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High Amperage Drain on Lead Acid Batteries 1700 watt kettle vs 400 watt coffee maker

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  • Sunking
    replied
    Originally posted by Mike90250 View Post
    Sir, you jest !! (good to hear your advice here) 300A @ 12V only requires 0000ga cable (0.46")
    Not so much jest. While I understand the tables you posted and where it came from has little to do with any wiring you can use and comply with NEC. The tables are true for equipment manufactures chassis wiring and utility power transmission with complete disregard for voltage loss incurred at low 12 volts. That has nothing to do with residential power and lighting wiring. That comes from NEC Table 310.15(B) using 90 degree cable insulation would require a minim 350 MCM which still does not address any voltage loss at 12 volt distribution. So I may be guilty of stretching it a bit, the point is the insanity of using low voltage for high wattage loads. Not only is it incredible expensive, but extremely dangerous doing so with that much current. Joe DIY Home Owner does not have the tooling, training, or experience to terminate such sized conductors properly.

    As you properly stated the first clue is something is not right is none of the 12 volt toy inverters sold can terminate a proper sized conductor or a massive surface area required. Example in any high current panels for commercial or industrial application using 350 amps would use Tongue type 2-hole terminals using Grade 5 3/8-inch hardware to secure it with proper torque.

    ?u=https%3A%2F%2Fi.ebayimg.com%2Fthumbs%2Fimages%2Fg%2FpccAAOSwkl5fPctd%2Fs-l200.jpg&f=1&nofb=1.jpg

    Here is a tip. If your application requires any wire to be larger than 6 AWG, your voltage is likely too low.

    .
    Last edited by Sunking; 09-15-2020, 04:23 PM.

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  • Mike90250
    replied
    Originally posted by Sunking View Post
    ............................
    As for cable size as big as your wrist called a 750 MCM that cost $9 per foot.
    Sir, you jest !! (good to hear your advice here) 300A @ 12V only requires 0000ga cable (0.46")

    Of course, none of the inverters use heavy 300A cable inside them which is why you need a good insurance policy
    And you need a gorilla to bend the cable to fit, along with the giant tool to crimp the ends onto the cable. And it's a very special inverter that has terminals of the proper size to connect that big wire to.


    Here's some unnamed chart from the internet that supports my theory.

    amp chart table.jpg

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  • Sunking
    replied
    Originally posted by chrisski View Post

    Another forum I was on, I was saying I plan to get an 1000 watt or 1500 watt inverter, and was even told a 3000 watt inverter for 12 volts is standard.
    They are fools and stupid.

    Originally posted by chrisski View Post
    The fuse I came up with was a 350 amp fuse if I ran my inverter at the 10.5 VDC voltage cutoff rate for the 2 KW inverter, I’d wonder what the fuse would be for a 3 KW inverter along with how thick the wiring would be..
    OK great you need to make sure you have two more things in place because you will need them sooner or later.

    1. Make sure you homeowners insurance policy is up to date and covers fire damage.
    2. Make sure you have plenty of smoke detectors and fire extinguishers.

    As for cable size as big as your wrist called a 750 MCM that cost $9 per foot.
    Last edited by Sunking; 09-15-2020, 01:56 PM.

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  • Ampster
    replied
    Originally posted by chrisski View Post
    .....
    Another site I'm on I'm told 3000 watts can be standard for a 12 volt system, but I don't think I would safely make one that strong at this point. One user talked of a professionally installed 12 VDC, 4000 watt continuous, 12000 watt surge inverter. ......
    I am on a number of sites and the variety of opinions, beliefs and "standard' facts vary exponentially as the number of sites visited increases.Some sites have more RV users, some have more off grid users, some are more geared toward DIY, some are referral sites for lead generation or sponsored by a retailer or manufacturer. I use each for the value it brings to the question at hand.

    It all depends on where you are standing.
    Last edited by Ampster; 09-05-2020, 03:06 PM.

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  • chrisski
    replied
    With the triangle and Ohms law, and measuring watt hours and amp hours, that gives you some data. But now turn on a high amperage draw device like the tea kettle and you use it for say, 500 watt hours, and you use your calculations you did at the 20 hour rate for your lead acid battery that mat have 2000 watt hours, and because that amperage draw is so high, you shouldn't have 1500 watt hours left in your battery if you drew that at the 1 hour rate, it should be more like 500 watt hours left, which would leave you well below the 50% DoD. Only taking into account the Peukert effect, those high amperage draws at the one hour rate chew up 3X the juice in the battery than the lower amperage ones using it at the 20 hour rate.

    For me I can really watch the load draws and be careful, but I've got three others that will be in this RV and not all of them will be that careful (Think teenagers). We'll see what its like the first time we use the RV. Could be about two or three weeks from completion.

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  • bob-n
    replied
    Just to make things more confusing, many heating devices contain a thermostatic switch that keeps the vessel or fluid at a desired temperature by turning power on and off. Last time I disected an old coffee maker, I found a thermostatic switch inside. This means that a 1700 watt coffee heater with a thermostatic switch will not use 1700 watts continuously. If you really want to know the energy it uses, use a Kill-A-Watt or other integrating energy meter.

