What do I need to know about solar

[Roger]

I think I've read every topic in the solar thread above. Running the worksheets I discovered I am not using that much battery at all, I just need a trickle, I use about 10-15 amps a day in the spring/summer. More in the winter, but I don't see the sun much in the winter so, no dice there. I figure, on a nice sunny spring/summer day I can get about, at least one, five hour day.

So I ordered a 40 watt solar panel (~2.5amps), a 6 amp controller and 30 foot of 10 AWG cable (overkill). The panel is about two foot by two foot by 2 inches. Needless to say I am thinking of a movable unit, not something screwed to the top. I figure I can park in the shade move my solar panel out into the sun and let 'er rip. Maybe I can add another one later and boost up to 80 watts and 5 amps? Who knows.

I am thinking of attaching a back center leg so I can lean the it back into the sun and a very small wire to secure it to the local tree/post/rock/nasty dog.

So what is my reality check here? Do-able or forget it?

Thanks!

Roger

 

[JohnF]

Doable.

There are pictures of folks with similar setups. (You probably saw them.)

I would use an outdoor extension cord for the leads. It would be much more durable and 14 gauge (seems to be most common) is not too expensive.

Go all hollywood and put a trolling motor connector/plug at the camper interface.

[Edit]

Oh yeah, and dogs tend to stick to shady spots...

 

[dsrtrat]

In my opinion 40 watts is a little small,you will have some gain, the trick is to maximize what you are working with.
Voltage drop with a small gauge wire is be a problem if you use a long cable. You are working with lower voltages than normal house power (about 1/10 as much) so using 10 gauge cable for your cord isn't overkill. You can measure the voltage drop by reading the output from the panel at the connections and then at the end of the connecting cable. You want as little drop as possible to take advantage of the small size of your panel. Maximum output of panels is under ideal lab conditions, likely never obtained in the field. There will also be losses in the controller and connecting wire in the camper. It all adds up. Try it and see how it works, panel prices are coming down all the time and you can always add to or buy a larger panel if you think it's necessary. I use a 80 watt panel and it keeps up with my Engel fridge in the sunny southwest where I live. It won't recharge a 50% depleted battery AND run the fridge in a day but brings the battery up to about 75-80% in addition to running the fridge.
Dsrtrat


quote name='Roger' date='30 September 2011 - 05:53 AM' timestamp='1317358424' post='57040']
I think I've read every topic in the solar thread above. Running the worksheets I discovered I am not using that much battery at all, I just need a trickle, I use about 10-15 amps a day in the spring/summer. More in the winter, but I don't see the sun much in the winter so, no dice there. I figure, on a nice sunny spring/summer day I can get about, at least one, five hour day.

So I ordered a 40 watt solar panel (~2.5amps), a 6 amp controller and 30 foot of 10 AWG cable (overkill). The panel is about two foot by two foot by 2 inches. Needless to say I am thinking of a movable unit, not something screwed to the top. I figure I can park in the shade move my solar panel out into the sun and let 'er rip. Maybe I can add another one later and boost up to 80 watts and 5 amps? Who knows.

I am thinking of attaching a back center leg so I can lean the it back into the sun and a very small wire to secure it to the local tree/post/rock/nasty dog.

So what is my reality check here? Do-able or forget it?

Thanks!

Roger
[/quote]

 

[pods8]

High as gauge as practical is always good for keeping drop low. In this case with only dealing with 2.5amps and 30' of cable a 14ga would see about 2.2% drop, 10ga would see about .9%

 

[JohnF]

Some good info there. I guess my main point was that I would use a drop cord rather than cable that is meant to be installed and left alone.

Reducing loss is important. Pods8's numbers seem to be calculated at 12V. I think most solar panels put out about 17 volts?

using pods8's calcs and assuming 12V:

A 2.2% drop (.264V at 2.5 amps) means .66 watts.
A .9% drop (.108V at 2.5 amps) means .27 watts.

The loss in watts is the same no matter what the input voltage is. (resistance is fixed and we fixed amperage)
I thought looking at the numbers in terms of power might be helpful. If Pods8 used a different voltage then I'll be off a bit.

1ft distance and it won't matter.
If you are fixating on 2% loss and you will place it a football field away then you need 2 gauge.

In summary, 10 gauge is not necessarily overkill, and I would use a drop cord.

 

[pods8]

I used 18V for those numbers since it was closest to what a solar panel would be on the calculator I was using.

