House Battery and Energy Management
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rebelrider.mike
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Thanks guys! That helps. I thought I read all the FAQs, but somehow I missed that one. So I'm looking for a hybrid inverter to handle an "energy pool" type of setup. I was thinking hybrid to mean multiple alternative sources, like solar + wind.
So is that what the PIP inverters do that everyone talks about? Manages the routing of power depending on the battery and grid status? Are there other products that do the same thing? I'm happy to read up on all these things if I could only find out what things to read up on! LOL. I will search for hybrid inverters and see what I come up with.
I feel like I'm really close to deciding on a chemistry for the battery. Lithium seems to be tied for now. The thing I keep coming back to is that lead/acid and Ni-Fehave high internal resistance and self discharge rates vs the lithium chemistries. If I were doing a backup battery only, I'd probably go with lead/acid. But as these cells will be used on a daily basis, I'm thinking lithium is the way to go.
I know I'm taking the long way round to make the same decisions everyone else has already made. But I think it's important to really understand this stuff. So I appreciate y'all putting up with me.
So is that what the PIP inverters do that everyone talks about? Manages the routing of power depending on the battery and grid status? Are there other products that do the same thing? I'm happy to read up on all these things if I could only find out what things to read up on! LOL. I will search for hybrid inverters and see what I come up with.
I feel like I'm really close to deciding on a chemistry for the battery. Lithium seems to be tied for now. The thing I keep coming back to is that lead/acid and Ni-Fehave high internal resistance and self discharge rates vs the lithium chemistries. If I were doing a backup battery only, I'd probably go with lead/acid. But as these cells will be used on a daily basis, I'm thinking lithium is the way to go.
I know I'm taking the long way round to make the same decisions everyone else has already made. But I think it's important to really understand this stuff. So I appreciate y'all putting up with me.
Korishan
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For me, lithium is pretty much a no brainer. Setting costs of new cells/batteries aside, Lithium is far superior. You can store for more energy in a smaller space with a lot less weight without the risk of hydrogen explosion or corrosive acid anywhere.
Lithium does need to have a bit more hands on work than LA's do, but the pros far outweighs the cons easily.
But in any light, it's good to know that you are educating yourself well, before making any lasting decisions. This will help drastically minimize any mishaps down the road
Lithium does need to have a bit more hands on work than LA's do, but the pros far outweighs the cons easily.
But in any light, it's good to know that you are educating yourself well, before making any lasting decisions. This will help drastically minimize any mishaps down the road
daromer
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Lithium is generally better for cycling yes.
PIP inverters are Offgrid. Atleast the grey ones The MPI and V series from MPP like the one I got is Hybrid. Beware that a Hybrid system is not as efficient and have rather high idle load. Like all Online UPS systems have
PIP inverters are Offgrid. Atleast the grey ones
The MPI and V series from MPP like the one I got is Hybrid. Beware that a Hybrid system is not as efficient and have rather high idle load. Like all Online UPS systems have
rebelrider.mike
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Searching for Hybrid inverters yielded tons of results! Also, thanks to all the folks on the Facebook group, I've been able to gather a list of several inverters that will work with split-phase. Contenders for today are:
Looks like real progress will be pretty slow for quite a while. I'm going to be away for several months this year, and won't be able to do much home improvement. It also looks like what improvement I am able to do will be focused on leak control. I've got an old leaky chimney that needs to be re-capped and re-sealed, as well as several leaks in the basement to take care of. That's going to require both inside and outside work to convince rain water that it's easier to stay outside, LOL.
Almost started sharing my math today, but I think I've discovered yet another huge flaw that I need to correct. They say if you can't explain it to someone else, you don't really know what you're talking about. That is me and math today!
- Outback
- SMA
- Schneider
- Selectronic
- Aims
Looks like real progress will be pretty slow for quite a while. I'm going to be away for several months this year, and won't be able to do much home improvement. It also looks like what improvement I am able to do will be focused on leak control. I've got an old leaky chimney that needs to be re-capped and re-sealed, as well as several leaks in the basement to take care of. That's going to require both inside and outside work to convince rain water that it's easier to stay outside, LOL.
Almost started sharing my math today, but I think I've discovered yet another huge flaw that I need to correct. They say if you can't explain it to someone else, you don't really know what you're talking about. That is me and math today!
Hey Rebelrider
I'm doing something of an experiment. My small (small for now) system is at first meant to run a few of my house AC circuits that I have drawing juice all day - namely a radio system that is connected to other radio systems over internet links.
I'm using LA batteries now but am in the process of an 18650 battery system.
I'm doing this because I got a free 2500W inverter and so I figure I'll run 2500Watts worth of stuff off grid. This means I'm taking a couple circuits out of my
main service panel and running them off batteries and inverter olny - no grid tie, no islanding issues etc...
If your power company has a good program of net metering or similar grid tie might be great. Some power companies actually pay YOU for excess energy - rare in USA but common in many provinces in Canada
Just a thought for you if you don't have the resources to go "all in" at once with a big 10kw inverter system etc...
Good luck
I'm doing something of an experiment. My small (small for now) system is at first meant to run a few of my house AC circuits that I have drawing juice all day - namely a radio system that is connected to other radio systems over internet links.
I'm using LA batteries now but am in the process of an 18650 battery system.
I'm doing this because I got a free 2500W inverter and so I figure I'll run 2500Watts worth of stuff off grid. This means I'm taking a couple circuits out of my
main service panel and running them off batteries and inverter olny - no grid tie, no islanding issues etc...
If your power company has a good program of net metering or similar grid tie might be great. Some power companies actually pay YOU for excess energy - rare in USA but common in many provinces in Canada
Just a thought for you if you don't have the resources to go "all in" at once with a big 10kw inverter system etc...
Good luck
rebelrider.mike
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Heyrtgunner, free is a hard price to beat!
I'll be working with my power provider to see exactly what they will be willing to help me with. Fortunately, they're pro solar, and are happy to buy any extra electricity I produce. In fact, they have a big solar array on their headquarters property which the community paid into. Every month, my power bill gets a tiny credit from the power it produces.
I'm not sure if my electronic meter is two-way ready, but they will swap it out if need be. At some point I will have to get together with them and figure out details. I do know they have a list of contractors who do PV installations. I've got tons of reading and preparation to do yet before I'm ready to begin installation though. Hopefully, hybrid systems will be more common by then.
I've been working on energy usage calculations to figure out what kind of battery I will need to run my house. I've got some assumptions that I've had to make in order to figure it all out, so I'll start with those.
Assuming all that is true, or close enough, I can translate my power bill into something I can use to design a house battery. So far as I can see, my peak power usage at any one point has been 5,600W. This is from looking back at the data collected from the electronic meter that got installed last year. I figure I'll round that up to 6,000W just to be safe. Drawing 6,000W from a 48V battery would mean drawing 125A.
