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Thread: Small cooking appliances off an inverter?

  1. #1

    Default Small cooking appliances off an inverter?

    In playing with my Kill-A-Watt EZ meter, I'm curious what puts more strain on a small off-grid system with inverter.

    1. Small grill that draws a constant 5 amps, used 5-7 minutes.

    Or

    2. A typical 1000 watt microwave which draws 12-13 amps on "High" but may only be used for a minute at a time.

    Are these the type of appliances that don't play well with a MSW-type inverter?

    Thank you

  2. #2

    Default Re: Small cooking appliances off an inverter?

    It isn't the inverter that will strain: an inverter capable of 'X' Watts is capable of 'X' Watts.
    The batteries, on the other hand ...

    The (nearly) constant draw of the resistance heating element may add up to a greater number of Watt hours than the larger draw of the microwave over a shorter time. What's really important is when you use it. If the batteries are at "Float" stage and the panels are "idle" you can use lots of power and still have your over-night reserve (providing the system is designed properly). This is when I run the water pump, digester, and yes microwave.

    When we talk about electric heating being not very efficient, we mean in relation to an alternate energy source such as propane.

    Some microwaves don't like MSW. It messes up their control mechanisms. I guess they don't put very good power supplies in them. Also, the magnatron will use greater power and function less efficiently on MSW. Been there, done that, bought the True Sine Wave!
    1220 Watts of PV, OB MX60, 232 Amp hrs, OB 3524, Honda eu2000.

    Ohm's Law: Amps = Volts / Ohms
    Power Formula: Watts = Volts * Amps

  3. #3

    Default Re: Small cooking appliances off an inverter?

    Forgot to mention: microwaves get rated by their cooking power. Don't get fooled into thinking this is their consumption. A "700 Watt" oven may well draw 1000 Watts or more. If you can get both numbers and compare them you'll see that some are more efficient than others (input power closer to cooking power).
    1220 Watts of PV, OB MX60, 232 Amp hrs, OB 3524, Honda eu2000.

    Ohm's Law: Amps = Volts / Ohms
    Power Formula: Watts = Volts * Amps

  4. #4
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    Default Re: Small cooking appliances off an inverter?

    Also, battery drain current (and charging current) changes efficiency too (more so for flooded cell batteries--the "spread" in efficiency for AGMs are less). The larger the current draw, the less useful capacity the battery bank appears to have.

    My observations of recommend Current/Capacity Ratios (flooded cell lead acid storage batteries):

    • C/20 (5%) of 20 Hour rate--pretty close to the typical off-grid system's average load/charging current. Also the minimum charging current rule of thumb.
    • C/8 (~12.5-13%)--Normally the maximum long term max continuous charging/discharging current. Higher rates can overheat Flooded Cell batteries and also shows reduced battery "efficiency" (Peukert's law).
    • C/2.5 (40%) is about the maximum surge current capability before you crash a charged battery's ability to provide current (battery "falls out of regulation").

    Note: The above is for flooded cell battery recommendations. Some AGM's have a Peukert Factor much closer to 1. There is at least one brand that has a max charging (and discharging?) ability of C*4 (400%) of the 20 Hour rate. At those current levels, it just becomes very difficult to attach enough copper to the battery to carry the current.

    -Bill
    20x BP 4175B panels (replacement) + Xantrex GT 3.3 inverter for 3kW Grid Tied system + Honda eu2000i Inverter/Generator for emergency backup.

  5. #5

    Default Re: Small cooking appliances off an inverter?

    If I am understanding you correcty, it is your experience that drawing current from battery bank should not exceed the 5-13% charging level of the bank's 20 amp hour rating( for flooded cell batteries)?

  6. #6
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    Default Re: Small cooking appliances off an inverter?

    As a continuous load for a standard flooded cell battery (say 225 AH battery):

    • 225 amps*hours (20 Hour Rate) * 0.13 = 29.25 amps

    Of course, we recommend discharging to a maximum of 50% capacity... The above is based on 20 Hour capacity, so we would only drain to 1/2 the 20 Hour capacity (~225AH/2=112.5 AH).

