KWH to AH Battery
SolarSailor
Solar Expert Posts: 49 ✭
If one knows their usage in KWh, how is that translated to Battery Amp-Hour requirements?
Does it require knowing the expected Amperage draw of the sum of all devices?
Or, if I were to say I want it to power a _circuit_ which has a capacity of 15-Amps and, say, 2500 Watts. Is that sufficient?
Does it require knowing the expected Amperage draw of the sum of all devices?
Or, if I were to say I want it to power a _circuit_ which has a capacity of 15-Amps and, say, 2500 Watts. Is that sufficient?
Comments
-
Re: KWH to AH Battery
Volts times Amps = watts Add hours to it, and you have watt hours
12V x 20 A = 240W
12V battery, 80AH capacity = 960wh (but you should only consume 50% to not destroy your battery) 480wh
2, 60 w bulbs = 120W for 3 hours = 360wh
2, 60 w bulbs = 120W for 4 hours = 480whPowerfab top of pole PV mount | Listeroid 6/1 w/st5 gen head | XW6048 inverter/chgr | Iota 48V/15A charger | Morningstar 60A MPPT | 48V, 800A NiFe Battery (in series)| 15, Evergreen 205w "12V" PV array on pole | Midnight ePanel | Grundfos 10 SO5-9 with 3 wire Franklin Electric motor (1/2hp 240V 1ph ) on a timer for 3 hr noontime run - Runs off PV ||
|| Midnight Classic 200 | 10, Evergreen 200w in a 160VOC array ||
|| VEC1093 12V Charger | Maha C401 aa/aaa Charger | SureSine | Sunsaver MPPT 15A
solar: http://tinyurl.com/LMR-Solar
gen: http://tinyurl.com/LMR-Lister , -
Re: KWH to AH Battery
Need battery system voltage.
kWH is amps per hour times voltage.
Battery A-H is capacity usually based on 20 hour discharge rate. 5 amps load on battery for 20 hours would be a 100 A-H battery.
AH capability is less at higher current drawn from battery. For a 1 to 2 hour discharge rate it may be 50% to 65% of what AH capability is at 20 hour rated AH. -
Re: KWH to AH Battery
- Power = Volts * Amps (rate--like miles per hour, gallons per minute)
- Energy = Volts * Amps * Hours (time) (total value, like miles or gallons)
- Amp*Hours= Energy/Time = Volts * Amps * Hours / Volts
For batteries--When dealing with Amp*Hours vs Watt*Hours, it is sort of nice to use Amp*Hours because Lead Acid Batteries are nearly 100% efficient (Amp*Hours in = Amp*Hours out) whereas Watt*Hours is around 80-90 percent efficient.
And it is easy to see why for a typical Lead Acid Battery (assuming amps and hours are the same for charging/discharging for this example):- Vbatt ~ 12.5 volts when discharging
- Vbatt ~ 14.4 volts when charging
- Eout/Ein = V*A*H/V*A*H ~ Vout / Vin = 12.5v/14.4 = 0.87 = 87% energy efficiency
But, that requires that the system be setup correctly (i.e., "12 volt panels" which are really Vmp~17.5 volts or so on a PWM controller). If we used 24 volt panels (vmp~35 volts) on a 12 volt battery bank with a PWM controller (which cannot down convert energy efficiently, unlike a PWM controller)--then we have to take those "design violations" into account because (in this example) we loose 50% of our power just in one improper design Panel/Charge controller setup.
Sort of make sense?
-BillNear San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset -
Re: KWH to AH Battery
Thanks, was thinking it not that easy.
BUT, really, one should expect to know the A-H draw on an hourly, or at least daily basis to get it right, I presume. Not a monthly KWH average. -
Re: KWH to AH Battery
The reason(s) for the confusing units--I like to use the PV Watts website to estimate how much energy XXXX watts of solar panels will generator using batteries (or grid tied) for an end load...
The PV Watts program is written/intended for estimating Grid Tied setups (solar panels + GT Inverter tied to utility power)--And the average person uses their monthly power bill (in kWH) to size their system.
For Off Grid installations, we can use the same data and a different derating factor (GT=0.77, Off-Grid=0.52 for flooded cell batters+inverter losses).
And since Off-Grid folks only can store a few days of load (we use a 3 day rule of thumb and 50% maximum discharge for a battery bank)--Off-Grid'ers tend to think in daily loads (I just divide the PV Results by 30 days per month).