6,000W / 48V = 125A
Then my maximum Amp draw will be 125A. This is the first critical bit of data I will need for the battery. And also the Voltage of the battery, which I just decided would be 48V, because all the cool people are doing it that way. And I want to be cool too!
Next is the average power usage. This one is a little tricky, because my power bill fluctuates wildly throughout the year. Ultimately, I decided just to average all the monthly totals for the last year together, because I don't know a better way to figure it. Then I needed to reduce it down to an hourly figure, as I'm calculating for a battery that will run for some number of hours. Here's what I came up with:
Average power => 3,008.333kWh/month x 1000Wh/kWh => 3,008,333Wh/month / 30days/month => 100,278Wh/day / 24h/day => 4,178Wh/h => 4,178W
Still with me so far? I hope so. Knowing the average Watts that will be used per hour, I can find the average Amps per hour too. Everything is calculated based on nominal voltage by the way. Different chemistries are going to have different nominal voltages, even when connected in series, but they all seem to be able to be set up to around 48V so I'm sticking with that.
4,178W / 48V = 87A
Next I've got to pick how long the battery should last, powering the house unassisted. It seems to be the general consensus that 3 days is the ideal. But I'm starting small, and plan to expand later with 3+ days being a goal to work toward. For now, I'm using a nice safe number like, say, 3 hours. (Long enough for a typical power outage in my area.) The minimum capacity I'll need can be calculated by the average Amps and how many hours I want to draw the Amps for.
87A x 3h = 261Ah
The minimum energy I'll need is capacity times Voltage:
261Ah x 48V = 12,535Wh
So now I have all the properties I need to design a battery that can power my house. Here's a summary of these properties:
I actually had a power outage yesterday during a morning windstorm. There were 4 substations out, along with about 12,000 families. The power company was able to get the vast majority of us connected back up within a couple of hours. In the meantime, I turned off the furnace thermostat, water heater, and a few other breakers in the hope that my local transformer wouldn't blow up when the power got restored. (This happens a lot where I live. Lights turn on, followed by several KABOOMs!)
Anyway, I was reminded that even using a backup battery for power outages, lots of things can be done to keep the battery from discharging as fast. The biggest would be shutting off the furnace and water heater. Those are the biggest power users in my house. In fact, during the coldest part of winter, I used around 5,500kWh of power. The mildest month I used only 1,000kWh. I figure the difference was my furnace. That's 4,500kWh difference!
There is something to be gained by doing without power for a few hours. It cleared the place out! My wife and son both fled in search of TV and Wi-Fi. I on the other hand, sat on the couch with a blanket and a good book, and had a nice quiet house for a while, all to myself.
I'll be working with my power provider to see exactly what they will be willing to help me with. Fortunately, they're pro solar, and are happy to buy any extra electricity I produce. In fact, they have a big solar array on their headquarters property which the community paid into. Every month, my power bill gets a tiny credit from the power it produces.
I'm not sure if my electronic meter is two-way ready, but they will swap it out if need be. At some point I will have to get together with them and figure out details. I do know they have a list of contractors who do PV installations. I've got tons of reading and preparation to do yet before I'm ready to begin installation though. Hopefully, hybrid systems will be more common by then.
I've been working on energy usage calculations to figure out what kind of battery I will need to run my house. I've got some assumptions that I've had to make in order to figure it all out, so I'll start with those.
- So energy usage is measured (for billing purposes) in kWh, which apparently accumulate over time. I'm assuming that can be expressed as kWh/month.
- I'm also a little confused as to what is energy vs. power. If I understand correctly, Watts are a measurement of power, and Watt-hours are a measurement of energy. Is kWh/month energy or power? Does it even matter?
- I'm also assuming that 1Wh/hour is the same as 1 Watt. Since the hours would cancel each other.
- Lastly, I'm assuming that DC Watts are close enough to AC Watts (orVolt-Amps?) in our application to figure if I use 1 Watt of power at 240VAC, that it's the same as 1 Watt of power at 48VDC.
Assuming all that is true, or close enough, I can translate my power bill into something I can use to design a house battery. So far as I can see, my peak power usage at any one point has been 5,600W. This is from looking back at the data collected from the electronic meter that got installed last year. I figure I'll round that up to 6,000W just to be safe. Drawing 6,000W from a 48V battery would mean drawing 125A.
6,000W / 48V = 125A
Then my maximum Amp draw will be 125A. This is the first critical bit of data I will need for the battery. And also the Voltage of the battery, which I just decided would be 48V, because all the cool people are doing it that way. And I want to be cool too!
Next is the average power usage. This one is a little tricky, because my power bill fluctuates wildly throughout the year. Ultimately, I decided just to average all the monthly totals for the last year together, because I don't know a better way to figure it. Then I needed to reduce it down to an hourly figure, as I'm calculating for a battery that will run for some number of hours. Here's what I came up with:
Average power => 3,008.333kWh/month x 1000Wh/kWh => 3,008,333Wh/month / 30days/month => 100,278Wh/day / 24h/day => 4,178Wh/h => 4,178W
Still with me so far? I hope so. Knowing the average Watts that will be used per hour, I can find the average Amps per hour too. Everything is calculated based on nominal voltage by the way. Different chemistries are going to have different nominal voltages, even when connected in series, but they all seem to be able to be set up to around 48V so I'm sticking with that.
4,178W / 48V = 87A
Next I've got to pick how long the battery should last, powering the house unassisted. It seems to be the general consensus that 3 days is the ideal. But I'm starting small, and plan to expand later with 3+ days being a goal to work toward. For now, I'm using a nice safe number like, say, 3 hours. (Long enough for a typical power outage in my area.) The minimum capacity I'll need can be calculated by the average Amps and how many hours I want to draw the Amps for.
87A x 3h = 261Ah
The minimum energy I'll need is capacity times Voltage:
261Ah x 48V = 12,535Wh
So now I have all the properties I need to design a battery that can power my house. Here's a summary of these properties:
- Nominal Voltage: 48V
- Maximum Current: 125A
- Maximum Power: 6,000W
- Average Energy Consumption: 4,178Wh
- Average Current: 87A
- Continuous Unassisted Runtime: 3 hours
- Minimum Capacity Needed: 261Ah
- Minimum Energy Storage Needed: 12,535Wh
I actually had a power outage yesterday during a morning windstorm. There were 4 substations out, along with about 12,000 families. The power company was able to get the vast majority of us connected back up within a couple of hours. In the meantime, I turned off the furnace thermostat, water heater, and a few other breakers in the hope that my local transformer wouldn't blow up when the power got restored. (This happens a lot where I live. Lights turn on, followed by several KABOOMs!)