    And, the 10 Hour capacity (C/8 Rate) is a bit less than the 20 Hour Capacity--so you would be less...

    For example a 6 volt T-105 "golf cart" type battery from Trojan (PDF):

    T-105:
    5HR Rate 185 AH
    20 HR Rate 225 AH
    100 HR Rate 250 AH

    So, call the "8 Hour Rate" to be round 195 Amp*Hours

    • 195 AH * 0.50 max discharge * 1/29.25 Amps = 3.33 hour recommended discharge time

    The available Watt*Hours for two of these 6 volt batteries in series (12 volts):

    • 29.25 amps * 3.33 hours * 12 volts = 1,170 WH to 50% state of charge

    Now, in reality, batteries are within 20% of their rated capacity (maybe 195 AH or it may be 0.8*195AH=156AH minimum)... The extra decimal points are just so you can check my math.

    Also, Lead Acid Batteries will increase their capacity from new as they are exercised over the first 10+ charging cycles.

    -Bill
    20x BP 4175B panels (replacement) + Xantrex GT 3.3 inverter for 3kW Grid Tied system + Honda eu2000i Inverter/Generator for emergency backup.

  7. #7

    Default Re: Small cooking appliances off an inverter?

    I was with you for the first two statements, and then was lost after that. Is there a way to determine the 50% discharge by measuring the volts? The Amp hours vs. Amps / Watt hours vs. Watts is still confusing to me.

  8. #8
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    Default Re: Small cooking appliances off an inverter?

    You can measure the battery resting voltage (couple hours of no loads and no charging).

    http://www.windsun.com/Batteries/Battery_FAQ.htm

    Other wise, measuring specific gravity or installing a battery monitor are two easier/more accurate.

    http://store.solar-electric.com/metersmonitors.html

    Watts and amps are rates--like miles per hour or gallons per hour.

    Wattxhours and ampxhours are an amount--like miles and gallons.

    AH and WH are about the same.

    AH x Volts = Watt x Hours

    With AH, we keep having to ask what voltage battery bank you are using,,, With watts and WH, we don't need to know anything more.

    -Bill
    20x BP 4175B panels (replacement) + Xantrex GT 3.3 inverter for 3kW Grid Tied system + Honda eu2000i Inverter/Generator for emergency backup.

  9. #9

    Default Re: Small cooking appliances off an inverter?

    Quote Originally Posted by bmet View Post
    I was with you for the first two statements, and then was lost after that. Is there a way to determine the 50% discharge by measuring the volts?
    Not accurately. The reason being that as a battery is discharged and its Voltage goes down the current (Amps) goes up to supply the same amount of Watts (over-all power). The closer the battery gets to "dead" the faster the rate of decline. Inversely, when you're recharging its the last "little bit" of power that takes the longest to put back. Batteries are not linear.

    The Amp hours vs. Amps / Watt hours vs. Watts is still confusing to me.
    A Watt is 1 Volt * 1 Amp. Use this much over an hour and you have 1 Watt hour.
    An Amp hour is basically a unit of storage capacity. If you had a 1 Volt, 1 Amp hour battery it could supply that 1 Watt hour. But not in real life, because you can never drain the full capacity. That's why there's that "arbitrary" 50% Depth Of Discharge limitation; you want to be able to bring the battery back to "full" afterward.

    So:

    12 Volt, 100 Amp hour battery can supply roughly; 50 Amp hours @ 12 Volts = 600 Watt hours. In the real world efficiency losses in the system can use up a surprisingly large amount of power themselves, so you won't necessarily have 100% of that power for the "end use device".

