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? 
Good job. You put it all together. Just what I needed. <br>Thanks.
Thanks a lot. I am glad you like it.<br><br>I am putting another method toghether. Wich is a power balance method. In a couple of weeks it will be up.<br><br>Thanks again.
Dear Supernatural. Today is 5th february 2012. Nobody know your method. Have wrote on other paper?<br>Please let us know your method!
Estimado Eliseo Sebasti&aacute;n, no entiendo su pregunta. El otro m&eacute;todo lo publiqu&eacute; hace ya un tiempo en otro instructable.<br><br>https://www.instructables.com/id/Hybrid-Design-PhotovoltaicEolic-for-rural-commun/<br><br>Fue publicado en noviembre de 2011.
&quot;MJ / 3,6 = Kw&quot;<br><br>Noupt, actually 3,6MJ = 1kWh (not kW!).
Yes, you whre right, I am cheking it now, and putting the full formula. But for &quot;easy to solve&quot; ways, if you have your mesurements in MJ and divide that by 3,6 you get the Peak Hours you need.<br><br>I am putting the correct formula:<br>So lets recap:<br>W = A x V<br>MJ / 3,6 = Peak Hours (HSP) in further formulas referred as &quot;H&quot;<br><br>The full formula looks like this:<br>1 HSP = [(1000 W x 1 h ) / m2] x [3600 s / 1 h] x [1 J/s / 1 W] = 3,6 MJ/m2<br><br>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.<br><br>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 &quot;agressive&quot; 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 ;).<br><br>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.
Dear NightLord, don't worry.<br><br>I still think your comment where right, and help me to improve my instructable.<br><br>I know, sometimes energy and power are confusing, and I try to keep it simple, that is wy I make some mistakes.<br><br>Have a good day.<br><br>Pablo
Supernatural, <br> <br>A couple of suggestions, since this is your first instructable: <br> <br>1. Your English, while not perfect, is good enough to convey your message and is much better than most of our spanish. Muy Bueno. <br>2. I think it would help if you could add an english translation to all of the photo captions, either yourself or with help from others. Some of the captions are already written with both english and spanish, but a few are not. <br> <br>Thank you for your effort, with this instructable and with your english write up.
Hi nanosec12.<br><br>Thanks a lot for your words. I have put all the translations in the pictures, that was a rookie mistake, since I didn't realize that the pictures are the same on all the instructables I use them. So I chanhe the captions for my spanish version, thinking that a copy of the english version is what I already have.<br><br>I have cheked that, and think is better now.<br><br>I hope this won't happend again.<br><br>Saludos,
This is a nice writeup, with decent guidance for the beginner on how to deal with medium-scale solar installations. I do have some minor comments on your units, though.<br><br>Joules are a unit of energy; watts are a unit of power, which is energy divided by time. When you write, &quot;MJ/3,6 = kW&quot;, you really should write, &quot;MJ / 3.6 [1000 s] = kW&quot;, since you have to divide joules by seconds to get watts.<br><br>I'm confused by your irradiance explanation. &quot;1 kW m<sup>2</sup>/day&quot; appears to have too many units of time in the denominator.<br><br>The SI symbol for &quot;kilo&quot; is &quot;k&quot; in lower case, not &quot;K&quot;.<br><br>I'm a little confused by your Ohm's law calculation. You write &quot;V = 2 x Length (in meters) x Current (in amps) x Cable Resistance (on &Omega;/meters)&quot;. Obviously length &times; resistivity is just resistance, so you get V = IR. But where does the extra factor of two arise? You don't need to deal with the return cable from the load back to the solar panel.
Thanks a lot for your comment. It took me a lot of hard work, since my language is spanish, and its an efford to pull this of in English.<br> <br> I will try to answer your comments in order:<br> We all agree, Joules are energy, Watts are power.<br> I just go right for the simplified formula, without dealing with units.<br> Here you have the right formula, in spanish, sorry, couldn't find it in English:<br> http://es.wikipedia.org/wiki/Hora_solar_pico<br> Since I teach to students with basic knowledge on mathematics, sometimes I have to forget all the units, and put just the real deal. You are right about it, but the simplified formula is a choice here, not a mistake.<br> <br> The irradiance is measured in our maps like this: MJ&nbsp;m2/day<br> This is, a device that measures how much power strikes the cell (The megajoules) on a particular surface area (the square meter) during an entire day. I made a mistake by converting that to kW.<br> <br> I have to say here, that :<br> V = 2 x Length (in meters) x Current (in amps) x Cable Resistance (on &Omega;/meters)<br> It's OK, since you are going to take each meter of cable as a resistance, you will have to multiply that by the number of meters the cable has, here is &Omega;/meter multiplied by the meters of lenght. Is not resistivity, is resistances in series, one for each meter. And regarding the cable back, it has to be taken into account, since you can't have a circuit with just one cable. You need two and a load. And if you don't have a circuit, you don't have current. I can't translate the Kirschoff' Laws, but there you will find that if you have a current comming in, you must have a current going out. Is really hard for me to do this in english, I am so sorry if it sounds too bad.<br> <br> I hope I haven't make a mess here with my explanations.<br> Thanks a lot for posting your comment.<br> Saludos, Pablo.
