New 100 Ah battery only lasting 90 minutes on 300w discharge
- Thread starter Tariq0101
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Hedges
I See Electromagnetic Fields!
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Do Peukert and 50% discharge limit apply at the same time?
What if I have a 100 Ah battery and discharge at say 50A for one hour, at which point it is 100% discharged (low voltage cutout) according to Peukert.
Then I stop draining it and let it sit. Is it now just 50% discharged, and acceptable SoC for reasonable cycle life?
To the extent the reduced available Ah at high current is due to simple IR drop causing reduced voltage output, obviously the chemistry is unaffected. But maybe there is something in chemicals and ion transport also occurring that I don't know about.
Not explicitly, but when someone is using a battery in as consistent a way as the OP, it's pretty sensible to use it that way. 22A vs. 5A is going to be harder on the battery, so maybe you're only discharging to 55 or 60%, but you're trying to offset reduced longevity associated with the more aggressive use.
While there's capacity recovery, it's not 1 for 1. The higher current over a shorter period of time isn't as efficient as a lower current.
One of these days, I'm going to test reserve capacity and then see how much is left at the 20 hr rate for a similarly sized battery.
rin67630
Solar Enthusiast
How could a norm cheat? This is the official reference for solar batteries. Cranking batteries are rated at C/20.
And it makes sense to dimension a solar battery at C/100 in Europe since you never can be sure to have enough sun to recharge next day, so you must have enough capacity for 100hours at least.
IEEE 1562:2007 states the minimum battery autonomy should be 5-7 days for non-critical loads and areas for high solar insolation, and 7-14 days for critical loads or areas with low solar insolation.
That brings you around C/330 for the last condition.
P.S. I just have seen that DIN EN 61427 is verbatim with IEC 61427 Secondary cells and batteries for photovoltaic energy systems (PVES) - General requirements and methods of test (IEC 61427:2005)
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rin67630
Solar Enthusiast
Just to get an idea: how heavy is your "100Ah" battery?
I'd say 30-35kg as far as I remember, I couldn't lift it with one hand
RCinFLA
Solar Wizard
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At 25 amp load, battery terminal voltage will drop to 10.5v in about 2.6 hours. That's about 65 AH of capacity.
If your battery cable voltage drop is too much I can understand inverter cutting out in 90 mins.
At 25 amps battery terminal voltage will drop to 12.0 vdc in about 30 minutes.
Attached is actual data on sixteen different 105 AH lead-acid batteries. This was just a short load time quality check.
If your battery cable voltage drop is too much I can understand inverter cutting out in 90 mins.
At 25 amps battery terminal voltage will drop to 12.0 vdc in about 30 minutes.
Attached is actual data on sixteen different 105 AH lead-acid batteries. This was just a short load time quality check.
| Battery Voltage at Load test of 25 Amps | 30 min. Return | |||||||
| 30 min rest | ||||||||
| Start | 30 sec | 1 min | 5 min | 10 min | 15 min | 30 min | Recovery | |
| 12.83 | 11.92 | 11.96 | 12.07 | 12.06 | 12.04 | 11.98 | 12.66 | |
| 12.82 | 11.96 | 11.99 | 12.11 | 12.11 | 12.10 | 12.05 | 12.68 | |
| 12.82 | 11.93 | 11.98 | 12.08 | 12.07 | 12.05 | 11.99 | 12.66 | |
| 12.83 | 11.98 | 12.02 | 12.13 | 12.13 | 12.11 | 12.05 | 12.69 | |
| 12.80 | 11.95 | 11.99 | 12.10 | 12.10 | 12.09 | 12.03 | 12.67 | |
| 12.80 | 11.92 | 11.97 | 12.08 | 12.07 | 12.05 | 11.99 | 12.66 | |
| 12.80 | 11.95 | 12.00 | 12.10 | 12.09 | 12.08 | 12.00 | 12.67 | |
| 12.80 | 11.95 | 11.99 | 12.10 | 12.09 | 12.07 | 12.00 | 12.67 | |
| 12.93 | 12.06 | 12.07 | 12.16 | 12.15 | 12.13 | 12.07 | 12.75 | |
| 12.92 | 12.04 | 12.07 | 12.16 | 12.15 | 12.13 | 12.07 | 12.74 | |
| 12.96 | 12.09 | 12.08 | 12.18 | 12.17 | 12.15 | 12.09 | 12.78 | |
| 12.89 | 12.03 | 12.05 | 12.15 | 12.14 | 12.12 | 12.06 | 12.73 | |
| 12.93 | 12.05 | 12.07 | 12.16 | 12.15 | 12.14 | 12.08 | 12.76 | |
| 12.94 | 12.07 | 12.09 | 12.19 | 12.18 | 12.16 | 12.11 | 12.78 | |
| 12.89 | 12.04 | 12.07 | 12.17 | 12.16 | 12.15 | 12.08 | 12.74 | |
| 12.91 | 12.05 | 12.07 | 12.16 | 12.15 | 12.13 | 12.08 | 12.74 |
Do you know why the voltage drops first and then recovers after 5 mins? I'm just curious
chrisski
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This is the best write up I've seen for the Puerkert Effect:
As I interpret this, if you discharge the 100 ah battery at 20 amps, you would only have 60 amps of usable power, so the battery would be at 50% in 1.5 hours.
Batteries 4 Dummies. The Peukert Lesson - Solar Panels - Solar Panels Forum
Batteries 4 Dummies. The Peukert Lesson In today's lesson we are going to discuss Peukert's Law. Specifically as it applies to Flooded Lead Acid batteries (FLA).
www.solarpaneltalk.com
As I interpret this, if you discharge the 100 ah battery at 20 amps, you would only have 60 amps of usable power, so the battery would be at 50% in 1.5 hours.