    There's a Voltage vs. State Of Charge chart in the Battery FAQS: http://www.windsun.com/Batteries/Battery_FAQ.htm

    But this is an approximation, not an absolute. With an operating range of 10.5 to 12.7 Volts, you can see what happens to the current draw as the battery discharges while supplying a steady load:
    100 Watts @ 12.7 Volts draws 7.8 Amps. 100 Watts @ 12.0 Volts it's up to 8.3 Amps. By the time the inverter would shut down @ 10.5 Volts the draw would be 9.5 Amps. If you compare that to how the Amp hour capacity is diminished by that drain you can see the capacity falls off faster the lower the battery is. The average draw would be 8.8 Amps over the full range so you would expect your 50 Amp hours to supply 100 Watts for 5.6 hours, or 560 Watt hours. Notice how that falls short of the simpler calculation's expected 600 Watt hours.

    In reality it's not that simple either.
    1220 Watts of PV, OB MX60, 232 Amp hrs, OB 3524, Honda eu2000.

    Ohm's Law: Amps = Volts / Ohms
    Power Formula: Watts = Volts * Amps

  10. #10

    Default Re: Small cooking appliances off an inverter?

    Quote Originally Posted by Cariboocoot View Post
    Not accurately. The reason being that as a battery is discharged and its Voltage goes down the current (Amps) goes up to supply the same amount of Watts (over-all power). The closer the battery gets to "dead" the faster the rate of decline. Inversely, when you're recharging its the last "little bit" of power that takes the longest to put back. Batteries are not linear.
    I get that part.


    ....That's why there's that "arbitrary" 50% Depth Of Discharge limitation; you want to be able to bring the battery back to "full" afterward.

    So:

    12 Volt, 100 Amp hour battery can supply roughly; 50 Amp hours @ 12 Volts = 600 Watt hours. In the real world efficiency losses in the system can use up a surprisingly large amount of power themselves, so you won't necessarily have 100% of that power for the "end use device".
    It was easier for me to divide the Amp hour rate by 20 to get the steady draw. I've been eyeballing 105Ah batteries because they seem to be a good match for a 135 watt panel. 105Ah divided by 20 = 5.25 Amps steady draw. In an all perfect world I would not want to discharge more than 50% of that, or 2.62 Amps.

    Here's where my math gets even weaker. 12V x 2.62 Amps = 31.5 Watts steady draw for 20 hours. Or 63 Watts steady draw for 10 hours.

    In another thread I measured <50 watts total draw for all the devices I would want to run at one time. By keeping a nice even number, 50, would I then be able to expect to power these devices for 10 hours? I know that my inverter would draw some power, but I'm thinking it would not exceed 13 watts (63-50=13).


    With an operating range of 10.5 to 12.7 Volts, you can see what happens to the current draw as the battery discharges while supplying a steady load:
    100 Watts @ 12.7 Volts draws 7.8 Amps. 100 Watts @ 12.0 Volts it's up to 8.3 Amps. By the time the inverter would shut down @ 10.5 Volts the draw would be 9.5 Amps. If you compare that to how the Amp hour capacity is diminished by that drain you can see the capacity falls off faster the lower the battery is.
    Okay. For my example that would be:
    50 Watts divided by 12.7V = 3.93 Amps
    50 / 12.0 = 4.16A
    50 / 10.5 = 4.76A

    If I add the full charge state with the 100% discharge state, could I approximate the 50% depth of discharge? If 12.7 were the 'upper' voltage and 10.5 is the 100% discharge, the the average between the two would be 3.93 + 4.76 = 4.345 A

    My usage, 50 watts, divided by 4.345A = 11.5 volts as 50% depth of discharge?

    Or have I completely missed the boat?

    I understand the real world includes wire losses, dynamic loads, and inverter needs, but I'm getting a little closer to knowing practical limitations. This is going to be a remote power pack, and will not be used more than a couple of hours a week.

    My next consideration will be, whether to invest in the 600TSW or 1000W TSW. It would be nice to use a small power tool, or maybe an electric weed eater around the shed. How many amps would a small motor need for a minute or two?

    Extrapolating my numbers from above (50%) would be 63 Watts for 10 hours. This would also get.

    504 Watts for 1.25 Hours

    or

    1008 Watts for 37.5 minutes

    Would a weed eater need more than this?

    My George Foreman grill only need 6 amps for 6 minutes to cook two burgers

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