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This question already has an answer here:

I'm going on a road trip and will be sleeping in the bed of the truck at night. I want to be able to have a fan running during this time. I have a couple old car batteries sitting around, and I was thinking I could charge them up before I leave home, and then use them to power the fan at night. The engine will not be running when I am using the fan, and I don't want to drain the battery that's used for starting the engine. The thought is that I get a power inverter so that I get 120 VAC out to power typical AC desk fans (the smaller the better).

I want to calculate how long a battery can power such a fan.

Let's say the fan is a 120 V, and on the "Low" setting consumes 30 W. Let's say the battery is a 600 CCA, 12 V, lead-acid car battery, with a 50 Ah capacity (20 Ah Rate).

How can I calculate the time that the battery will last? I know that the power rating of the fan can give the current used (P=IV), and the capacity of the battery can tell me how long at whatever current (Capacity = Amps*hours). But I don't know which voltage to use (12 or 120), or how to incorporate the inverter or the part about the 20 Ah Rate.

I'm also open to other ideas for running a fan with the engine off.

Another additional thought is to connect the battery in parallel to the truck's battery and that way it would be charging when I drive during the day, but the batteries are different capacities, and I don't know if there would need to be other hardware to prevent my truck battery from being drained. If this can't be done then I will just drain the battery and when it's dead, I'll just have to wait to get back home to charge it.

Thanks!

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marked as duplicate by winny, Harry Svensson, Voltage Spike, Dmitry Grigoryev, MCG Jun 1 '18 at 10:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    \$\begingroup\$ Going from 12V to 120V cost energy. Most efficient would be to use a 12V fan. (Or more then one). The first one I checked uses 100mA. (that is 1.2W) \$\endgroup\$ – Oldfart May 16 '18 at 17:33
  • \$\begingroup\$ There are certainly 12V fans and blowers used in cars, a bit more powerful than the one @Oldfart found. \$\endgroup\$ – Brian Drummond May 16 '18 at 18:22
  • \$\begingroup\$ Any idea about calculating battery life time though? \$\endgroup\$ – mrkevelev May 16 '18 at 19:17
  • \$\begingroup\$ 30 watts sounds high for a 12V desk fan. I'm thinking 10-15 would be more like it. \$\endgroup\$ – Robert Endl May 16 '18 at 21:49
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@PDuarte's answer is not quite correct. However, let's try another tack.

Your 30 watt fan will draw about 2.5 amps at 12 volts (2.5 times 12 is 30). Assuming 90% efficiency from the inverter means that the inverter will draw 2.5/.9, or 2.78 amps. A 50 Ahr battery will last 50/2.78 or 18 hours.

Why is PDuarte's answer different? Because a battery does not lose voltage linearly with time. Here is a site which addresses the issue. A "typical" lead-acid battery will have a discharge curve which looks more or less like enter image description here for a 20 Ahr unit. Since you are talking about a 50 Ahr battery, the 0.6 A curve is closer to correct. Note that the battery voltage will remain above 11 volts pretty much to the end.

However, you need to be aware that, as the site says,

Don't expect battery to give you 100% of capacity and even don't try to be close to that.

enter image description here

Granted, 8 hours times 2.78 gives 22.2 hours, which is less than a 50% discharge for your 50 Ahr battery, so you're unlikely to experience problems, but it's something to keep in mind in any future endeavors

There are such things as "deep discharge" lead-acids available which will tolerate heavy use, but they are expensive, and car batteries are not made to that standard.

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Rough calculation following....

Fan is drawing 30W. Assuming an inverter of 80% efficiency, this means that the power consumed by the battery is 37.5W (37.5 x 0.8 = 30)

Assuming a nominal value of 12V from the battery in order to not complicate things, this means the current consumed @ 37.5W is 3.125 A. This is the current value we use for calculating the time.

For a 20Ah battery, this means a 6.4 hour (20 / 3.125) lifetime.

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Quick calculation that could estimate the best case (i.e. don’t expect to see the fan working longer than that:

  1. Your battery energy: 50Ah (that doesn’t actually means it will have this charge, generally is less than that)
  2. Your inverter efficiency: a fantastic circuit could reach around 90%.
  3. Your load: 30W
  4. Let’s not consider the battery voltage drop during time and estimate 8V average until it looses all energy

So: 50 x 8 = 400 W (this is 400W power during 1 hour) For your fan, 400 / 30 ~ 13.3 hr Considering losses: 12 hours of total fan if battery is fully charged.

THIS IS THE BEST CASE you’ll get, normally it’s less. Also, I’m not considering that your inverter won’t be able to use all the battery’s energy since it will stop working after the input voltage is less than a specific threshold depending on the manufacture’s design.

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  • \$\begingroup\$ Why are you doing 50Ah x 8V? I'm confused. \$\endgroup\$ – mrkevelev May 17 '18 at 16:25
  • \$\begingroup\$ To find the total Wh of your battery. I used 8V as an approximation of the average voltage you’ll experience during the battery discharge. 50 x 12 is not realistic. \$\endgroup\$ – PDuarte May 17 '18 at 16:57

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