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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!
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!
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Signing UpStep 1Basic 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
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
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http://www.instructables.com/id/Current-method-for-Photovoltaic-Calculations/
My Task is part of my coursework and in all truth I am completely naive of the rudiments, as we have not been taught a single thing on how to do a similar work, so I most heartily seek your assistance.
I really need to learn the procedure, as well as excel in this course work, that is why in confidence that you can solve this problem, I most heartily entreat you to help me design one or put me through how I can go about it.
Could you help?
Thanks in anticipation.
My email is nonstop4heaven@yahoo.com
This is the task:
Renewable Energy Technologies Task
Objectives
• To identify and assess the scale of renewable energy resources;
• To use appropriate techniques to assess the feasibility of renewable energy resources, and
• To use appropriate software to help assess renewable energy systems.
The task
You have been asked to investigate the potential for renewable energy for a small dwelling in the Scottish Highlands. The owner, a writer, plans to refurbish it and use it mainly as a retreat up to four weeks at a time, to ‘get away from it all’ so that he can concentrate on his writing. He envisages occasional weekend visits by friends, up to four people at a time, and expects that the total use will be about six months per year.
The house could be connected to the electricity grid, but connection is expensive, and the unit price of electricity will be high. Therefore the owner wants to know whether it might be possible to develop a stand-alone renewable energy system to provide for his needs.
You should consider the energy requirements for the dwelling (heating, lighting, cooking, etc.) for a single person, with occasional use by up to four people, as described above. Write a report for the owner which to outline the technical possibilities for a renewable energy system for the dwelling.
You are required to:
a) Outline briefly a possible energy strategy for this building. For each of the available renewable energy resources, estimate the size of the resource (see below for data), discuss the characteristics (whether it can provide heat or electricity (or both), seasonal variability, storage requirements, etc.), and hence determine which resources are most likely to be most useful for this application (say, 800 words, 25% of the marks).
b) Analyse the heating or electricity requirements for the house in some detail, and, considering one of your principal energy sources identified in (a) above, discuss in detail how you think it could be incorporated into your plans (say, 1000 words, 30% of the marks). Note: for a ‘real’ report, you would normally expect to analyse the all the energy requirements and all the potential energy resources – this full analysis is not required for the purposes of this coursework.
c) Use the RETScreen software to analyse the contribution that your chosen energy resource (from (b) above) could provide. Write a commentary on the most important features of your analysis (45% of the marks).
Note that this is principally a technical exercise and you are not required to investigate financial details, but the costs of any scheme should fall within ‘reasonable’ limits. The primary function is to provide for the energy requirements of the dwelling, and not to generate heat to electricity for export, or to act as a test-bed for experimental technologies. This means that the scale of your scheme and the technologies used should be in keeping with the size of the dwelling and its requirements.
Data
Use the following data. You may also seek out other sources of weather data. Make reasonable estimates and assumptions for any information not given, but make sure that you state clearly any estimates and assumptions you make. You may work in MJ or kWh, but you should ensure that the report is written using consistent units.
• The house is in an unspecified location in the Scottish Highlands. You may assume that the principal sources of renewable energy are: solar, wind, hydro and woodfuel, as given below.
• The house has a plan area of 45m2, with the main façade facing almost due South. After applying insulation, double-glazing and draught-stripping, the building will have an overall heat loss coefficient of 160 W/°C, i.e., for every °C temperature difference between the inside and outside, the heat lost will be 160W.
• Weather data may be accessed via the RETScreen software. Note: the wind data given in RETScreen for Scotland is low – you may use a mean average windspeed of 7.2 m/s.
• You may assume that the owner of the house has access to 1.5 hectares of woodland, which can provide 6m3 of timber per hectare per year, with a calorific value of 11 GJ/m3.
• The house has access to a small hillside burn, with a catchment area of 0.8 km2. It may be possible to make use of a fall of about 8m to provide a source for hydroelectric generation.
Marking criteria
You are expected to:
• outline the relative size, characteristics and usefulness of the available resources;
• generate a specification of energy requirements;
• choose, and justify your choice of, a principal energy system for supplying the perceived requirements;
• use the RETScreen software to size a renewable energy system, including storage requirements;
• discuss the validity of your analysis and your assumptions.
A mark of 55% - 65% will be awarded for a report which addresses the above points and appears to be largely correct in its calculations.
A mark of over 70% will be awarded for a piece of work which demonstrates an in-depth understanding of the issues involved, perhaps by incorporating more than one energy resource, or by assessing and comparing more than one alternative system.
A mark of less than 50% will be awarded for a piece of work which shows only a basic understanding of the problem, has significant errors in the calculations and makes unwarranted or unreasonable assumptions.