2.2% drop at 18V isn't something I would personally worry that much about if the panel is getting good voltage output, that is where the "practical" part comes in. Heavier never hurts but no sense going lots heavier than needed if the costs and availability of supplies go up too.

 

[JohnF]

I used 18V for those numbers since it was closest to what a solar panel would be on the calculator I was using.


Ah, I poked my finger at a handy distance chart and tried to guess. Your method is much more accurate.

So, the new numbers are (in case someone is wondering):

2.2% drop = .396V * 2.5 amps = .99 watts
.9% drop = .162 * 2.5 amps = .405 watts

And I agree with your "practical" assessment.

Roger, I didn't do all the math available here. Some things to consider now that I've sucked myself deeper into the subject.

Your numbers (40 watts and 2.5 amps) equates to 16 volts.
If it really puts out 18 volts then the amperage is 2.222 amps.
These small differences matter because your situation is right on the edge.

If we assume you'll loose 20% of your power due to sun angle/line loss/conversion loss/ etc. (and again, I do not know what is typically) then you have 33 watts to put into your battery.

You said you need a 15 amp day. let's stick the voltage to 14.2V for battery charging (low end of absorption stage - constant voltage - and I don't know what your controller does). Your system can then supply 2.34 amps to the battery which is 6.41 hours of sunlight.

if you assume 33% power loss then it comes out to 7.89 hours. (27 watts)

Keep in mind that losses are cardinal and constant so a jump in panel size won't change the loss any but will increase the power supplied to the battery.

7.89 hours works for me here this time of year.

 

[Roger]

Wow,
Thanks for the help. I received my panel, controller and the wire. Whoops, wire should have been plural. Duh. Well it will be another week before I can plug it in, but that will give me some time to find a good place for the controller and to get all of the interior wiring squared away.
I know the panel is kinda small, but I really don't use that much of the battery so I think I got the right size. Good thing I bought the larger wire, I kinda see now why it is so important.


Thanks!

Roger

 

[pods8]

JohnF, you're making some incorrect assumptions for the typical solar setup. You can't take the panel amps & voltage and convert to watts and then divide by the charge voltage to get the charge amps (which is an increase in amps) because most affordable charge controllers that people use are pulse width modulation (PWM) and they don't do a voltage conversion. They "chop it" to drop the voltage. So whatever amperage goes into the controller at that higher voltage is basically the same amperage you get out at the charge voltage. You'd need a more expensive Multipoint Power Tracking (MPPT) controller to actually harness all the wattage out of your panel. However it wouldn't make any sense to spend the money on an MPPT controller for a 40watt panel so it's pretty save to assume we're taking PWM here.

So in your example above if you had 18V @ 2.22amps (40watts) going into the PWM controller and it's output was 14.2V you'd still only have 2.22amps (at 31.5watts) coming out assuming no efficiency loss in terms of amps in the controller (I believe the remaining 8.5watts are dissipated as heat in the controller).

 

[Roger]

JohnF, However it wouldn't make any sense to spend the money on an MPPT controller for a 40watt panel so it's pretty save to assume we're taking PWM here.

**** right, no loss of energy, lost to heat...***


Yeah it is a PWM controller. Morning Star Sun Saver 6-12V to be exact.
Roger

 

[JohnF]

JohnF, you're making some incorrect assumptions for the typical solar setup.

That's not hard to believe. The last time I did any real(ish) solar calculations it was for a buddies house many years ago to show him that it would never pay off. That system converted to AC and did a stepup to line voltage at the meter. So thanks for correction.

So in your example above if you had 18V @ 2.22amps (40watts) going into the PWM controller and it's output was 14.2V you'd still only have 2.22amps (at 31.5watts) coming out assuming no efficiency loss in terms of amps in the controller (I believe the remaining 8.5watts are dissipated as heat in the controller).


Ah. Knowing this - use 14 gauge. The effect of the voltage drop at the controller is that it will have to dissipate less power as heat. So there is a ton of room to be lossy in transmission. (8.5 watts - (there is some required conversion cost))

And we know 2.22 amps. So, a 15 amp/day is 6.88 hours of charge time.
Don't have to guess the effeciency, it's dialed into the controller and it's linear. (Yuch) So, at 40 watts you lose 8.5 watts. Double the input current and lose double the power - 17 watts.

[edit]

I'm back. I thought I would back up the "use 14 gauge" argument a bit.

From "Amateur Radio Relay Handbook" 1985 I find resistance in 14 AWG is 2.575 ohms / 1000ft.

