My first project - Solar power computer....I need your help

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  • KellezKellez Registered Users Posts: 21
    Re: My first project - Solar power computer....I need your help

    Guys i am aware of what you are talking about, chinese parts will brake eventually, good quality products last a lot longer and they do have warranty, but what you are suggesting (a good charger + samlex inverter) costs about $250, if you consider morning star inverter then this goes up to $350. What i am suggesting is only $50. By comparing the price it is obvious that there is big difference in quality, however $50 is nothing and please keep in mind what i am doing is only a test, this is just for learning experience nothing more. I may as well just use the system for a couple of days just to see how it works and use a watt meter to monitor the system
  • BB.BB. Super Moderators, Administrators Posts: 32,319 admin
    Re: My first project - Solar power computer....I need your help

    Not arguing--Just agreeing that these are the choices you are making.

    We tend to help folks with off grid cabins/homes and such--So we are generally after long life, reliability, and the ability to run some medium sized equipment (refrigerator, well pump, etc.).

    As long as you are OK with experimentation--And the good vs bad that can happen--That is perfectly fine with us.

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • KellezKellez Registered Users Posts: 21
    Re: My first project - Solar power computer....I need your help

    For sure we are not arguing, i understand your concern and i really appreciate it, you have already helped me lots. I totally agree with you and if i was to power some serious equipment i wouldn't go with $50 charger and inverter. As you can see i have already changed my mind about powering my desktop computer. I wouldn't dare to use such cheap equipment with my computer even if my computer can handle the MSW inverter.
  • JohannJohann Solar Expert Posts: 245 ✭✭✭
    Re: My first project - Solar power computer....I need your help
    BB. wrote: »
    Welcome to the forum Kellez,

    Your calculations and numbers are correct in the "ideal" world. In the real world, there are some issues.

    First, you want to discharge the battery at a C/2 hour rate (fully discharging the battery to "dead" in 2 hours)--Flooded cell batteries really only like to be discharged at a C/8 hour rate, or at most, a C/5 discharge rate.

    Also, the faster you discharge a battery, the less apparent capacity (a battery discharged at C/20 rate may appear to have a 100 AH capacity, a lead acid deep cycle storage battery discharged at C/2 rate may have an apparent capacity of 60 AH or less).

    So--You have two choices, either get a larger battery to drop the discharge rate to C/8 - C/20, or change the battery type. AGM type Lead Acid batteries can be discharged at a much faster rate... Or looking at the various types of Lithium Ion chemistry type batteries (lots of choices, and you get into the whole "Battery Management System" electronics--Some/most Li-Ion battery types need a per cell voltage monitoring/charging adjustment circuit to prevent over/undercharging and damage/battery failure).


    -Bill

    WHAT IS C /2 , C/8 etc etc.
    What does it mean etc etc.
  • CariboocootCariboocoot Banned Posts: 17,615 ✭✭
    Re: My first project - Solar power computer....I need your help
    Johann wrote: »
    WHAT IS C /2 , C/8 etc etc.
    What does it mean etc etc.

    Charge (or discharge) rate where 'C' is the battery's capacity in Amp hours (at the '20 hour rate'). So for a 100 Amp hour battery C/10 = 10 Amps, C/8 = 12.5 Amps, C/5 = 20 Amps et cetera.
  • BB.BB. Super Moderators, Administrators Posts: 32,319 admin
    Re: My first project - Solar power computer....I need your help

    C = Amp*Hour Capacity at C20 hour discharge rate (i.e., Battery discharging at 5 amps * 20 hours until dead = 100 AH capacity).

    There are other discharge rates (C100, C24, C8, etc...). For a flooded cell lead acid battery, the "optimum" discharge rate is >C8... The faster you discharge the battery, the more energy that is lost as heat, and the less apparent capacity.

    For example, a Trojan "golf cart" 6 volt 225 AH battery:

    5-Hr Rate = 185 AH
    20-Hr Rate = 225 AH
    100-Hr Rate = 250 AH

    Different batteries will have different discharge "curves"... A good quality Concorde AGM may look something like this:

    2 Hour 8 Hour 24 Hour 48 Hour 120 Hour
    172 194 212 235 253 (Amp*Hours)

    Notice that the AGM is rated down to C2 discharge rate---And you can pull upwards of C*4 hour discharge rate (kill the battery in 15 minutes) for some AGM batteries.

