Are all amp hrs equal?

[Gjac]

Say you have 230 amp hrs in 2 6vGC batteries and you discharge at a low amp rate say 1 amp so if you discharge to 50 percent SOC you should have used 115 amps in 115 hrs. If you then discharge at a higher rate say 10 amps would it then take 11.5 hrs to get to 50 percent SOC? Or does the higher discharge rate create more heat and loss occurs during discharge causing batteries to have less amp hrs?

[BFL13]

Re: previous post.

That is why Trimetric does not use Peukert for its AH counter. It all works out "close enough" for the usual mix of high and low amp draws while RVing.

Having done this type of mix many times over the years, I can confirm that you can use the Trimetic AH counter with confidence for both discharging and recharging.

[Searching_Ut]

To add to your confusion on lead acid battery "Capacity", you also have to remember that temperature, and the Peukert effect are only applicable if the current conditions remain the same, ie temperature or discharge rate. If you warm up a battery that has only been able to deliver a small amount of current, it's ability to deliver more current at a higher voltage will rise. Peukert effect only applies if you maintain the current discharge rate. If you discharge a small battery bank at a high current rate for say running a microwave off an inverter, you will quickly encounter a low voltage cutoff on your inverter after having only pulled a few amp hours. The battery voltage will quickly recover however after you remove the load, and if you then apply a load at say the 20 hour rate, you will still get close to the same amp hour total when you add the amp hours you pulled at the lower rate to the amp hours pulled at the high discharge rate.

[BFL13]

Edit I swapped this over to the res fridge thread--

Hard to compare yourself with another RVer. You don't know if he chooses to use an electric kettle or a pot on the propane stove to boil water, eg.

The fridge is a killer all by itself. Say 8 amps to run it (if a big fridge---my 3.2 cu ft electric one in the TC draws 4.5ish.) with 65% on time out of 24 hrs. That is 125 AH a day right there.

We use about 70AH a day in the 5er in summer, but that gets close to 200AH a day in winter with more lights-on time, more TV time, and way more furnace time, which can hit 100AH a day by itself if it is cold out (say 35-40F).

So I carry 6 batteries in the winter and four in the summer. Hate to think how it would work with a res fridge in the winter. 300AH a day? Out of the bank of 6 batts to 50% can do, so now to recharge every day back to 90%.

That is going to take some big amps of charging with a big generator and take at least 3 hours. (Takes me two hours and 15 minutes to restore 200 AH as a 50-90 using 150amp charger and a 3000w gen to run the chargers)

Don't forget when it is cold your bank's AH capacity is much reduced from as rated at 25C/77F, so you are already hurting for battery.

[Gjac]

DrewE wrote:

Gjac wrote:
Thanks for all the explanations. I have a friend with a 2012 Pheaton with 6 6v GC batteries in it. He has a residential refer and a lot of electronic devices inside. On the other hand I have a 1996 GBM with 2 6 v GC batteries but very low amp usage water pump and a few lights. When dry camping I go 7 days before drawing down to 50 percent SOC on the other hand he has to run his genset 2 hrs a day to stay above 50 percent. I never could figure out why before I thought his batteries were bad.

The explanation there has much more to do with simple ratios of battery capacity to energy usage. His battery capacity is about three times yours, but his electric usage rather much more than three times your usage. The residential fridge uses somewhere roughly around ten times the electrical energy of your absorption fridge's control board (but, of course, no propane).

I was thinking 2 batteries would take care of the refer, and 2 would take care of the tv and more advanced electronics in a newer MH and the other 2 would put him equal with me but I was thinking wrong.

[DrewE]

Gjac wrote:
Thanks for all the explanations. I have a friend with a 2012 Pheaton with 6 6v GC batteries in it. He has a residential refer and a lot of electronic devices inside. On the other hand I have a 1996 GBM with 2 6 v GC batteries but very low amp usage water pump and a few lights. When dry camping I go 7 days before drawing down to 50 percent SOC on the other hand he has to run his genset 2 hrs a day to stay above 50 percent. I never could figure out why before I thought his batteries were bad.

The explanation there has much more to do with simple ratios of battery capacity to energy usage. His battery capacity is about three times yours, but his electric usage rather much more than three times your usage. The residential fridge uses somewhere roughly around ten times the electrical energy of your absorption fridge's control board (but, of course, no propane).

[Gjac]

Thanks for all the explanations. I have a friend with a 2012 Pheaton with 6 6v GC batteries in it. He has a residential refer and a lot of electronic devices inside. On the other hand I have a 1996 GBM with 2 6 v GC batteries but very low amp usage water pump and a few lights. When dry camping I go 7 days before drawing down to 50 percent SOC on the other hand he has to run his genset 2 hrs a day to stay above 50 percent. I never could figure out why before I thought his batteries were bad.

