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So I have a Bosch L5077 180A 12v deep cycle sealed Pb battery for my solar setup. I bought it recently (2 weeks ago) but since the weather was rainy the last 2 weeks and the solar panels didn't produce any significant power, I keep charging it on-grid with a 10A 3-stage smart charger.

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

So, in 6 hours of usage which usually reaches the threshold in which the inverter shuts down, I'm using a 43'' 50w LEDTV, one 32'' 35w LEDTV, a laptop charging (60w/hour) and few home led lights opening now and then.

With a quick calculation we have :

43'' LEDTV 50w x 6 hours = 300W

32'' LEDTV 35w x 6 hours = 210W

Laptop charger 60w x 6 hours = 360W

Inverter 1500W consumption, let's say 30w x 6 hours = 180W

And the lights which shouldn't be more than 100watts total in those 6 hours as I don't use much the lights.

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

Where did the rest of 85 amps went to ? The charger is always charging fully the battery, when the battery is fresh charged it will show a value of 13.6V in PWM charger, while 2 hours later it will show between 13.4-13.5V so I guess the battery can't be damaged so soon.

Any ideas ?

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    \$\begingroup\$ Have you accounted for DC/AC conversion efficiency of the inverter under the load of your use case? \$\endgroup\$ – vicatcu Nov 21 '19 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\$ – kostas parlapas Nov 21 '19 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 '19 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\$ – Kevin White Nov 22 '19 at 1:41
  • \$\begingroup\$ take a gander at this e-education.psu.edu/eme812/node/738 \$\endgroup\$ – vicatcu Nov 22 '19 at 1:43
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The battery's capacity is specified at 20h, so 9A. That means that if you pull 9A, its capacity is 180Ah.

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

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

This is probably where the bulk of your capacity "loss" is. There are also charging and conversion losses, and the cut-off voltage to consider.

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