    Anyway, we use some generic rules of thumbs here for Flooded Cell Lead Acid batteries--Basically, how much current you can pull from the batteries in different situations (and recharging)--We use the C20 capacity for our rules of thumbs. Other vendors may used different Cxx rates.
    • C/20 rate = Run 5 hours per night, two nights and 50% maximum discharge (i.e., 20 hour discharge rate for an off grid home--Charge during the day, discharge in the evening).
    • C/8 rate = Basically running the battery flat out (more or less efficiently) until dead. You don't want to exceed this rate (total discharge) without thermal management (can warp plates, etc.).
    • C/5 rate = Short term heavy discharge (run for tens of minutes to an hour or so.
    • C/2.5 Rate = Maximum surge current... Starting a well pump, few minutes on a microwave etc...
    We (I) use these numbers when planning out an off grid system with standard usage patterns. The idea is that if you do not exceed these numbers, then the system should support the desired loads under typical conditions (50%+ State of charge, nominal temperature range, over the life of the battery bank).

    Of course, you can run the batteries "harder" when new and fully charged--But most people need to have their systems work well over the years... So you want to design a system conservatively.

    If you have special needs, perhaps an AGM battery with higher discharge capabilities (running an large pump for relatively short times a day). The new Lithium chemistry type rechargeable batteries are starting to look very interesting--But it is on the "bleeding edge" of the technology right now.

    Some light reading:

    http://www.windsun.com/Batteries/Battery_FAQ.htm
    http://www.batteryfaq.org/
    http://batteryuniversity.com/

    And, if you are really interested about battery charging/discharging characteristics (such as the Peukert Factor), the Smart Gauge website has a lot more information:

    http://www.smartgauge.co.uk/technical1.html

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • KellezKellez Registered Users Posts: 21
    Re: My first project - Solar power computer....I need your help

    I started reading the thread from the beginning and i have one question

    For the system below

    140W panel at 18V
    12V battery

    which is the best way to calculate the Amperes?

    140W/18V = 7.7A

    or

    140W/12V = 11.6A


    I am really confused on when do we use the 12V of the battery or the 18V of the panel in the calculations.

    In this forum you guys use the 17.5Vmp of the panel when calculating the size of the controller but in a book i have seen they use the 12V of the battery
  • CariboocootCariboocoot Banned Posts: 17,615 ✭✭
    Re: My first project - Solar power computer....I need your help

    The Imp of a 140 Watt panel with a Vmp of 18 is 140/18 = 7.7 Amps.
    We use 17.5 Volts as a 'default' Vmp for "12 Volt" panels as it is a sort of generic number found on many panels in that range. Some are lower, some are higher.

    If you are using a PWM controller the current will be the Imp of the panel as calculated above. If you are using an MPPT controller the current formula is a bit different as it can down-convert higher panel Voltage into greater charge current. There is no exact way of telling this in advance, but it is approximately calculated as:

    Maximum current = (Watts * 0.77 typical efficiency) / nominal system Voltage.

    The use of nominal system Voltage for this is because that is approximately 50% SOC of the battery and where you are likely to need/see the maximum Voltage. Some people use Absorb Voltage but this is in my opinion incorrect because you will not necessarily need/see maximum current at that level (because the battery is already partially recharged). Likewise you can't expect it for load supply because that would involve knowing both the loads @ Battery Voltage and the array output at the same time; you may as well try to pick the lottery numbers. So I go with straightforward charging as an estimate of maximum current when it is most likely needed.
  • BB.BB. Super Moderators, Administrators Posts: 32,319 admin
    Re: My first project - Solar power computer....I need your help

    It depends... On the type of charge controller, the ambient temperatures, typical/best case/worse case numbers, etc...

