Help: Calculating battery sizes and inverter sizes?

RobertMfromLI · September 2011

Hello.

In trying to figure out how big a battery bank we need to power our place for 12 hours, I found (ugh, first one) a really crappy battery size calculator that made things seem oh so wonderful. Now, I'm rechecking all of my math, as it was off by quite a bit. Found others, which seem to agree, and used some of the formulas posted elsewhere, but I'd be very appreciative if someone far more knowledgeable than me on this stuff rechecked my figures.

First, here are the details of the genny and motor we're looking at. We plan on adding solar to this project as soon as possible (which may be a year - but still should be factored in to the design).

We use a minimum of 3,500W. I'd "estimate" we use about 5,000-6,000W (we've metered it at various times). I've based my calculations off 5KWh.

So far, my math is as follows:

At 12hr, 60KWh/day from batteries
5000Ah battery capacity (at 12V... we'll be using 48V) to "full" discharge
10,000Ah to 50% discharge (again, using 12V for the math)
Forty batteries to ensure no lower than 50% discharge
Batteries at 250Ah
??? Can two inverter chargers recharge them in 12 hours? Currently looking at the Magnum MS4448PAE recommended in the other thread.

During the off time (ie: not using the batts), power to charge will be provided by the generator (and eventually the genny and solar). The genny puts out 11KW, leaving minimum of 5KW available for charging. At most times, as much as 7KW will be available. I'd *like* to limit the genny (motor actually) to 7KW output (more efficient fuel range on the motor), but that leaves between 2KW and 3KW for charging - usually 3KW.

About 2KW of the load cannot be turned off (it has a minor variance of 1800W to 2200W), as it's 24/7 servers and network equipment.

The deeper I dig, the more lost I seem to be getting...

Best,
Rob

Cariboocoot · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

It's easy to get lost in the calculations. Especially if you try to switch between one system Voltage and another.

So, start with the power usage from the batteries in one day (never mind the hours).

If this number is 5 kW hours:

5000 / 48 Volts = 104 Amp hours used. @ 50% DOD = 208 Amp hour 48 Volt battery bank minimum.

If you were to recharge this from solar, you'd be looking at this rule-of-thumb formula:
10% Amp hour rate = 21 Amps @ charging Voltage of 59.2 = 1243 less 77% derating = 1615 Watt array. That would be minimal for recharging the bank at that DOD. Look at the "harvest" potential:

1243 "average Watts" * hours of equivalent good sun (4 usually as a minimum) = 4972 Watt hours.

This can be adapted to generator charging, but you have to consider the power factor of the charger. Some inverter chargers aren't very good, but the Magnum's is close to 1.0.

You're not going to find a 208 Amp hour battery. You would probably go up to 320 Amp hour L16's, and raise the charge rate correspondingly.

There should be no trouble charging this bank in one day with either PV or generator. We do this sort of thing all the time using the same rule-of-thumb formuli above.

RobertMfromLI · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

Hmmm... ok, so I just need to multiply by 12 then? (ie: 12 hours at 5K)

5000W/hr = 5KWh x 12h = 60KWh / 48 = 1250Ah battery capacity *2 = 2500AH capacity (?)

I'll take a look for the 320Ah you recommend. I'd found some 220 and some 250 (at 12V)

RobertMfromLI · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

RobertMfromLI wrote: »

Hmmm... ok, so I just need to multiply by 12 then? (ie: 12 hours at 5K)

5000W/hr = 5KWh x 12h = 60KWh / 48 = 1250Ah battery capacity *2 = 2500AH capacity (?)

I'll take a look for the 320Ah you recommend. I'd found some 220 and some 250 (at 12V)

Ugh... wait, this means that using a 48V setup, since the batteries are 12V, I'd need 10,000AH worth of 12V batteries? That's about where I got last time I figured this out.

BB. · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

You are (partially) confusing yourself over units...

Watts (and kWatts) is a rate (like miles per hour, gallons per minute).

Watt*Hours and kWatt*Hours is an amount (like miles driven, gallons pumped).

So, when you type and equation like this:

5000W/hr = 5KWh x 12h = 60KWh / 48 = 1250Ah battery capacity *2 = 2500AH capacity (?)

It gets even more confusing...

