hanaparts.com ## System Capacity and Calculating Storage of DC current.

 The storage capacity of your system and the generator output is a critical factor to how long your batteries will last. A balance has to be maintained for optimum performance and longevity of any DC storage system.     Charging the battery or batteries has to be calculated to make up for the usage in time or the system will become problematic and fail. A simple example is a car or truck battery that goes dead because the driver starts it up and drives just 5 minutes to work and shuts it off and then drives 5 minutes home and parks the vehicle. If that person repeats that daily his or her battery will fail. Not because the alternator or generator failed but because it took more DC current to start the vehicle than the charging system could make up in the time it took before shutting the engine off. So there has to be an equal exchange or the system will fail.     Here we will begin with a simple 6 cell 12 Vdc battery that has the optimum charge @ 13.2 Vdc, our charged battery has the surface charge removed and drained and is maintaining 12.8 Vdc for 10 hours with no load applied (current drawn).   NOTE: As a battery is charging it maintains a high static charge from the compression of electrons inside and the higher voltage charging rate. A regular 12 Vdc light bulb (1156) should be connected to the battery and turned on and ran for a minimum of 15 seconds and shut off. The battery voltage becomes nominal after a period of 1 minute of small current draw, but you don't need to test your system like that unless it is failing to produce current for the expected period in time. Electrical current (Watts) is calculated by the hour i.e. a light using 100W for 1 hour uses 100 X 1 = 100 Watts. Calculating current draw is not the same as calculating consumed watts when calculating usage, wattage is a total based on the hour (consumed) and the current is based by the second (current Amps). The Amps (per hour) is the factor that determines the size of your charging system. The total consumed watts is simply how long in hours did the generator have to run at the calculated Amp Capacity. That is what costs more fuel, less energy production in kWh (kilo Watt hours) and maintenance in the power generating fields ending up as inefficiency in the system.
 Calculating current draw in Amps to size your generator or alternator is made by figuring out what electrical things you are going to use at one time together. For example 2 light bulbs rated @ 100 Watts each (per hour) added to the fridge @ 200 Watts added to electric heater @ 2500 Watts. Added together 2900 Watts per hour or 2900 Watts divided by system voltage 120 Vac would equal 24.17 Amps. So the generator now can be sized to charge a battery or bank of batteries at it's optimum performance. Optimum performance relates to producing more for less. Running any equipment or machinery out of it's optimum rating equals more ongoing costs to the operators of them. In the machine design field we know that horsepower is written in foot pounds, and that one horsepower is required to generate 746 Watts (electrical horsepower is called true horsepower and is a direct conversion factor). 746 Watts running @ 120 Vac would equal 6.21 Amps and @ 12 Vdc would equal volts times Amps or 12 X 6.21 = 74.52 watts. So the Watt is the factor here for conversion in Vdc into Vac and inverse Vac to Vdc. When you size your generating system and storage needs the horsepower is needed primarily to set a parameter. Knowing that we need 2900 Watts at one time or (2900 divided by 746) a 3.89 horsepower engine. Manufacturers don't make that size of small engine, there are 5 horse engines which are slightly to large by 1.11 horsepower  We will have to purchase a generator with a larger horsepower by 20% (that percentage is a general calculation by machine manufacturers). We would then need an engine that had 4.67 horsepower so the 5 horse engine would be the optimum choice.     When calculating system size you have to go over 20% because no system lives very long at 100% output. So our system requires 4.67 horsepower and at this point you know we need a generator with a horsepower greater than 4.67 horsepower but not larger than the next size available or you will use more fuel and endure more maintenance cost. Sizing in a percentage is important to good design. A good design will work as anticipated for more than 90% of it's design life with no additional costs for the consumer. A slap up sloppy design will always cost in anguish from not performing, and deliver intermittent clean electricity or charging capabilities. There are many out there putting up and installing PV panels and chargers with little electrical engineering knowledge, companies want to sell things and will put them in service when the customer keeps on paying, like a yo-yo system up for a few weeks then down for a few weeks. Just because written on their truck or business front the words solar does not mean they have adequate experience to design a system that works well.       