    The details. the rating of a solar panel is given by Pmp=Vmp*Imp at STC (standard test conditions---Basically full noon time sun equivalent of 1,000 Watts/m2 and ~75F cell temperature). Note that Vmp actually falls as solar cell temperature increases. Upwards of 20% drop in Vmp on a hot day under full sun (~worst case 30C/54F temperature rise under full sun, little wind). So Pmp=Vmp*Imp*~0.80 on a hot day/roof.

    Next, based on the charge controller. A PWM charge controller is nothing more than an On/Off switch. On, the panel is charging the battery. Off, the panel is not charging the battery. A solar panel is a (mostly) a current source. It puts out ~Imp (current at rated sun) at typical battery voltages. So the energy going into the battery is:
    140W panel at 18V
    12V battery

    which is the best way to calculate the Amperes?

    140W/18V = 7.7A

    or

    140W/12V = 11.6A
    • Pbatt=Vbatt*Imp*some losses~ 14.5 volts charging * 7.7 Amps = 111.65 Watts into battery (near full battery)
    • Pbatt=Vbatt*Imp*some losses~ 12.0 volts charging * 7.7 Amps = 92.4 Watts into battery (near dead battery)
    Now, for a (more expensive) MPPT (maximum power point tracking) charge controller... These are constant power converters, they can take high voltage/low current from a solar array and efficiently (~95%) and down convert (think sort of like the DC equivalent of a variable transformer) to low voltage/high current needed to charge the battery bank.
    • Pmp panel = Vmp*Imp = 18 volts * 7.7 amps = 138.6 watts (round up to 140 watts)
    • Pbatt=Vmp*Imp*some losses~ 140 Watts panel * 0.81 hot panel losses * 0.95 controller eff = 108 Watts charging battery
    • 108 Watts / 14.5 volts battery charging = 7.44 Amps into battery (near fully charged battery)
    • 108 Watts / 12.0 volts battery charging = 9 Amps into battery (near dead battery)
    For different reasons, we end up using ~0.77 derating of panel "marketing wattage" as the available charging energy (PWM "matching losses", MPPT conversion losses, Hot panel losses, dust on panel losses, aging losses, battery voltage variations, etc.).

    An MPPT charge controller is a "switching power supply" (typically a Buck mode switch mode down converter).

    Buck converter

    There are lots of variations and "adjustments" we can make--But for a back of the envelople calculation, the way we calculate the overall charging energy is "close enough" for solar worth. Anything within ~+/-10% power calculations is pretty much "dead on" when all of the variables (weather, measurement, equipment selection, how the system is operated) that most people will get approximatly the numbers we use here over the life of the system.

    So, if you have a 140 Watt panel and want ~5% to 13% rate of charge into the battery bank:
    • 140 Watts * 0.77 panel+controller derating * 1/14.5 volts charging * 1/0.05 rate of charge = 149 AH battery minimum charging specification.
    • 140 Watts * 0.77 panel+controller derating * 1/14.5 volts charging * 1/0.10 rate of charge = 74 AH battery nominal charging current
    • 140 Watts * 0.77 panel+controller derating * 1/14.5 volts charging * 1/0.13 rate of charge = 57 AH battery maximum "cost effective" charging rate
    And based on a typical sunny day, minimum available sun for ~9 months of the year in much of the US:
    • 140 Watt panel * 0.52 end to end system efficiency * 4 hours minimum sun per day = 291 Watt*Hours per day of AC power
    Based on charging during the day and using the power during the evening/night (panel losses*controller losses*battery charging losses*AC inverter losses = ~0.52 end to end system efficiency).

    Yes, about 1/2 of the "marketing number" of the panel wattage actually makes it out to your AC appliance. Get an AGM battery, use DC load--You will get a bit better end to end efficiency.

    We use these numbers to give you a conservative output energy. I use 14.5 volts battery voltage--I think Marc/Cariboocoot uses 12.0 volts in the MPPT charging calculations.

    Mine is a bit more "pessimistic"--either is pretty close to what you will see in actual operation.

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • KellezKellez Registered Users Posts: 21
    Re: My first project - Solar power computer....I need your help

    Thank you guys for the great explanation

    BB. wrote: »
    One thing to watch with "12 volt car" adapters and battery clips...