First, you had typed 5,000W, 5000WH, and 5000W/H -- And I think you actually are talking about the same thing... A maximum (or average?) load of 5,000 Watts (or 5 kWatts).

That is a huge load for any off-grid/emergency power system--but not unheard of when looking at starting a well pump, an Air Conditioner, etc... And at that power level (again, rate of power usage), you would be looking at a 48 volt battery bank.

The next question is a combination of what does 5,000 Watts really mean--is that your peak usage (average load plus starting the well pump), or is that truly average power usage over a period of time. And is that period of time 1 hour per day, 10 hours per day, 24 hours per day, or what?

Next, in your equation, you typed 12 hours--So that appears to mean that you are looking at an average of 5,000 Watts @ 12 hours per day.

Next, when sizing the battery bank, we need to ask how much "oversize" do you want to run the battery... Typically, we say you should be able to run 1-3 days "without sun/generator use"... The numbers are rules of thumb, but after many long and confusing threads :roll:, it turns out for many reasons smaller than 1 day, the battery is not large enough to handle surge current. And over 3 days, the battery can be difficult to recharge and may set below ~75% state of charge for periods of time--which means that the battery bank will tend to sulfate (can cut years off the life of a battery bank, depending how long and how deeply discharged).

The other "rule of thumb" is to try and avoid discharging a deep cycle battery below 50% state of charge in normal fixed use (you can violate the rule for weekend cabins and RV's where space is critical and the actual number of cycles per year are not that great--as long as the battery is quickly recharged the next day)...

So, the way I would work the math:

5000W/hr = 5KWh x 12h = 60KWh / 48 = 1250Ah battery capacity *2 = 2500AH capacity (?) these are not "=" signs, they are more like process signs "=>" indicating steps... I try to avoid this.

So, instead:

5,000 Watts * 12 hours per day * 1/48 volt battery bank * 2 days of "no sun" * 1/0.50 maximum discharge * 1/0.85 inverter efficiency = 5,882 AH battery bank @ 48 volts

That is a huge battery bank... As an aside, we also need to check the sizing of the battery bank for surge loads (recommend around C/0.4 as maximum surge current). For a 6kW inverter, assume 12kW surge, on a 48 volt battery bank:

12,000 Watt * 1/48 volt battery bank * 1/0.4 maximum surge current = 625 AH @ 48 volt battery bank

Obviously, your battery bank requirement for 2 days of battery only use and 50% maximum discharge is 10x the surge requirement.

Now, lets take a look at the recharging requirements... For solar, roughly 5% to 13% of battery AH requirements is a good range (below 5% is too low and above 13% is very expensive for most solar systems and "overkill"). 10% is a good medium (but pretty expensive for many people in solar panel/charge controller costs)--but I will use it here to make the math less confusing.

Note that for generator, you would use the same 5% to 13% range--although you could go as high as 25% for a generator.

5,882 AH battery bank * 59 volts charging * 1/0.77 derating for solar array+charger * 0.10 rate of charge = 44,610 watt solar array

Or roughly a 50-100 kW genset + charger.

The above is based on recommended charging current... To size the array for hours of sun--Lets assume that this is near Long Beach California. And that you want to run from solar for ~9 months of the year (use a genset during bad weather). On average, a fixed array will get around 5 hours of "full sun" equivalent per day in February:

5,000 W * 12 hours per day * 1/0.52 end to end efficiency * 1/5 hours of sun per day = 23,077 Watt array

So--based on the above numbers, the solar array should be around 23-45(+ a bit) kWatt array to run your load off grid for ~9 months of the year with backup generator for the other 3 months of the year.

I made a whole bunch of assumptions--Take the above and plug in your own assumptions/needs and see if it makes sense to you.

It is not that this cannot be done--but you are looking at spending some serious money to do this sort of system. (maybe $200,000 to nearly $1,000,000 depending on the details).

-Bill

Cariboocoot · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

RobertMfromLI wrote: »

I've based my calculations off 5KWh.

See, that's what I was going from.
As Bill points out, there's a big difference between 5 kilowatt hours and using 5 kilowatts for 12 hours, which is 60 kilowatt hours.

