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Copanas
May 30th, 2006, 8:45 PDT
Hi,
I was wondering if anyone could help me out?* I am helping a friend design and install a two small systems for two off-grid yurts she has on her property in Vacaville, CA.

The first yurt will use PV direct to run a solar chill evaporative cooler which uses 35-50 Watts.
We have been thinking of using a Sharp 80 Watt panel Vmax=17.1V and Imax=4.67A with a Linear Current Booster rated
at 7A and 12V.

It looks like we might have to place the panel a minimum of 100ft from the yurt.* At a distance of 200ft in wire length.* It seems like this would cause a large voltage drop and require that we use #2 gauge wire.

Cost wise, as well as performance, would we be better off using a higher voltage panel (for example a 165WAtt, Vmax 34.6, Imax 4.77A) with a 24V evaporative cooler and 24V rated Linear Current Booster to cut down on our voltage drop?

Also, any suggestions on brands of Linear Current Boosters?

System #2
This is a larger yurt that would have a battery back-up for a small refrigerator, laptop, cell phone charger, one compact flourescent and if possible an evaporative cooler.*

It seems like too large of a load to have the evaporative cooler and refrigerator on the same battery system, so we were thinking of running the evaporative cooler direct separate from the batteries.

Summer (Vacaville is very hot)
The Load:
DC
Sundanzer chest refrigerator* 12V 168WH/day

AC
1 compact fluorescent* 15W *4 = 60 WH/day
Cell phone (5W *12 hours * 4 days)/7days = 34.29WH/day
Laptop (50*3hours * 4days)/7 days = 85.714 WH/day
Cd/player/Radio 15W * 3 hours = 45 WH/day*

Total AC = 225WH/day x 1.15 to correct for inverter loss = 259WH/day

To Calculate number of panels:
AC + DC = 427WH/day
427 / 12 = 36 Amp Hous/day
Multiply by 1.2 to compensate for loss from battery charge/discharge = 43AH/day

43AH/day* divided by average sun hours per day 5.1 = 8.37A (the total amp hours required)

8.37A divided by panel peak amps (sharp 80W Vmax 17.1, Imax 4.67A) = 1.79 panels = 2 panels in parallel

Battery Bank Calculations:
43AH/day x 3days = 129Amp hours
129Amp hours / .5 discharge = 250Amp hours

A Golf cart 6V with 220 Amp hours
250/220 = 1.13 batteries
Can I use one battery if it will only give me a 3.8% charge?
PLus will the refrigerator load, since it is summer and over 90 degrees bring that 3.8% down even lower? Any way to bring this up* to the 5-10%?

This system will have some of the same wire distance issues as the direct system for the smaller yurt mentioned above, as we are working to see if we can get it closer, but it may be 100 feet away as well.
Should we switch to a 24V refrigerator and change the panel voltages to match?* Also, given the hot climate and a small system, should we use a charge controller with MPPT? Any suggestions for battery chargers?

This is a lot of questions I know!* But, any help would be greatly appreciated!*

Thank you!
I really appreciate this discussion forum, it is very helpful!
Beth

BB.
May 31st, 2006, 15:51 PDT
You have pretty much the basics down for doing the load and panel calculations from what I can see... However, before trying to put hard numbers against everything, a few questions:

Vacaville can be pretty humid--is it dry enough that an evaporative cooler will help--instead of just creating humidity/mold problems?

Using solar for fan/cooling is great because there is usually bright sun with high temperatures. Placing a battery in the system with the cooler can allow you to shift some of the cooling into the afternoon and evening--may be worth while.

Also, for the Yurt where folks will be living(?). You talked about it having battery backup... Does this mean that you will have AC and you are providing for stormy/hot weather grid failures, or are just looking to be "more green" and reduce the use of utility power? Is this a fully occupied Yurt, or only just 2-3 seasons a year?

The reason I ask is that very few folks can live year in and year out using only solar power if you have constant power needs (like a fridge and/or computer for work). For me, I am grid-tied so that I can average my power over the entire year, and have excess generated power with respect to the utility meter. However, in winter, my power is severely curtailed. (3.5 kW array only produced 134 kWatts all December. May, I have produced about 500+ kWatts). So, there will be choices to make, backup generator, utility power, larger array, etc.

Lastly, you want to mount the solar panels 100' away from the Yurt. If you do this, you will want to keep the voltages much higher to keep the wire size small. Roughly, if you double the voltage, you can drop the wire gauge by 3 sizes (12-24v, instead of 2 awg, you can go to at least 4 awg).