    A typical "cigarette lighter" plug is good for around 10 amps... 10 amps * 12 volts = 120 Watts--Not the 300+ watts the inverter is rated for.

    Similar for battery clips. They are not really good enough to run your inverter at higher current levels... How high? You should be planning on:

    The inverter i suggested (link below) does not have an attached cigarette lighter plug but bolted connections

    http://www.amazon.co.uk/Inverter-inverter-notebook-emergency-MRI3013J2/dp/B008HOECP0/ref=sr_1_6?s=electronics&ie=UTF8&qid=1398783755&sr=1-6&keywords=bestek+inverter

    BB. wrote: »

    300 Watts * 1/0.85 inverter eff * 1/10.5 battery cutoff * 1.25 derating for wiring/fuses/breakers = 42 amps minimum (40 amp internal inverter fuse supplied)

    -Bill

    Can you please better explain the result of this calculation? and what is the purpose.

    i see that 42 amps > 40 amps of inverter, so what does this mean?
  • BB.BB. Super Moderators, Administrators Posts: 32,319 admin
    Re: My first project - Solar power computer....I need your help
    Kellez wrote: »

    It appears to come with two cables--One clips, the other a cig plug. As long as you hard wire it with heavy gauge cable--you should be fine.
    300 Watts * 1/0.85 inverter eff * 1/10.5 battery cutoff * 1.25 derating for wiring/fuses/breakers = 42 amps minimum (40 amp internal inverter fuse supplied)
    Can you please better explain the result of this calculation? and what is the purpose.

    i see that 42 amps > 40 amps of inverter, so what does this mean?

    Many times, we do not have the specifications of the AC inverters (paper design phase, sometimes the manual does not state the operational specifications). The above calculations are "close enough" that we can "ball park" the operational requirements without going into the specifications for a 1/2 dozen AC inverters.

    AC inverters have conversion losses (typically 85% efficient). And they are constant power devices. XYZ Watts out to AC load, the Pin=Vin*In -- Looking at the equation, the maximum input current to the inverter is going to happen at minimum input voltage (which causes maximum input current). The 1.25 derating is a United States building/electrical code derating for ciruit breakers and wiring (and how "we" rate breakers and fuses--Other countries do not need to follow these customs, and some specialty fuses/breakers are rated differently even in the US--We have "trip curves" that are used to design for specific applications--slow blow, fast blow, etc. protective devices)... I.e., the breakers and wiring need to be 1.25x larger than the maximum continuous current rating of the load (circuit breakers/fuses are supposed to "blow" at 100% of rated current).

    So, my generic calculations show that a 42 amp minimum branch circuit would support a 300 watt AC inverter running at 300 Watts AC loads. Typically, there are no 42 amp fuses, so the mfg. either rounds down to 40 amps or up to 50 amps--In this case, either is fine (close enough). And perhaps their inverter runs at better than 85% efficiency or not quite down to 10.5 volts input--Anyway--It was just a conformation that you need wiring+any fuses+battery connections capable of ~40 amps minimum for reliable operation (you don't want false trips on your fuses/breakers or overheated wiring+connections).

    Also, very few people will run a 300 watt AC inverter at exactly 300 watts for hour on end--So 40 amp internal fusing will be just fine. If you choose the run the inverter at >300 watts for minutes-hours--You will probably have the internal fuses trip and/or the inverter overheat.

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • CariboocootCariboocoot Banned Posts: 17,615 ✭✭
    Re: My first project - Solar power computer....I need your help
    Kellez wrote: »
    Can you please better explain the result of this calculation? and what is the purpose.

    i see that 42 amps > 40 amps of inverter, so what does this mean?

    300 Watts AC output of the inverter. Inverter is 85% efficient so that 300 Watts requires 353 Watts from the batteries. 10.5 Volts is the lowest Voltage the inverter will operate at (too low, in fact, to be reasonable). So 353 Watts / 10.5 Volts = 33.6 Amps from the batteries. (Note: this is omitting the inverter consumption which must be included). NEC regulations require 80% derating, so the current is multiplied by 1.25 to get the 42 Amps the circuit is design to handle. The fuse is 40 Amps, so it should pop before the wire fries.
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