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  • Battery Sizing for Surge Loads

    I somewhat understand the peukert effect and I understand sizing the batteries for the expected load. What I am trying to get my head around is sizing for seldom usage of say a table saw. I have a woodworking shop that I use for a hobby and I would rather not have to way oversize my batteries to support my woodworking habit. Can 1000AH 48v FLA batteries handle the startup draw of a 3hp table saw?

    I am still trying to figure out what I don't know so I can figure out what to ask

  • #2
    You may be better off finding a 3 phase motor for your table saw and then installing a VF drive on it so you have a slow startup. A VF drive can cut your load surge down from 3 or 4 times nameplate current to not much more then nameplate. I wish some company would come up with low cost single phase output VF drives but the market seems to be three phase. Three phase motors are a lot more rugged design than any single phase motor.

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    • #3
      The issue is the battery Internal Resistance or Ri. As current flows through resistance, a voltage is developed. Voltage = Current x Resistance.

      So for Flooded Lead Acid batteries a typical True Deep Cycle 12 volt 100 AH battery when fully charged has an Ri of roughly .019 Ohm's which does not sound like much, but is quite significant. Significan enough that if you draw C/8 current (100 AH / 8 H = 12.5 amps), your Open Circuit Voltage (OCV) drops from 12.6 volts down to 12.6 - (12.5 amps x .019 Ohms) = 12.36 volts or roughly 3% of you power and voltage is lost on the wiring. That means a 12 volt 100 AH battery can only handle a 150 watt maximum continuous load.

      But now let's say that 150 wat load is a SAW motor which has a start up surge of 6X or 900 watts. At 900 watts requires 90 amps of current. Use that same battery and your voltage drops from 12.6 fully charged down to 12.6 - (90 amps x .019 Ohms) = 10.89 volts at the battery and your .2 volt loss on the wire between the battery and Inverter you are down to 10.6 volts. Your Inverter trips at 11 volts. So when you start the say motor you hear a lound Hum and a CLICK from your Inverter shutting down from Low Voltage Disconnect when nothin gis wrong except the equipment selection is not compatible.

      What do you do in that situation. Well there is three answers and they all cost the same. Replace FLA battery with twice the capacity, or AGM or LFP of same 100AH capacity as they have much lower Ri than FLA.

      So to answer your question a 48 volt 1000 AH battery has a peak power of roughly 10,000 watts and carry a 5000 watt continuous load for about 4 to 5 hours before completely discharged.
      MSEE, PE

      Comment


      • #4
        One more thing to consider is exactly how the inverter and battery system react to surge load such as a motor starting.
        If the voltage out of the inverter sags significantly below nominal voltage, that will reduce the surge current to the motor. It will have less starting torque and will accelerate more slowly, but may manage to start just fine.
        If the inverter cuts off instead of delivering a reduced voltage OR if the battery voltage drops so low under load that the inverter cuts out on low input voltage then the motor will not start.
        The key, following what SunKing said, is what the battery terminal voltage will be based on internal resistance and whether that drops below the LVCO of the inverter. (Well, allow for voltage drop in the wiring to the inverter too.)
        SunnyBoy 3000 US, 18 BP Solar 175B panels.

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        • #5
          Originally posted by inetdog View Post
          The key, following what SunKing said, is what the battery terminal voltage will be based on internal resistance and whether that drops below the LVCO of the inverter. (Well, allow for voltage drop in the wiring to the inverter too.)
          Well in a properly designed system you want to keep voltage losses to 2 or 3% or less. On a 12 volt toy system that is only .24 to .36 volts. As you stated you have to voltage drops.

          1. Battery Ri

          +

          2. Wire loss between battery Term Post and Inverter Input Terminals.

          What is funny, or at least I find funny. most folks under size the wire between Battery and Inverter, and that wire alone can exceed 3%. Couple that with over loading the battery by 500% and I get a good laugh when they come here and ask what is wrong and what it takes to fix it. My answer is real and real expensive. Only from a proper design can it be made to work, and if that was donew from th estart, I would say 90% would give up once they realize the cost involved.

          They then understand just how dirt cheap and plentiful electricity really is from the utility.
          MSEE, PE

          Comment


          • #6
            Originally posted by Sunking View Post
            The issue is the battery Internal Resistance or Ri. As current flows through resistance, a voltage is developed. Voltage = Current x Resistance.

            So for Flooded Lead Acid batteries a typical True Deep Cycle 12 volt 100 AH battery when fully charged has an Ri of roughly .019 Ohm's which does not sound like much, but is quite significant. Significan enough that if you draw C/8 current (100 AH / 8 H = 12.5 amps), your Open Circuit Voltage (OCV) drops from 12.6 volts down to 12.6 - (12.5 amps x .019 Ohms) = 12.36 volts or roughly 3% of you power and voltage is lost on the wiring. That means a 12 volt 100 AH battery can only handle a 150 watt maximum continuous load.
            Great info, that helps a ton.