Another consideration you may want to make--much of your use in the second Yurt can be AC (if not all). If you are keeping the solar/battery box/etc. a hundred feet away, you may wish to think about going 120 VAC from the solar/battery shed to the Yurt. You won't lose that much in the conversions and you may be able to save some money on the fridge (some folks have experimented with a small AC chest freezer using a separate thermostat to keep the interior at refrigerator temperatures--I would probably place a small fan in there too that turns on when the compressor is running to keep the temperature more stable and uniform). The differences these days in energy usage between a 12/24 vdc fridge and an AC fridge is not that great anymore.

Without backup power (generator/utility) the panel sizing is probably OK for six months of the year. If you have backup power, you can always add more panels later to reduce your generator runtime/utility costs.

If you have AC utility power, then consider using a Grid-Tied system--all the advantages of an unlimited battery for only $6/month (plus any excess utility power charges you used over the year).

The MPPT controller can help you get more power in cooler weather. Where it can help you a lot is in system placement. For example, you want a 12 vdc system with the controllers and batteries in the Yurt, but the panels 100' away. If this was a non-MPPT controller, you would basically be sizing the 100' of wire to pass the 12 volts and 10 amp current. With a MPPT controller like the Outback MX 60 with the panels wired to give you, say 80 volts or so at 1.5 amps of current--much smaller gauge of wires required from the panels to the Yurt/controller/battery box. Also, with a MPPT controller you can look at more brands/models of solar panels for the best price per watt. The controller will, very efficiently, converter any reasonable input voltage into the correct battery voltage (a non-MPPT controller will not down convert any excess voltage over that needed to charge the battery into energy--i.e., a 120 watt 12 volt, 10 amp panel will supply the same "power" to the battery as a 240 watt, 24 volt, 10 amp panel-120 watts. A MPPT controller will transfer almost 240 watts to the battery with the higher voltage panel--less a few 10's of watts for the conversion electronics).

For small systems with panels/batteries/loads close together, an MPPT will not generate much "extra" power over a non-MPPT (and may actually give less because of the 10-20 watts of internal power consumption).

In your case, a 160 watt system PV system is probably a close call (you will have to look at the different brands/models to match one to your needs--you don't want to pick too large of MPPT controller if the amount of solar PV watts is low).

At this point, I probably have more questions for you than answers... Does this help?

-Bill

Beth
June 3rd, 2006, 14:22 PDT
Hi Bill,
Thank you for your response! I really appreciate it! Your questions/suggestions are most helpful. Sorry, about the delay on my end, I was visiting with my grandmother, who is sick and does not have internet access.

The yurts are located on a small organic farm and are occupied full-time by farm interns from at least March through November, although occasionally someone might be there year round. The idea is to provide them with a means of cooling during the hot summer and to equip the larger yurt (they are about 150 feet apart) with a small refrigerator, a light or two, cell phone and laptop charging capabilities. There is no grid tie in available. The interns do have access to another refrigerator on the property as well as other electrical hook-ups, however those are located far away.
The yurts are located far (1/4 mile) from the main property utility, which is why my friend would like to utilize solar, as well as wanting to take a green approach. The winter load seems a lost less to me given that the evap cooler won't be necessary and the SunDanzer chest refrigerator we were thinking of using, or any refrigerator for that matter, will have a lower ambient temperature to deal with. Any AC chest refrig. suggestions? The ones I have come across have been 12V or 24V only.

I don't actually live in Vacaville, but the owner of the farm utilizes a evaporative cooler in her mobile home which seems to work well for her, but maybe we should just use a solar fan like you suggest?

The distance from the yurts will be better determined tomorrow when I can get out there with a solar pathfinder to figure out the ideal placement for the panels. Based on an initial glance we were thinking the panels would have to be 100' away from the yurt with batteries, controller and loads. We were thinking of having a well ventilated battery box for housing the batteries under the yurt platform to help keep them cooler, but I am concerned about safety, since they would be under the yurt living space if we did that. I am interested in your suggestions to go all AC from the panels to the yurt, since I thought that we would lose more to conversion from DC-AC, given that the cooler, refrigerator and compact fluorescents can all be purchased as DC items and are pretty efficient. But, if I am understanding correctly, if it is a long run, converting to 120VAC before running to the yurt will make up for the voltage drops that would be experienced using a 12V or 24V system? I have to spend more time researching and calculating this one out (the cost of the refrig and coolers versus, wire, panel and battery costs).