Also, statistically, weather is highly variable (and location specific)--so it is easier to deal with a 1 month average than to deal with each day (PV Watts does have an extended data output with 365 days of "average days" (in 1 hour increments). It would drive me crazy to discuss the differences between February 3'rd and February 28th--and what about Feb. 29th? )... It is just easier to put it in 1 month (or even 3 month seasonal) power estimates when talking with people needing to size their system.
PV Watts has a nice "help file" (click on one of the links next to the explanation of variables) that describes more of the details/assumptions behind the numbers.
The kWH vs WH is just a factor of 1,000 difference (1,000 WH = 1 kWH). Many times we use Watts because it is "voltage/current" independent--We don't have to ask is this 12 volts, 24, 48, 120, 230, 240 etc.)...
A 100 watt bulb running for 10 hours is 1,000 WH or 1 kWH. In Amp*Hours (ignoring efficiencies of conversion):- 1,000 WH / 240 VAC = 4.2 AH @ 240 VAC
- 1,000 WH / 120 VAC = 8.3 AH
- 1,000 WH / 48 VDC = 21 AH
- 1,000 WH / 12 VDC = 83 AH
-BillNear San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset -
Re: KWH to AH Battery
Thanks again. That all makes perfect sense.
However, I was kind of going from ground up, how much do I need (to then size the collectors) rather than determining what I can get (collection results) and designing around that. Both ways should arrive near the same point, but if I design from a usage/capacity standpoint it seems more logical to me.
Then it's what will one panel do, divided into desired capacity as to how many panels are required.
he,he, prolly by the time I multi cost, I'll back up and reconsider again. -
Re: KWH to AH Battery
This is sort of how I do it--Say you are near Tulsa OK, using flooded cell batteries and an inverter (0.52 derating). The controller is either a well designed PWM system or MPPT with fixed solar panels pointing south at latitude:"Station Identification"
"City:","Tulsa"
"State:","Oklahoma"
"Lat (deg N):", 36.20
"Long (deg W):", 95.90
"Elev (m): ", 206
"PV System Specifications"
"DC Rating:"," 1.0 kW"
"DC to AC Derate Factor:"," 0.520"
"AC Rating:"," 0.5 kW"
"Array Type: Fixed Tilt"
"Array Tilt:"," 36.2"
"Array Azimuth:","180.0"
"Energy Specifications"
"Cost of Electricity:"," 7.7 cents/kWh"
"Results"
"Month", "Solar Radiation (kWh/m^2/day)", "AC Energy (kWh)", "Energy Value ($)"
1, 4.01, 64, 4.93
2, 4.46, 63, 4.85
3, 5.33, 81, 6.24
4, 5.88, 85, 6.54
5, 5.61, 80, 6.16
6, 5.77, 78, 6.01
7, 6.06, 83, 6.39
8, 5.95, 82, 6.31
9, 4.99, 69, 5.31
10, 5.41, 81, 6.24
11, 4.11, 61, 4.70
12, 3.70, 58, 4.47
"Year", 5.11, 884, 68.07
Say that you would like to use solar panels only for ~9 months of the year, and for the 3 months of winter you will use a backup genset. That gives us the 4th month as January and 64 kWH per month usable AC power (out the inverter outlet) per 1,000 watts of solar panels:- 64 kWhr per month * 1/30 days = 2.1 kWHrs or 2,100 Watt*Hours per day base load
- 2,100 WH / 12 volts = 175 AH @ 12 volts
- (80kWH-64kWH) * 1/30 days = 0.533 KWH or 533 WH per day extra power (summer)
- 533 WH excess per summer day / 300 watts Sanyo on low = 1.77 hours per 1kW of solar panels
I try not to get too generic in laying out a solution--it can create lots of confusion... Generally, it works better if you describe your needs and we try to work out possible answers based on your requirements.
And remember that these are all estimates based on ~20 year weather history--Some places have very consistent weather and other have highly variable weather from year to year--So any estimate we work out while generally be really good if we get within 10% of your actual power available/usage.
I try to be a bit on the conservative side when making estimates so as not to surprise you (in a bad way) when you have to run the genset much more than your original plans.
-BillNear San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset -
Re: KWH to AH Battery
Thanks for all that.