In particular, note that the software part of the assignment is valued at 45% of the total marks. Therefore it is imperative that you work to understand how to use the software, and make sure that the data input matches your own calculations, and that your commentary discusses the results in the context of your own estimates.
Note on referencing and presentation: You are required to cite references for your work. Although the coursework will be marked on content, and no mark is given specifically for references or presentation, where poor presentation is detrimental to understanding your work, or where references are clearly missing, marks may be deducted (up to 10 marks in extreme cases). Equally, where presentation particularly helps the understanding of the work, additional marks may be awarded. However, it will be rare for a mark to be increased above 70% or decreased below 40% principally on the basis of its physical presentation.
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,
I am really grateful, i have seen it, its a whole lot of work you did, i will apply your directive accordingly, and will get back to you if need be, or when I am through.
Meanwhile, thanks a lot for your time and seasoned intellectual resources.
Ajiero, Ikenna Reginald
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.
Many thanks.
Ajiero, Ikenna R.
Thanks.
I am putting another method toghether. Wich is a power balance method. In a couple of weeks it will be up.
Thanks again.
Please let us know your method!
http://www.instructables.com/id/Hybrid-Design-PhotovoltaicEolic-for-rural-commun/
Fue publicado en noviembre de 2011.
Noupt, actually 3,6MJ = 1kWh (not kW!).
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.
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.
I still think your comment where right, and help me to improve my instructable.
I know, sometimes energy and power are confusing, and I try to keep it simple, that is wy I make some mistakes.
Have a good day.
Pablo
A couple of suggestions, since this is your first instructable:
1. Your English, while not perfect, is good enough to convey your message and is much better than most of our spanish. Muy Bueno.
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.
Thank you for your effort, with this instructable and with your english write up.
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.
I have cheked that, and think is better now.
I hope this won't happend again.
Saludos,
Joules are a unit of energy; watts are a unit of power, which is energy divided by time. When you write, "MJ/3,6 = kW", you really should write, "MJ / 3.6 [1000 s] = kW", since you have to divide joules by seconds to get watts.
I'm confused by your irradiance explanation. "1 kW m2/day" appears to have too many units of time in the denominator.
The SI symbol for "kilo" is "k" in lower case, not "K".
I'm a little confused by your Ohm's law calculation. You write "V = 2 x Length (in meters) x Current (in amps) x Cable Resistance (on Ω/meters)". Obviously length × 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.
I will try to answer your comments in order:
We all agree, Joules are energy, Watts are power.
I just go right for the simplified formula, without dealing with units.
Here you have the right formula, in spanish, sorry, couldn't find it in English:
http://es.wikipedia.org/wiki/Hora_solar_pico
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.
The irradiance is measured in our maps like this: MJ m2/day
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.
I have to say here, that :
V = 2 x Length (in meters) x Current (in amps) x Cable Resistance (on Ω/meters)
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 Ω/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.
I hope I haven't make a mess here with my explanations.
Thanks a lot for posting your comment.
Saludos, Pablo.
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.
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 discussion group where members interested in multilingual or non-English I'bles have some topics going.
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).
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×L×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.
You can also think of it this way. You could 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 -> panel -> load -> earth, with the two (possibly distant) earth locations connected implicitly by dirt, ground water, whatever.
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.
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, "hands on the job". 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.
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 ρ (Rho) wich stand on the copper like this: 1/57 value in [Ω 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.
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.
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 "conductor" they say "you mean the cable?"
The: earth -> panel -> load -> cable
Is like this: panel ->conductor -> load -> conductor.
So you have: 2 conductors.
Look at here:
http://www.mikeholt.com/technical.php?id=technicalvoltagedrop1
Near the bottom you wil found this sentence:
VD = 2 wires x 12.9 ohms x 44 amperes x 160 feet/26,240 Circular Mils
That is what I am using, a similar actually formula.
Here it is again:
http://www.elec-toolbox.com/Formulas/Useful/formulas.htm
Nice to have this talk! I am practicyng with my English ; )
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.
Thanks for your help, it has given me the confidence to go ahead with my project.
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.
I am really glad you found it usefull.
I am working on another one, with other calculation system known as "Energy Balance" and used in a paper of the Universidad del Nordeste. Is good for hybrid systems too.
Also, I have some nice pictures of the course I am running here in argentina. You could take a look here:
http://www.facebook.com/energiasolarfotovoltaica
And, if you run over my site:
http://instalaciones.pablomaril.com.ar
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.
Once again, thanks a lot for your words, and good luck on your project, I hope it works well.
Don't forget to make an "Instructable" with your results. Even if you have only pictures. That way we help the growing community of renewable users.