We originally wanted to put a battery with the cooler to allow for shifting some of the cooling into the afternoon and evening like you suggest, but for the yurt with the refrigerator we were concerned about the size of the summer load with both the refrig. and cooler having large power demands. The summer load seems like the largest, especially if you include the evap cooler or cooling fan with the refrigerator on the same battery system. I couldn't find data on Vacaville insolation specifically, so I have been going by the Davis yearly average of 5.1, 6 in the summer and 3 in the winter. With just the refrigerator is it possible to utilize solar year round or do you think they need a back-up generator? You mentioned six months for the current calculations, is that due to the lower solar insolation in winter? Even with the refrigerator working less hard due to ambient temperature will we still fall short?

I guess we are just trying to figure out what the most cost effective option or options is for each yurt, especially the larger which involves the refrig. and batteries. If I up the panel wattage and battery bank, are we going to have overkill during the winter if there's less load? I guess I need to work with these numbers a little more.

Thanks for your help! Any other suggestions or corrections would also be appreciated.
Thanks again,
Beth

BB.
June 3rd, 2006, 22:53 PDT
One of the reasons I was suggesting "full AC" was that AC appliances are, in general, much cheaper than their DC equivalents--and for small things, like laptops and cell phone chargers--every one probably has the AC adapter--fewer would have the DC equivalents.

Money, that you can put into an inverter, batteries, and/or extra solar panels...

Here is the link to one of the "chest freezer" conversion threads...

http://www.wind-sun.com/smf/index.php?PHPSESSID=8374c031bedaa27dab793febdb1dd317&topic=1073.0

By the way, the fan I was speaking of was for use inside of the "chest fridge" conversion--to mix the air when the compressor is running (otherwise, you may end up with unequal temperatures in the fridge). And if the owner is happily using an evaporative cooler (and there is sufficient water on had to run it)--sounds like a winner to me.

The reason that I was saying 6 month of the year was because of clouds/storms during our "rainy" season can dramatically reduce the output of the solar panels. For my system, Mid-November 2005 through about end of January 2006, there a few times of 3-4 days of sub 1kWhr per day production (full sun was about 10+ kWhr/day for a few days during that time). During the worst, 1 month period (3weeks Dec+ 1 week Jan) I averaged about 3 kWhrs per day with a system that had 3,500 watts of panels (3 kW peak rated using CA more conservative numbers).

Will next winter be better or worst? Will your area be better or worst? I don't know. But as an example, that same system for May 2006 averaged 17.6 kWhrs per day. So, eventhough the Solar Radiation (RedBook) ratios for San Francisco are May/Dec 6.4/3.4=1.8:1, my real numbers were 546/134 = 4:1 between those two periods.

So, what is one supposed to do? Buy 2-4 times the panels to run the winter load than will be needed for the summer load? Or, do you purchase a small generator that you run a few hours every day or three to charge the batteries--Or, do you assume that the Yurts folks may not reliably use the fridge/computer and such larger loads for roughly Mid-November through the end of January (no generators). Will my numbers apply to your area--hmmmm maybe. But each year will, of course, be variable wrt the year before.

At this point, you probably have enough information to start building a couple "paper systems" and can begin to estimate the costs between the various options.

Usually, what I would do is start with a web site like www.wind-sun.com (shameless plug for host of this forum) and just pencil a rough diagram and start placing some componets. A DC 8cuft fridge here ($1,000), an AC 8cu "chest fridge" there ($500), no inverter here, a xxx or x,xxx watt (sine wave preferred) inverter over there ($500-$1,800), golf cart or AGM batteries,... Back calculate power efficiencies and size the solar panels accordingly. Then find the charge controller (MPPT or not) that fits your size requirements (again, with meters or not--suggest meters/logging are very helpful to keep track of the system--but a waste of money if nobody is ever going to look at them).

The problem is that you will have to make a decision if you are going for lowest costs (wire yourself, rig your own chest refregerator conversion, modified sine wave inverter, no generator provisions), or ease of operation (full sinewave inverter with 2 watt automatic AC search/standby mode, with built in AC battery charger/generator connection, i.e., Xantrex SW series) and such.

It will seem pretty daunting at first... But just go step by step. Like you already have, start adding up loads (and loads by season), starting loads (if things like well pumps, or small inverter powering AC fridge), etc... And as you do this, you will gain confidence and start understanding better the trade-offs between price, quality, and reliability. There is no one right answer here...