In the latter case you're looking at massive battery banks made with this type of 2 Volt cell:
http://www.solar-electric.com/repoba2vo25a.html

And you would need to use a couple of Xantrex XW 6048 units because their chargers are 100 Amps, and you'd need all that and more for such a huge bank.

For solar, you'd be looking at a 20 kW array. Nothing that $50,000 couldn't buy (uninstalled).

So let's make triple certain we're working with the right Watt hour demand.

RobertMfromLI · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

Hmmm... ok, I see how I confused the situation. Let me try to do the breakdown in a proper fashion:

Our average load is about 4KW
Our "it's all on, no start surge since it's on" load is just shy of 5KW
Our peak load is about 6KW (theoretically, if all the "normally on/used" stuff is on, and all the surge load stuff is turned on at once (ie: fridge compressor starts at the same time as the single AC compressor and at the same time I decide to do something that takes the laserprinter out of sleep mode (use it once a month maybe)))
We have selected a generator that can handle this load (11KW) with surplus, including theoretical surge
We hope to achieve 12 hours on batteries... though achieving 8hrs would be fine
No need for 2-3 day "on battery" time - just 8 to 12 hours (we've got grid power to fall back on - usually, and the genny can be run 24/7 for a month off the tank we already have (minus weekly maintenance))
Our minimal load is 3.5KW (there are always servers and an AC unit running which draw 2KW)
Heat (water, house) is natural gas, so not in the equation. The stove is a mix... rangetop is gas, inside oven is electric.

Now... the battery AH part of the math is where I've been getting confused with what I've seen online (and various calculators, where when the same figures are entered, come up with radically different results). The way I'd been doing it was "cheating" by converting everything to watt hours. So (as a simple example), if the place used 100Wh for 10 hours, (assuming for simplicity, not caring about deep discharging) we'd need a battery bank that at C/10 were rated 8.3Ah for 12V, 4.16Ah for 24V and so on.

Now, from there, since the batteries come in 12V (and smaller voltages, which I hadn't been looking at), I used that for the equations... thus, if I wanted an hour at 5KWh, I'd need (for 50% discharge) double 417Ah worth of batteries. (?)

At a C/10 rating for Ah on the batteries, for 10 hours, I figured I'd need 10 times the batteries... I know the drawable amps are different at C/10 and C/20, etc... and thus in reality, figuring out only an hour wouldn't be done the way I did above, but since I'm using C/10 figures and 10 hours desired run-time, I figured the math works for my needs.

So for ten hours, I figure I need, if I haven't totally screwed something up, 2085Ah at 48V for 50% discharge.

And right about there (getting the figures for 48V and then selecting proper Ah batteries) is where I get confused... what I figured was 8,340Ah worth of 12V batteries, (in the correct quantity of course for balanced strings) set up as a 48V bank would simply be 48V @ 2085Ah (minus efficiency differences, etc, etc). Which is how I was trying to select... or about 36 250Ah batteries (giving it a little surplus) at 4 strings in series of 9 parallel batteries...

BB. · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

C/10 is a fairly stiff rate of discharge (typical flooded cell lead acid batteries max out around C/8 without forced air cooling).

Pretty much, you have it now--But, I think, you missed/are confusing an extra "h" in this paragraph:

So (as a simple example), if the place used 100Wh for 10 hours, (assuming for simplicity, not caring about deep discharging) we'd need a battery bank that at C/10 were rated 8.3Ah for 12V, 4.16Ah for 24V and so on.

Are you after:

100 Watt*Hours?
100 Watts * 10 Hours = 1,000 Watt*Hours
100 Watt*Hours / 10 Hours = 10 Watt Average Load

Battery size (assuming 50% maximum discharge):

100 Watt*Hours * 1/12 volts * 1/0.50 max discharge = 16.7 Amp*Hour battery bank @ 12 volts
1,000 WH * 1/12 volts * 1/0.50 = 167 AH 12 volt battery bank

Backing up a moment... If you have grid power, and a back up genset, etc... Are you trying to create a Battery backup (say XX hours, then start genset) for a Grid Tied system? (high reliability/uptime?)

Generally, for such a system, you would run from the grid 99% of the time, and only run from battery bank/genset during power outages... Much more efficient (less conversion losses, less cycling on the battery bank, etc.). Simply, utility power (at least in North America) is pretty much cheaper than you can generate from solar/wind/genset--when going through batteries.