Anyway, I was reminded that even using a backup battery for power outages, lots of things can be done to keep the battery from discharging as fast. The biggest would be shutting off the furnace and water heater. Those are the biggest power users in my house. In fact, during the coldest part of winter, I used around 5,500kWh of power. The mildest month I used only 1,000kWh. I figure the difference was my furnace. That's 4,500kWh difference!
There is something to be gained by doing without power for a few hours. It cleared the place out! My wife and son both fled in search of TV and Wi-Fi. I on the other hand, sat on the couch with a blanket and a good book, and had a nice quiet house for a while, all to myself.
Korishan
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You have most of your information correct. Just a little hazy in some areas:
You got your math right here. 125A max draw from your packs is what you'd expect. Well, sorta. There will be power losses, so assume 125 + 10-15% = ~150A to be on the safe side.
The best way to calculate the amount you need, is to base it off your max per year. So, in Winter is when you use the most energy, for you. Down here in Florida, Summer is when we use the most energy. So you'd determine based off averages from Dec - Feb. Assuming these are the most energy hungry months for you.
So, if you had 3008 kWh/month, divide by 30d = 100.3Wh/day. divide by 24h = 4.17Wh. The rest you got
rebelrider.mike said:
- Wh (Watt-Hours) is a measure of energy consumption over time. kWh is 1000Wh. For billing purposes, this is usually equated to a monthly amount, usually greater than 20kWh for most US users
- Energy and Power is practically synonymous here. We usually equate energy as the amount of power stored in a device, and power as how much energy was consumed. But they are basically the same. It doesn't really matter. It just can get annoying if someone goes back and forth between the two terms through out their chat
- There is no such thing as 1Wh/hour, unless you are trying to determine how many Watts a device is rated at by knowing how much it consumed over time. If your water heater used 7500Wh over a period of 5 hours, then you'd take 7500Wh/5hrs = 1500Watts
- Watts is Watts is Watts. This is calculated by Amps * Volts. It doesn't matter if it's AC or DC. Watts is the usage of power/energy. Just like Horsepower is the same in Gas engine vs diesel engine.
rebelrider.mike said:
You got your math right here. 125A max draw from your packs is what you'd expect. Well, sorta. There will be power losses, so assume 125 + 10-15% = ~150A to be on the safe side.
The best way to calculate the amount you need, is to base it off your max per year. So, in Winter is when you use the most energy, for you. Down here in Florida, Summer is when we use the most energy. So you'd determine based off averages from Dec - Feb. Assuming these are the most energy hungry months for you.
So, if you had 3008 kWh/month, divide by 30d = 100.3Wh/day. divide by 24h = 4.17Wh. The rest you got
daromer
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As Korishan said you have alot of power losses. Atleast 10% or even 20% or more in worst case.
Also if your inverter is 6kW nominal you need to calculate 2x that on surge so the system wont die because you strart something heave. Motor or stuff draws easy 5x the nominal power so a 250w fridge can easy take 1500w to start.
Also if your inverter is 6kW nominal you need to calculate 2x that on surge so the system wont die because you strart something heave. Motor or stuff draws easy 5x the nominal power so a 250w fridge can easy take 1500w to start.
rebelrider.mike
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Korishan said:You got your math right here. 125A max draw from your packs is what you'd expect. Well, sorta. There will be power losses, so assume 125 + 10-15% = ~150A to be on the safe side.
The best way to calculate the amount you need, is to base it off your max per year. So, in Winter is when you use the most energy, for you. Down here in Florida, Summer is when we use the most energy. So you'd determine based off averages from Dec - Feb. Assuming these are the most energy hungry months for you.
So, if you had 3008 kWh/month, divide by 30d = 100.3Wh/day. divide by 24h = 4.17Wh. The rest you got
My peak power usage was 5,600W, so 5,600W / 48V = 117A, but I went ahead and rounded up to 6,000W for a safety margin to get the 125A. Do you think I need anothersafety margin on top of this?
Also, you went from 3,008kwh/month to 100.3Wh/day. Do I not need to convert from kWh to Wh at some point?
My wife has just shown me a whole other area of our power app with tons more information than I've been working with! All these numbers are now outdated, LOL. But as long as the math is solid, the numbers are easy to change.
daromer said:
I will keep that in mind for looking at inverter specs. The cells I end up using should also have a peak draw as well as a maximum continuous current. I will take that under consideration as well.
I appreciate the feedback everyone!
I'm learning a huge amount of stuff in a short time, but it's all slowly getting less confusing.
Korishan
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k = 1000, you don't need to convert from kWh to Wh as it's just the same with extra 000's on it. You just need to make sure you do your divisions right accordingly. We both came up with similar numbers (I did some rounding), so either way was fine I just prefer to minimize the math to as least steps as possible. This drove my classmates in Algebra crazy when I did equations on the board that normally took 8-10 steps and I did them in 4
I just prefer to minimize the math to as least steps as possible. This drove my classmates in Algebra crazy when I did equations on the board that normally took 8-10 steps and I did them in 4 
not2bme
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rebelrider.mike said:My peak power usage was 5,600W, so 5,600W / 48V = 117A, but I went ahead and rounded up to 6,000W for a safety margin to get the 125A. Do you think I need anothersafety margin on top of this?
Also, you went from 3,008kwh/month to 100.3Wh/day. Do I not need to convert from kWh to Wh at some point?
My wife has just shown me a whole other area of our power app with tons more information than I've been working with! All these numbers are now outdated, LOL. But as long as the math is solid, the numbers are easy to change.
daromer said:
I will keep that in mind for looking at inverter specs. The cells I end up using should also have a peak draw as well as a maximum continuous current. I will take that under consideration as well.
I appreciate the feedback everyone!
Korishan said:You got your math right here. 125A max draw from your packs is what you'd expect. Well, sorta. There will be power losses, so assume 125 + 10-15% = ~150A to be on the safe side.
The best way to calculate the amount you need, is to base it off your max per year. So, in Winter is when you use the most energy, for you. Down here in Florida, Summer is when we use the most energy. So you'd determine based off averages from Dec - Feb. Assuming these are the most energy hungry months for you.
So, if you had 3008 kWh/month, divide by 30d = 100.3Wh/day. divide by 24h = 4.17Wh. The rest you got
I think Korishan may be off on his kWh and Wh
. the 3008kWh/30 = 100.3kWh/day. Divide by 24hr = 4.17kWh. So you would use 4.17kWh per hour. But if I recall, you're the guy with the electric furnace. So your biggest use is probably that furnace/cooling system. I won't be surprised if that unit uses 10-20kw when running. It'll probably be ambitious to put that system on your inverter. Not that it's not possible to run through inverter if sized properly, but it's definitely more power hungry, with high start up loads etc. By taking that system out of the equation you might find an easier time to calculate your actual load.