You can even go by the hardware store and find out the costs of the 100' of buried cable at different gauges (heavy for DC low voltage system, lighter gauge for AC or high voltage DC system).

Once you have an equipment list (and prices)... It should be pretty easy to check back here and ask specific questions about your planned system.

My suggestions for looking at AC over DC only system are just that--until you specify and price the solutions, and look at their issues, you won't know which will be best for your situation. For a specific Chest Freezer suggestion--I would simply look at the Energy Tag (should list kWhrs/Year) and price. Any high volume production manufacturer should work just fine (and if one dies early, they are cheap enough to buy a second brand later). There were a couple of suggestions in the link above. Also, watch the size of these guys... Turns out that, for example a Crosley 12 cuft model uses only 10% more energy than an 8 cuft model (and the retail price is lower). The smaller fridges (under 7cuft) seem to use almost the same power as the medium sizes (7-15cuft).... Just look around.

One other suggestion I would make--if you can, power both Yurts from the same panel/battery system (assuming that the panel/battery pack can be located where you want). This will allow you to use the batteries for both Yurts and allow you to use the extra panels for the cooling only summer Yurt to give you extra capacity to take you deeper into the winter months.

Another question, are you remote locating the solar/batteries because of safety? Proper design and venting (with agm much less hydrogen) should keep everyone safe. Or, build a "shed" Yurt/outbuilding very close to the "living" Yurt... You can also put the fridge in it too--give more space to the interns in the living Yurt (less noise from fridge too). You will 1/2 the wiring and wiring losses wrt your first post (with two solar systems 100' from their respective Yurts?).

Am I giving you useful answers to your questions? Or are you looking for Xantrex SW2424 with x 165 Unisolar panels, xx T105 batteries, etc.???

-Bill

BB.
June 5th, 2006, 12:15 PDT
Beth,

Speaking as an engineer and not as a person who has purchased and ran an evaporative cooler off of solar panels... I am going to put a lot of math and numbers into the stuff below--may or may not be interesting to you--the results were intereting for me--in that it showed (again, to me) that going through all the math gives you about the same numbers as using simple rules of thumb. And, remember, solar is a game of estimates anyway. Weather, seasons, local climate, installation, wind, shading, temperature, etc. will probably affect your solar output more than worrying about the exact Vtoc for a panel and how it will affect today's (June 5th, 2006) exact power output.

DC motors in general, use more current the slower they turn--and they turn more slowly when the voltage drops. So, in an environment where you have varying power available, it is helpful to supply "match" the output of the solar panels, in this case, with the needs of the load, DC motor...

A solar panel's current is, more or less, proportional to the amount of sun (and panel temperature), and the voltage will simply drop if you try to pull more current... So, remembering Power=Current*Voltage, if you can slightly drop the amount of current used, the voltage will dramatically rise. In the above example, if you have 2 amp (morning sun) and voltage = 2 volts, you are only getting 4 watts. But, with the "current booster", if you get 1.8 amps but 12 volts, you now have 21.6 watts. So, now you have almost 10x the amount of power available from the solar panel and can do something with it...

Your DC cooler load--mid-morning sun and the motor is just trying to turn. The motor is rated for 12 volts and 52 watts, or P=IV ; I=P/V=52/12= 4.33 amps at full speed. However, at starting, the motor make take 2x or more current... So, you get a steady operation with the motor spinning slowly at 2 volts and 2 amps (only 4 watts of power available). Place the current booster in between the solar panel and the DC motor load... The electronics now (this is a guess on my part--never seen the insides of the booster) finds the best power point (just like the MPPT charge controllers) to be 12 volts at 1.8 amps (21.6 watts) by "dropping the current" and watching the voltage rise to the peak power (should be close to the solar panel's rated voltage) (actual algorithm is probably different, and there are several ways I can think it could be done--details don't matter for our discussion). Then, internally is a circuit (again, probably very similar to a MPPT charge controller) that converts the power of 21.6watts = 2v*2a to, for example, 4 amps at 5.4 volts (the P in P=IV is constant, change the I and V output to match the load)--for the motor, it now sees close to 20 watts of power to turn the fan instead of the original 4 watts of energy.

