ah or kwh

To a degree--The change from AH to kWH can be laid to the changes in our loads...

In the "olden days", much of the loads are (more or less) constant Amp loads such as filament lamp loads. And lead acid batteries are (relatively) close to 100% efficient in charging when measuring Amp*Hours. You take out 50 AH, it takes around 50 AH to recharge--With only the final bit of charging being less efficient (final charging, EQ charging, etc. when gassing).

I first started seeing batteries rated in Watt*Hours for UPS systems. AC inverters are constant power devices. The inverter supplying 300 Watts at 120 VAC constant output, but the input current is also dependent on DC bus voltage:

• 300 Watts / 12.7 volts charged battery = 23.6 Amps
• 300 Watts / 10.5 volts near dead battery = 28.5 Amps

So, as the battery discharges, the current draw goes up... If you only looked at AH capacity of the battery, it is less accurate at estimating how much energy (WH) it can support.

Of course, the higher the current draw (C/20 vs C/5 discharge rate--I.e., 20 hour discharge vs 5 hour discharge to dead rate), the lower the apparent capacity of the battery--So you will still have a different capacity in WH rating depending on rate of discharge.

And for non-technical folks, 300 Watt discharge at 120 VAC vs 300 Watt discharge at 12 VDC is still 300 Watts. And why we here suggest people do all of their load calculations in Watts and Watt*Hours to prevent the confusion when doing Amp and AH calculations at 240 VAC vs 120 VAC vs 48 VDC vs 24 VDC vs 12 VDC. Again a difference when boats/cars/etc. only at native DC loads vs today's usages (cabin, RV) that have 12/24 VDC loads and 120/240 VAC loads--Because of the AC inverters and DC to DC converters that easily change voltage as needed.

So when talking about a battery bank in WH and kWH storage capacity--It does remove a bit of confusion. We do run into that question frequently here--Somebody tells us they have a 200 AH battery bank for their system, but we always have to ask them what the battery bus voltage is to start the load/charging design questions.

Watts and Watt*Hours are "mathematically complete" descriptions of loads and charging systems. With Amp and Amp*Hours, we are always asking at what voltage (120 VAC and 10 Amps vs 12 VDC and 10 amps--10x more power at 120 VAC):

• Power = Voltage * Current
• 12 VDC * 10 amps = 120 Watts
• 120 VDC * 10 amps = 1,200 Watts
• Energy (Watt*Hours) = Voltage * Current * Time
• 12 volts * 10 amps * 5 hours = 600 Watt*Hours
• 120 volts * 10 amps * 5 hours = 6,000 Watt*Hours
• Amp*hours = Current * Time
• AH = 10 amps * 5 hours = 50 AH (at what voltage?)

Anyway--That are my guesses. I don't disagree with you a marking guy is looking a 100 AH (at 12 volts) number "looks smaller" than 1,200 Watt*Hours--The bigger number looks better on the marketing glossies.

Regarding home power--Watts is a "moment in time" value... For some examples, say an average energy efficient North American home uses 300 kWH per month:

• 300 kWH month / 30 days per month = 10 kWH per day
• 10 kWH per day / 24 hours per day = 0.417 kWH = 417 Watt constant load over 1 month

A typical USA refrigerator may use 120 Watts running, and only run about 50% of the time in "nominal" room temperaters... And 600 Watts starting and something like 500 Watts running a defrost heater... Long term, may average:

• 50 Watts average * 24 hours per day = 1,200 WH per day
• 1,200 WH per day * 30 days per month =  36,000 WH per month = 36 kWH per month (or about 12% of average home load at 300 kWH per month bill)
• 1,200 WH per day * 365 days per year = 438,000 WH = 438 kWH per month on the Energy Star hang tag

The average load on a home may be around 417 Watts... But peak loads:

• 600 Watts fridge starting or defrosting heater
• 1,200 Watts microwave
• 200 Watts of lighting
• 100 Watts TV/radio/cell phone charging
• 30 Watts laptop computer
• 300 Watts desktop computer
• 4,000 Watts well pump
• home heating (central heat vs wall heater)
• cooking (gas/propane vs electric)
• hot water (gas vs electric)
• total = 6,230 Watts "typical worst case home loads"
  

Why I/we always say to "know your loads"... It is not unreasonable to see "worst case" loads at 10+ times that of average loads (above mythical home of 6,000 Watts worst case vs 400 Watts of average loading).

A reason why we always are asking is that 6,000 Watt (average loads) or 6,000 Watt*Hour (average energy usage per day)...

And why we push conservation... A 300 Watt desktop system vs a 30 Watt laptop--1/10th the energy usage for normal house hold usage (vs gaming, scientific, server usage).

-Bill

There is always the healing power of "and".  

-Bill

It all depends on the type of loads...