If you plan to grid-tie with the electric company, but also want backup in case of outages, then you probably want to get separate subpanel from the main panel and you will focus this subpanel to power your critical needs, like the lights, fridge, etc. This subpanel will be powered by your inverter. If you're planning to grid tie then you'll need to find out what units are supported or approved by the electrical company, such as the Outback Radian.
rebelrider.mike
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Yeah, I've got an old furnace that uses something like 4/5ths of my winter power usage. For now, I'm doing my calculations as things are, but I do want to get a new heat pump, as well as improving the efficiency of my air ducts. Anyway, check out my first post and you'll see all the stuff that can be improved so I can use less power.
There are some definite pros and cons to living in my area.
My local power company is very pro solar. Seems like everything here requires a permit, but no one ever seems to get one, and no one ever checks. (Though I generally do get permits for larger jobs.)
Some cons though; most batteries and plastics here are not accepted and have to be thrown out with the trash. There is a huge shortage of contractors in my area. The few that exist are outrageously expensive and they tend to do a terrible job. If you can even get them to show up! And it's overcast here a lot during winter. Though we do have some sunny days too. I'll likely always be using grid power in the winter unless I can augment the solar with something else.
Well, math has been put on hold for now, as I've just got a new batch of cells to test, and I got my testing rig repared. I figure I'll need to collect about 2,300 cells to make a "starter" battery for my house. So I've only got about 2,000 more to go. I gotta quit using them up on other projects! LOL
There are some definite pros and cons to living in my area.
My local power company is very pro solar. Seems like everything here requires a permit, but no one ever seems to get one, and no one ever checks. (Though I generally do get permits for larger jobs.)
Some cons though; most batteries and plastics here are not accepted and have to be thrown out with the trash. There is a huge shortage of contractors in my area. The few that exist are outrageously expensive and they tend to do a terrible job. If you can even get them to show up! And it's overcast here a lot during winter. Though we do have some sunny days too. I'll likely always be using grid power in the winter unless I can augment the solar with something else.
Well, math has been put on hold for now, as I've just got a new batch of cells to test, and I got my testing rig repared. I figure I'll need to collect about 2,300 cells to make a "starter" battery for my house. So I've only got about 2,000 more to go. I gotta quit using them up on other projects! LOL
rebelrider.mike
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Finally had a chance to gather some more numbers and do some more math.
I've already shared the math used for the house power requirements, so I'll simply list my results with the latest numbers. This won't be the final characteristics, just a baseline for building the system. The idea is to meet or exceed these specs where appropriate. One idea I'm toying with is using a 52V battery instead of a 48V battery. Seems that most people are using 14s 18650 configurations (51.8V) in their batteries, and most hardware is able to deal with that, so why not? More efficient right?
House battery Voltage: 52V
Maximum continuous current: 240A (this is a calculated value plus 20%)
Maximum power: 12,480W
Average energy consumption 6,103Wh (based on hourly usage over the three highest consumption months of the year)
Average continuous current: 117A
Independent battery run time: 3h (Ideally 36h, but I'll work up to that later.)
Minimum capacity: 352Ah
Minimum energy: 18,310Wh
I've been comparing different chemistries, but the math is the same for each. So I'll use 18650s as an example. Now, not all 18650s are equal, so I've chosen the MoliICR18650H as a readily available and low-cost option (as of now) on eBay, in order to get specific properties. Here they are:
Full charge: 4.2V
Nominal charge: 3.7V
Maximum DoD: 2.8V
Total capacity: 2.2Ah
Recommended DoD: 100% (important because other chemistries differ)
Usable capacity: 2.2Ah (again, this makes more sense when comparing chemistries)
Useable energy: 8.41Wh
Maximum discharge current: 4.4A (if I were to use old laptop cells, I'd limit this to 0.5A)
Peak current draw: 5A (important because lots of devices have a large initial current draw.)
With those properties, I can figure out how many cells I need, and it what configuration.
Voltage is easy. 52V / 3.7V (nominal) = 14.05 cells. Let's just say 14.
The minimum power needed in the battery will be 18,310Wh as listed above. And each cell has 8.14Wh.
18,310Wh / 8.14Wh = 2,249.35 cells. Let's round that to 2,250 cells.
I've found calculating cells needed in parallel to be a little tricky. Sometimes you need more in parallel so as to share current across the cells. And sometimes you need more cells in parallel to get enough overall energy in the battery, since adding more cells in series would affect the Voltage. So I calculate both, and simply choose the larger number.
Cells needed in parallel to get the necessary Wh: 2,250 cells / 14s = 160.71p Let's round that to 160 cells for the sake of building actual packs.
Cells needed in parallel to keep current per cell low enough: 240A / 4.4A = 54.5p. Let's round that up to 55 cells.
Of those two calculations, 160p is the higher number so I'll go with that. This means I will need a battery configured as 14s160p. This of course, can be broken down into packs for easier maintenance and installation. A person could build 14 packs of 160 cells in parallel, and install each in series, or 28 packs of 80 cells in parallel, and make a 14s2p configuration with those. Anyway, the important thing is that now the exact number of cells is known. 14 x 160 = 2,240 cells. This is a little less than the calculated 2,250 cells, but it's pretty darn close. And the actual properties of the final battery design can now be calculated, and compared with the theoretical values above.
Using the same math (so I won't repeat it), here are the values of the battery I'd need to build if I were making one today with the theoretical values in parentheses:
Full Voltage: 58.8V
Nominal Voltage: 51.8V (52V)
Discharged Voltage: 39.2V
Usable capacity: 352Ah (352Ah)
Total energy: 18,234Wh (18,310)
Total power: 12,432W (12,480W)
Maximum continuous current: 704A (117A)
Peak momentary current: 800A (240A)
Independent runtime: 3h (3h)
It seems the power and energy is a little low (I think this is because I chose to round down to 160p). But the rest of the specs are as good or better than what I need. I will probably redo these calcs based on laptop cells instead of the Moli's because I doubt I'll be able to get 2,240 identical Moli cells. So these numbers will change. But the important thing is that I've got the math down. (Or at least I hope I do.)
The next step is to figure out installation cost, and cost over time. And later, I'll share the same results but with different chemistries.But for now, I'll let those who are interested review what I've done, and maybe offer feedback.
I've already shared the math used for the house power requirements, so I'll simply list my results with the latest numbers. This won't be the final characteristics, just a baseline for building the system. The idea is to meet or exceed these specs where appropriate. One idea I'm toying with is using a 52V battery instead of a 48V battery. Seems that most people are using 14s 18650 configurations (51.8V) in their batteries, and most hardware is able to deal with that, so why not? More efficient right?
