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I am planning to set up an off-grid solar system for my small home. Based on my research, I now have some level of understanding/clue about the devices used in the system but need clarifications as I was given two very different setup/offer.

WHAT I HAVE IN MY HOME:

  • LCD TV: 21” (51cm) 100Watt
  • Small Fridge: 300Watt
  • Bulb: 10W

If I run all simultaneously, total is 410Watt. Simultaneous usage will occur only for 6 hours in a day.

ENVIRONMENT WHERE MY HOME IS:

  • Sun light from 6am till 8pm and average is 25C.

OFFER 1 (Is this overkill?)

  • Solar Panel: 200W Monocrystalline
  • Controller: 20A MPPT
  • Inverter: 2000W Pure Sign Wave
  • Battery: 2 x 12V 100A Deep Cycle Gel

OFFER 2 (This seems to be what I need but I might be wrong!!!)

  • Solar Panel: 150W Monocrystalline
  • Controller: 15A MPPT
  • Inverter: 1000W Pure Sign Wave
  • Battery: 12V 100A Deep Cycle Gel

FINDINGS

I learned that:

  • Solar Panel Short Circuit Current Amp must not exceed controller capacity. Formulas:

    1. Solar Panel Watt divided by Battery Volt.
    2. Solar Panel Short Circuit Current Amp multiplied by 1.56.
  • Total watt used by home appliances should not exceed inverter’s watt. In my case: 100W+300W+10W = 410W which is at least below 1000W.

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closed as off-topic by duskwuff, Voltage Spike, Enric Blanco, PeterJ, Wesley Lee Jun 19 '17 at 2:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Questions on the use of electronic devices are off-topic as this site is intended specifically for questions on electronics design." – duskwuff, Enric Blanco, PeterJ, Wesley Lee
If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ realize that you're not going to get 200W from such a panel, except in sunny mid-july at noon; avg is about 1/3rd of rating for "daylight", 1/6th for all-day avg. avg in: 816wh, avg use: 2460wh, so you need about 4X more than 200w for true uninterrupted power, less if you can live without the power at times (winter/night). \$\endgroup\$ – dandavis Jun 16 '17 at 14:25
  • \$\begingroup\$ On the other hand, a 300W fridge doesn't always run continuously. \$\endgroup\$ – pjc50 Jun 16 '17 at 14:52
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Your fridge draws 300W. Have you measured this? From my experience when I put an energy meter on my fridge out of curiosity, you might be surprised.

  • Starting the motor will draw a rather large transient current, which might trip your UPS.
  • If there is one, the resistor which heats the door to prevent it from freezing shut draws a surprising amount of power when it is switched on.

So, get an energy meter like a Killawatt and check measurements.

Now, you're missing the big picture: energy.

Your 12V 100Ah (not 100A) deep cycle battery can contains 12*100=1200 Wh or watts*hours... very optimistically.

If your fridge runs for 1 hour at 300W power, it will consume 300 Wh of energy. The battery stores 1200 Wh, so it can power the fridge for 4 hours. If it runs for 4 hours in the night because the weather is hot, then... next morning batteries will be discharged.

So, use an energy meter to measure how much actual energy the fridge uses in a day. Same for other loads. Watts*hours gives Wh which is energy, and you should have enough energy stored to run your stuff.

Putting warm stuff (like the groceries you just bought) in the fridge will make it use more energy, so also test that.

This will give you an idea about the battery capacity you require.

Note that you can also store cold, by putting lots of water bottles (or phase change material) into your fridge and only running it during the day when the solar panels produce electricity, which reduces your battery needs.

Now, lets talk about the panels. Once you know how much energy the loads will use in a day, then you need enough panels to produce that energy in a day.

Remember the power specification of a solar panels is for direct solar exposure with best orientation on a sunny day! You're not going to use a solar tracker, so the sunlight will not arrive at the optimal angle. Overcast or rainy days will massively reduce power ouput.

If you still want a cold fridge on an overcast or rainy day, or in winter, you will actually need quite a bit more solar panels.

Ultimately, it depends on the value of what's in the fridge. If it's just beer, you can drink it warm... okay. But if your meat starts to rot then it's a different problem!