The real purpose of this exercise (since I cannot afford the $20-$30K to actually do an install today, or even $15K after tax credits) is for my own understanding. So, I'll be going through the calcs in enough ways and times to get it down. Then, when I do install, I won't be starting from scratch.
In the process, I'll probably start small, with a single panel of some kind, a 12-volt or pair for 24v battery, an inverter, et al, so I can physically go through it again. Then, I'll know what all the parts are and do.
I know my process is painfully slow and intricate, probably expensive, but it's the only way I learn it in a way which sticks.
I figure I'm the one who'll have to maintain the system once it's in place.
At least, know what's wrong when it doesn't work. -
Re: KWH to AH Battery
re: "it is sort of nice to use Amp*Hours because Lead Acid Batteries are nearly 100% efficient (Amp*Hours in = Amp*Hours out) whereas Watt*Hours is around 80-90 percent efficient."
whoops! The units you use makes no difference in efficiency.
The proper unit for power is watts and for energy it is watt hours.
Amp hours as a measure of energy assumes a voltage and that is only convenient if the voltage of all components is the same. Assumptions are seldom safe unless carefully constrained. (they can make an ass[of]u[and]me as an old engineer once said - like in the efficiency comment quoted above) -
Re: KWH to AH Battery
Bryan,
Actually, for a first approximation, Amp*Hours to/from a battery bank is different than Watt*Hours to/from the same bank...
Very roughly, we are charging a flooded cell battery bank at 14.5 volts and discharging it at 12.5 volts or:- 12.5 v/ 14.5 volts = 0.86 or 86% efficiency (load/charging)
Granted, the closer you get to 100% charge, the less efficient in both AH and WH a battery bank becomes--As well as when equalizing (gassing) which approaches 0% efficiency.
The battery bank tends to be more efficient at lower states of charge where the "bulk" charging voltage is less than the "absorb" voltage.
For an AGM battery where you cannot really do an equalization/higher voltage charging without damaging the cells, they have roughly 98% coulombic efficiency.
In the end, we are talking about rough rules of thumb and 14% losses are not that much by themselves--However, every little bit of losses adds up and we end up with the ~52% end to end losses.
I used to go into great detail in my first posts about losses and assumptions--However, over the years I have stopped doing that... First it gave me tunnel carpel syndrome and second it was really not much use because nobody has that much information available for a new system and the rules of thumb were "good enough" to get the reader a system that worked well for their needs without a lot of confusion.
Mike90045 posted a nice little report about efficiency assumptions and what happens if you operate a lead acid bank only in the top 20% of its capacity and the fact that the "system" becomes much less efficient than the 80% battery rule of thumb we (I) typically throw around here.
There is risk in all assumptions/simplifications--But they usually work out well for most users and I always try to tell people that rules of thumbs are starting places--not an endpoint in themselves.
-BillNear San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset -
Re: KWH to AH Battery
re: "Actually, for a first approximation, Amp*Hours to/from a battery bank is different than Watt*Hours to/from the same bank..."
The only way to make this work is to play games with the voltage. (i.e. simplify the measurement in ways that can be 'adjusted' to suit whatever point one wants to make)
Lead acid batteries are inherently inefficient and that efficiency has nothing to do with voltage differences between charging and discharging but rather in internal resistances and state changes involved in the actual energy storage and retrieval process.
Watts are defined as volts times amps. Therefore watt hours is, by definition, the same as amp volt hours. If voltage changes, both measures of energy will need appropriate integration.
Amp-hours is really only a meaningful measure within a very small scope, such as battery capacity within carefully defined limits over a period of time for the use of a referent. That is why there are measures such as 'reserve minutes' or other AH measures.
Whenever you take energy from a battery or apply it to a battery, the power you use will make a difference in the efficiency you achieve. This is the essence of the Peukert effect.
With the Peukert effect, temperature, and other variables that produce significant variation in available battert energy capacity, trying to make a case for high precision in battery capacity measures is not a fruitful exercise.
Where this goes is towards a proper discussion of the accuracy of the 'electron counting' type of battery status meters. These are getting smarter but still loose accuracy due to the use profile because, in part, they do not measure the actual state of charge of the battery. Conductance testers and battery behavior analyzers are attempts to do that.
