Complete rookie needs help with first project

[Jokeaccount]

Hi,

I'm a student in electrical engineering school and I have undertaken the task to power a 40W LED (or two 20W) using a small solar panel for a front yard. I'm pretty new to this stuff and even though i've taken some classes that were relevant to the use of solar panels, I don't feel that knowledge is really enough for my first hands on experience with such a thing. I have spotted a store that sells panels and this sort of stuff but before I go I want to know what exactly I'm looking for, what to look out for, what my options are, cost effectiveness etc. The only thing you should assume is that I can understand most of what you're talking about when it comes to circuits and electrical stuff but I'm really just a neophyte beyond theory. Also, I do not have any of the basic equipment like a multimeter and such tools which I guess I'm gonna need.

Anyway what I'm looking for is either your personal guidance or a direction to some good and reliable sources (either in this forum or generally on the internet. Please don't tell me to read a 500 page book ) regarding what panels to use, what type of LED (I think they need it for night lighting), what to look out for when I do the connections etc.

Thanks in advance

[Sunking]

You have not specified any defined parameters like how many watt hours per day, time of year use, and location. Once you sort that out it is 5th grade math. Just read this Sticky and it will tell you exactly what to do.

[Jokeaccount]

Thanks for the sticky seems to cover lots of stuff

For anyone who cares: He location is sunny most of the time of the year. I'm guessing its gonna be 40*10 Wh all year.

[SunEagle]

Thanks for the sticky seems to cover lots of stuff

For anyone who cares: He location is sunny most of the time of the year. I'm guessing its gonna be 40*10 Wh all year.

Just because it is sunny does not mean you are getting enough sunlight.

Your "location" determines the number for useable hours of sunlight a day to generate electricity from your panels.

In good locations during the summer time it gets you about 5 hours of "usable" sunlight. In winter it can go down to 2 hours of "usable" sunlight.

[Jokeaccount]

Yeah i remember we did that in school. I suppose i don't have to calculate sunlight in the specific location right? I remember using a billion weird equations for radiation calculation etc but don't those numbers exist already with the meteorological service?

[Naptown]

You don't have to do any calculations at all
Everything you need to know is in that sticky along with a calculator will do everything for you.
If 12v with no inverter set inverter efficiency to 100% to take it out of the equation. You will however have to factor in PF if less than 100% on the LED's

[SunEagle]

Yeah i remember we did that in school. I suppose i don't have to calculate sunlight in the specific location right? I remember using a billion weird equations for radiation calculation etc but don't those numbers exist already with the meteorological service?

There is a link (which unfortunately i don't have but may be in one of the stickies) that tells your solar isolation (# of hours of usable sunlight all year long) based on either a city location or long / lat location anywhere in the world.

[Jokeaccount]

SunEagle: Thanks i'll look out for that link

Naptown: PF means?? English is not my native language thus not knowing all the terms yet (much less abbreviations).

[Naptown]

Power factor

[Sunking]

Yeah i remember we did that in school. I suppose i don't have to calculate sunlight in the specific location right?

You had better or else it will fail. Most importantly is the winter Solar Insolation Sun Hours.

It also makes you realize the huge cost. For example 400 Wh/day with 3 Sun Hours requires a 200 watt panel.

[inetdog]

SunEagle: Thanks i'll look out for that link

Just Google "PVWatts" for a start.
PF = Power Factor. It is a number less than 1.0 which tells you how much "useless" current is going through your wiring in addition to the part that is actually delivering power. An inductor (motor) , a capacitor (CFL) and a power supply all can have a low Power Factor. Which makes an inverter work harder to deliver the same amount of actual power.

[Sunking]

PF = Power Factor. It is a number less than 1.0

Horse feathers. I have for sale a unit that contains an electric motor, alternator, and a PF device with a PF of 2. All you have to do is get the motor spinning to get the alternator generating electricity an you have free power for life.

[inetdog]

Horse feathers. I have for sale a unit that contains an electric motor, alternator, and a PF device with a PF of 2. All you have to do is get the motor spinning to get the alternator generating electricity an you have free power for life.

I'll be looking for it on Kickstarter, so I can invest and maybe even become a dealer.

[Sunking]

I'll be looking for it on Kickstarter, so I can invest and maybe even become a dealer.

Tell you what. I am feeling generous today. I will cut you in for 30% ownership of the company. Send a check of $10,000 good faith deposit and then we will talk.

[Jokeaccount]

Ok i've checked the tutorial on the sticky and I'm now looking to get the numbers down. I found this link for sun hours per day: http://pveducation.org/pvcdrom/prope...lar-insolation

The thing is, the graph calculates 3 things. Incident power, power on the horizontal and module power. It says that module power is when the panel faces south with a 45 degree angle. Is that the number I'm looking for? I tested to see if Tucson (which the tutorial says has 5.6 kwh/m2/day) with latitude of 32 gives the same via this calculator and it gives a 6.1 (close to 5.6) on the module graph whereas Seattle gives 1.2 on the horizontal graph, so now I'm confused. I'm guessing the Module graph is the one i need as long as I put the panel in a 45 degree angle but still I'd like to be sure

THanks