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 Author Topic:   Calculating Battery Capacity in Ampere-Hours brainstormer posted 09-21-2010 04:48 PM ET (US)         How do you [calculate] how much battery is used up in X hours of use? [A particular] machine uses 12-Volts at 4.2-amperes, which it presently gets from a transformer from 120-VAC . A good sized battery would stand how many hours of 4.2-ampere use? Hoosier posted 09-21-2010 05:12 PM ET (US)             Divide the battery capacity in ampere-hours by 4.2, then divide by 2. You don't want to use more than 50% of your capacity. If that doesn't get you where you want to go then hook up a second, identical, battery in parallel so you double the amp-hour capacity, but keep the system at 12 V. For this application I'd get good deep cycle batteries. Chuck Tribolet posted 09-21-2010 05:31 PM ET (US)             There are three (at least) types of inverters:Square wave. This is either full positive, then full negative, the full positive. Never zero except as the signal swings by zero. The area under the curve is greater thana true sine wave. These are unfriendly to electronics, butfortunately very rare today.Modified sine wave: This is full positive, then zero for a little bit, then full negative, then zero for a little bit,then full positive. The area under the curve is the same asa true sine wave. These are friendly to most electronics, and the most common today.True sine wave. This is a graceful curve from full positiveto full negative and back to full positive. It is friendlyto electronics. It is becoming more common as it gets incorporated into chipsets.Chuck AZdave posted 09-21-2010 06:02 PM ET (US)             I'm probably displaying my ignorance, but could you simply make up a line to run the machine off your existing 12-Volt battery? That would save losses in the inverter and transformer. jimh posted 09-21-2010 08:22 PM ET (US)             You can run a 12-volt DC device from a 12-volt battery. This is much more efficient than using a battery to create 120-VAC via an inverter, then running an AC-operated power supply to make 12-volt DC. To do that would be very inefficient.The capacity of a battery is specified in its ampere-hour rating. The ampere-hour rating is specified at a particular discharge rate. Generally the smaller the discharge rate the longer the battery will last. This looks like a good article on the topic: http://www.windpowerunlimited.com/batteries/Amp_Hours.htm jimh posted 09-21-2010 08:48 PM ET (US)             An example:A battery is rated at 42-ampere-hours (Ah). How many hours can it run a device that draws 4.2-amperes?A battery of 42-Ah could theoretically supply 4.2-amperes for 10-hours. In actual practice, the performance would depend on other factors, including the temperature.The 42-Ah rating assumes the discharge rate is for 20-hours, typically. Therefore, to reach the 42-Ah battery rating, we would have to only draw enough current to force a 20-hour discharge. Such a current would therefore be42-ampere-hours / 20hours = 2.1-amperesSince the load is double the rating, we would find that the battery would not provide its rated 42-Ah performance. A load of 4.2-amperes will discharge a 42-Ah battery in less than 10-hours. Exactly how much less would be determined from a curve provided by the battery manufacturer. jimh posted 09-21-2010 08:54 PM ET (US)             If asking about battery capacity while the battery is being charged by an outboard motor, most outboard motors have charging current capacities greater than 4.2-amperes. Thus you could run a machine that consumed 4.2-amperes from a 12-volt battery as long as you wanted while the outboard motor was running and charging the battery. David Pendleton posted 09-22-2010 11:26 PM ET (US)             Also keep in mind that AC appliances draw ten times their rated amps when running on 12V (using an inverter). brainstormer posted 09-24-2010 11:21 AM ET (US)             OK, the machine in use is a Cpap machine, helps momma breathe and needs to last 8-10 hours every night. I have an Optima battery rated @ 55ah . I could tie in another battery if necessary. The machine right now, in the house, uses a transformer from 110v to 12v (its not here, so I can't tell if its AC or DC) @ 4.2 amps of useage. On the boat, I can recharge either/both batteries during the day, but it'll be needed every night. I just want to make sure (not much of a math person) that it'll do the job without killing machine or batteries. jimh posted 09-25-2010 10:08 AM ET (US)             If you want to be certain to be able to operate for 10-hours from battery power a machine that draws 4.2-Amperes at 12-volts, you should use a battery with about double the calculated ampere hours needed.The calculated ampere hours are4.2 Amperes x 10 hours = 42-Ampere-hoursA battery with double that rating would have2 x 42-Ampere-hours = 84-Ampere-hoursA battery with a rated capacity of 84-Ampere-hours