The current booster, basically, changes the output of the solar panel into something that better matches the load of the motor. A solar panel is, more or less, a constant voltage source. Batteries are also a constant voltage device and you can connect the two together without a MPPT controller and get pretty good results. A motor is not a constant current/voltage load, so something that can "match" the solar panels output with the motor's input requirements. (a 12 battery cannot be charged with 5 volts--however a 12 volt motor can run at 5 volts if there is enough current). So, install the current booster as close to the motor load as possible for the least amount of loss energy (lower current on the solar-booster input connection, higher current on the booster-motor wires).

Choosing panels for the DC motor application... I will use my grid-tie system as an example...because I have some real numbers and I am not too far from your proposed location. If you decide on direct connect through a linear current booster and not batteries--then you will need to choose how much power you will want to run the cooler. Assuming 52 watts at high speed/full load, you will need to look at the solar panel specifications. While the current is pretty much proportional to the amount of sun's energy on the panels, the peak power (Vpp or Vpmax from the solar panel spec.) is proportional to the temperature of the solar panel. Most panels are tested under standard conditions of full sun and 70 degrees F. However, if you put your hand on a dark surface in full sun and no wind, you know that it is much hotter than 70F--130F or even 160F is more likely.

California has its own rating system that seems to be much closer to actual power... Xantrex has a nice sizing tool for their Grid-Tie systems, but they also list most of the solar panels available and their specifications too...

http://www.xantrex.com/support/gtsizing/index.asp?lang=eng

For example, I have BP 4175 (175 watt panels at STC ratings) and the Xantrex specs. show:

175.0 watts STC rating
155.2 watts PTC rating
35.7 Vdc Vmp (max power)
44.0 Vdc Voc (open circuit)
-0.16 V/C Vtoc (voltage temp. coefficient) or -0.2 V/F (note: this number is also proportional to panel's voltage rating)
4.9 amps Adc (max current)

From my system, I have seen voltages around 29-34 volts and 4.95 amps (per panel, I have 20 panels PTC rating of 3,500 watts, about 100 watts is used by the inverter electronics--95% efficiency at full load of 3,000 watts max). The most power I have seen is 2,999 watts (150 watts per panel, morning bright sun, ~55F) on a cool morning and most of the time my high power is around 2,500 to 2,700 watts during the middle of the day (125-135 watts) on a 80F-70F day. Notice that the power goes down as the temperature increases... In the late afternoons, I am probably around 2,000 watts (100 watts per panel, warm later afternoon)...

Hmm... what temperature are my panels. Assume 2,500+100 watts panel + hidden inverter power, full sun, warm 80F day 9.6 amps (130 watts per panel)...

P=IV=I*(Vmp + Vtoc * (temp-70F)) ... After a bit of algebra =>
temp= [(P/I - Vmp) / Vtoc] +70F = [(130w/4.9a - 35.7v) / -0.2 V/F] + 70F = 115.8F calculated panel temperature

115.8F is probably not a bad estimate of the temperature of my roof mounted solar panels. If you have pole mounted panels (with good air circulation), your panel temperatures will be lower, and the voltage (and therefore power) output will be higher than mine.

After all was said and done, the thing to notice is that my ratio of actual output to rated output on a warm day was in the range of 130/175 = .75 : 1 ... So, if you want to run your cooler on a hot day at full power, your panels will need to be, at least 1/.75 = 1.33x rated or, in your case, 52*1.33=70 watts minimum for full power on a warm day. Assume that the current booster is only 80% efficient (just an educated guess of 80% efficiency for the appropriately sized booster)), then you are looking at 70watts/0.80 = 88 watt panel PTC rated panel power (assuming mono or poly crystalline solar panels).

If you wanted to do the same calculations for the Uni-Solar US-64 panel, looking at Xantrex site again, the Vtoc is -0.07378 V/C or -0.092225 V/C -- it is less than 1/2 affected by temperature increases--a good thing for a system running in the hot summer sun. Using my temp of 116F and the rest of the numbers from the Xantrex work sheet:

P=IV=I*(Vmp + Vtoc * (temp-70F)) = 3.88a * [16.5v + -0.092225 V/F * (116F-70F)] = 47.6 watts US-64 on a warm day.

US - 64 derating @ 116F = 47.6 / 64 = 0.75

To compare with a Sharp 80 watt panel... There appear to be at least 3 types of 80 watt panels from Sharp. I will pick two. One appears to be a less expensive one (with larger Vtoc). Example NE-80EJE (remember, I am dividing Vtoc in C by 0.8 to get Vtoc in F):

P=IV=I*(Vmp + Vtoc * (temp-70F)) = 4.67a * [17.1v + -0.09 V/F * (116F-70F)] = 60.5 watts on a warm day.