Muchas gracias, Pablo. Mi lengua primera esta ingles, y mi espanol esta pobre.<br><br>I am very grateful that you will take the time to work through the English for me, and I am sorry that I don't know enough Spanish to properly return the favor.<br><br>Have you communicated with any of our other Spanish-speaking members? A few of the ones I know are Rimar2000, ricardoruizo, M.C.Langer, but there are many others. There's also a <a href="https://www.instructables.com/community?categoryGroup=forums&category=esl">discussion group</a> where members interested in multilingual or non-English I'bles have some topics going.<br><br>I am a physicist, so for me the units are very important. I appreciate wanting to keep things simple (and thank you for clarifying that you made deliberate decisions, not mistakes! :-), but my own training and teaching of students suggests that with proper units, you can tell quickly if you have the wrong formula (the units don't work out correctly).<br><br>I still think the factor of two in your resistance calculation is wrong. Here's why: What you're computing is the voltage drop from the output of your solar panel to your load. That's V = I&times;L&times;r (here r = resistivity, ohm/m). The load itself will introduce some additional voltage drop due to its own resistance (or impedance). The return path is necessarily dropping all the way to ground (by construction), so the voltage drop due to the return cable is irrelevant.<br><br>You can also think of it this way. You <i>could</i> set up your system with just an outgoing cable! Connect the negative side of the solar panel to a true earth ground, run a single wire from the panel to your load, and connect your load to a true earth ground at its location. The current flow will be earth -&gt; panel -&gt; load -&gt; earth, with the two (possibly distant) earth locations connected implicitly by dirt, ground water, whatever.
I'm not sure about the voltage drop point of view, but the return leg resistance is just as important to keep low as the other side of the circuit as all of the current charging the battery must flow through it. <br><br>I don't think your earth example would work as there would be too much resistance from the ground/earth/soil water, so almost no current would flow.
Hi again! I haven't comunicated with the spanish members yet, this is my first instructable, and I think my next will be this same subject, but in spanish.<br><br>I understand what you say about units, and for people with some degree it's easier. I teach in vocational training Centers, where you teach people that sometimes don't have more than elementary school, and in a small course with only 4 months. Sometimes, it's easier to show them without units, in practical matters. I mean, &quot;hands on the job&quot;. I have show them the right formula, the one I put you in the earlier post, and they all look at me like I was going mad... so I drop it and go for the straight-forward-system.<br><br>With the two factor, it's ok what you say, I need the voltage drop on the load side. I think I am not too clear with resistance and resistivity. Here Resistivity is &rho; (Rho) wich stand on the copper like this: 1/57 value in [&Omega; mm2 / m], and is related to the section of the cable, while resistance is the equivalent value of an electrical resistance on each meter of lenght.<br><br>If you build the equivalent circuit, it will be the Generator (module or panel) with a resistance which value will be the one given for your cable + (plus becouse is in series) the internal resistance of the load, and here we don't use a particular value for that + the resistance value of the return cable. If you use earth, the last one will differ from the first one, but if you use the same cable, you have 2 resistances with the same value.<br><br>As you say in the last paragraph, you could set up the system as you stand, and the earth will be the second conductor. I think I should say conductor instead of just cable, but conductor is som how a difficult term to use, I don't know why, but people when you say &quot;conductor&quot; they say &quot;you mean the cable?&quot;<br><br>The: earth -&gt; panel -&gt; load -&gt; cable<br>Is like this: panel -&gt;conductor -&gt; load -&gt; conductor.<br>So you have: 2 conductors.<br><br>Look at here:<br>http://www.mikeholt.com/technical.php?id=technicalvoltagedrop1<br>Near the bottom you wil found this sentence:<br><br>VD = 2 wires x 12.9 ohms x 44 amperes x 160 feet/26,240 Circular Mils<br><br>That is what I am using, a similar actually formula.<br><br>Here it is again:<br>http://www.elec-toolbox.com/Formulas/Useful/formulas.htm<br><br>Nice to have this talk! I am practicyng with my English ; )
Hi, I am currently investigating how to install Solar power in my home to supplement the mains supply. If you are a complete novice then this can be quite confusing. <br> <br>I would like to say that this instructable is probably the easiest one I have come across to date. I congratulate you on your easy to understand expainations. <br> <br>Thanks for your help, it has given me the confidence to go ahead with my project.
Hi!, thanks a lot for your encouraging words!<br><br>Has been quiate a ride to make this in English, you know. I can manage reading and seeing movies in plain English (without subtitles) but writing a wole instructable was a challenge.<br><br>I am really glad you found it usefull.<br><br>I am working on another one, with other calculation system known as &quot;Energy Balance&quot; and used in a paper of the Universidad del Nordeste. Is good for hybrid systems too.<br><br>Also, I have some nice pictures of the course I am running here in argentina. You could take a look here:<br>https://www.facebook.com/energiasolarfotovoltaica<br>And, if you run over my site:<br>http://instalaciones.pablomaril.com.ar<br>You could found a link to a bunck of documents I have collected over the years on renewables. Many are in spanish, but most of them are in English.<br><br>Once again, thanks a lot for your words, and good luck on your project, I hope it works well.<br><br>Don't forget to make an &quot;Instructable&quot; with your results. Even if you have only pictures. That way we help the growing community of renewable users.

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More by Supernatural:Hybrid Design: Photovoltaic/Eolic for rural communities Metodo de Calculo Sistemas Fotovoltaicos: Metodo de Corrientes (Version en Espa�ol) Current method for Photovoltaic Calculations 
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