. Look for data on your weather service on a year basis. Usually 3 to 5 days are good numbers.

The formula here is like this:
Bb = (Iday x Cdays) / DoDWhere Bb is, battery bank
Iday, is as we have seen, again, our energy need.Cdays is our cloudy days.
And DoD is or “Depth of Discharge”
Here we have to talk to the battery manufacturer. Since each technology and kind of battery has an unique DoD.

We are talking here of 12 V batteries, to simplify. But you could form your bank from single-cell batteries, that are very powerfull.

For deep cycle, lead-acid batteries, a DoD of 0,7 is quite right. AGM is 0,5. look for specifications on the web.

So:
Bb = (16,5 Ah/day x 5 days) / 0,7 = > Bb = 117,86 Ah

I choose a 110 Ah /12 V AUTOBAT SOLAR Battery. 

TIP: Remember, serial connections increments voltage, but maintains same current.Parallel connections maintains same voltage, but increments the current.

So, I could use two 12 V / 60 Ah batteries in parallel, or two 6 V / 110 Ah in series, and we always have the same energy. 12 V / 120 Ah in the first case, and 12 V / 110 Ah in the second.

Step 6: Choose Your Regulator

The charge regulator has to manage all the currents involved on the circuit, and I would recommend to pick up one with at least:
• Reverse polarity protection.
• LVD, which stands for Low Voltage Disconnect. This will cut your energy on the load side if the battery is too low, and wont let you reconnect until is half charged.
• Battery type selection: Gel, AGM, sealed or classic lead acid.
• LEDs to check the state of charge.

Those are really basic features, and you have lots of regulators, with displays, remote controller and monitoring, USB connections.

Just don't get too picky, and choose one reliable, proved on the field.

A small advice on connections:
The Universal technical standard recommends some cable diameters and voltage drops.

• For the Panel-to-Regulator length: 2,5 mm^2 (AWG 10) and 3% voltage drop.
• For the Regulator-to-Battery length: 4 mm^2 (AWG 6) and 1% voltage drop.
• For the Regulator-to-Load length: Just 5% voltage drop, and choose the wire that don't let the voltage go below that.

How to calculate voltage drop?
Well, if your system is 12 V, this is your 100%.
For 3%: (3 x 12 V) / 100 = 0,36 V
For 1%: (1 x 12 V) / 100 = 0,12 V
For 5%: (5 x 12 V) / 100 = 0,60 V

Now, you have to use this formula to test if your cable run is not too long or too thin.
V = 2 x Length (in meters) x Current (in amps) x Cable Resistance (on Ω/meters)

You probably have the resistance in Ω/Km. So divide that number by 1000.

So in each cable run, you have to make that calculation and check that the result is less than the spected V (1, 3 or 5 %).

For a run from panel to regulator of 4 meters, with 6,9 Amps (two 3,5 amps modules in parallel) and you whant a 0,36 V voltage drop tops, goes like this:
V = 2 x 4 m x 6,9 A x 0,0095 Ω/m = 0,52 V

As you can see, is above 0,36 V, so you will have to go to a thicker cable, since that is the resistance of 2,5 mm cable. Probably with 4mm will works.

So, protections now:
You will have to keep an interrupter, dual pole, that cuts the connection from the modules to the regulator, so if there is a electric storm, or you have to do some mantenaince, you could cut there.
A fuse must be installed on the positive of the battery. You have to use a DC fuse. So choose it right. Remember that batteries could push a lot of power if a short circuit appears.
Also, you have to put a surge protection, or fuses on the load side, to disconnect and protection to short circuits.

So, this is a resume of all you need to calculate all the materials needed, with one method.
Hope this will be found useful for someone.
Thanks.

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20 Comments

I am glad you found the Instructable usefull.

If you check my site, you could see the actual spreadsheet (in spanish) with the formula.

http://www.pablomaril.com.ar

Also, in my site I have put a library on-line, is there. Just choose the logo on the right "pablomaril" and below "instalaciones y renovables".

If you have some particular question, write me an e-mail using the direct message feature of Instructables.

Good luck.

Dear Ikenna:

You are asking for a full time job here!

if you follow the instructable, and build a spreadsheet with the step-by-step formula you will reach your goal, but I can't do all this job for you.

If this instructable is too simple, you could check the other I have put (for solar-eolic systems).

And I have to say thar RETScreen is free and has it's own course on-line, and you could run it for free.

Sorry I can't help you more, but is too much work and I don't have the time.

Good luck,

Good job. You put it all together. Just what I needed.
Thanks.

Thanks a lot. I am glad you like it.

I am putting another method toghether. Wich is a power balance method. In a couple of weeks it will be up.

Thanks again.

Dear Supernatural. Today is 5th february 2012. Nobody know your method. Have wrote on other paper?
Please let us know your method!

AAA.jpg

Estimado Eliseo Sebastián, no entiendo su pregunta. El otro método lo publiqué hace ya un tiempo en otro instructable.

https://www.instructables.com/id/Hybrid-Design-PhotovoltaicEolic-for-rural-commun/

Fue publicado en noviembre de 2011.

"MJ / 3,6 = Kw"

Noupt, actually 3,6MJ = 1kWh (not kW!).

Yes, you whre right, I am cheking it now, and putting the full formula. But for "easy to solve" ways, if you have your mesurements in MJ and divide that by 3,6 you get the Peak Hours you need.

I am putting the correct formula:
So lets recap:
W = A x V
MJ / 3,6 = Peak Hours (HSP) in further formulas referred as "H"

The full formula looks like this:
1 HSP = [(1000 W x 1 h ) / m2] x [3600 s / 1 h] x [1 J/s / 1 W] = 3,6 MJ/m2

Thanks for your advice, but remember, I sometimes forget on porpouse of the units just to give to the students an easy way to make the calculations.

And dometimes I make mistakes, like you point to me here. Once again, thanks for the advice. I am always looking to improve.

Sorry for such "agressive" comment, your article is still great and I knoy you've put a lot of work into it. I wasn't in a really good mood when I wrote it ;).

I just really don't like confusion between kW and kWh, because this is really common and nobody understands even so simply physics as difference between energy and power, and I didn't really check your formula in details, to see what you were really calculating. Sorry again.