UPS system

[PG1995]

Thank you for pointing this out although I knew that Ah rating is dependent upon the current drawn.

UPS run-time calcs:

**broken link removed**

It seems as if you need to consult the manufcturer's data.

You have a non-linear amp-hour capacity depending on discharge rate. See here: https://www.batteryweb.com/pdf/inverter_battery_sizing_faq.pdf
Amp-hours is an estimate only and might not even be close.

Let's talk about how Ah rating is affected by the current drawn.

Q1:
We will use this datasheet for lead-acid AGM 6V, 7Ah battery:
https://drive.google.com/file/d/0B_XrsbDdR9NEZWRyRzB6bndodUE/view?usp=sharing

i: In the table below, it can be seen that if the current drawn is 0.35A then the battery gives full 7Ah and when closed circuit voltage falls to 5.25V then the battery has fully discharged. Could you please confirm the information underlined?

On the opposite if the current drawn is 7A then it only gives 3.97Ah and the closes circuit voltage indicates the battery has fully discharged.

ii: I had though that "C" in the first column stands for 'capacity' but I don't think that it's correct. It should have something to with time.
7A*1C=3.97Ah
=> C = 34 minutes

Do you agree that "C" stands for 34 minutes

Q2:
Is the table below using voltage for a single lead-acid battery cell which has voltage of around 2.3V because voltages shown in the top row, like 1.85V, 1.8V, 1.75V, etc., are too low for a 6V lead-acid battery which consists of 3 cells?

This curve shows charging profile of a single cell of lead-acid battery.

Please help me with the queries. Thanks.

Regards
PG

 

[KeepItSimpleStupid]

Q1: C is capacity at the one hour rate. C=7 A/Hr.

the "20 hour rate" is 20 C. Later they didn't do a unit switch, but sort of did. You will also see charig rates specified as C/10 etc.

My guess for UPS capacity, one would want to use the E rate (Constant power). https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&uact=8&ved=0CDMQFjAC&url=https://mit.edu/evt/summary_battery_specifications.pdf&ei=8XjuVKeTL4GFgwTz14GYAQ&usg=AFQjCNH4N6qmO_B03zcVKo1BJLRs0sZgVw&sig2=3l_AXKTDEH4mv_Edbfs12w&bvm=bv.86956481,d.eXY

Your data for Q2 shows the E-rate (constant power).


Q2: Not as easy s Q1

 

[KeepItSimpleStupid]

Q2:See table 1: https://www.google.com/url?sa=t&rct...4xGXIhrpiqkwVAw&bvm=bv.86956481,d.cWc&cad=rja

 

[MrAl]

Hi,

In one of those links they specify watts and surge as before. In anther link they are stating a VA rating but "at" a certain power factor load, so they are really stating that 'watts' is the only way to specify the inverter. That's the way we did it way back when i worked in the industry.

 

[KeepItSimpleStupid]

Watts and "Surge Watts for x seconds" makes a lot of sense. "Surge" is too simplistic.

 

[MrAl]

Hello,

Well sometimes they want to limit the information and keep it simpler. For example they dont want to have to tell us the wire diameter to use on the output. Some however will tell the wire gauge on the INPUT, or at least suggest that it should be heavy with a larger inverter.

 

[KeepItSimpleStupid]

Hello,

Well sometimes they want to limit the information and keep it simpler. For example they dont want to have to tell us the wire diameter to use on the output. Some however will tell the wire gauge on the INPUT, or at least suggest that it should be heavy with a larger inverter.

They can't. It depends on the length or 2x the length depending on how you think.

Some "Bright" Electrical Engineering major in my class asked me if he could jump a car with a standard, say 18 AWG extension cord.

 

[MrAl]

They can't. It depends on the length or 2x the length depending on how you think.

Some "Bright" Electrical Engineering major in my class asked me if he could jump a car with a standard, say 18 AWG extension cord.

Hello again,


Sorry, but i think you are just getting too critical here. It is good to think critically in engineering, but you have to be able to moderate it. In your last post you seem to assume that nobody else knows anything so you assume they are both wrong: the manufacturer was wrong because they cant possibly know the wire size, and the bright engineer was wrong because you believe that a car can not be jumped with 18 gauge wire.

In all actuality the manufacturer can give the wire size if they provide a table that includes run lengths, just for one example.
And the size of the wire required to jump a car depends on how 'dead' the battery is and how much time you've got because even a smaller current can jump a car if you give the battery enough time to charge, as long as the battery is not totally beat. In some cases you have to do this anyway where you let the jumpers stay connected for a few minutes while you let the battery charge up.