The other Sharp I will use is the Sharp NE-80U1:

P=IV=I*(Vmp + Vtoc * (temp-70F)) = 4.67a * [17.1v + -0.0025 V/C * (116F-70F)] = 79 watts on a warm day.

Sharp 80EJE derating at 116F = 60.5/80 = 0.76
Sharp 80U1 derating at 116F = 79/80 = 0.99

Hmmm... I am not convinced that the Sharp 80U1 temp/coeff is correct (personally, I think it is dead wrong)... I would tend to believe that there is a typo in the Xantrex data--and I cannot, with a quick web search, find any further information... Take that 0.002 number with a grain of salt.

Anyway.... After a lot of boring math (I had fun doing it though...), I would simply use the rule-of thumb of take what ever load you want, and use the following "fudge" factors to calculate your ratings:

Panel Watts = Load Power * Temp derating * 1/Charger/Booster Eff * 1/Inverter Eff * etc....

Warm Weather derating ~ 1.33 (hot desert weather--run your numbers at 150F--cold weather, you will get more than your rating)
Charger Eff (see specs) ~80-95%
Inverter Eff (see specs) ~75-90%
Battery Charging Lead Acid ~80%
Battery Charging AGM ~98%

Run a typical lead acid battery system for efficiencies:

Panel Factor = 1.33 * 1/0.93 charger * 1/0.80 lead acid charge * 1/0.85 inverter * 1/0.97 wiring loss = 2.2

So, for all of the losses, to get power to your 7 watt night light, you will need a panel that can power 2.2*7w = 15.4 watts

Each piece of loss is not that bad in itself, but when all the losses are multiplied together, the power required is much larger than it first seems.

And, generally, if the system is properly designed, having "too many solar panels" will not damage anything, it will just give you more power that can help you ride through bad weather, power more than you planned, and not get surprised by running out of power when you need it most. But, those extra panels are not cheap...

Also, when you look at solar panels, they age over time (exposure to sun). The Mono and Poly Crystaline are usually rated at to 80% power at 25 years (and I have read folks that say 10-20 year old panels still meet new specs). The Amorphous panels can degrade faster (especially the older panels)... Check the warranty and ask questions.

To be honest, on a properly installed cooler--nobody will probably even notice a 1% drop in energy available from the solar panel from year to year... And even a 20% drop (after 25 years) is not even noticeable for a blower application. In general, where light, fans, and such are used, anything that is less than 1/2 (50%) less power (like 80% of power, or 20% power drop) is, usually, hardly noticeable by the average human. So, while I would recommend a 90+watt panel for your cooling system to run at full power--nobody would be able to tell the difference unless they were using test equipment--and it certainly would not be worth purchasing two 80 watt panels to power the load--except to run the fan faster in the late afternoon (may be very much worth while to spend the money on that feature). Using the Current Booster should, more than likely, noticeably improve the performance of the cooler on very hot days and in less than full sunlight conditions.

Long way around to your short questions--I would consider:

80 watts (or 90+ if cooling is important) for the evap. cooler (by the way, does that include a pump for the water flow?). If not using batteries, I would very much consider the current booster (because it matches peak power curve from the solar panel with the best power delievered to the motor at times other than solar noon).

Get fewer, larger panels than many smaller panels (wiring/mounting/more things to go wrong with lots of small panels). I would also, if possible, keep all of the solar/equipment together, and, if possible, wire up the panels so that you can through a switch that will move the panels from summer cooling use to winter battery charging.

Just from a paper check, the UniSolar 64 module is probably a bit on the small size--However, if you could get two UN 64 or one Sharp 80 watt for the same price--I would consider the two UN-64 units (more power at same price). With the current booster, the exact current/voltage ratings are a little less important as the booster should adjust for optimum power.

Regarding your question of running DC/Battery/Charging/etc. directly... Hmmm. I am not sure exactly what you are asking--but see if I can explain it in another way...

Think of your batteries as a pond. Your solar panels as a water faucit. And you loads as a hose to somewhere else.

You can hook the hose directly to the faucet, however, if the water flow is low, or off (dark), then no water. If you connect the hose to the pond, you can use as much or as little water anytime you wish--within the limits of the pond (it will run out and you will suck-up mud--etc.).