But definitions are just that and it really doesn't help much, IMHO, to try to distort them. Instead, we need to better learn what it is we are actually measuring. -
Re: KWH to AH Battery... what it is we are actually measuring.
Joules, maybe? I am considering replacing my TriMetric "electron counter" with the SmartGauge. That site has been linked to many times on this forum for battery bank wiring schemes. It also has a few good physics discussions regarding lead acid batteries. It's not easy to navigate the site. But I did see some mention of the Watt being defined as 1 Joule per second. Then, some simple dimensional analysis makes clear why we use Watt*Hours or Ampere*Hours.
K
I had to pull out my 1981 copy of Halliday & Resnick. -
Re: KWH to AH Battery
re: "Joules, maybe?" -- that would be 'proper' but it is not so convenient with things electrical where most are rated in terms of the power they use in watts and electrical energy is bought and sold as watt hours.
The Trimetric is extremely popular for those who want to monitor DC energy in and out of their battery. It has improved over the years to accommodate such things as Peukert phenomena but still needs proper programming and careful interpretation.
I am curious about the Smartgauge. I haven't seen one in operation. Its method of determining battery state of charge by an analysis of battery behavior seems to me to be a more direct measure. The details are in the implementation and what I read on the website for technical information indicates that should be top notch.
This thread got me wondering about the measurements to determine battery amp hour specifications. Most are made with a battery voltage change of 20% or so occurring within the test. Is the test done keeping the load constant or the current constant?
More than anything else, though, I think the message is that we need to be careful to keep the limitations of battery measure in mind when we look at specifications and ratings. It does not appear to me that you can pin down a battery energy capacity measure to the sorts of precision often seen in the numbers for it. -
For BB / Bill,....in trying to follow one of your answers above, I am no electrical guy in the least bit when it comes to calculating kWH to Ah...ie: i am figuring the solar generation/battery storage needs for a class B motor home by determining electrical demands of all appliances in Amp hour usages ... all based of course on approximately how long each will be used a day to properly size the solar generation/battery storage.Since I don't 'get it', when reading energy consumption on a spec sheet, I get confused when appliances are different....for example a Dometic CFX 75DZW Portable Refrigerator reads:
- Energy consumption (DC@5/32degC) 1.309 kWh/24h... ...the following is my presumption....
1.309 kWh = 1309 Wh divided by 24 hours = 54.5 Amps/dayAnd if the fridge runs 16 hrs/day, then my total amps used for a 16 hour day would be 54.5 A/day divided by 24 hours/day x 16 hours use = 36.36 A/dayAm I presuming correctly?Thank you in a major way !!Lanse
-
1.309 kWh = 1309 Wh divided by 24 hours = 54.5 Amps/day
If you want to get to amps define the voltage and divide. 54.5 watts per hour at 12 volts is 54.5/12= 4.5 amps per hours at 12 volts.
Home system 4000 watt (Evergreen) array standing, with 2 Midnite Classic Lites, Midnite E-panel, Magnum MS4024, Prosine 1800(now backup) and Exeltech 1100(former backup...lol), 660 ah 24v Forklift battery(now 10 years old). Off grid for 20 years (if I include 8 months on a bicycle).
- Assorted other systems, pieces and to many panels in the closet to not do more projects.
Categories
- All Categories
- 222 Forum & Website
- 130 Solar Forum News and Announcements
- 1.3K Solar News, Reviews, & Product Announcements
- 192 Solar Information links & sources, event announcements
- 888 Solar Product Reviews & Opinions
- 254 Solar Skeptics, Hype, & Scams Corner
- 22.4K Solar Electric Power, Wind Power & Balance of System
- 3.5K General Solar Power Topics
- 6.7K Solar Beginners Corner
- 1K PV Installers Forum - NEC, Wiring, Installation
- 2K Advanced Solar Electric Technical Forum
- 5.5K Off Grid Solar & Battery Systems
- 425 Caravan, Recreational Vehicle, and Marine Power Systems
- 1.1K Grid Tie and Grid Interactive Systems
- 651 Solar Water Pumping
- 815 Wind Power Generation
- 624 Energy Use & Conservation
- 611 Discussion Forums/Café
- 304 In the Weeds--Member's Choice
- 75 Construction
- 124 New Battery Technologies
- 108 Old Battery Tech Discussions
- 3.8K Solar News - Automatic Feed
- 3.8K Solar Energy News RSS Feed