2
\$\begingroup\$

I have a Bosch L5077 180 Ah, 12 V deep-cycle sealed Pb battery for my solar setup. I bought it recently (two weeks ago) but since the weather was rainy the last two weeks, and the solar panels didn't produce any significant power, I keep charging it on-grid with a 10 A, 3-stage smart charger.

Assuming that this battery has a capacity of 180 Ah which is around 2 kWh of energy, the battery is almost fully drained (11.3 V) before even consuming 1 kWh or so.

So, in six hours of usage which usually reaches the threshold in which the inverter shuts down, I'm using a 43", 50 W LED TV, one 32" 35 W LED TV, a laptop charging (60 W/hour) and a few home LED lights now and then.

With a quick calculation we have:

  • 43" LED TV 50 W x 6 hours = 300 W
  • 32" LED TV 35 W x 6 hours = 210 W
  • Laptop charger 60 W x 6 hours = 360 W
  • Inverter 1500 W consumption, let's say 30 W x 6 hours = 180 W.
  • Lights, which shouldn't be more than 100 W total in those 6 hours, as I don't use the lights much.

The result is 1050 W of energy consumed and the battery is fully drained.

Where did the rest of the 85 Ah go?

The charger is always charging the battery fully; when the battery is freshly charged it will show a value of 13.6 V on the PWM charger; 2 hours later it shows between 13.4-13.5 V so I guess the battery can't be damaged so soon.

Any ideas?

\$\endgroup\$
7
  • 6
    \$\begingroup\$ Have you accounted for DC/AC conversion efficiency of the inverter under the load of your use case? \$\endgroup\$
    – vicatcu
    Nov 21, 2019 at 23:23
  • \$\begingroup\$ How to do that ? I have no idea how much the inverter draws in standby or under load. It mentions nothing also in its manual. \$\endgroup\$ Nov 21, 2019 at 23:37
  • \$\begingroup\$ the battery is probably rated at 9 A over a 20 hour discharge interval ... for example, if you draw 45 A, then the battery will last less than 4 hours ... refer to the datasheet, if you can find one (i could not find it) \$\endgroup\$
    – jsotola
    Nov 22, 2019 at 1:08
  • \$\begingroup\$ How do you know the various loads? Especially the power for TVs will vary considerably depending upon usage - in particular the backlight intensity setting. I don't think you can conclude much unless you actually measure the current being taken from the battery. \$\endgroup\$ Nov 22, 2019 at 1:41
  • \$\begingroup\$ take a gander at this e-education.psu.edu/eme812/node/738 \$\endgroup\$
    – vicatcu
    Nov 22, 2019 at 1:43

1 Answer 1

4
\$\begingroup\$

The battery's capacity is specified for 20 hours of use, so 9 A. That means they specify that when you pull 9 A, the battery's capacity is 180 Ah.

Unfortunately, that does not mean you can pull 18 A for 10 hours, or 36 A for 5 hours; for higher currents, the battery's capacity is (much) lower.

You say 1050 Wh in 6 hours, so 175 W on average. At roughly 11.5 V, that means the average current is a little over 15 A; well over the 9 A at which the capacity is specified.

This is probably where the bulk of your capacity "loss" is. Some batteries' data sheets show different capacities at different discharge currents, so maybe you can look this up.

There are also charging and conversion losses, and the cut-off voltage (i.e. how far you discharge, which determines what percentage of the capacity is actually used) to consider.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ All correct. And even if they state 9 A for 20 h is possible, it already must be considered to be a "deep-cycle" and the number of such cycles is limited (few hundred, if specified at all). Also deep-cycle batteries live longer if only discharged down to 50 or 30 % SOC. So if calculating the Wh also the changing voltage and the lowest SOC should go in. \$\endgroup\$
    – datenheim
    Jan 25 at 8:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.