Capacity calculations     Calculating storage capacity for any type of equipment or battery bank system is the most important and first step for producing a well designed system.     For example: There are two types of electrical devices, IR (current resistance) like a conventional light bulb or space heater or  induction devices which are like motors, fans, air conditioning and anything else operating on electrical current that produces a magnetic field to make useable power. An induction device rarely operates continuously it is intermittent and has a duty cycle written in time or percentage. A 50% duty cycle motor can only operate by shutting off an equal amount of time than it has been operating in time. So the motor is designed to operate at it's load capacity for a period that will not overheat the motor windings. A motor has wires with fragile insulation paint on them and once this melts the motor shorts out and burns up the windings inside the motor. Most all electric motors have a built in thermal switch that shut's off the electrical current to the windings after the circuit reaches a set upper limit.     Now you count all of your IR devices and add up all the current requirements. In this example we will calculate 4 100Watt light bulbs 400W total if all four are turned on at the same time. Then add the number of hours per day (24 hours) those lights could possibly run and then add in your 20% over. Four lights operating 4 hours per day would require watts per hour X 1 = 400 watts per hour divided by 1000 = .4 KWhr (kilowatt hour).     It is really difficult for me as a mechanic to understand how the companies rated the total watts in the above formula. I looked that up in a watt calculator program on the net and the odd thing is that a light bulb requiring a 1 amp load is using one amp each second it is lit up by electrical current. So the formula should be 1 Amp per second or 60 total Amps for one minute. At 12 volts that would be 12 watts per second or 720 total watts per minute or X 60 = 43200 total watts per hour and that is 43.2 KW/hr. (43.2 kilowatts per hour). It is not calculated like that for some reason when AC and watts are calculated.     So household appliances using 120 Vac or 240 Vac have a rated wattage. A conventional refrigerator is rated @200 watts over a period of an hour so 24 hours of usage would equal 24 X 200 = 4,800 total watts or 4.8 Kilowatt hours. If the cost of electrical current is 48 cents per kilowatt hour (Hana average) then that fridge will cost  \$2.30 per day or \$69.12 per month.     So in simple if we only want to run a refrigerator for one day without charging a battery we need a capacity that will run the load as well maintain operating voltage for proper operation of the fridge. A DC battery is rated in Amps for capacity and every time you invert meaning DC to AC there is a small power loss factor in watts total and when you use a converter it takes AC and makes it DC that also takes a small amount of total energy away.     At this point we need a battery capacity of 200 watts per hour, a 100 Amp hour battery will produce 12 Vdc X 10 = 120 watts per hour for 10 hours with a volt drop limit to 9 Vdc. Charging a 100 Amp hour lead acid battery will require 10 Amps of charge for 10 hours and will give the battery a theoretical 100% charge with balanced electrolyte. Balanced electrolyte is one fundamental target goal for your DC storage systems. To keep a system balanced you never want to exceed the Amp hour charging foundation 1/10th the battery Amp hour rating is optimum. A 12 Vdc battery that has 200 Amp hour rating is supposed to last 10 hours @ 20 Amp current draw. Important is the lower limit of discharge which would be 9 Vdc lower limit or damage from excessive system heating could result. So in percentages is how we figure the good design, a 12 Vdc battery has the potential of 13.2 Vdc so we take minimum @ 9 and divide it my optimum calculated and get 32%. Each battery in series adds to voltage and each battery in parallel adds capacity to the system.     In watts consumed if something is rated at 200 watts with a 120 Vac potential then at 240 Vac that device will draw 400 watts (almost instantly burn out). When designing your system the conventionally used is 12, 24, 48 Vdc (Volts dc). A 12Vdc system has 6 cells in series, a 24 Vdc system has 12 cells in series, a 48 Vdc system has 24 cells in series. Using batteries in series only adds voltage to the system. The 48 Vdc systems have an optimum of 24 X 2.2 = 52.8 Vdc and a minimum lower limit of 52.8 X .32 = 16.9 - 52.8 - 16.9 = 35.90 Vdc. A one cell battery @2.2 Vdc would have a lower limit of 1.5 Vdc. This number can be applied to all the DC voltage storage systems using lead acid batteries. Storage capacity of your system should be calculated based on these foundational numbers. The difference between the 35.90 Vdc calculation and the single 2.2 Vdc cell @ 36 Vdc is only .1 Vdc apart.      