So, if you use the pond for everything--the hose to take water when needed. The solar panels (through the charge controller) to fill the pond when water (sun) is available... It is probably easier to think about running your load (light, laptop, etc.) with the, more or less, constantly available water source (pond/battery) than it is to figureout how to deal with a variable source (like a faucet/solar panel where voltages, currents, and power are constantly changing).

It can be done, but it is much more of an engineering issue. For example, your DC Motor can run from zero to 18 volts. However, your battery needs about 12 to 14.6 volts. How do I connect the two together? If the solar panel is not putting out enough voltage, the battery will begin to discharge and run the motor. If the solar panel is bigger than the motor load, the voltage will increase so much that the battery will overcharge and get ruined in just a few days or weeks.

The laptop 12vdc adaptor is probably designed to work from 10.5 volts to 15 volts. Below 10.5 volts it will turn off to save the battery, above 15 it will can get destroyed. So, connecting to a solar panels with Vmax of 21 volts will kill the laptop adaptor. At the beginning of the day the solar panel slowly receives sun, the voltage and current starts to rise--when the voltage goes over 10.5 volts, the laptop charger turns on and tries to take 50 watts of power (about 5 amps) to charge the laptop battery. The solar panel can't supply this, the voltage falls under 10.5 volts, the fan slows down, the current drops, the solar panel voltage begins to rise again above 10.5 volts, and the whole cycle begins again. You may here the fan motor buzz, the laptop adapter may begin to overheat, and things just are not playing well together. (kind of like when you are taking a hot shower and somebody turns on the kitchen sink and your shower turns cold--you re-adjust the water temp. Then they turn off the kitchen sink and your shower becomes scalding hot...).

So, yes you can make a purpose built solar panel / motor system (cooling, water pumping, etc.) for when the sun is out and save some money on batteries, charge controllers and loss efficiencies of charging...

But, when mixing many different uses together--it is much easier to assume a fix source (battery) and figure all of your loads and charging sources with respect to that battery. And only do Solar/Load for special cases.

I will stop here--probably too much typing... I hope that it has been helpful for you to better understand your problem and design. Try to keep things simple in your mind. The Solar Panel, the Charge Controller, the battery, the loads (each with its own needs). Converters to change 24vdc to 12vdc. Or 12vdc to 120vac. It is easier to have higher voltages and lower currents to transmit the same amount of power:

P=IV=240 watts=20amps * 12 volts = 10 amps * 24vdc = 4 amps * 48vdc = 2 amps * 120vac
Uggg... Put brain back in gear...
P=IV=240 watts=20amps * 12 volts = 10 amps * 24vdc = 5 amps * 48vdc = 2 amps * 120vac

Power Loss (for our discuss here)= I^2*R... As current goes up, heat goes up at the square of the current--2x current 4x power loss).

And safety... Fuses and circuit breakers are rated differently between AC and DC (DC is, surprisingly, much more difficult for CB's and fuses than AC). You can certainly do 90% of the sizing/costing of a system. But you need somebody with experience (electrician and/or solar installation experience) to put everything together safely and with the right protection against fumes and short circuits...

And remember, I am discussing some of the above from the point of view of engineering theory... Others here have much more experience in the day to day items, such as DC evaporative coolers and current boosters than I. I am just trying to suggest several approaches to your problem that, hopefully, can help you get a cost effective solution for your friend that they will be happy with.

-Bill

BB.
June 6th, 2006, 1:34 PDT
Beth,

I might as well add one more option... Using Grid Power and Grid-Tied Solar Panels...

Find out how much it would cost to run (overhead or buried) power from the main buildings out to the Yurts (just one 20 amp 120 VAC service) from an electrician. And just use energy efficient items (fridge/evap. cooler/laptops/CPF lighting) to keep the loads/costs low.

Then, at the main house, install a Grid-Tied solar PV system. You will get the reliability of utility power, and the owner gets to go green for as little (two-Yurts' worth), or as much as they wish (up to a whole farm's worth of solar panels). And this system would be virtually no maintenance, no batteries, no fumes, no running out of power, and rebates/tax credits would be available from the state (and even the Feds if this is a business) (the Yurts may not have California rebates available).

They could also sign the farm up for net metering (basically, a 1 year energy bank--use more power during the winter and get generating credits in the summer. At the end of one year, either pay the balance (if consumed more than generated) or Zero-Out the balance (if generated more than consumed).