House battery Voltage: 52V
Maximum continuous current: 240A (this is a calculated value plus 20%)
Maximum power: 12,480W
Average energy consumption 6,103Wh (based on hourly usage over the three highest consumption months of the year)
Average continuous current: 117A
Independent battery run time: 3h (Ideally 36h, but I'll work up to that later.)
Minimum capacity: 352Ah
Minimum energy: 18,310Wh
I've been comparing different chemistries, but the math is the same for each. So I'll use 18650s as an example. Now, not all 18650s are equal, so I've chosen the MoliICR18650H as a readily available and low-cost option (as of now) on eBay, in order to get specific properties. Here they are:
Full charge: 4.2V
Nominal charge: 3.7V
Maximum DoD: 2.8V
Total capacity: 2.2Ah
Recommended DoD: 100% (important because other chemistries differ)
Usable capacity: 2.2Ah (again, this makes more sense when comparing chemistries)
Useable energy: 8.41Wh
Maximum discharge current: 4.4A (if I were to use old laptop cells, I'd limit this to 0.5A)
Peak current draw: 5A (important because lots of devices have a large initial current draw.)
With those properties, I can figure out how many cells I need, and it what configuration.
Voltage is easy. 52V / 3.7V (nominal) = 14.05 cells. Let's just say 14.
The minimum power needed in the battery will be 18,310Wh as listed above. And each cell has 8.14Wh.
18,310Wh / 8.14Wh = 2,249.35 cells. Let's round that to 2,250 cells.
I've found calculating cells needed in parallel to be a little tricky. Sometimes you need more in parallel so as to share current across the cells. And sometimes you need more cells in parallel to get enough overall energy in the battery, since adding more cells in series would affect the Voltage. So I calculate both, and simply choose the larger number.
Cells needed in parallel to get the necessary Wh: 2,250 cells / 14s = 160.71p Let's round that to 160 cells for the sake of building actual packs.
Cells needed in parallel to keep current per cell low enough: 240A / 4.4A = 54.5p. Let's round that up to 55 cells.
Of those two calculations, 160p is the higher number so I'll go with that. This means I will need a battery configured as 14s160p. This of course, can be broken down into packs for easier maintenance and installation. A person could build 14 packs of 160 cells in parallel, and install each in series, or 28 packs of 80 cells in parallel, and make a 14s2p configuration with those. Anyway, the important thing is that now the exact number of cells is known. 14 x 160 = 2,240 cells. This is a little less than the calculated 2,250 cells, but it's pretty darn close. And the actual properties of the final battery design can now be calculated, and compared with the theoretical values above.
Using the same math (so I won't repeat it), here are the values of the battery I'd need to build if I were making one today with the theoretical values in parentheses:
Full Voltage: 58.8V
Nominal Voltage: 51.8V (52V)
Discharged Voltage: 39.2V
Usable capacity: 352Ah (352Ah)
Total energy: 18,234Wh (18,310)
Total power: 12,432W (12,480W)
Maximum continuous current: 704A (117A)
Peak momentary current: 800A (240A)
Independent runtime: 3h (3h)
It seems the power and energy is a little low (I think this is because I chose to round down to 160p). But the rest of the specs are as good or better than what I need. I will probably redo these calcs based on laptop cells instead of the Moli's because I doubt I'll be able to get 2,240 identical Moli cells. So these numbers will change. But the important thing is that I've got the math down. (Or at least I hope I do.)
The next step is to figure out installation cost, and cost over time. And later, I'll share the same results but with different chemistries.But for now, I'll let those who are interested review what I've done, and maybe offer feedback.
rebelrider.mike
Member
- Joined
- May 25, 2017
- Messages
- 554
Finally approaching the end of the maths! Today I'm going to attempt to figure out how to calculate the cost of building a battery. The installation cost is fairly straightforward, but I'm not sure about long term cost. Especially when it comes to used 18650s. Ok, here goes:
Sticking with my example cells, the Moli 18650s I saw on eBay, I can set the price per cell at $1.40. This is all in USD by the way. Assuming I can get exactly 2,240 cells. That's $1.40 x 2,240 = $3,136. Knowing that the battery will have 18kWh (rounded) as calculated from the earlier post, that works out to: $3136 / 18kwh = $171.99/kwh. This number doesn't really mean anything yet, but it will when I start comparing other chemistries.
A more difficult number to calculate will be the cost over time. Lots of variables here, like actual DoD vs. recommended DoD, EoL vs. using cells past EoL. Basically, trying to predict how long each cell will last, and how often will it need to be changed. So I'm going to have to make a few assumptions in order to proceed.
First, cycle life. Searching the database, and also around the internet, it seems that 18650s have a cycle life of around 300-600 cycles before they are considered End of Life. It also seems to me that the general consensus is that EoL is when the cell is only able to retain 80% of its original rated capacity. (For example, the 9,000mAh cells you find on eBay are EoL before they're even done being manufactured, LOL.)
Depth of Discharge also has a huge impact on the life of a cell. Li-Ions seem to be meant to use their entire capacity between charges, compared to say, a LiFePO4 which is recommended to use only 80% DoD, or a lead/acid which is recommended at 50% DoD. Consequently, keeping an 18650 above 100% DoD, will increase its life expectancy. Not sure how much. I'm guessing its linear, because I don't know how do do math if its logarithmic.
Also, what is considered 100% DoD? There is no consensus on this. It depends on who you ask. The more popular numbers I've read are 2.5V, 2.8V, 3V, and 3.2V. This could be because people are talking about the cell's Voltage under load. Cells I test recover their Voltage when the load is removed, to around 3.2-3.6V.But how much load? This will make a difference in how "fast" a cell will reach its full discharge Voltage.
So again, I've had to make more assumptions. I'm going with 100% DoD = 2.8V @ 0.5A. And I'm assuming that the Moli cells will show their rated mAh if I were to run them in a test like this. Now, several datasheets for other chemistries have given some of these values so I don't have to guess. Unfortunately, not all manufacturers are as forthcoming with their specs.
Ok, now I have my values either read, calculated, or just guessed, time to figure this out. So the cycle life of the Moli cells as given on the spec. sheet, is 300. Let's say I decide that rather than going with the standard 80% capacity, I'll triple that to 40% capacity before I remove them (or reuse them as weaker packs). So losing 20%, then 40%, then 60% of their capacity would mean I'll get 900 cycles instead of just 300. (Again, assuming linear degradation.)
Also, I'm going to assume I'll be using the cells at less than 100% DoD per cycle. This means (maybe) that I'll double the life spanto something like 1,800 cycles. And let's say I do this every day. That's 1,800cycles / 365days/year = 4.93 years. I may have skipped a step there, but that's about 5 years per cell. So if I have to spend $3,136 every 5 years, that works out to $635.91/year, as cost over time.