In some regions, people use batteries + AC battery chargers to carry loads through the "normal" 8 hour afternoon power outages (some Caribbean islands, etc.).

Battery backed power, solar/wind power, etc. are expensive forms of power. As is running a backup genset for long periods (particularly if operated at less than 50% of rated load)...

I guess I lost/never understood "why"?

-Bill

RobertMfromLI · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

Hi Bill, here's what we worked out... running 24/7 off the generator, currently would be cheaper than grid power. Yes, we pay that much. Our area fights for the top spot when it comes to highest price (and often makes it, as I am sure we will after the upcoming rate increase). Currently, we're behind by under a penny per KWh. Buying fuel, we'd save about $100/mo.

But, the idea was to use the surplus the generator is making to charge batteries... once at 100%, kill the generator (which feeds through inverters), at which point the inverters would switch to the batteries to 50% discharge, then start the genny.

At minimum, we create about 3.2KWh surplus on the generator. Worst case, with it "dialed down" to about 7.5KW. At best, we generate over 7KW surplus.

Hence, where the batteries came into the equation... store the surplus, use it, then restart the generator. Later, generator run-time could be limited even further with solar and additional batteries (or so the plan goes).

At half day runtime on the genny, we save over $350/month... minus of course, the initial expense in batteries, inverters, chargers, etc.

And of course, it allows for maintenance intervals for the generator... grid power would be used as a last resort in the event the genny or the battery/inverter system needed more than their regular "maintenance" (ie: battery checking/filling/etc, oil change on the generator, etc).

Best,
Rob

stephendv · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

A diesel generator is fuel efficient between 40% to 90% load. If your minimum load is 3.5kW and your maximum is 6kW, then as long as the genset is about 7-8kW you'll be running efficiently on diesel alone.

The only time you'll have any "surplus" power to speak of is if the generator is massively oversized for those loads. So simply adding batteries into the mix isn't going to save you a whole lot as you'd now be running the generator harder for a shorter amount of time - and using roughly the same amount of fuel.

Adding batteries would help if there was a large disparity between the minimum and maximum power requirements, which 3.5-6kW isn't.

I think you'll only start seeing savings if you add solar into the mix.

Cariboocoot · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

4 kW average load is massive for an off-grid home. What sort of industry are you running there?

That works out as 4000 Watts * 24 hours = 48 kW hours per day. Spending money getting that number down will save you more than any battery/solar scheme. I don't know what your electric rate is, but it's hard to imagine it exceeds $1.00 per kW hour which is about the cost of solar electric. In Hawaii it's only around $0.50 per kW hour and that's all generator driven with imported fuel.

At any rate, 4kW over 12 hours would be 24kW hours. 24000 / 48 Volts = 500 Amp hours, times two for a minimum 1,000 Amp hour 48 Volt battery bank. Again the 2 Volt cells would be better: http://www.solar-electric.com/repoba2vo10a.html You could use two Magnum http://www.solar-electric.com/maenms4444wa.html inverters on that @ 80 Amps peak each for 160 Amps total current capacity. Assuming that any given full loading doesn't exceed 8kW at any time. For solar you'd be looking at an array size around 7.8 kW. From the generator it would need around 6kW peak for battery charging, plus whatever loads are running. Using the 4kW average that would come to a maximum of 10kW, so the 11kW gen would be fully loaded at maximum power usage.

I still think you're better off spending money on reducing consumption.

mike95490 · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

Cycling batteries to 50% is pretty hard on them, and will shorten their lifetimes a LOT. Better check your lifetime/cycle depth chart to see when you will be buying batteries again, and factor that into the costs.

Running servers- there are now some lower power servers out, and I think google was looking at running some of their data centers at high ambient airflow, no chillers. Conserve, conserve, conserve !

RobertMfromLI · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

stephendv wrote: »

A diesel generator is fuel efficient between 40% to 90% load. If your minimum load is 3.5kW and your maximum is 6kW, then as long as the genset is about 7-8kW you'll be running efficiently on diesel alone.

The only time you'll have any "surplus" power to speak of is if the generator is massively oversized for those loads. So simply adding batteries into the mix isn't going to save you a whole lot as you'd now be running the generator harder for a shorter amount of time - and using roughly the same amount of fuel.