will typically be in the battery size Group-24 or Group-27.Of course, you will want to use a battery designated as a deep-cycle battery. Each overnight 10-hour period of discharge will be pulling the charge on the battery to 50-percent or lower. A battery designed for deep-cycle use will tolerate this sort of charge-discharge cycle much better than a starting battery or a marine battery.Another good source of information on deep-cycle battery operation can be found in this FAQ:Deep Cycle Battery FAQ http://www.windsun.com/Batteries/Battery_FAQ.htm If you are intending to recharge the battery from an outboard motor, you must consider the time needed to fully recharge the battery. In my own use of my boat, I often see that my daily time underway with the engine running is often only three hours. Using three hours as a typical running time, we will need to re-charge the battery completely during that time. Since we used 42-Ampere-hours of charge, we must restore all of that charge. Because we will only have three hours of charging, the average charging current must be42-Ampere-hours / 3-hours = 14-AmperesIt is typical that the efficiency of the battery at accepting a charge is not 100-percent. The precise factor for efficiency of charging will be determined by many factors, such as the battery itself, the temperature, the charging voltages, and so on. However, we should consider that the efficiency might only be 70-percent, that is, for every 10-Ampere-hours of charging current, the actual charge on the battery increases only 7-Ampere-hours. We apply this to our average charging current for the three hours of charging we will have:14-amperes / 0.7 = 20-AmperesWe now have to consider that in most situations the charging current will not be uniform during the charging period. Typically as the battery charge begins to reach full-charge level, the charging current will taper off. The exact curve of charging current as a function of battery state of charge is not known, and it will, again, depend on several factors such as the battery, the charger, the temperature, and so on. We should assume that we will need a much higher charging current during the initial phase of charging if we are going to maintain an average charging current of 20-Amperes over the three hours. Here we will use a factor of two, and we will need a source of charging current of 40-Amperes if we want to expect to be able to restore all of the charge to the battery.A charging current of 40-Amperes is a substantial charging current, and it is more than most outboard motors can supply. Many popular outboard motors can only supply a net charging current output of 10-Amperes or less. It would not be reasonable to expect that an outboard motor with limited charging current will be able to replace the charge lost in 42-Ampere-hours of discharge current in only three hours. jimh posted 09-25-2010 10:19 AM ET (US)             Another consideration in charging batteries is the rate of charge. One source suggests that the rate of charge should be limited to the battery capacity at the 20-hour rate divided by 8. For our hypothetical 84-Ampere-hour battery, the recommend limit for the charge rate is therefore84/8 = 10.5-AmperesIf we limit the charging current to 10.5-Amperes, we can assume that this will taper over time, giving an average of only 5.25-Amperes charing. If we reduce the charge by our factor of 70-percent efficiency, we will obtain only a net average charge of 5.25 x 0.7 = 3.675-amperesThus to restore 42-Ampere-hours of charge at a net average rate of 3.675-Ampere, we should plan to have the battery on the charger for42-Ampere-hours / 3.675-Ampere = 11.4-hoursThat is a lot of time underway with the outboard motor running on a daily basis. Cf.: http://www.windsun.com/Batteries/Battery_FAQ.htm#Battery%20Charging jimh posted 09-25-2010 10:21 AM ET (US)             If this machine uses 4.2-Amperes from a 120-Vac source, it would not be reasonable to operate this machine for 10-hours from a battery. You would need something like 840-Ampere-hours of battery capacity. A battery of that capacity would be so large and have so much weight that it would not be practical to install in a small boat. Hoosier posted 09-25-2010 03:08 PM ET (US)             According to this info from Interstate batteries all deep cycle/cranking 12v batteries can do the 5 amp load for 10 hours. It would have been more useful if they had listed the same parameters for each battery. brainstormer posted 10-05-2010 09:14 PM ET (US)             sorry for taking so long...doctor visit again. I have no problem hooking 3 or 4 55ah batteries or better together to secure the discharge and once in dock, I can put the batteries on a land based charger. wouldn't that fairly well cover the few overnight trips? I appreciate all the knowledge sharing Jim, I'm just hoping not to have to get a generator and ruin the night.

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