If we use a 100ft drop cord (that ought to be plenty, eh?) then we will be adding .2575 ohms resistance to each of the positive and negative leads for a total resistance of .515 ohms.

That drops our 18V input to (uhm calculator...) (6.35% loss) (uhm...) 16.86V.

Since we are constant current and assuming the 16.86V meets the cost requirements of the controller, we end up with the same 2.22 amps to the battery and need to shed only 5.9 watts as heat.

And since the controller is so ineffecient, I could even argue that 14 gauge is the right choice. It will let the controller run cooler.

Keep in mind that there are other losses in other places too. Hook everything together and test at the controller input. If it meets minimum voltage requirement for the controller then you are good-to-go.

 

[Roger]

I'm back. I thought I would back up the "use 14 gauge" argument a bit.

From "Amateur Radio Relay Handbook" 1985 I find resistance in 14 AWG is 2.575 ohms / 1000ft.

-snip-

That drops our 18V input to (uhm calculator...) (6.35% loss) (uhm...) 16.86V.

-snip-

And since the controller is so ineffecient, I could even argue that 14 gauge is the right choice. It will let the controller run cooler.

-snip-
[/quote]

Umm, most of that went over my head , but is not 14 AWG "thinner" than 10 AWG? I know this may sound dumb, I am so good at dumb, but would 10 AWG actually be not as good as it may possibly cook the converter because it delivers too much power?

And speaking of converters: I am thinking of mounting it above the battery in the battery storage area. Would this be a poor choice as there is not a lot of cooling? Another location would be under the couch so I could monitor it's little green light.
What AMP fuse would be good? I am thinking 10 amp between the controller and the battery, would that be sufficient? I noticed that FWC had 30 amp fuses on the pos and neg side of the battery.
Also what is the max distance I can mount the controller from the battery? I am thinking 2 feet would be the maximum before I had a lot of loss??
Thanks for all of your help.

Roger

 

[pods8]

I wouldn't worry about giving the controller too much as long as you're within it's rated parameters, it's designed to deal with it.

Maximum distance again is dependant on the gauge wire, however here your lines losses become more important since they drop the voltage going into the battery. But we're still talking low amps so you have lots of room to work with while still being in moderate gauge wire.

 

[JohnF]

would 10 AWG actually be not as good as it may possibly cook the converter because it delivers too much power?

Oh gosh, my bad. No, 10 AWG is perfectly fine. It's just overkill, like you said.

All of that stuff I wrote really just speaks to the really wide margins the designers put into the system.

 

[Roger]

Good evening!

I finally received the my other 30 foot cable today and I had the panel installed about an hour later.
I used a 10 amp fuse, but I probably should drop down to a 5 amp just to keep it under the 6 amp rating of the unit.(?)
I had some good clear sun and a battery that was a bit discharged.
When I plugged the unit in, even in the shade, the charge control LED glowed very dimly. When I put it in the full sun it glowed bright green and my battery charge indicator noted I had a "full" charge! SWEET!
Oh yeah, I didn't detect any voltage drop from the panel to the charge controller, but that could very well be a function of my cheap multimeter.

Some of the things I learned.
1. There is no "local" source of MC4 connectors.
2. It is cheaper to get one 100 foot MC4 cable then two 30 foot cables. You only need one cable for positive and one for negative. (duh moment) You just chop that 100 foot cable in half anyway.
3. Wander the west rocks.
4. If this panel is not enough I can add another panel on easily.

Again,
Thanks to everyone who responded to this tread.

Roger

 

[craig333]

Good to hear. Solar is on my short list.

 

[Den]

If you use a multimeter to measure output at the panel, should you be reading amps, watts, volts ?

 

[pods8]

If you use a multimeter to measure output at the panel, should you be reading amps, watts, volts ?


You're only going to be able to read volts with most cheap multimeters, most are only rated to measure up to .2amps. You'd need a rated amp meter of on sort or another if you want to find out how many amps or watts (watts = voltage X amps) you're getting.

 

[Roger]

If you use a multimeter to measure output at the panel, should you be reading amps, watts, volts ?


I am using a "Watt's up" between my solar controller and the battery to "meter" my amps and amp hours.

My link

I'll check back in the new year, after my winter trip, to let you know how it works.

Roger

 

[MarkBC]

I am using a "Watt's up" between my solar controller and the battery to "meter" my amps and amp hours.

My link

This is the same meter that I've been using in my camper solar system for over a year, and I've been very happy with it!

From "My Post-Pork-Rally Eastern California Trip" report: