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  • Working out the figures for a small scale solar system.

    Hello all, I

  • #2
    OK this one is pretty easy based on what info you have provided except your location.
    The brick (AC adapter) ratings are meaningless as is your laptop power input requirements All that tells you is the maximum power input/output the device is rated for. It has very little to do with how much energy your laptop used over a period of time. The only useful bit of data is the laptop minimum input requirements.

    What you need to know is how many watt hours the laptop uses over a given amount of TIME, and your location...

    I can tell you from personal experience a laptop uses about 30 watts average over a period of time. So you want to run the laptop 10 hours per day right? If correct your laptop will use roughly 30 watts x 10 hours = 300 watt hours. For calculation we need to adjust for system losses to determine the panel size. More than likely you will have a dual conversion from panels to a 12 volt battery. That will power a inverter to charge yet another battery in your laptop.

    So to account for all th eloses I assume only a 50% efficiency which means your solar panels need to generate 300 wh / 2 = 600 watt hours.

    So to determine how large of a solar panel wattage you need , we need to know your location, and your December Solar Insolation because that is the worse case for a battery system.

    I do not know your location, but the formula is simple. All you need to do is factor out the time element to find wattage. The formula is simple w = wh/h. We know wh = 600, now al we need to know is your winter insolation is in hours. Let's assume it is 3 for now. Then 600 wh / 3 h = 200 watts..

    As for the battery multiply daily adjusted watt hours by 5, then divide by the battery voltage. So [5 x 600 wh] / 12 volts = 250 Amp Hours @ 12 volts.

    Charge controller minimum amperage needed is panel wattage divided by the battery voltage. So in this example 200 w / 12 v = 17 amps. So you would need a 20 amp charge controller in this example.
    MSEE, PE

    Comment


    • #3
      To have usable amounts of power required for equipment you really need to measure power over something like 24 hours or longer. In many cases a Kill A Watt device or a simple electric meter with leads on it can supply that.

      Gets a bit more difficult with anything not plugged in but is still possible.

      The numbers on the back of the case are the max ratings (as Sunking pointed out) and would lead you to grossly oversize a system.
      [SIGPIC][/SIGPIC]

      Comment


      • #4
        Originally posted by Sunking View Post
        OK this one is pretty easy based on what info you have provided except your location.
        The brick (AC adapter) ratings are meaningless as is your laptop power input requirements All that tells you is the maximum power input/output the device is rated for. It has very little to do with how much energy your laptop used over a period of time. The only useful bit of data is the laptop minimum input requirements.

        What you need to know is how many watt hours the laptop uses over a given amount of TIME, and your location...

        I can tell you from personal experience a laptop uses about 30 watts average over a period of time. So you want to run the laptop 10 hours per day right? If correct your laptop will use roughly 30 watts x 10 hours = 300 watt hours. For calculation we need to adjust for system losses to determine the panel size. More than likely you will have a dual conversion from panels to a 12 volt battery. That will power a inverter to charge yet another battery in your laptop.

        So to account for all th eloses I assume only a 50% efficiency which means your solar panels need to generate 300 wh / 2 = 600 watt hours.

        So to determine how large of a solar panel wattage you need , we need to know your location, and your December Solar Insolation because that is the worse case for a battery system.

        I do not know your location, but the formula is simple. All you need to do is factor out the time element to find wattage. The formula is simple w = wh/h. We know wh = 600, now al we need to know is your winter insolation is in hours. Let's assume it is 3 for now. Then 600 wh / 3 h = 200 watts..

        As for the battery multiply daily adjusted watt hours by 5, then divide by the battery voltage. So [5 x 600 wh] / 12 volts = 250 Amp Hours @ 12 volts.

        Charge controller minimum amperage needed is panel wattage divided by the battery voltage. So in this example 200 w / 12 v = 17 amps. So you would need a 20 amp charge controller in this example.
        Thank you for the reply, being a newbie, the info you’ve given me has lead me to ask as many question as you’ve answered.

        Taking that the power ratings on my laptop and laptop PSU are maximums, I understand that I will have to use a power consumption meter over a 24Hr period to get a true estimate of my laptop power use. It is a large and hefty 17” laptop, or it was five years ago.

        Until I can work out the actual usage, just to clarify how it’s calculated I’ll use your recommended estimate of 30W AC.

        240V x 0.125A = 30W AC

        For example I’d use the following inverter, JayCar 380W 12V to 240V pure sine wave inerter.
        http://www.jaycar.co.nz/productView....e&form=KEYWORD

        So now...

        240V x 0.125A = 30W AC
        ...becomes...
        12V x 2.5A = 30W DC
        30W x 10Hr = 300Wh

        I’m with you so far. I don’t understand what you mean when you say “More than likely you will have a dual conversion from panels to a 12 volt battery. That will power a inverter to charge yet another battery in your laptop” Lets say I’m not using the laptops battery just to remove it from the equation for the meanwhile.

        I’d need to generate 300Wh a day to use 30W for 10Hr a day @ 100% efficiency.
        With an overall system efficiency @ 50%, I’d need a set up that could generate 600Wh @100% efficiency to get the needed 300Wh.

        Still with you, but my first question is how did you estimate an efficiency figure to calculate with. I’m sure your experience in the field more than accounts for accuracy but as I’m starting out, could you explain how efficiency is worked out?

        Solar insolation for Wellington, New Zealand is as follows:
        http://www.apricus.com/html/insolation_levels_asiap.htm

        Country - NZ
        City - Wellington
        Latitude - 41' 17" S
        Longitude - 174' 47" E
        Jan - 6.27
        Feb - 5.31
        Mar - 4.17
        Apr - 3
        May - 1.95
        Jun - 1.54
        Jul - 1.74
        Aug - 2.46
        Sep - 3.66
        Oct - 4.7
        Nov -5.73
        Dec - 6.01
        Year Avg - 3.88

        Our midwinter insolation in June is 1.54kWh/m2/day with an annual average of 3.88kWh/m2/day.

        So my ideal panel wattage in mid winter would be 600Wh x 1.54kWh/m2/day = 390W panel. By this I understand I would need at least 3 x 130W panels to generate 600Wh on a daily basis to run a 30W laptop for 10H a day, on a system running @ 50% efficiency.

        ...then a panel suited to our annual average would be 600Wh x 3.88kWh/m2/day = 155W panel generating 600Wh a day? Would this be a boom and bust form of generation? Fine midsummer, virtually nothing in midwinter?

        I’m hanging in there just, but you’ve lost me when it comes to battery size. I’m not sure I understand what you mean by multiply daily adjusted watt hours by 5. What do you mean by “daily adjusted” and why by 5?

        Also (5 x 600Wh) / 12V = 250Ah @ 12V, then 250Ah x 12V = 3000Wh. So I would only be using 10% of my battery's capacity on a daily basis?

        Does this also mean it would take the midwinter ideal size panel of 390W, 15.4H to charge it completely?
        (390W / 3000Wh) * 2 = 15.4H @ 50% efficiency?

        Would then recharging the 300Wh use by the laptop take the 390W panel only 2.6H?
        (390W / 300Wh) * 2 = 2.6H @ 50% efficiency?

        Again, many thanks for your assistance.

        PS, this site might provide any missing info on daylight hours in Wellington, NZ.
        http://www.gaisma.com/en/location/wellington.html
        http://www.apricus.com/html/insolation_levels_asiap.htm

        Comment


        • #5
          [QUOTE=sbonham;13438]I
          MSEE, PE

          Comment


          • #6
            I brought a power consumption meter today to test out my laptops usage.

            Fairly surprising...

            With the battery attached it draws a peak wattage of nearly the full 90W, but runs at about 50W+ otherwise.

            Removing the battery, which to be honest, I rarely use, the peak wattage dropped to 65W and it ran at about 35W to 55W otherwise.

            One things for sure, the laptops battery is going back in it's case from now on.

            Comment


            • #7
              The Li ion battery from what I read - for the longest life - you do not want to keep fully charged or allow to fully discharge but to allow it to float in between maybe 70 and 100%.

              So maybe take it out of the case weekly for a day each week or 10 days.
              [SIGPIC][/SIGPIC]

              Comment


              • #8
                I do not advise giving up on the laptop battery just yet. I am not sure how long you ran the test, but I suspect not near long enough to have useful data.

                I would bet when you ran the test the battery was in a discharged state and you never allowed it to charge up at which point all the power going to it would have went to 0.

                Your test should be done over the course of several days measuring watt hours consumed. So for example lets say you run the laptop for 120 hours, and in that 120 hours it consumes 4800 watt hours or 4.8 Kwh you would know the laptop averages 40 watts and consumes 400 watt hours in 10 hours which is your goal.

                So my advice is run a long term test to get accurate data, it will pay off being patient, and could save you quite a bit of money.
                MSEE, PE

                Comment


                • #9
                  OK, I think I’m starting to understand the basics of solar panels now. I’ve been working it out and with a lot of help from good old Excel, I think I’m nearly there.

                  So, one laptop with a PSU rated at 240V x 1.5A = 360W AC max, though the laptop actually uses more like approx. 40W.

                  40W / 240V = 0.16A AC
                  40W / 12V = 3.33A DC

                  40W for a period of 10H a day gives a Watt Hour requirement of 400Wh then factoring the efficiency of the solar system @ 50% gives a 800Wh daily requirement.

                  40W x 10H = 400Wh * 2 = 800Wh

                  Wellington’s insolation is quite poor in our midwinter. Insolation is a bit tricky for a newbie to grasp but as I understand it if our June insolation is 1.54kWh/m2/day, then no matter if the day has 4, 8, or 12 hours of daylight it’s still only the equivalent of approx. 1.54 hours of direct full sun on the panel in June. However our annual average is a bit better at 3.88kWh/m2/day or 3.88 hours.

                  800Wh / 1.54H = 519W Panel = 1 day in midwinter.
                  800Wh / 3.88H = 206W Panel = 1 day over annual average.

                  So a 519W panel would take one day to generate 800Wh in midwinter (June@1.54kWh/m2/day) and similarly a panel suited to our annual average, 206W would take 2.52 days in midwinter to generate the same 800Wh.

                  206W * 1.54H = 317Wh (483Wh short of required 800Wh @ midwinter.)
                  519W * 3.88H = 2013Wh (1213Wh over required 800Wh over annual average.)

                  A battery suitable for a daily use of 400Wh works out at 165Ah. I figured out that the multiply by 5 was to allow for the daily watt hour usage to equate to about 20% of the batteries total capacity.

                  400Wh / 12V = 33Ah * 5 = 165Ah
                  165Ah * 12V = 2000Wh approx.

                  So 400Wh from a 33Ah battery would empty it daily and would need to be recharged 100% daily.
                  Making the daily requirement of 400Wh only 20% of the batteries capacity gives room to manoeuvre with regards to charging and actual daily usage.

                  40W x 10H = 400Wh * 2 = 800Wh
                  519W * 1.54kWh/m2/day = 800Wh
                  400Wh / 12V =33Ah
                  33Ah * 12V = 400Wh approx.

                  400Wh * 5 = 2000Wh / 12V = 165Ah approx.

                  A charge controller that could handle the Amps from the panel would be:

                  206W / 12V = 17.17A (20A controller, annual average insolation.)
                  519W / 12V = 43.25A (45A+ controller, midwinter insolation.)

                  So to wrap it up, using a laptop drawing 40W for 10 hours, 400Wh a day would ideally need a solar panel of at least 24V and 519W to generate 800Wh over midwinter, a charge controller that could handle 45A+ and a battery of at least 165Ah @ 12V. This system would replace the daily used watt hours over a day even in midwinter.

                  Equally, a laptop using 40W for 10H, 400Wh a day with a solar panel suited to our annual average insolation would need to be at least 24 V and 206W to generate 800Wh a day, a charge controller that could handle 20A and again a battery of at least 165Ah @ 12V. This system would replace the daily used watt hours over a day for 6 months of the year but would need nearly three days over midwinter.

                  Now the big question...

                  Have I worked this out right?

                  If I have, then I estimate a midwinter compliant system for Wellington would cost as follows:

                  519W Solar panel = 3 * 175W PowerTech Monocrystalline panels = $3417 NZD (14.5KG each)
                  1 * 45A Morning Star MPPT Charge Controller = $699 NZD = (4.2KG)
                  1 * 12V 200Ah AGM AA Champion Battery = $599 NZD (61KG!)
                  1 * 180W 12V PowerTech Pure Sine Wave Inverter = $239 NZD = (1KG)

                  I’m uncertain how much to allow for wiring and other bits that would be needed.

                  Total $4954NZD and weighing in at 110KG!

                  I currently pay 25c a kWh so at the above system cost I would have to use my laptop over 19,816kWh before I broken even off the grid.

                  $4954 / $0.25 = 19,816kWh
                  19,816kWh / 400Wh = 49,540 Days or 136 years!

                  Wow!

                  Comment


                  • #10
                    Originally posted by sbonham View Post
                    So, one laptop with a PSU rated at 240V x 1.5A = 360W AC max, though the laptop actually uses more like approx. 40W.

                    40W / 240V = 0.16A AC
                    40W / 12V = 3.33A DC
                    OK this information is not of any real use except for sizing the inverter to run the Laptop Brick. It tells you will need at least a 360 watt inverter. 500 watt is the mark.

                    Originally posted by sbonham View Post
                    40W x 10H = 400Wh * 2 = 800Wh
                    Correct if you are going the dual conversion method

                    Originally posted by sbonham View Post
                    Wellington’s insolation is quite poor in our midwinter. Insolation is a bit tricky for a newbie to grasp but as I understand it if our June insolation is 1.54kWh/m2/day, then no matter if the day has 4, 8, or 12 hours of daylight it’s still only the equivalent of approx. 1.54 hours of direct full sun on the panel in June. However our annual average is a bit better at 3.88kWh/m2/day or 3.88 hours.
                    This fact escapes most newbies. Newbs assume a 40 watt panel generates 40 watts as ling as the sun is shinning. Unfortunately that is far from the truth. A panel will never generate its rated power. The rating is a lab rating using a special high output lamp, at 25 degree C, 0% humidity. with the lamp directly facing into the panel You will never experience those conditions except maybe on a freezing cold December 21 at 12 noon for a few brief minutes down under. When the sun rises the panel will only generate a few watts. as the sun gets higher, and more direct angle to the panel the wattage slowly increases to a peak at solar noon, then the process reverses itself until dusk.

                    So on a June day if you have a 100 watt panel, at the end of the day in June the panel generates 100 watts x 1.54 hours = 154 watt hours at the terminals.

                    For a battery system you design for worse case, and for you that is June If you designed a battery system for the yearly average of summer months, You would be dark and in trouble for the whole winter. FWIW Grid Tied systems are designed for your yearly average.

                    Originally posted by sbonham View Post
                    800Wh / 1.54H = 519W Panel = 1 day in midwinter.
                    800Wh / 3.88H = 206W Panel = 1 day over annual average.
                    You got it, if you design for 3.88 hours you are doomed

                    Originally posted by sbonham View Post
                    A battery suitable for a daily use of 400Wh works out at 165Ah. I figured out that the multiply by 5 was to allow for the daily watt hour usage to equate to about 20% of the batteries total capacity.

                    400Wh / 12V = 33Ah * 5 = 165Ah
                    165Ah * 12V = 2000Wh approx.

                    So 400Wh from a 33Ah battery would empty it daily and would need to be recharged 100% daily.
                    Making the daily requirement of 400Wh only 20% of the batteries capacity gives room to manoeuvre with regards to charging and actual daily usage.

                    40W x 10H = 400Wh * 2 = 800Wh
                    519W * 1.54kWh/m2/day = 800Wh
                    400Wh / 12V =33Ah
                    33Ah * 12V = 400Wh approx.

                    400Wh * 5 = 2000Wh / 12V = 165Ah approx.
                    If you discharge your battery 100% each day, you will be replacing that battery every couple of months. There is an error in your calculation which I did not address. To calculate the battery AH capacity assuming daily 20%, you have to adjust the daily watt hours by multiplying by 1.5 first. So 400 wh x 1.5 = 600 wh. Now you can calculate the battery by multiplying the adjusted number by 5 and dividing by the battery voltage. AH = [WH x 5] / Battery Voltage = [600 wh x 5] / 12 v = 250 AH

                    Originally posted by sbonham View Post
                    A charge controller that could handle the Amps from the panel would be:

                    206W / 12V = 17.17A (20A controller, annual average insolation.)
                    519W / 12V = 43.25A (45A+ controller, midwinter insolation.)
                    Again something we have not covered. The answer depends on which type of charge controller you use being either an inexpensive Shunt/PWM or MPPT.

                    If using a shunt type the input current = output current. Knowing this the input current will be the maximum current the solar panels will supply. So lets say you have a 500 watt rated at 18 volt VMP @ 27 amp. (500 w / 18) Sou you would shop for a 30 amp controller.

                    For an MPPT type controller the formula is panel wattage / battery voltage So 500 watts / 12 volts = 42 amps. So you would look for a 50 amp MPPT controller.

                    Take note here, MPPT controllers are fairly expensive, so you may be inclined to pinch money and go with a Shunt controller. However you will get snake bit doing that. Remember I said for a Shunt type input current = output current? Well here is the snake bite. So you have a 500 watt panel. It outputs 18 volts @ 27.7 amps (500 watts) into the charge controller. On the output of the charge controller is 12 volts @ 27.7 amps. What is wrong with this picture? Do the math and see if you spot the problem? 500 watts into the controller, 333 watts output. You just lost 33% or 167 watts of your power. You now need another panel to make up for the losses.

                    A good quality MPPT controller is 95% efficient. 500 watts in, 475 watts output. With a MPPT controller input current is not equal to output current.

                    Originally posted by sbonham View Post
                    Have I worked this out right?