These numbers still don't have any particular value (other than the effect on my budget) until I start comparing them to other chemistries. I also ran the same calculations for random laptop cells, and came up with some very significant differences! One last assumption. I'm going with the idea that these cells will be used every day. In the beginning, that may not be the case as I won't have enough storage capacity to be off-grid all night even if my solar panels can charge the battery full every day with excess power that isn't being used onmy house. But being semi-off-grid is the ideal I'm working toward, eventually.
Any flaws in logic, math, or maybe bad assumptions? I'd appreciate feedback. I've certainly gotten a lot already, which has been a huge help! Otherwise, I'll compare the cost of various chemistries next and maybe even decide some things.
Sticking with my example cells, the Moli 18650s I saw on eBay, I can set the price per cell at $1.40. This is all in USD by the way. Assuming I can get exactly 2,240 cells. That's $1.40 x 2,240 = $3,136. Knowing that the battery will have 18kWh (rounded) as calculated from the earlier post, that works out to: $3136 / 18kwh = $171.99/kwh. This number doesn't really mean anything yet, but it will when I start comparing other chemistries.
A more difficult number to calculate will be the cost over time. Lots of variables here, like actual DoD vs. recommended DoD, EoL vs. using cells past EoL. Basically, trying to predict how long each cell will last, and how often will it need to be changed. So I'm going to have to make a few assumptions in order to proceed.
First, cycle life. Searching the database, and also around the internet, it seems that 18650s have a cycle life of around 300-600 cycles before they are considered End of Life. It also seems to me that the general consensus is that EoL is when the cell is only able to retain 80% of its original rated capacity. (For example, the 9,000mAh cells you find on eBay are EoL before they're even done being manufactured, LOL.)
Depth of Discharge also has a huge impact on the life of a cell. Li-Ions seem to be meant to use their entire capacity between charges, compared to say, a LiFePO4 which is recommended to use only 80% DoD, or a lead/acid which is recommended at 50% DoD. Consequently, keeping an 18650 above 100% DoD, will increase its life expectancy. Not sure how much. I'm guessing its linear, because I don't know how do do math if its logarithmic.
Also, what is considered 100% DoD? There is no consensus on this. It depends on who you ask. The more popular numbers I've read are 2.5V, 2.8V, 3V, and 3.2V. This could be because people are talking about the cell's Voltage under load. Cells I test recover their Voltage when the load is removed, to around 3.2-3.6V.But how much load? This will make a difference in how "fast" a cell will reach its full discharge Voltage.
So again, I've had to make more assumptions. I'm going with 100% DoD = 2.8V @ 0.5A. And I'm assuming that the Moli cells will show their rated mAh if I were to run them in a test like this. Now, several datasheets for other chemistries have given some of these values so I don't have to guess. Unfortunately, not all manufacturers are as forthcoming with their specs.
Ok, now I have my values either read, calculated, or just guessed, time to figure this out. So the cycle life of the Moli cells as given on the spec. sheet, is 300. Let's say I decide that rather than going with the standard 80% capacity, I'll triple that to 40% capacity before I remove them (or reuse them as weaker packs). So losing 20%, then 40%, then 60% of their capacity would mean I'll get 900 cycles instead of just 300. (Again, assuming linear degradation.)
Also, I'm going to assume I'll be using the cells at less than 100% DoD per cycle. This means (maybe) that I'll double the life spanto something like 1,800 cycles. And let's say I do this every day. That's 1,800cycles / 365days/year = 4.93 years. I may have skipped a step there, but that's about 5 years per cell. So if I have to spend $3,136 every 5 years, that works out to $635.91/year, as cost over time.
These numbers still don't have any particular value (other than the effect on my budget) until I start comparing them to other chemistries. I also ran the same calculations for random laptop cells, and came up with some very significant differences! One last assumption. I'm going with the idea that these cells will be used every day. In the beginning, that may not be the case as I won't have enough storage capacity to be off-grid all night even if my solar panels can charge the battery full every day with excess power that isn't being used onmy house. But being semi-off-grid is the ideal I'm working toward, eventually.
Any flaws in logic, math, or maybe bad assumptions? I'd appreciate feedback. I've certainly gotten a lot already, which has been a huge help!
Otherwise, I'll compare the cost of various chemistries next and maybe even decide some things.
rebelrider.mike
Member
- Joined
- May 25, 2017
- Messages
- 554
It's finally time to compare chemistries!
What I've done is to put all the variables I can think of on a spreadsheet and try to organize them in some fashion that makes sense. From there, I can duplicate the math for each chemistry I'm considering, and also easily make updates as I get better information. Here's the sheet I made for lead/acid:
The first section is the power requirements that my battery will need in order to run my house for 3 hours. This will stay the same across all the chemistries I'll be looking at. By the way, the 240A was calculated from 52V not 48V like it says in the margin. I also added in a 20% safety margin for things like efficiency losses and appliance surges.
The second section lists properties of the cells (or batteries in the case of lead/acid) that are available for me to purchase. So I based my calculations on what I could actually get my hands on.
The third section with green highlights is figuring out the necessary battery configuration. Series is straightforward enough. Parallel is a little more tricky because there is a minimum amount of capacity needed, but also each cell has a maximum amp draw. Both situations have to be satisfied, so I ended up just listing both and picking the larger number. In the case of lead/acid, 4 batteries in parallel will provide the needed capacity, but I'd need 13 in parallel to spread the Amps out.
The last section is what the actual battery properties would be. So this can be compared to the house requirements in the first section to double check that it will work ok. Also, at the bottom is an attempt to predict the installation price and the price per year. Installation is pretty straightforward too. SxP shows the total number of cells or batteries needed, and multiply that with the cost, including tax and shipping, and you get the total installation price. I also listed price per kWh, as that was pretty easy to do. Cost per year is not so easy. I wrote down the advertized life expectancy of each chemistry in cycles, and assumed that the end of life is considered reached at 80% of the original capacity. I also figured I could let the cells degrade further on to say, 40% capacity so as to use them 3x longer. I also assumed that putting each cell through half a cycle instead of a full cycle would double its lifespan. Maybe that's true, maybe it isn't. But I know that most chemistries benefit greatly from getting used at a partial cycle rather than a full cycle. From there I figured how many years of service I might get from each cell based on all these assumptions.
For me though, the really important number is cost for installation. And so I highlighted that in red.
For Li-Ion, I did two different calculations. One is based on a large number of similar cells available from a reputable supplier on eBay. The other is based on continuing to collect and test cells from old laptop batteries.