Adding batteries would help if there was a large disparity between the minimum and maximum power requirements, which 3.5-6kW isn't.

I think you'll only start seeing savings if you add solar into the mix.

Sorry for any confusion above... yeah, I know the efficiency range... it's an 11KW generator, hence the surplus. We're gonna drop it (the diesel engine) down in RPMs so it can handle roughly 7KW (a little over 7.5KW) if we cannot store the surplus - it's a bit better on fuel consumption at that range. We've got pulleys designed for just such a purpose... drop 30% RPM, switch to the proper pulley, still generating dual 120VAC @ 60Hz - the only differences being the diesel at that RPM can't handle the 11KW load (rated at about just shy of 8KW) but consumes less fuel at the lower RPM.

Since our load remains the same (and within range of using either pulley/RPM combo), dropping the RPM to keep load within the diesel's efficiency range seems to make sense since it drops fuel consumption.

Best,
Rob

Edit:
We're planning on about 8.5KW-9KW available from the generator if we use batteries for surplus. I think I screwed that up earlier on in one of my earlier posts...

mike95490 · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

Don't forget to keep your genset loaded to about 50-80%, or it "wet stacks" and does not run hot enough to prevent carboning up.

RobertMfromLI · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

mike90045 wrote: »

Don't forget to keep your genset loaded to about 50-80%, or it "wet stacks" and does not run hot enough to prevent carboning up.

Will do... also, fuel efficiency drops dramatically outside of that range. Fortunately, the office/server room is a continuous load that barely fluctuates (about 15% change, max), which will help ensure that with the other loads, it will stay in that range. And when it gets nearer to 100%, the governor/fuel rack rides way too far out for my comfort - so, I'm definitely planning it out to avoid getting near that. We should be sitting in the 50-70% range, with rare surges of coincidence pushing it to 80% for a few seconds (if even that long or at all, due to the inertia built up in the flywheels).

BB. · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

Do you have an approximate fuel usage figure for the genset (gallons per hours at X kW output)?

Very roughly, I have seen numbers of around 7kW diesel genset running at 1 gph... Assuming ~$3 per gallon for untaxed diesel (no road taxes):

$3 per gallon * 1 gallon per hour * 1/7kW*1 Hours = ~$0.43 per kWH

No capital/maintenance costs... In my area of California, residential for >~900 kWH per month is ~$0.34 per kWH (top rate). My Time of Use/Seasonal maximum rate for Mon-Fri, noon-6pm rate is ~$0.52 per kWH.

If you can swing a Grid Tied system--You may actually save money.

In California, it is the only "legal" way to generate your own power is Grid Tied (solar/wind)... Assuming your weather is similar to New York City... Your average sun is 4.56 hours per day over a year. Assuming 4kW * 12 hours a day average load (7 days a week), and 0.77 solar power system derating:

4kW * 12 hours per day * 1/0.77 derating * 1/4.56 hours of sun per day = 13.7 kW array = 13,700 Watt Array.

That is actually a pretty doable system... Assuming $6 per watt installed with a 30% federal tax credit:

13,700 Watt * $6 per Watt * 0.70 after fed tax rebate = ~$57,540 system cost

The per kWH price, assuming 20 year life and $7,000 maintenance costs over 20 years (I am a pessimistic engineer):

($57,540+$7,000)/(4kW*12 hours per day * 365 days * 20 years) = $0.184 per kWH

That will be a real savings over almost anything else you can do (short of moving to a low cost state)...

Of course, this assumes you have:

a utility that will allow Grid Tied Solar
utility supports net metering (ideal 1 year net metering)
utility supports retail priced net metering
you are a residence (power plans for commercial usually get way more complex for ROI calculations)

Anyway, just some back of the envelope calculations.

Also, pure grid tied does not provide for battery backup power... You have several ways of providing backup power--but it will add to the costs:

Traditional UPS systems for computers + AC generator backup
Hybrid Grid Tied / Off Grid solar+battery system.

Either system has its plus/minuses... And will add significantly to the cost of power (for a Hybrid GT system, you are probably looking at, very roughly, ~$0.45 range for cost of power). Pure Off Grid Solar Power is probably more in the range of $1-$2+ per kWH.

Depending on your billing plans, you may not have to offset 100% of your power costs... For example, my power plan is ~$0.09-$0.12 per kWH for less than 300 kWH per month usage... And can be as high as ~$0.35-$0.52 per kWH (depending on plans/usage)... So, here, we can save quite a bit of power by shaving off any usage over ~900 kWH per month.

-Bill

RobertMfromLI · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

BB. wrote: »

Do you have an approximate fuel usage figure for the genset (gallons per hours at X kW output)?

Very roughly, I have seen numbers of around 7kW diesel genset running at 1 gph... Assuming ~$3 per gallon for untaxed diesel (no road taxes):

Heh... ready for this? At 5.5kW load, roughly 5.33... ummm... hours per gallon (yes, hours per gallon - not gallons per hour). I literally spent a week digging and digging to make sure these figures were correct, and found others who'd charted it.

BB. wrote: »

$3 per gallon * 1 gallon per hour * 1/7kW*1 Hours = ~$0.43 per kWH

No capital/maintenance costs... In my area of California, residential for >~900 kWH per month is ~$0.34 per kWH (top rate). My Time of Use/Seasonal maximum rate for Mon-Fri, noon-6pm rate is ~$0.52 per kWH.

Top rate here is third in the country, only behind Connecticut and Hawaii - and only by under a penny at that... soon to be rectified by the rate increase coming up in a couple months.

Now, I won't mention the abundance of veggie fuel in the area, which this thing is well suited to run with no mods...

BB. wrote: »

If you can swing a Grid Tied system--You may actually save money.

In California, it is the only "legal" way to generate your own power is Grid Tied (solar/wind)... Assuming your weather is similar to New York City... Your average sun is 4.56 hours per day over a year. Assuming 4kW * 12 hours a day average load (7 days a week), and 0.77 solar power system derating:

(yes, very similar to NYC sunlight - minus some of the pollution)

Still waiting on the back and forth with the various installers and such... it's been a year of nonsense. Cutting three trees we'd rather not... building a solar array stand where we get the most sun, re-testing for exposure, moving this, re-testing, moving that... not sure if we qualify for rebates; change this... re-testing... use these cheapo, low power, high price cells (yeah, right), and on and on.

BB. wrote: »

That will be a real savings over almost anything else you can do (short of moving to a low cost state)...

Or running the genny 24/7 through a decent power conditioner, and with whatever amount of backup capacity we can afford via batteries to allow maintenance without power interruption (and without the need to switch to grid for maintenance)...

BB. wrote: »

Of course, this assumes you have:
a utility that will allow Grid Tied Solar

utility supports net metering (ideal 1 year net metering)

utility supports retail priced net metering

you are a residence (power plans for commercial usually get way more complex for ROI calculations)

Anyway, just some back of the envelope calculations.

Also, pure grid tied does not provide for battery backup power... You have several ways of providing backup power--but it will add to the costs:
Traditional UPS systems for computers + AC generator backup

Hybrid Grid Tied / Off Grid solar+battery system.

Either system has its plus/minuses... And will add significantly to the cost of power (for a Hybrid GT system, you are probably looking at, very roughly, ~$0.45 range for cost of power). Pure Off Grid Solar Power is probably more in the range of $1-$2+ per kWH.

Depending on your billing plans, you may not have to offset 100% of your power costs... For example, my power plan is ~$0.09-$0.12 per kWH for less than 300 kWH per month usage... And can be as high as ~$0.35-$0.52 per kWH (depending on plans/usage)... So, here, we can save quite a bit of power by shaving off any usage over ~900 kWH per month.

-Bill

Buyback rate (last I checked, which was almost 2 years ago) is retardedly low. A relatively small fraction of what they sell at. But, it's better than Baltimore 10+ years ago, when they'd passed a law forcing the electric company (BGE) to buy back power... which they claimed they couldn't do, because the bill did not specify how - so until the govt decided to pass a bill that stated how they should pay people, they refused... for at least 5 years (then I gave up keeping track of what was going on). But then again, BGE is the most corrupt, idiotic power company I know of. We went off grid, had the power lines physically cut (by THEM no less) between the pole connection and house, had the meter disconnected and went off grid, entirely generator based (and small inverter/battery setup for supplemental lights)... and they billed us based off claimed meter readings month after month after month. Took a year of fighting and them insisting that's what our meter was saying each and every month - and the only thing that resolved it was a 30 year supervisor who we managed to drag out there to confirm there was no mains power, couldn't be, the meter was and had been disconnected, matched all that to their work orders, and fought for us. Three months later, the bills were fixed. What's that... 1.5-2 years of that crap? They did that to the Baltimore City Govt as well - for another building THEY disconnected - to the tune of a quarter million dollars. The city of course fought and won very quickly (and it made the papers)... but that's how stupid and corrupt they are.

Ugh, sorry, off tangent. LIPA, while not nearly as corrupt, does have ridiculous electricity pricing sadly... makes almost any other power alternative cheaper.

Best,
Rob

BB. · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

Yea, I was getting to the point of believing a good quality diesel generator was getting to be cheaper to run than utility power in California (if you have heavy loads)...

I still wonder about the fuel consumption... I would be closer to 1/2 gallon of diesel per hour for a 5kW load from looking around the web. But I certainly could be wrong. Worst case:

$4 per gallon * 1/2 gallon / 5 kWH = $0.40 per kWH

In California, they are pretty much killing the use of diesel engines for off road / fixed applications unless they meet new air pollution standards (not cheap). They have even gone as far as having parts stores log the name/vehicle/engine information from customers back to the state for enforcement action (all based on bad science from a diploma mill Phd state employee--another issue).

In California--We actually have to pay the Utility Company if a home/business goes "off grid" to pay for "utility investments" based on assumed power usage / stranding of generator capacity (solar/wind has an exemption for small generators--typically under 10kW).

-Bill

RobertMfromLI · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

BB. wrote: »

Yea, I was getting to the point of believing a good quality diesel generator was getting to be cheaper to run than utility power in California (if you have heavy loads)...

I still wonder about the fuel consumption... I would be closer to 1/2 gallon of diesel per hour for a 5kW load from looking around the web. But I certainly could be wrong. Worst case:
$4 per gallon * 1/2 gallon / 5 kWH = $0.40 per kWH

In California, they are pretty much killing the use of diesel engines for off road / fixed applications unless they meet new air pollution standards (not cheap). They have even gone as far as having parts stores log the name/vehicle/engine information from customers back to the state for enforcement action (all based on bad science from a diploma mill Phd state employee--another issue).

In California--We actually have to pay the Utility Company if a home/business goes "off grid" to pay for "utility investments" based on assumed power usage / stranding of generator capacity (solar/wind has an exemption for small generators--typically under 10kW).

-Bill

Perkins makes one that's about 1/3 gallon at 5kWH. The lower speed an higher horsepower on the Lister Type engines end up using under 2/10th a gallon an hour... (I think Perkins has one rated at .29 gph - not drastically more than the Lister's .2 gph). The lower revolutions normally cause issues with surge variances, which is compensated for by two massive flywheels and the governor (the job of which is to deal with any extended surges before inertia is overcome). The engine with compression valves open (ie: no compression), takes up to a minute (depending on what RPMs it's set to) to spin down just based on inertia. I used a little less than 5gph for my math and came up with a savings of $136 a month buying fuel at $4/gal.

solar_dave · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

So assume you save $136 a month, what is your ROI on the equipment and hookup. Better figure diesel costs will rise in the future as well, unless you can secure a source of fuel at a fix cost.

RobertMfromLI · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

solar_dave wrote: »

So assume you save $136 a month, what is your ROI on the equipment and hookup. Better figure diesel costs will rise in the future as well, unless you can secure a source of fuel at a fix cost.

Veggie oil, waste oil and such (as noted someplace buried above). Even without using that, if we can eventually manage 12 hours of battery time, we'd save $450/month at current prices... WELL over $700 a month on veggie oil and waste oil. The initial cost would be high, but at $9,000+ saved a year, 4 years would cover it, leave 3-6 years for saving for another set of batteries, which is still far cheaper than what we currently pay for electricity.

solar_dave · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

WOW $9000 at year i would be looking for alternatives as well, when my bill hit $4500 a year I did the grid tie solar deal and the conservation routine. My ROI is similar.