                    519W Solar panel = 3 * 175W PowerTech Monocrystalline panels = $3417 NZD (14.5KG each)
                    1 * 45A Morning Star MPPT Charge Controller = $699 NZD = (4.2KG)
                    1 * 12V 200Ah AGM AA Champion Battery = $599 NZD (61KG!)
                    1 * 180W 12V PowerTech Pure Sine Wave Inverter = $239 NZD = (1KG)

                    I’m uncertain how much to allow for wiring and other bits that would be needed.

                    Total $4954NZD and weighing in at 110KG!

                    I currently pay 25c a kWh so at the above system cost I would have to use my laptop over 19,816kWh before I broken even off the grid.

                    $4954 / $0.25 = 19,816kWh
                    19,816kWh / 400Wh = 49,540 Days or 136 years!
                    You are very close.

                    You need a 500 watt inverter. Allow $400 for materials.

                    Now here is the big thing you missed. You are assuming the battery last forever. Wrong, about 5 years and it will need replaced. So the break even point is never. With a battery system you voluntarily elected to pay 10 to 15 times more for electricity for the rest of your life. If this was a business, your competitors would eat your lunch and force you out of business. You would never be able to compete.

                    Well I think now maybe you understand why I always say never ever use solar battery systems if you have commercial power available to you. You will get taken to the cleaners.
                    MSEE, PE

                    Comment


                    • #11
                      Originally posted by Sunking View Post
                      Correct if you are going the dual conversion method.
                      With the laptop battery removed, so the laptop is only pulling power for itself and not topping up it's battery, what would the systems efficiency estimate as?

                      60%?

                      Comment


                      • #12
                        So for a laptop using 40W for 10H a day I would need,

                        Laptop PSU requires max:
                        240V * 1.5A = 360W

                        Laptop requires:
                        240V * 0.17A = 40W AC
                        12V * 3.33A = 40W DC

                        To run the laptop 10 hours a day requires:
                        40W * 10H = 400Wh

                        Allowing for 60% system efficiency, a solar panel that can generate 667Wh a day is require:
                        400Wh / 60 * 100 (60%) = 667Wh

                        Allowing for midwinter insolation, a panel that can generate the required Wh would be:
                        667Wh / 1. 54H (kWh/m2/day) = 434W

                        A battery to support a device requiring 400Wh would need to be:
                        400Wh * 1.5 = 600Wh * 5 /12V = 250Ah

                        A device using 400Wh of a 250Ah battery is using approx. 13% of the battery’s capacity a day, while a 434W solar panel @ midwinter insolation would be recharging 22% of the battery’s capcity.

                        An MPPT charge controller to support the solar panel would need to be rated about 40A.
                        434W / 12V = 36.16A (40A)

                        The pure sine wave inverter required for the laptop PSU would need to be a minimum of 360W

                        This system would allow me to run a 40W laptop for 10 hours a day off the grid, taking a day to recharge the battery in our midwinter and only 0.25 of a day in our midsummer.

                        So the recosting would be:

                        1 * 434W solar panel = 3 * 180W 24V (540W) AA SolarTech panels = $2397 NZD (15.5KG each)
                        1 * MPPT Solar Controller = 1 * Morning Star Tristar MPPT 45A = $699 NZD (1.5KG)
                        1 * 12V 250Ah battery = 1 * AGM AA Champion Deep Cycle = $759 NZD (70KG!)
                        1 * Pure Sine Wave Inverter = 1 * PowerTech 380W (650W Surge) Inverter = $325 NZD (1.1KG)

                        Not including materials.

                        Total $4180 and weighing in at 120KG!

                        I currently pay 25c a kWh so at the above system cost I would have to use my laptop over 16,720kWh before I broken even off the grid.

                        16,720kWh / 400Wh = 41,800 Days or 115 years!

                        Still... wow!

                        It's clear efficiency is the key and the draw back to solar technology. If taking a laptop off the grid is this prohibitively expense, I wonder how much a system to charge a mobile phone would cost?

                        Comment


                        • #13
                          Charging laptop

                          We had discussed long time ago, charging laptop using just a solar electric panel connected to a car adaptor for the laptop.
                          Hopefully it can be found with searching back.

                          Comment


                          • #14
                            Originally posted by sbonham View Post
                            It's clear efficiency is the key and the draw back to solar technology. If taking a laptop off the grid is this prohibitively expense, I wonder how much a system to charge a mobile phone would cost?
                            No not really if it were 100% you are are stillo looking at no ROI, you have to replace the battery every 5 years. Going green means wrecking your economy and going broke.

                            FWIW I pay $0.096/Kwh about 1/3 what you do. I would never ever consider solar for my own use.
                            MSEE, PE

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