Although testing laptop cells is cheaper per cell, I'll need a lot more of them to make a house battery. Buying large numbers of new-ish pre-tested cells on eBay looks to be more economical. Of course, there's no reason not to use both. For example, using the better cells in "primary" modules as the minimum to run the house, and adding laptop cells as I go in "secondary" modules that just expand the batterie's capabilities later.
Another chemistry I though to use was LiFePO4.
These would certainly be more manageable. Eighty vs. 3,000 of the 18650s. And the life span is something like 10x Li-Ion.But it would cost 4x more to install these cells.
The last chemistry I seriously considered was Ni-Fe. The advantage of these is that with proper maintenance, These cells would last longer than me! However, the price and number of cells I'd have to get makes these completely infeasible.
Just for fun, I did one more chemistry. I wondered what the numbers would say about using NiMH D-cells. The numbers were not good, LOL!
So it seems lithium chemistry wins out. I was quitesurprised, as I hear often that lead/acid is cheaper than lithium. From the math I've done, even brand-new LiFePO4 cells are cheaper than lead/acid, and are in fact, superior to them in every way that I could find. Perhaps this because lithium has been decreasing in price while lead/acid has remained stable? Anyway, looks like I'll be going with 18650s of one sort or another. Even having to buy them rather than collect them for free, they're still the cheapest solution.
Looks like the next phase for me is to start collecting a metric-sh_t-ton of 18650s and also tackling some of the home improvement projects to get my power usage down.
What I've done is to put all the variables I can think of on a spreadsheet and try to organize them in some fashion that makes sense. From there, I can duplicate the math for each chemistry I'm considering, and also easily make updates as I get better information. Here's the sheet I made for lead/acid:
The first section is the power requirements that my battery will need in order to run my house for 3 hours. This will stay the same across all the chemistries I'll be looking at. By the way, the 240A was calculated from 52V not 48V like it says in the margin. I also added in a 20% safety margin for things like efficiency losses and appliance surges.
The second section lists properties of the cells (or batteries in the case of lead/acid) that are available for me to purchase. So I based my calculations on what I could actually get my hands on.
The third section with green highlights is figuring out the necessary battery configuration. Series is straightforward enough. Parallel is a little more tricky because there is a minimum amount of capacity needed, but also each cell has a maximum amp draw. Both situations have to be satisfied, so I ended up just listing both and picking the larger number. In the case of lead/acid, 4 batteries in parallel will provide the needed capacity, but I'd need 13 in parallel to spread the Amps out.
The last section is what the actual battery properties would be. So this can be compared to the house requirements in the first section to double check that it will work ok. Also, at the bottom is an attempt to predict the installation price and the price per year. Installation is pretty straightforward too. SxP shows the total number of cells or batteries needed, and multiply that with the cost, including tax and shipping, and you get the total installation price. I also listed price per kWh, as that was pretty easy to do. Cost per year is not so easy. I wrote down the advertized life expectancy of each chemistry in cycles, and assumed that the end of life is considered reached at 80% of the original capacity. I also figured I could let the cells degrade further on to say, 40% capacity so as to use them 3x longer. I also assumed that putting each cell through half a cycle instead of a full cycle would double its lifespan. Maybe that's true, maybe it isn't. But I know that most chemistries benefit greatly from getting used at a partial cycle rather than a full cycle. From there I figured how many years of service I might get from each cell based on all these assumptions.
For me though, the really important number is cost for installation. And so I highlighted that in red.
For Li-Ion, I did two different calculations. One is based on a large number of similar cells available from a reputable supplier on eBay. The other is based on continuing to collect and test cells from old laptop batteries.
Although testing laptop cells is cheaper per cell, I'll need a lot more of them to make a house battery. Buying large numbers of new-ish pre-tested cells on eBay looks to be more economical. Of course, there's no reason not to use both. For example, using the better cells in "primary" modules as the minimum to run the house, and adding laptop cells as I go in "secondary" modules that just expand the batterie's capabilities later.
Another chemistry I though to use was LiFePO4.
These would certainly be more manageable. Eighty vs. 3,000 of the 18650s. And the life span is something like 10x Li-Ion.But it would cost 4x more to install these cells.
The last chemistry I seriously considered was Ni-Fe. The advantage of these is that with proper maintenance, These cells would last longer than me! However, the price and number of cells I'd have to get makes these completely infeasible.
Just for fun, I did one more chemistry. I wondered what the numbers would say about using NiMH D-cells. The numbers were not good, LOL!
So it seems lithium chemistry wins out. I was quitesurprised, as I hear often that lead/acid is cheaper than lithium. From the math I've done, even brand-new LiFePO4 cells are cheaper than lead/acid, and are in fact, superior to them in every way that I could find. Perhaps this because lithium has been decreasing in price while lead/acid has remained stable? Anyway, looks like I'll be going with 18650s of one sort or another. Even having to buy them rather than collect them for free, they're still the cheapest solution.
Looks like the next phase for me is to start collecting a metric-sh_t-ton of 18650s and also tackling some of the home improvement projects to get my power usage down.
Crimp Daddy
Member
- Joined
- Feb 21, 2018
- Messages
- 973
Nice cost analysis!
We did some home improvement projects to make the house more efficient and reduced the energy use in lighting by 80%, which really left large appliances. We even used things like motion detector / occupancy sensor switches to take lighting control to that next level.
In your installed price, I presume the labor is not accounted for as this is all done by you? One thing to mention is that while the 18650 route is cheapest, you will often times save in kWh raw costs, but end up spending in consumables to finish the project. Time, testing, and other things should be taken into consideration when buying new cells vs second live.
For me personally, I looked to larger lithium second life cells to balance between labor/time and material costs since I found collecting 40 kWh+ 2000 mah at a time was a bit time consuming with a higher maintenance lifecycle.
We did some home improvement projects to make the house more efficient and reduced the energy use in lighting by 80%, which really left large appliances. We even used things like motion detector / occupancy sensor switches to take lighting control to that next level.
In your installed price, I presume the labor is not accounted for as this is all done by you? One thing to mention is that while the 18650 route is cheapest, you will often times save in kWh raw costs, but end up spending in consumables to finish the project. Time, testing, and other things should be taken into consideration when buying new cells vs second live.
For me personally, I looked to larger lithium second life cells to balance between labor/time and material costs since I found collecting 40 kWh+ 2000 mah at a time was a bit time consuming with a higher maintenance lifecycle.
DK100
Member
- Joined
- Feb 18, 2018
- Messages
- 63
I belive you could reduce cost by reducing consumption first. Change all lights to LED's. How much are fridge and frezer using? Per watt insulation are cheaper than batteries.... A heatpump would off course give you the highest return on investment?