[MrWizard]

Gjac wrote:

jharrell wrote:

Gjac wrote:
I assumed this loss in amp hr capacity was due to heat. If this is true, can one also assume that these higher discharge rates producing more heat also reduce battery life every thing else being equal (still only discharging to 50 percent)?

No in fact Peukert's law does not take into account temperature change from self heating. Temperature does effect capacity but that it is somewhat independent of discharge rate. That is if you keep the batteries cooled to a constant temperature during discharge, capacity will still be reduced according to Peukert's law.

. If the reduced AHs due to the higher discharge rates is not caused by heat what then actualy causes this reduction?

chemical reaction efficiency is reduced and increased plate sulfation, which in turn decreases exposed plate surface area and weakens the electrolyte at the same time, reducing over all capacity

think of it like fuel mileage,
the faster you go the less miles per gallon, and the less miles per tank of fuel, there is a sweet spot that gives you maximum fuel mileage, but you won't drive all day 30 mph
and you won't use your batteries at 3 amps either
although each of those examples will give you more mpg and more ampHrs from your battery bank

[grizzzman]

Gjac wrote:
Say you have 230 amp hrs in 2 6vGC batteries and you discharge at a low amp rate say 1 amp so if you discharge to 50 percent SOC you should have used 115 amps in 115 hrs. If you then discharge at a higher rate say 10 amps would it then take 11.5 hrs to get to 50 percent SOC? Or does the higher discharge rate create more heat and loss occurs during discharge causing batteries to have less amp hrs?

"The Peukert Number is directly related to the internal resistance of the battery. Higher currents mean more losses and less available capacity"

That being said, resistance during charge/discharge will create heat.
Remember it is a electro/chemical action that takes time.

[Gjac]

jharrell wrote:

Gjac wrote:
I assumed this loss in amp hr capacity was due to heat. If this is true, can one also assume that these higher discharge rates producing more heat also reduce battery life every thing else being equal (still only discharging to 50 percent)?

No in fact Peukert's law does not take into account temperature change from self heating. Temperature does effect capacity but that it is somewhat independent of discharge rate. That is if you keep the batteries cooled to a constant temperature during discharge, capacity will still be reduced according to Peukert's law.

. If the reduced AHs due to the higher discharge rates is not caused by heat what then actualy causes this reduction?

[DrewE]

BFL13 wrote:
"An amp hr at 12v is 12 watt hrs , an amp hr at 13v is 13 watt hrs"

Wait a sec! That is counting the volts twice or something tricky?

A Watt-hour is (volts x amps) x hour while an amp-hour has no voltage. Or something. 🙂

They're measuring different things, even though we in the RV world tend to treat them as equivalent measures. We likewise tend to do the same thing with amperes (current) and watts (power).

An amp-hour is a measure of charge (a certain number of electrons, if you like). A watt-hour, on the other hand, is a measure of energy or work--the amount of useful (or useless, I guess) activity that can be achieved by the charge. An amp-hour at 12V is less energy than an amp-hour at 24V (or any other greater voltage).

If the voltage is constant, then the charge is exactly proportional to energy and current proportional to power and our simplifications work out okay in practice. If there are differing voltages involved, it can get rather confused rather quickly.

A water analogy might be helpful here. An amp-hour (or other unit of charge) would be equivalent to a volume of water--say a gallon. Voltage is analogous to pressure. A gallon of water working a turbine at a pressure of 12 psi will do less work than a gallon at a pressure of 24 psi. (The equivalent of the ampere in this analogy would be the rate of flow of the water, gallons per hour or something like that.)

[jharrell]

Gjac wrote:
I assumed this loss in amp hr capacity was due to heat. If this is true, can one also assume that these higher discharge rates producing more heat also reduce battery life every thing else being equal (still only discharging to 50 percent)?

No in fact Peukert's law does not take into account temperature change from self heating. Temperature does effect capacity but that it is somewhat independent of discharge rate. That is if you keep the batteries cooled to a constant temperature during discharge, capacity will still be reduced according to Peukert's law.

[BFL13]

"An amp hr at 12v is 12 watt hrs , an amp hr at 13v is 13 watt hrs"

Wait a sec! That is counting the volts twice or something tricky?

A Watt-hour is (volts x amps) x hour while an amp-hour has no voltage. Or something. 🙂

[MrWizard]

An amp hr at 12v is 12 watt hrs , an amp hr at 13v is 13 watt hrs
No they are not equal
But more important is the chemical reaction efficiency of the battery

It's related to lead plate thickness and density vs electrolyte
Thick solid plates react more slowly to produce electrical current
A slow low amp draw is very efficient
A higher amp faster draw is less efficient , warmer temps actually speed up reaction and are more efficient until the temp is to high and starts causing battery damage

Porus grid plates expose more surface area aka start batteries, and produce more current faster, but do no recover well from deep discharge, and shed more plate material, and loose capacity

[time2roll]

Capacity chart based on load

Amp-hour chart based on time