            Comment


            • #7
              Originally posted by Sunking View Post
              The issue is the battery Internal Resistance or Ri. As current flows through resistance, a voltage is developed. Voltage = Current x Resistance.

              So for Flooded Lead Acid batteries a typical True Deep Cycle 12 volt 100 AH battery when fully charged has an Ri of roughly .019 Ohm's which does not sound like much, but is quite significant. Significan enough that if you draw C/8 current (100 AH / 8 H = 12.5 amps), your Open Circuit Voltage (OCV) drops from 12.6 volts down to 12.6 - (12.5 amps x .019 Ohms) = 12.36 volts or roughly 3% of you power and voltage is lost on the wiring. That means a 12 volt 100 AH battery can only handle a 150 watt maximum continuous load.

              But now let's say that 150 wat load is a SAW motor which has a start up surge of 6X or 900 watts. At 900 watts requires 90 amps of current. Use that same battery and your voltage drops from 12.6 fully charged down to 12.6 - (90 amps x .019 Ohms) = 10.89 volts at the battery and your .2 volt loss on the wire between the battery and Inverter you are down to 10.6 volts. Your Inverter trips at 11 volts. So when you start the say motor you hear a lound Hum and a CLICK from your Inverter shutting down from Low Voltage Disconnect when nothin gis wrong except the equipment selection is not compatible.

              What do you do in that situation. Well there is three answers and they all cost the same. Replace FLA battery with twice the capacity, or AGM or LFP of same 100AH capacity as they have much lower Ri than FLA.

              So to answer your question a 48 volt 1000 AH battery has a peak power of roughly 10,000 watts and carry a 5000 watt continuous load for about 4 to 5 hours before completely discharged.
              I have been checking my potential load balance with my battery and inverter with:

              1) That the inverter can handle the sustained load.
              2) That the inverter can handle the surge.
              3) The number of hours the battery will last at load.
              4) The the battery is being charged fast enough (eg C/10)

              But I have not been checking whether or not the battery can handle the surge. From what I
              read here, a battery does not have any surge capacity.

              So how about this:

              4 Costco golf car batteries @ 207 AH for 24v system (207 X 24 = 4968 watts)

              402 watts possible if all loads active at the same time.
              503 watts momentary surge for refrigerator compressor.
              ===
              905 watts

              4968 battery watts / 905 load watts at surge = C/5.5

              I know that golf batteries are hybrid and so can take a higher surge, and if so, what is it? But for a battery that has C/8, will this be a problem?

              Comment


              • #8
                Originally posted by Sunking View Post
                The issue is the battery Internal Resistance or Ri. As current flows through resistance, a voltage is developed. Voltage = Current x Resistance.

                So for Flooded Lead Acid batteries a typical True Deep Cycle 12 volt 100 AH battery when fully charged has an Ri of roughly .019 Ohm's which does not sound like much, but is quite significant. Significan enough that if you draw C/8 current (100 AH / 8 H = 12.5 amps), your Open Circuit Voltage (OCV) drops from 12.6 volts down to 12.6 - (12.5 amps x .019 Ohms) = 12.36 volts or roughly 3% of you power and voltage is lost on the wiring. That means a 12 volt 100 AH battery can only handle a 150 watt maximum continuous load.

                But now let's say that 150 wat load is a SAW motor which has a start up surge of 6X or 900 watts. At 900 watts requires 90 amps of current. Use that same battery and your voltage drops from 12.6 fully charged down to 12.6 - (90 amps x .019 Ohms) = 10.89 volts at the battery and your .2 volt loss on the wire between the battery and Inverter you are down to 10.6 volts. Your Inverter trips at 11 volts. So when you start the say motor you hear a lound Hum and a CLICK from your Inverter shutting down from Low Voltage Disconnect when nothin gis wrong except the equipment selection is not compatible.

                What do you do in that situation. Well there is three answers and they all cost the same. Replace FLA battery with twice the capacity, or AGM or LFP of same 100AH capacity as they have much lower Ri than FLA.

                So to answer your question a 48 volt 1000 AH battery has a peak power of roughly 10,000 watts and carry a 5000 watt continuous load for about 4 to 5 hours before completely discharged.
                I'm assuming that Ri adds in series just like normal resistance so that for your example a 48v battery would have an Ri of 0.076. Also I'm assuming the Ri goes down as the AH capacity of the battery goes up. If that is true do you have any kind of formula?

                Thanks.

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