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When considering a solar system, there are several efficiency factors that must be taken into account. A common mistake is to simply look at the "name plate" power information of the solar panel itself. This is not a sufficient analysis.

Solar panels are rated in watts by the manufacturer. These are the watts that can be achieved under ideal, direct overhead sun conditions. But these conditions do not exist for most of the day, not all days are sunny, and the angle of the sun changes seasonally. To account for these real world conditions, a factor known as PV solar radiation must be included. This factor is specific to the geographic location of the panel and is based on historic solar radiation information gathered from that location. An example of this data for the United States can be found here. Your solar supplier should be asked to quote the annual energy (kilowatt hours) output of the solar panel(s) based on your annual PV solar radiation for your location.

Another important factor to be considered is the inefficiency of the charge controller and inverter system. In the case of an off the grid system, the inefficiency occurs twice - first as the controller charges the battery from the solar panels and again as the controller converts the energy stored in the battery to alternating current. There are also losses in the solar panel wiring and in the battery itself. The combined losses can be estimated at 80% if your cannot obtain specific data for your controller, battery, and site wiring plan.

A conservative plan for an off the grid solar system takes into account the possibility that the battery may be nearly fully drained at the same time there are loads from the household requiring energy. This can occur after a period of inclement weather, for example. So the system would be sized to fully charge a depleted battery during a single day of sun while supplying the energy to the attached loads at the same time.

Your refrigerator, while rated at a maximum power consumption of 300 watts is not the information you need to size your solar system. The real question is how much energy per month or year does your refrigerator consume. This will be far less than 300 watts * 24 hours * 365 days/year. Depending upon your location and the age of the refrigerator, it may include the estimated annual energy consumption on its nameplate. If not, you may be able to look it up here. Failing that, you will need to measure the consumption of your refrigerator over say a one week period. There are various inexpensive devices available on the Internet that perform this function. Here is an example of such a device.

Another factor that must be considered with any motor powered appliance like a refrigerator is the startup current that is required. This is typically much higher than the running current of the refrigerator. So while this has little effect on the overall energy requirements of your panels, the battery and inverter combination must be able to handle this "blip" of higher power consumption while keeping all other connected devices powered. If you are not able to measure the startup current or power drawn by your refrigerator, a conservative estimate is that is will take 3 times the run time current or power. Your battery and inverter must be able to supply this power requirement combined with all other connected loads. This sets the minimum power specifications for your battery and inverter combination.

There are several factors that go into sizing the battery:

1.) The battery must supply some or all of the energy during inclement weather. This means that if you have two very cloudy days in a row, the battery may have to supply all of your energy needs for a 48 hour period.

2.) The inverter effectively reduces the capacity of the battery. A 10% loss from conversion efficiency is not uncommon.

3.) Battery amp hour ratings are based on a 20 hour discharge period. So a 100 AH battery is rated at 5 amps for 20 hours. If you attempt to draw 100 amps for 1 hour, the realized energy capacity of the battery will be significantly lower - probably in the order of 65 AH.

4.) The temperature of the battery has an effect on its realized energy capacity. For a typical AGM battery, freezing temperatures reduce its capacity while higher temperatures increase its capacity. Higher temperatures also reduce the overall lifetime of the battery.

5.) The inverter will drain your battery even when there is no load if you do not turn it off. This minimum load on the battery must be taken into account but do not duplicate inverter inefficiencies in this calculation.

6.) You should always build in some reserve capacity into the battery for unforeseen circumstances. Running the battery at the edge of its ratings is never a good strategy.

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  • \$\begingroup\$ Regarding to your 5th point, do all inverters do that or only cheap/bad ones? In terms of MMPT controller, they promise to prevent energy traveling towards panel from batteries in some particular cases I cannot remember what. \$\endgroup\$ – BentCoder Jun 17 '17 at 9:36
  • \$\begingroup\$ Yes, all inverters will continue to drain the battery if not switched off. Look for the term quiescent current or standby current. \$\endgroup\$ – Glenn W9IQ Jun 17 '17 at 13:03

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