Remember - the cheapest energy is the energy you dont use
Remember - the cheapest energy is the energy you dont use
completelycharged
Active member
- Joined
- Mar 7, 2018
- Messages
- 1,083
Having read through the posts I have one small point to make, which I am not sure that you have considered.
For any renewable power source you have (wind or solar) given your average winter power demand ove over 4kW you would need and output from your wind/solar to be in excess of 4kW BEFORE the surplus can charge your battery power pack. And a typical solar install would have to be at least 5kW to start with.
Say you install 10kW of solar, in winter this may provide a peak of 6-8kW for a few hours during the day and the excess of 2-4kW above your existing demand will only last quite a short time making the charge/discharge period very short and your battery pack will only be active for a 2-3 hour charge and then 16-20hr discharge. This is also marginal cost output from your solar, so the economics have to be factored in.
I would sugest that you create an hourly data set and add in your expected generation/demand and battery position per hour and see what the profile ends up like. How much generation do you expect to have to charge the battery pack ?
Convert your 1500W generator to CHP (recover the heat output) as this may make some sense.
Wind - if you can install a turbine and it is anything close to windy in the winter then install one rather than solar, you should get more value out of it..
For your situation given the very high level of demand 24 hours a day any storage is going to cycle very quickly, if you have a power source large enough to offset your existing demand first. The issue you may face is the high and short charge rate rather than the demand side.
4kW 24 hours a day seems like a very high consumption rate....
If you have any battery pack, because you have such a high and continuous load I would just buy a small 500W battery grid tie inverter that can set the discharge to say 300W. They cost around $100.
Say you have a battery pack of 10kWh and it is charged from solar in 4 hours in the middle of the day during the peak. You then have 20 hours to slowly discharge the battery pack against your existing load. This works out to be only 500W. Building for >100A and peak demand only makes sense if you need run off-grid but your battery would need to be very, very large kWh and does not make sense to me.
Put any money into wind first if an option, solar second and then think about a battery when you have enough generation. That would be my suggestion. Woodburner ????
With the numbers from your calculations...
FLA6 - with $268.31/kWh cost and the two scenarios of cycle life the energy throughput cost is $0.279 per kWh life (1200 cycles @ 80% DoD)and $0.186/kWh (3600 cycles @ 40% DoD). This is how much the battery setup costs you to put energy in and get it back out over the life of the cells.
LiIon eBay - $171.99/kWh cost. 300 cycles at 80% DoD works out to be $0.717/kWh cost and 900 cycles at 40% DoD works out to be $0.478/kWh. - Not viable for cycle life expectations and cost.
LiIon Laptop - $97.30/kWh cost. 300 Cycles at 80% DoD works out to be $0.405/kWh and 900 cycles at 40% DoD works out to be $0.27/kWh. - Not viable for cycle life expectations and cost.
LiFePO4 - $620.12/kWh cost. 3000 cycles at 80% DoD works out to $0.258/kWh. 9000 cycles at 40% DoD works out to be $0.172/kWh.
The LiPo cost you have is very high.... I would expect pricing closer to half and then you system could make sense.
This is working out the Cost of the battery in $per kWh and then dividing it by the expected throughput you are expecting, which is the cycles x DoD. e.g. if you have a 1kWh battery that will last 500 cycles at 80% DoD then the throughput is 500 x 0.8 x 1kWh = 400kWh.
If you battery cost is $150 for 1kWh then your cost per kWh in and out of the pack is $150 / 400kWh = $0.375 per kWh.
Cycle life expectations are key to the whole economics..... work the cost to put 1kWh into your battery and get it back out again...
For any renewable power source you have (wind or solar) given your average winter power demand ove over 4kW you would need and output from your wind/solar to be in excess of 4kW BEFORE the surplus can charge your battery power pack. And a typical solar install would have to be at least 5kW to start with.
Say you install 10kW of solar, in winter this may provide a peak of 6-8kW for a few hours during the day and the excess of 2-4kW above your existing demand will only last quite a short time making the charge/discharge period very short and your battery pack will only be active for a 2-3 hour charge and then 16-20hr discharge. This is also marginal cost output from your solar, so the economics have to be factored in.
I would sugest that you create an hourly data set and add in your expected generation/demand and battery position per hour and see what the profile ends up like. How much generation do you expect to have to charge the battery pack ?
Convert your 1500W generator to CHP (recover the heat output) as this may make some sense.
Wind - if you can install a turbine and it is anything close to windy in the winter then install one rather than solar, you should get more value out of it..
For your situation given the very high level of demand 24 hours a day any storage is going to cycle very quickly, if you have a power source large enough to offset your existing demand first. The issue you may face is the high and short charge rate rather than the demand side.
4kW 24 hours a day seems like a very high consumption rate....
If you have any battery pack, because you have such a high and continuous load I would just buy a small 500W battery grid tie inverter that can set the discharge to say 300W. They cost around $100.
Say you have a battery pack of 10kWh and it is charged from solar in 4 hours in the middle of the day during the peak. You then have 20 hours to slowly discharge the battery pack against your existing load. This works out to be only 500W. Building for >100A and peak demand only makes sense if you need run off-grid but your battery would need to be very, very large kWh and does not make sense to me.
Put any money into wind first if an option, solar second and then think about a battery when you have enough generation. That would be my suggestion. Woodburner ????
With the numbers from your calculations...
FLA6 - with $268.31/kWh cost and the two scenarios of cycle life the energy throughput cost is $0.279 per kWh life (1200 cycles @ 80% DoD)and $0.186/kWh (3600 cycles @ 40% DoD). This is how much the battery setup costs you to put energy in and get it back out over the life of the cells.
LiIon eBay - $171.99/kWh cost. 300 cycles at 80% DoD works out to be $0.717/kWh cost and 900 cycles at 40% DoD works out to be $0.478/kWh. - Not viable for cycle life expectations and cost.
LiIon Laptop - $97.30/kWh cost. 300 Cycles at 80% DoD works out to be $0.405/kWh and 900 cycles at 40% DoD works out to be $0.27/kWh. - Not viable for cycle life expectations and cost.
LiFePO4 - $620.12/kWh cost. 3000 cycles at 80% DoD works out to $0.258/kWh. 9000 cycles at 40% DoD works out to be $0.172/kWh.
The LiPo cost you have is very high.... I would expect pricing closer to half and then you system could make sense.
This is working out the Cost of the battery in $per kWh and then dividing it by the expected throughput you are expecting, which is the cycles x DoD. e.g. if you have a 1kWh battery that will last 500 cycles at 80% DoD then the throughput is 500 x 0.8 x 1kWh = 400kWh.
If you battery cost is $150 for 1kWh then your cost per kWh in and out of the pack is $150 / 400kWh = $0.375 per kWh.
Cycle life expectations are key to the whole economics..... work the cost to put 1kWh into your battery and get it back out again...