RobertMfromLI · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

solar_dave wrote: »

WOW $9000 at year i would be looking for alternatives as well, when my bill hit $4500 a year I did the grid tie solar deal and the conservation routine. My ROI is similar.

Yeah, Star Trek Phase 2 has been costing us a pretty penny for electricity... though we'd probably be at $6500 without it. LIPA is ridiculous (even moreso as they press for more rate increases - there's a new 1.9% coming, followed by a 2.5%). Current privatization plans are supposed to come with an even bigger increase. This, from the power company that's proud their JD Power rating went from last, to second to last, to third to last (16th of 18 ) for eastern seaboard power companies. And 6th worst in the country. Talk about spinning something bad in a big way.

BTW, the 21 points they brag about moving up? It's on a THOUSAND point scale. LoL!

Still better than BGE though. I'd rather get billed a lot for power I actually use (LIPA), than get billed for power I've never used (BGE).

Shane Jackson · September 2011

Re: Help: Calculating battery sizes and inverter sizes?

I think you are confused on the surplus of your generator. If your load is say 5KWH you will burn 5gph of fuel (just pulling numbers from the air for illustration purpose). If your load is 3KWH you will burn 3.7ghp. Add the battery bank and chargers and your load will go up to say 9KWH and use 8.6gph.

What I am trying to say is your fuel usage is directly related to the load applied to the generator. It's not like you are wasting the surplus as you truly are not generating it!

Hope that helps.

Shane

Danielkamoah · October 2018

Please help me calculate the number of 12v 200AH batteries do i need to run a load of 7000 watts for 15 hours. The system is 48v.

BB. · October 2018

Daniel,

Please start your own thread--However I will give you a quick example here...

Assume that you are using flooded cell lead acid deep cycle storage batteries. Normally, discharge to 50% state of charge (for longer battery life). And, if this was using solar panels for recharging, assume 2 days of energy storage (bad weather, use a genset on the 3rd day, if needed). The calculation would be:

7,000 Watts * 15 hours * 1/0.85 AC inverter efficiency = 123,529 Watt*Hours of storage required

Now convert that into Amp*Hours at 48 VDC battery bank:

123,529 Watt*Hours * 2 days of storage * 1/0.50 max discharge * 1/48 VDC battery bank = 10,294 AH battery bank @ 48 volts

Each battery you have stores 200 AH, so the number of Parallel strings would be:

10,294 AH / 200 AH per battery string = 51.47 ~ 52 parallel strings of 200 AH batteries

You have 12 volt batteries, so the number of batteries per string:

48 VDC Bus / 12 volts per battery = 4 batteries per string

And the total number of 12 volt @ 200 AH batteries would be:

52 parallel strings * 4 batteries per string = 208 Batteries total

I do not know what you intend this system to run, but I would highly suggest that you evaluate your loads (really 7,000 Watts * 15 hours per day?). Also, 12 volt @ 200 AH Batteries for such a large battery bank is not an ideal solution (lots of small batteries in series/parallel could be a maintenance nightmare).

Also, this is making assumptions that this is a solar powered system used every day... If this was for a battery backup system (rarely used, have Utility power and Gensets for normal operation), you could use a smaller battery bank (discharge to 20% state of charge, no 50% buffer for longer battery life). In this case, you use the batteries 2-4 times over a couple of years, and then replace the whole battery bank (UPS system batteries are typically replaced every 2 years whether used or not):

123,529 WH per cycle * 1/0.85 AC inverter eff * 1/0.80 depth of discharge * 1/48 volts = 3,785 AH @ 48 volts
3,785 AH / 200 AH per battery string = 18.9 ~ 19 parallel strings of batteries
19 parallel strings * 4 batteries per string = 76 Batteries total

This is why I suggest you start your own thread/discussion--We can discuss your needs without confusing those with the needs of the original poster in this thread.

-Bill

mike95490 · October 2018

Danielkamoah said:

Please help me calculate the number of 12v 200AH batteries do i need to run a load of 7000 watts for 15 hours. The system is 48v.

Ouch. Not likely to happen, unless you get a diesel electric submarine to hold the batteries you need.

Help: Calculating battery sizes and inverter sizes?

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