Small cooking appliances off an inverter?

bmet · October 2010

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

Cariboocoot · October 2010

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!

Cariboocoot · October 2010

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).

BB. · October 2010

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

bmet · October 2010

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)?

BB. · October 2010

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

bmet · October 2010

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.

BB. · October 2010

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

Cariboocoot · October 2010

Re: Small cooking appliances off an inverter?

bmet wrote: »

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.

bmet · October 2010

Re: Small cooking appliances off an inverter?

Cariboocoot wrote: »

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

JESSICA · October 2010

Re: Small cooking appliances off an inverter?

Cariboocoot wrote: »

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).

Also:

A microwave oven uses the same power, whether in "high" or in "medium".
If its magnetron draws 1,200 watts, it will use 1,200 watts as long as it is "on".
In the "medium" setting, for 1 minute, the magnetron will be on 30 seconds, and then will "rest" for 30 seconds, and so on, though the oven may seem to be working all the time. In "low" setting, the magnetron may work for 15 or 20 seconds out of every minute.

Cariboocoot · October 2010

Re: Small cooking appliances off an inverter?

For the most part, the batteries have 'X' Watt hours available and don't care how that quantity is used.
So an available 600 Watt hours can be 600 Watts for one hour or 60 Watts for ten hours or 30 Watts for 20 hours. Usually it's that last number we plan for: the long-term average. That would be about 2.5 Amps on a 12V system.
For the most part.
Always there are exceptions. :roll:

As in they might not be able to supply the full current at a given time without suffering sudden Voltage drop. In other words, if you try to pull 50 Amps from a 100 Amp hour battery in one go, the Voltage could fall to 10 V and the inverter will shut down. Most manufacturers recommend the discharge rate doesn't exceed the charge rate, or about C/10 for either.

That doesn't mean you can't draw more, but usually off-grid systems don't. But don't say I didn't warn you that the George Foreman grill may deliver a knock-out current draw that puts the inverter on the mat and down for the count (until the batteries 'recover').

The lower the battery is, the more likely it will be "overwhelmed" by a high current demand. So the grill may work when everything is fully charged and the panels are ready with some back-up current, but cooking bacon for breakfast after a night's discharge (however mild) may shut the system down.

Don't rely on a Voltage reading as a state-of-charge indicator. It works fairly well for a battery at rest (one that's been sitting for a day with nothing going in or out) but is unreliable for an in-use system. If you want to keep a close eye on it, get a battery monitor. http://store.solar-electric.com/metersmonitors.html These are particularly useful with AGM batteries as you haven't got the option of checking the specific gravity.

As for running a string trimmer ... you'd have to get a Kill-A-Watt meter and find out what it draws. You can't guess with motors, and you can't rely on the label info either.

techntrek · October 2010

Re: Small cooking appliances off an inverter?

I think you need to look at this spreadsheet:

http://www.smartgauge.co.uk/calcs/peukert.xls

Batteries - when charging or discharging - experience the "Peukert effect". The slower you discharge, the more total amps you can get from them. This spreadsheet takes several factors into account and will let you see how long you can run depending on the amps you need. Plug your battery's amps in near the top, and if you want to see the numbers for a specific amp draw, plug that number in the lower left. Keep in mind - this spreadsheet shows the numbers down to a complete discharge. Cut the numbers in half to see a 50% discharge.

So let's say you have a 100 amp battery (20 hour rate). That's a 5 amp load for 20 hours. So you might think a 10 amp load will only last 10 hours, and a 2.5 amp load would last 40 hours. But due to Peukert, 10 amps lasts 8.12 hours and 2.5 amps lasts 49.25. Keep in mind the numbers shown are to 100% DOD which you never want to do. So you should only run a 5 amp load for 10 hours which will be 50% DOD.

bmet · October 2010

Re: Small cooking appliances off an inverter?

JESSICA wrote: »

Also:

A microwave oven uses the same power, whether in "high" or in "medium".
If its magnetron draws 1,200 watts, it will use 1,200 watts as long as it is "on".
In the "medium" setting, for 1 minute, the magnetron will be on 30 seconds, and then will "rest" for 30 seconds, and so on, though the oven may seem to be working all the time. In "low" setting, the magnetron may work for 15 or 20 seconds out of every minute.

So is that why I sometimes hear an additional drumming noise(along with table motor and fan), attributable to when the klystron kicks in?

BB. · October 2010

Re: Small cooking appliances off an inverter?

If you are running less than 100% cooking power--you may "hear" the load and see the oven light dim a bit as the magnetron cycles on and off.

-Bill

JESSICA · October 2010

Re: Small cooking appliances off an inverter?

BB. wrote: »

If you are running less than 100% cooking power--you may "hear" the load and see the oven light dim a bit as the magnetron cycles on and off.

-Bill

Bill:

I really don't know if we are talking about the same thing. The microwave's magnetron cannot run at less than 100% power; it is either "on" or "off". Differents settings ("high", "medium", "low", or "90%", "70%", etc.) only alter the times, during a given cycle, that the magnetron works or is "iddle", so to speak. Thus, in my oven, "90%" means that if I use it to heat a cup of coffee for one minute, the magnetron will be on for just 54 seconds, though the light and rotating plate are "on" for the whole 60 seconds.

P.S.: I have read that some newer ovens can actually regulate the magnetron's power output level. But I don't know for sure.

BB. · October 2010

Re: Small cooking appliances off an inverter?

I was talking about less than 100% setting on the display... I agree that normally the magnetron just turns on for XX seconds and turns off for YY seconds depending on the exact power level programmed in (maybe a 15-20 second cycle on my oven).

-Bill

mike95490 · October 2010

Re: Small cooking appliances off an inverter?

Every microwave I've used, the tone of the fan and the internal light, cycle as the magnetron cycles on and off. (and they all cycle when you use less than 100% power)

waynefromnscanada · October 2010

Re: Small cooking appliances off an inverter?

Could be wrong, but have been given to understand that some of the higher end microwave ovens now have what they are calling an inverter power supply, that can be somehow modulated at a high enough speed, as to for all intents and purposes, control the magnetron in such a way as to make it appear infinitely variable, in the same way wall mounted light dimmer switches control light bulbs. Interesting, if true.

Small cooking appliances off an inverter?

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