RCinFLA
Solar Wizard
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On a lead acid battery there is often an initial dip in voltage when first loaded. This is particularly pronounced when a battery has been on float charge for a long time prior to discharging.
Lead-oxide coats the positive plate. It is what makes the positive plate positive polarity. If you have floated for a long time the lead oxide becomes thicker. Since lead oxide is not a good conductor it increases battery internal resistance a bit until the discharge burns off some of the lead oxide.
Beyond that, lead-acid batteries requires a lot of chemical transformations and a long path through electrolyte so the kinetic voltage required to create and transport ions is fairly high. The more charge or discharge current the greater the kinetic voltage overhead.
When the current demand stops the battery will slowly go back to equilibrium within 30 to 120 mins. and the kinetic voltage will go to zero.
Lead-oxide coats the positive plate. It is what makes the positive plate positive polarity. If you have floated for a long time the lead oxide becomes thicker. Since lead oxide is not a good conductor it increases battery internal resistance a bit until the discharge burns off some of the lead oxide.
Beyond that, lead-acid batteries requires a lot of chemical transformations and a long path through electrolyte so the kinetic voltage required to create and transport ions is fairly high. The more charge or discharge current the greater the kinetic voltage overhead.
When the current demand stops the battery will slowly go back to equilibrium within 30 to 120 mins. and the kinetic voltage will go to zero.
rin67630
Solar Enthusiast
you just eluded the interesting part of Tariq's question "...and then recovers after 5 mins".
That is indeed one of the mystery of battery chemistries.
I have the same problem upon trying to measure the battery's internal resistance by calculating the ratio between voltage drops and current increases: that ratio is not constant and changes noticeably when the drop is important, just to recover once the new situation is established.
No mystery at all. Answer is implicit. If a current causes a voltage drop, the removal of said current removes the voltage drop, so the voltage will rise. Same concept applies for charging. Apply a charge current, the voltage will rise. Remove charge current, the voltage will drop.
RCinFLA
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As a continuous discharge current that would be the appearance. If you stop the current drain periodically, and let it rest to equilibrium, the battery voltage will recover to some degree and there will be some available capacity still left in battery. On my 30 min., 25 amp discharge tests, the battery terminal voltage dropped below 12v but after 30 minutes of no load resting it recovered to near full charge voltage. Actual amount of discharged capacity was about 10%.
The actual capacity loss for high current is the kinetic voltage under load at the current demand for the amount of time loaded. For 25% C loading it is roughly about 0.5 to 0.7v for a six cell, 12v battery. Battery internal resistance which is a function of age and electrolyte acid concentation will add to the terminal voltage drop.
I am not a fan of Peukert's Law as it doesn't represent a normal useage profile of varying amount of loading current on battery. If you drain a battery at low current you will completely discharge capacity. If you drain it at higher current it will recover a significant amount of its capacity when allowed to rest for a while.
AGM have lower kinetic voltage drop because they have higher electrolyte acid concentration and shorter distance between positive and negative plates for ions to traverse. Rested voltage will be a bit higher because of higher electrolyte acid concentration.
For flooded lead-acid battery, sulphuric acid concentration is set to about 35% percent at full charge. It drops to about 10% concentration at full discharge. Specific gravity is used to measure concentration. Kinetic voltage gets greater as acid concentration decreases during discharge because it is harder to find available sulphuric acid molecules in electrolyte to combine with lead plates.
Acid concentration is a compromise between discharge and charge efficiency and effects battery longevity. Acid is needed for discharge, water is needed for charging.
| Battery open circuit 'rested' state voltage | |||||||
| Chrg State | Spec Grav 80 deg F | % Acid | Cell Voltage | 6v Battery | 12v Battery | 24v Battery | 48v Battery |
| 100 | 1.277 | 33.0% | 2.122 | 6.37 | 12.73 | 25.46 | 50.93 |
| 90 | 1.258 | 30.7% | 2.103 | 6.31 | 12.62 | 25.24 | 50.47 |
| 80 | 1.238 | 28.3% | 2.083 | 6.25 | 12.50 | 25.00 | 49.99 |
| 70 | 1.217 | 25.8% | 2.062 | 6.19 | 12.37 | 24.74 | 49.49 |
| 60 | 1.195 | 23.2% | 2.040 | 6.12 | 12.24 | 24.48 | 48.96 |
| 50 | 1.172 | 20.5% | 2.017 | 6.05 | 12.10 | 24.20 | 48.41 |
| 40 | 1.148 | 17.6% | 1.993 | 5.98 | 11.96 | 23.92 | 47.83 |
| 30 | 1.124 | 14.8% | 1.969 | 5.91 | 11.81 | 23.63 | 47.26 |
| 20 | 1.098 | 11.7% | 1.943 | 5.83 | 11.66 | 23.32 | 46.63 |
| 10 | 1.073 | 8.7% | 1.918 | 5.75 | 11.51 | 23.02 | 46.03 |
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rin67630
Solar Enthusiast
I'm afraid, you did not catch the question:
If you apply a sudden load, you will get a voltage drop and if you let the load connected, after a couple of minutes the voltage will partially recover. RCinFLA gave an answer to that phenomenon.
Delmar
Solar Addict
What are you using for an inverter? I suspect a UPS where you replaced the internal battery.
Most consumer UPS's will ignore the actual battery voltage and instead use a timer chart based on the AC power draw. You can substitute any larger battery and it will still shutdown at the same time. Leaned this the hard way.
No I'm using a 1000 watts inverter
GSXR1000
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you say 300W i don't think you included the inverter losses so more like 350w to 400w.
at 400w it would be 33 amps at 12v