• Filament bulbs=  tend to be constant current.
• Purely resistive loads such as a heater, current will drop as voltage drops.
• DC motors will generally drop current as voltage falls--However motor design and type of loads i.e., fan vs compressor will affect current consumed.
• Electronic loads (LED lighting, motors with electronic controls, AC inverters) are constant power and will generally increase current as voltage falls.

Since most cabin/home power systems these days are 120/240 VAC, these systems use AC inverters. The AC output voltage of the inverter is usually regulated to its specified output voltage as long as its DC input voltage is within specifications (for a 12 VDC input inverter, typicality around 10.5 to 16 volts or so). This means the AC inverter is required to maintain output AC voltage into the loads (constant power). This means the AC inverter must draw constant power from the DC input, regardless of input voltage.

Using math... We can setup a very easy equation (model) for an AC inverter:

• Power = Voltage * Current (ignoring losses)

From the equation, we see that if voltage falls, the current must increase to keep the AC output constant voltage and supply the "constant" AC loads. For example:

• Example: 10 amp load @ 120 VAC load
  
• Power = Voltage * Current = 120 VAC * 10 Amps = 1,200 Watts
• Assume DC battery bank voltage is from 15.0 volts (charging) to 10.5 volts (near dead):
• Power = Voltage * Current
• Current = Power / Voltage
• DC bus current = 1,200 Watts / 15.0 VDC = 80 Amps (12 volt battery bank charging)
• DC bus current = 1,200 Watts / 10.5 VDC = 114 Amps (12 volt battery bank under heavy load, near dead and/or voltage drop in DC bus wiring)

-Bill

Hi Marc,

No problem... I was just not sure I was clear in my earlier post--And I had some free time to (again) type way too long of answers.  

-Bill

One of the things I try to do here is avoid "generalities". It is hard to have a "feeling" of what 3.3 kWH per day vs 33 kWH per day is...

I may start with some generalities on how many WH/kWH per day an "average" home/cabin may use and then do some math to show what a generic system could look like (costs/size of system/complexity) because 1,900 Watts of solar panels and 12 "golf cart" batteries starts to make it "real" when trying to power a refrigerator and some lights and an RV water pump.

But then quickly try to figure out what their actual energy usage is--Or close too it (from electric bill, Kill-a-Watt meter, location, why/when need for power, etc.).

The problem with generalities is that hard right and wrong are not always clear. As we talked about here, AH @ xx volts vs kWH battery capacity. They both work and are useful. Loads types are specific to an install... When talking about generic systems--I tend to "error" on the conservative side. With off grid solar, one is always over designing the system capacity. We have to because having a system that does what we need when new on a sunny day--Have to allow to supply needed power on a stormy day 5 years down the road instead of leaving the owner in the dark.

-Bill

I think there is a misunderstanding of the question. The question is "falls" not "fails" (FALL vs FAiL).

And for that answer, the old power equation still works:

• Power = Voltage * Current
• Current = Power / Voltage
• Current = 1,000 Watts / 15.0 volts (battery charging) = 66.7 Amps
• Current = 1,000 Watts / 10.5 volts (battery near dead) = 95.2 Amps

The Amp*Hour rating for a battery bank is (from what I can tell) based on a "constant current draw" (filament lamps, some electric motors are examples of, roughly, constant current devices).

The Watt*Hour rating is based on a constant power draw (typical inverter is a "constant power" device).

A battery bank rated in Amp*Hours is "less accurate" in today's world of UPS (uninterruptible power supplies--Where I have seen batteries rated in "native Watt*Hours") vs the typical Amp*Hour rating method (constant current draw).

Here is a glossy that shows one way how to do kWH and AH calcuations for UPS battery banks:

https://www.solidstatecontrolsinc.com/knowledgecenter/~/media/85b8e51754c446bda1f38449f444471c.ashx

The problem is that batteries are discharge rate sensitive... So AH at C/20 (20 Hour discharge rate) is different that C/6 or C/1 discharge rates.

From Trojan (just a random pick):

https://www.trojanbattery.com/products/deep-cycle-flooded/solar-industrial-line-flooded/

For example, a 6 volt battery has 610 AH @ 100 hour rate and 3.66 kWH @ 100 hour rate... What is the "average" battery voltage?

• P=V*I; I=P/V
• 3,660 WH / 610 AH = 6.000 volt average voltage

Now is that 6.000 volt number "real" or just somebody pulling 6 volt nominal... 3 cells * 1.75 volt "dead" = 5.25 volts (much less than 6 volts for much of the time, and less Watt*Hours (V*I*Hours). Don't know how "accurate" this kWH number is (at least for Trojan).

Anyway, just trying to show that a FLA battery has significant "capacity" variation between high discharge rates and low discharge rates... "One magic number" does not really give useful (accurate) answers. kWH capacity will also vary based on rate of discharge.

-Bill