Working now with the conventional 48 Vdc storage systems as example, needs a 24 hour period between full charges. To run simply the refrigerator requires the 4,800 watts or 4.8 kilowatt hours. Amps times volts are watts, a 100 Amp hour capacity battery has the potential of 480 watts continuous in a good designed system. A motorcycle 12 volt battery is around 12 Amp hours, a conventional 4 cylinder automobile engine has about 85 Amp hour and a large heavy equipment battery around 200 Amp hours capacity.     For example now we are making a 48Vdc system knowing that the AC load capabilities are for a refrigerator 24 hours per day, 4 light bulbs 3 hours per day, a toaster oven running 4 hours a day, a space heater running 2 hours per day. Fridge 200 X 24 = 4,800 watts or 4.8 kWh per 24 hour period. Light bulbs 100 X 4 = 400 watts X 3 or 1.2 kWh per day, toaster oven 1000 X 4 = 4 kWh per day and a space heater 2500 X 2 = 5.0 kWh per day. Total consumption per day would total 15 kWh per day. The battery storage system then would require 15000 / 120 = 125 Amps The delivery rate on the lead acid battery should not be more than 1/10 of it's rated capacity in Amp hours. So if you purchased 200 Amp hour batteries to keep this system working for a 24 hour period would require 4 12 volt 200 Amp batteries in series to achieve the 48 Vdc potential system voltage (52.8 optimum) then those four in series would total a 48 volt DC system and 200 Amp hours capability or 20 Amps per hour delivery potential (well designed system). The rest is simple we need 105 more Amps capability @ 48 Vdc or 12 more 200 Amp hour 12 Vdc batteries for a bank where there are 16 total 12 Vdc 200 Amp hour batteries sets of 4 batteries in series and three more sets of four more sets paralleled together. This would require a minimum of two banks of 200 Amp hour batteries with 16 each bank or a total of 32 200 Amp Batteries to produce current for one day and based on the above scenario.     To make your system smaller or larger capacity makes no difference the facts are the same. Batteries are rated in Amp hours because they are a specific voltage 6, 12 and a 100 Amp hour battery can safely deliver 10 Amps for 10 hours and then need re charged @ 10 Amps for 10 hours to be fully charged.     NOTE: The optimum charge going into the battery cell from a charger is 1/10th of it's Amp hour rating for 10 hours. On a 12 volt battery a person needs to charge it after it's running voltage drops below 9 Vdc. The running voltage is the voltage meter reading as the loads are being operated. A good battery will recover voltage quicker than an unbalanced battery having cells that have lost a percentage of their storage capacity. A faster or slower charge rate is ok but not the recommended way for optimum battery life. The rate at which the battery is taking a charge is directly and proportionally related to it's cell temperature. A battery temperature of 115 degrees Farenheight is the upper charge limit. It does not matter what the charger indicates for Amps charge the temperature is the important factor. Softening the cell metal begins at around 120 degrees f and will affect the cells ability to hold a charge.
 The formulation for discharging should be equal to the charging rate and heat results in the battery cell if to much load is drawn to quickly exactly like the heat results from charging the battery to quickly. So when figuring the total capacity make sure the battery capacity is calculated for 20 hours minimum. For example a 100 Amp hour battery delivers current at the rated voltage and a minimum is set. The rating is in 1/10th the amp hour rating so for a 20 hour storage system that figure needs to be divided in half. Optimum current draw for a 100 Amp battery in a 20 hour storage system is 5 Amps @ system voltage or in the 12 volt system 5 X 12 = 60 watts.     Again we make these calculations on the most minimum system powering a simple refrigerator for 24 hours per day. Fridge draws 200 watts for 24 hours = 4,800 watts total or 4.8 kWh (kilowatt hour). The system wants to be designed 20% larger so minimum storage would require 4,800 X 1.2 (original number + 20%) would require storage of 5760 total watts. Each 10 Amp hour battery in a 20 hour system has the potential of 60 watts so the figures would be 5760 / 60 = 96 or /2 = 48 batteries 12 Vdc @ 200 Amp hours cabled up in parallel. This is a whole lot of expense when you figure battery cost versus wear on the batteries. No one likes to accept this fact so almost all storage systems become short lived and have less capacity stored than is required to run the calculated loads for a 24 hour period without charging.     The bottom line is that at the current costs here for AC power @ approximately 50 cents per kilowatt hour that fridge would cost around 2.50 USD per 24 hour period. 48 batteries 12 Vdc 200 Amp hours will cost around \$7200.00 and then a charging system is needed along with that. The initial costs are very high to set up a reliable 20 hour storage system for electrical consumption needs. You can get by with much less but when the charge is not available your system run time can be much less. Remember we are designing a good storage system that will last much longer than one that has less than calculated storage capabilities.