Also, depending on the energy use of the farm, may try Time of Use metering. If can avoid (for example) Noon-6pm Mon-Friday peak power use (when solar is generating a lot and powe is expensive), they can then use lower off-peak rates for cooking/processing/etc...

May not fly with owners, but probably would come in close, if not cheaper, than a stand-a-lone Yurt power system (using AC appliances, no batteries, controllers, current boosters, etc.). And very little maintenance costs with no dark day blackouts or generator requirements.

-Bill

PS: And, no expensive solar panels/batteries/controllers/inverters out in the south 40 that can be stolen/vandilized. Also, could run phone/network/cable/etc. lines out with the power lines too.

Patman3
June 8th, 2006, 22:50 PDT
BB Bill, I enjoyed reading your essay and was interested in the temperature calculations.



Hmm... what temperature are my panels. Assume 2,500+100 watts panel + hidden inverter power, full sun, warm 80F day 9.6 amps (130 watts per panel)...

P=IV=I*(Vmp + Vtoc * (temp-70F)) ... After a bit of algebra =>
temp= [(P/I - Vmp) / Vtoc] +70F = [(130w/4.9a - 35.7v) / -0.2 V/F] + 70F = 115.8F calculated panel temperature

115.8F is probably not a bad estimate of the temperature of my roof mounted solar panels. If you have pole mounted panels (with good air circulation), your panel temperatures will be lower, and the voltage (and therefore power) output will be higher than mine.

After all was said and done, the thing to notice is that my ratio of actual output to rated output on a warm day was in the range of 130/175 = .75 : 1 ... So, if you want to run your cooler on a hot day at full power, your panels will need to be, at least 1/.75 = 1.33x rated or, in your case, 52*1.33=70 watts minimum for full power on a warm day. Assume that the current booster is only 80% efficient (just an educated guess of 80% efficiency for the appropriately sized booster)), then you are looking at 70watts/0.80 = 88 watt panel PTC rated panel power (assuming mono or poly crystalline solar panels).

If you wanted to do the same calculations for the Uni-Solar US-64 panel, looking at Xantrex site again, the Vtoc is -0.07378 V/C or -0.092225 V/C -- it is less than 1/2 affected by temperature increases--a good thing for a system running in the hot summer sun. Using my temp of 116F and the rest of the numbers from the Xantrex work sheet:

P=IV=I*(Vmp + Vtoc * (temp-70F)) = 3.88a * [16.5v + -0.092225 V/F * (116F-70F)] = 47.6 watts US-64 on a warm day.

US - 64 derating @ 116F = 47.6 / 64 = 0.75

To compare with a Sharp 80 watt panel... There appear to be at least 3 types of 80 watt panels from Sharp. I will pick two. One appears to be a less expensive one (with larger Vtoc). Example NE-80EJE (remember, I am dividing Vtoc in C by 0.8 to get Vtoc in F):

P=IV=I*(Vmp + Vtoc * (temp-70F)) = 4.67a * [17.1v + -0.09 V/F * (116F-70F)] = 60.5 watts on a warm day.

The other Sharp I will use is the Sharp NE-80U1:

P=IV=I*(Vmp + Vtoc * (temp-70F)) = 4.67a * [17.1v + -0.0025 V/C * (116F-70F)] = 79 watts on a warm day.

Sharp 80EJE derating at 116F = 60.5/80 = 0.76
Sharp 80U1 derating at 116F = 79/80 = 0.99

Hmmm... I am not convinced that the Sharp 80U1 temp/coeff is correct (personally, I think it is dead wrong)... I would tend to believe that there is a typo in the Xantrex data--and I cannot, with a quick web search, find any further information... Take that 0.002 number with a grain of salt.


-Bill


All have a derating of around .75 (except for anomaly), so there is no real advantage to the US64 in warm weather.* Just as I thought, a marketing gimmick.* Thanks BB...

BB.
June 9th, 2006, 3:07 PDT
Thank you for the compliment Patman...

When I first was looking the Vtoc for the UniSolar vs the BP panels, it seemed that there was going to be a big difference in hot panel performace... Then I realized that the thermal derating (voltage drop per degree C/F) is per solar cell junction.

The UniSolar was a "17 volt" panel, and the BP was around a "34 volt" panel... So, it had twice the number of junctions (and twice the voltage), so its derating had to be 2x the UniSolar panel too... Basically, in power produced, this resulted in a wash for the temperature derating factor.

-Bill