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  • Mike90250
    replied
    Originally posted by chrisski View Post

    These IR losses are what we calculate in wire loss on a DC calculator? .....
    Yes, include also, the crimp and all bolted connection interfaces, You can measure them with a volt meter and running a known Amps thru them ( load your inverter so your cables are supplying 40A and you can caculate the resistance from the voltage drop across the battery terminal and the Lug.

    Volts divided by amps (I) = resistance

    EIR_elect_triangle.png

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  • chrisski
    replied
    Originally posted by Mike90250 View Post
    They all consume enough power to heat the water.

    The losses are many:
    IR loss in the cables
    These IR losses are what we calculate in wire loss on a DC calculator?

    I'm building a spreadsheet for my use that shows amperage draw based on amp hours of battery bank with ratings from 1 hour to 100 hours, based off one of the site battery stickies. It's just trying to show why its not a good idea to push a 12 volt system to 2000 watts or 3000 watts for an inverter. The 2000 watt inverter would be fused at 350 watts, and would need at least 2/0 cables. Also found the fuses only go up to 500 amps, at least for my ANL battery fuse.

    There's not a lot out there that I can find for formulas that explain the peukertt effect with lead acid batteries at higher amperage draws, so I based the formulas off the stickies.

    Another site I'm on I'm told 3000 watts can be standard for a 12 volt system, but I don't think I would safely make one that strong at this point. One user talked of a professionally installed 12 VDC, 4000 watt continuous, 12000 watt surge inverter. Not that this can't be done, but should not be the goal of a first time self install in your RV.

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  • Mike90250
    replied
    They all consume enough power to heat the water.

    The losses are many:
    IR loss in the cables
    Peukert's loss in the batteries
    Inverter standby losses & inversion losses

    All this data gets loaded into your spreadseet and an engineered project gets designed.

    The smallest 1 cup boiler (with internal heating element directly heating the water, brewing a cup at a time, no "keep warm" (that's what a thermos does ) element
    coupled with a 48V system, will be the most efficient. (Until you use well chosen specialty gear)

    Leave a comment:


  • Ampster
    replied
    Originally posted by PNPmacnab View Post
    ..... Half the rated voltage yields 1/4 of rated power.
    The operative question was which alternative would result in the least amount of Watt hour consumption.

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  • Mike90250
    replied
    Originally posted by PNPmacnab View Post
    You can always use a buck or auto transformer to reduce the voltage to the kettle for a lower wattage at the expense of time. It doesn't have to be a big drop as reducing the voltage has dramatic effects. Half the rated voltage yields 1/4 of rated power.
    But that's only for a restive element heater, not an induction (97% efficient) or a microwave (60% efficient)

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  • PNPmacnab
    replied
    You can always use a buck or auto transformer to reduce the voltage to the kettle for a lower wattage at the expense of time. It doesn't have to be a big drop as reducing the voltage has dramatic effects. Half the rated voltage yields 1/4 of rated power.

    Leave a comment:


  • gbynum
    replied
    Originally posted by Ampster View Post
    I believe inductive heating of water is more efficient than resistive heat.
    For a pan on an surface "hot plate or range type" element, I suspect you are correct. For a "submerged" heating element, conceptually like an electric water heater, I believe the resistance unit would be much better, over 99%. I think the coffee makers are effectively submerged.

    From Google, Stovetop or cooktop electric cooking allows only 65-70% of heat to reach food as opposed to induction's 90%. This results in your kitchen staying cooler with induction than it does with electric cooking.

    So to use modern marketing terminology (JOKE), 99% vs 90% is 10 times more efficient ... 10% loss ve 1% loss.

    Leave a comment:


  • chrisski
    replied
    So with this, I’m trying to put some numbers to why an 2Kw inverter on a system my size, 478 AH lead Acid Batteries, 600 watts of panels would not be a good idea, or at very least done carefully.

    Another forum I was on, I was saying I plan to get an 1000 watt or 1500 watt inverter, and was even told a 3000 watt inverter for 12 volts is standard.

    So with above, I tried to take into account the lead acid batteries discharging quicker under high load, and I think I did the math correctly, or at least as best as I could without the manufacturer publishing data.

    I came up with its not a good idea To discharge at the C1 rate because even wired with 2/0 wire for a short run. Since Trojan does not publish a max discharge rate for my battery, but publishes a 10 hour rate, I think it would be worst than that.

    The fuse I came up with was a 350 amp fuse if I ran my inverter at the 10.5 VDC voltage cutoff rate for the 2 KW inverter, I’d wonder what the fuse would be for a 3 KW inverter along with how thick the wiring would be.

    Calculating the amp hours a battery holds and loads is pretty straightforward, until you start using a high discharge rate, and the only reference I could find is Sunkings post that had the coeefficient based off some classes he taught and project he’s done.

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  • Ampster
    replied
    I don't know what kind of coffee pot you are using but the one in my home is induction and boils one cup in under a minute and I think less than 100 Watthours. I don't know how they compare to the Mr Coffee type which I believe use resistive heat. I believe inductive heating of water is more efficient than resistive heat.
    Last edited by Ampster; 08-30-2020, 12:13 PM.

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