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

[chrisski]

With a 478 amp hour 12 VDC flooded lead acid battery bank, if I brew 1700 ml of coffee, this is the percentage of my battery drained for the same amount of coffee:

1700 Watt electric kettle: 11.5255% drained
400 watt: 3.720951% drained

I'm posting this because on a separate forum I am on, I am being lead to me not needing to go up to 24 VDC to power a 2000 watt inverter with a 478 amp hour bank, and the batteries are strong enough to push a 3000 watt inverter. I'm not really like what I'm seeing when I run the numbers. I'm sure there's errors in the formulas, but its close enough to get the point across.

I have been curious how much better it is on an inverter system to run a single cup coffee maker to make an equivalent amount of coffee from 1700 watt kettle. I was pretty surprised at how much keeping the wattages low with the single cup saves you versus just doing it all at once with the electric kettle.

My system I'm building has flooded lead acid batteries with a C20 of 478 amp hours so I calculated how the wattage devices used up a battery quicker than the lower watt devices. The 1700 watt kettle used 141 amps when turned on, and that uses stored energy up at the neat C1 Rate, so the battery could produce 143 amps. The 400 watt coffee maker used energy at 33 amps, or the C12 rate of 430 amps or 35.85 amps per hour. So to determine the percentage of battery use, I divided amp hours used by total battery amp capacity. Then what I did with that percentage of battery used is multiply that percentage of the C20 rate, to see how much was pulled out at that rate. With my 36 amps I expect in ideal conditions with my system, I used this to see how long it would take to charge the battery.

For the 1700 watt kettle, it will make 1.7 liters of boiled water that you then add instant water to. To make the same amount of coffee, it's 7.2 cups of coffee.

Single use coffee maker
478 C20 Rate
400 Watts of coffee maker
12 VDC Batteries
33.33333333 Amps for coffee pot
12 Hour Rate (C rating)
430.2 Total Amps at the C12 Rate
35.85 Amps per hour at the C12 rate
4 Minutes to brew a cup
236 ml per cup
1700 electric kettle ML
7.203389831 # of cups to kettle
28.81355932 # of minutes to brew 7.20338983050847 cups
16.00753296 Amps hours used at C12 Rate
17.78614773 Equivalent Amp Hours at C20 Rate
3.720951408 % of Battery used
36 Charging Amps
0.494059659 Hours to recharge

1700 Watt 1700 ml Electric Kettle Coffee Maker
478 C20 Rate
1700 Watt electric kettle
12 VDC Batteries
141.6666667 Amps for coffee pot
1 Hour Rate (C rating)
143.4 Total Amps at the C1 Rate
143.4 Amps per hour at the C1 rate
7 Minutes to brew a cup
236 ml per cup
1700 electric kettle ML
1 # Kettles brewed
7 # of minutes to brew 1 cups
16.52777778 Amps hours used at C12 Rate
55.09259259 Equivalent Amp Hours at C20 Rate
11.52564699 % of Battery used
36 Charging Amps
1.530349794 Hours to recharge

The table below I got from @sunking ‘s https://www.solarpaneltalk.com/forum...teries-and-you

Discharge Rate Coefficient
C Hour Coef Ttll Amps
1 Hour 0.3 143.4
2 Hour 0.5 239
3 hour 0.6 286.8
4 hour 0.65 310.7
5 hour 0.7 334.6
6 hour 0.75 358.5
8 hour 0.8 382.4
10 hour 0.85 406.3
12 hour 0.9 430.2
16 hour 0.95 454.1
20 hour 1 478
24 hour 1.05 501.9
36 hour 1.1 525.8
72 hour 1.25 597.5
100 hr 1.3 621.4


Some other issues, on the negative side, there is no loss accounted for and what could make this better than posted is if the solar panels are producing a lot of charging amps, the batteries won't drain as much.

[Mike90250]

several short runs of heavy load, with idle period, is better than a long run of the same load & time total.

I'd go with the single cup, and maybe by cup # 2, solar power will be assisting the batteries.

[Ampster]

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.

[chrisski]

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.

[gbynum]

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.

[PNPmacnab]

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.

[Mike90250]

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)

[Ampster]

..... 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.

[Mike90250]

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)

[chrisski]

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.

[Mike90250]

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

[bob-n]

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.

[chrisski]

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.

[Ampster]

.....
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.

[Sunking]

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.

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.