Calculating UPS Run Time for 12V 7Ah Battery & 220VAC 1A Device: Using Formulas & Assumptions

Hello.

it is known that when you have a 12V 7Ah battery and when you hook it up with a 12V device that gets 1A, it is known that theoretically it comes on this battery for 7 hours.
And what if we have a 12V / 220V UPS with a 12V 7Ah battery and we attach a 220VAC device to it, ofcours what gets 1A, how long does such a device come from on this battery? Is there any formula to count it ??

Thank you in advance for your answer.
Greetings.

First of all, at 220V, the 1A draw device will eat P = U * I = 220W.
At 12V and 1A P = U * I = 12W.
It can be easily calculated that the device consuming 1A at 12V will last almost 20 times longer.
The UPS is a voltage converting device - the power remains constant.
In addition, the efficiency of such a device is in the range of 50-70% (possibly even less - maybe someone has data on this subject). So, for example, if the UPS consumes 1000W, the usable power will be 500-700W.

And also the discharge characteristics of the battery. It has 7Ah with a 20-hour current (350mA) - the higher the current, the less energy can be drawn from it, e.g. to a current of 7A, it will be half, i.e. it comes from 0.5 hour

Added after 16 [minutes]:

In the case of downloading 12W "directly" from the battery I get 6 hours.
When using a UPS with 60% efficiency and a load of 220V 12W 4 hours

you should deduct heat losses, low current conversion to high current and UPS own power supply with a value of about 0.5A

elektronik321 wrote:

you should deduct heat losses, low current conversion to high current and UPS own power supply with a value of about 0.5A

yeah, everything is correct, it only interests us specifically how many of these losses are. I also met the 50-70% range given by Mrrudzin and I would stick to that.

In concrete terms, they can only be determined empirically - connect wattmeters to the input and output and divide the value at the output by the value at the input.
I suspect that the efficiency will depend heavily on the load.

forestx wrote:

In the case of downloading 12W "directly" from the battery I get 6 hours.
When using a UPS with 60% efficiency and a load of 220V 12W 4 hours

okay, but unfortunately at 220V the device will take about 200W, not 12W.

the second thing, if you count on watts, how many of these watts can UPS give away? If we take, for example, a 600W UPS, i.e. if we connect a device to it (for 220V, consuming 1A) = 220W, it means that what will keep it powered for less than 3 hours ...?
Somehow I can't figure it out.

And looking from the other side:

mrrudzin wrote:

First of all, at 220V, the 1A draw device will eat P = U * I = 220W.
At 12V and 1A P = U * I = 12W.

so at 12V and 1A P = U * I = 12W but these 12W are for an hour? or how? because theoretically a device for 12V 1A comes for 7h, i.e. 12W * 7 = 84W and what is the number and what do I need it for? how does this relate to the device that must download 220W ?? no, I don't know anymore
How to count how long a given device will be supported by a given UPS?

Thank you in advance for your answer.
Greetings.

Quote:

okay, but unfortunately at 220V the device will take about 200W, not 12W

Read carefully:
Both devices charge 1A

Quote:

the second thing, if you count on watts, how many of these watts can UPS give away? Them

In the case of the UPS and other electrical devices, the rated power is given.
The UPS has a given maximum load.
The second thing is the capacity of the batteries.
Generally: the 1000W UPS will not power the electric locomotive for 1s (example) - simply the converter will not allow it (read: it will burn out).

Quote:

so at 12V and 1A P = U * I = 12W but these 12W are for an hour? or how?

Another thing that is defined is energy (W = U * I * t) where time is still involved.
For example, if you have a 100W bulb and it stays on for 2 hours, it will use 200Wh.

Okay, then how to count how long a given device will be supported by a given UPS?

Let's assume I have a 600W upsa with one 12V 7Ah battery and connect a 220V 1A device to it, which is 220W.
What formulas do you need to use to count it?

Thank you in advance for your answer.
Greetings.

The energy that the battery will theoretically give away W = 12 * 7 = 84Wh
For this, let's assume that the converter efficiency (and all losses) are 60%.
So the energy that we are able to receive on the 220V side is 85Wh * 0.6 = 50.4 Wh
A 1A 220V device eats 220Wh
Therefore, for example, proportions could be written
220Wh - 1h
50.4Wh - x

Hence we calculate that X = 50.4 / 220 = 0.23h which approximately gives us 14 minutes of work.

Of course, all this is an estimate.

Oh gosh, THANKS FOR YOUR HELP, and sorry you had to write so much, sorry mrrudzin I already had an explanation in your first post "UPS is a voltage converting device - the power remains constant." somehow I did not argue it: / sorry and thanks a lot for your help and patience because it was your last post that struck me. :)

greetings

Buddy mrrudzin, this unit of energy is strange. Probably a simple oversight, but it's worth correcting ...

Well, maybe a little but after a little thought, it becomes understandable. I would calculate it like this: 220W at the output is 366W from the battery (60% efficiency). And that is 30.5A at a voltage of 12V. I would compare the current of 30.5A with the capacity of 7Ah, so we get the current of 4.35C. And then a chart from this pdf http://www.edw.com.pl/pdf/k01/10_15.pdf " target="_blank" rel="nofollow noopener ugc" class="postlink inline" title="" > http://www.edw.com.pl/pdf/k01/10_15.pdf and unfortunately we have 4 minutes. Of course, all this is an estimate .

I am sorry to stick to this current, but here you can clearly see how the "capacity" changes depending on the current. And not everyone knows it-remembers it.

I think that it is enough to close windows and save the MA thesis in the word and that's it!

Quote:

Buddy mrrudzin, this unit of energy is strange. Probably a simple oversight, but it's worth correcting ...

That's right. I don't know where this macabre idea came from. I corrected the error of course - thanks for paying attention.

Quote:

I am sorry to stick to this current, but here you can clearly see how the "capacity" changes depending on the current. And not everyone knows it-remembers it.

Frankly speaking, I did not mention this matter - without a graph it is difficult to "estimate" something in this matter.

Quote:

I think that it is enough to close windows and save the MA thesis in the word and that's it!

Whoever lost even 5 pages of work or any document does not need UPS to make frequent copies and save work progress

I think we explained the subject "roughly". Of course, the whole thing is based on a rough assessment of efficiency, and on the capacity of the battery, which does not necessarily have as much Ah as it writes on it (e.g. aging of the gel tubes). And there is also a matter of reactive power, which, tangling between the UPS and the receiver, can "add" to it. And now, out of curiosity, I will ask why the author needs this time. Taking into account the power (220W), I suspect the normal use of a UPS - i.e. for a computer. And because I am leaning on the APC350 with my left leg (7Ah battery, but bitten by a bit of time), I have just measured: the computer + monitor that downloads 170W from the network turns off my UPS "and after 3 minutes 41 seconds :D

greetings

I have a question, I am not electro, but my underfloor heating pump consumes 62W of power and is connected to a 230V socket, does it make sense to buy a UPS with a 12V 7ah power supply - how long will this solution last?
This is the only heating I have, and there was no electricity a few times , the floor heating must run for at least 2 hours to keep it warm.
Thanks to everyone for the answers and best regards

I will step on because there are a million topics, and this one fits.
I have a power supply as in the link http://www.salicru.com/files/documentacion/ficha-pdf-5641aa34cb4df_ingles.pdf
With the following information: (Typical back up time (3) - 10 minutes, (3) At 75% of load), can I calculate the backup time for loads of 30%, 50%, 100%?

It will definitely not be a linear function. The greater the load, the faster it rains. At 100%, it will not be 13.3 minutes, but less.