But in any case we have found out how manufacturers rate their inverters, and there is some variation here and there so we have to be careful when we go to pick out a unit for a particular purpose. When i worked in the industry we made units that went up to 30000 watts, and they were always rated by the watts, and the limit for any load would be found from the current that a 30000 watt load could draw. That seemed to make sense because the internal parts are all rated by current.
Funny though, about a year ago i purchased a 200 watt unit that was on sale for something like $12 USD. It actually ran my soldering gun which was what i was hoping for because i needed to do some soldering on the car and didnt want to have to run an extension cord. Didnt try it on much else yet though as it puts out a single pulse per half cycle not a sine wave.

 

[KeepItSimpleStupid]

Basically what I said.

I too have a small inverter, but I forget how its rated. It's rated as Peak power 400 W and constant power 140 W so it fails.

The laptop supply has real ratings at: 100 VAC 1.5 A and 19V at 4.74A, so it cuts out when running the laptop - As expected.
150 W in and 90 W Out. Care to guess what 90/160 is? The magic number of 0.6.

I too use the inverter for a soldering iron when I decide I want to go to the park and set up shop on a picnic table.
I use an iGo charger for the car. I actually have two, but parts for the first are getting scarce.

The newer version is designed differently with a standard power cords for DC in and AC in. Only the output connectors are custom with a make before break contact. Two resistors in the connector set the voltage and current limits. Thus the charger can be moved from laptop to laptop theoretically. The newer supply may shutdown when the DC output connector comes apart. The older one won't. They can also feed a small wide range DC-DC converter simultaneously to drop the voltage to 5V with multiple cell phone type connectors available. The supplies support Aircraft power too.

Aircraft power does use a break before make contact that generally turns on the socket only when the plug is fully inserted or power is removed as your taking out the plug so there is little chance of a spark.


FWIW:
I HAD to change the lighter plug on my inverter to a European version. e.g. https://www.amazon.co.uk/Durite-Cigarette-Lighter-Plug/dp/B0051OC7UQ

They are FAR superior to the US versions, but they have really strange fuses inside. The ring comes off and is used in a more secure socket.

==

So hey, I got the 400 W inverter that won't power a 100 W laptop. (You know what I'm trying to say).

 

[PG1995]

Thank you, KISS.

This post is continuation of discussion from previous posts #21, #22, and #23.

Q1:

Q1: C is capacity at the one hour rate. C=7 A/Hr.

the "20 hour rate" is 20 C.

Don't you think that it should be C/20 instead? Please have a look here.

Q2:

Your data for Q2 shows the E-rate (constant power).

Theoretically, a 6V, 7Ah battery should be able to supply 1A of current continuously for one hour which means 42W. But looking at the table below it can supply 42W for only 30 minutes.

Q3:
We have missed one important point so far that a lead-acid battery shouldn't be discharged below 50% of its capacity. I believe that at cut-off voltage a lead-acid battery is considered to be 100% discharged and it will greatly its lifespan. I believe that the data posted in post #21 assumes that the battery is fully discharged to its cut-off voltage. Do you think this too?

Note to self: For a lead-acid battery cut-off voltage for a cell could as low as 1.6V (when load applied).

Please help me with the queries. Thanks.

Regards
PG

Helpful links:
1: http://web.mit.edu/evt/summary_battery_specifications.pdf (contains very good definitions)
2: **broken link removed** (is quite helpful)
3: **broken link removed** (detailed discussion of chemistry of lead-acid battery)

 

[KeepItSimpleStupid]

Q1: Good catch
Q2: One of the references I posted and I was hoping that you would catch is that 50% discharge is DEPENDENT on the current drawn. That's why the ultra-low cell voltages.
Q3: "50% capacity" cell voltage is defined based on the current drawn.

 

[tronitech]

Why has this post got overly complicated?

The quoted power factor of 0.6 must be for the UPS.

It's rated at 1000VA so if used on a PC with a good quality power supply which will have a power factor approaching unity then it will deliver its quoted capacity of 1000VA or 1000W.

If it's not used for a PC then the power factor (if known) of the intended load will have to be taken into account.

I'm of the opinion that a UPS should only be used has an emergency power source. They are not designed for long term use and for example to provide power to a PC/Server to shut down safety in the event of a power outage.

 

[MrAl]

I too use the inverter for a soldering iron when I decide I want to go to the park and set up shop on a picnic table.

Hi,

That's a very very interesting idea. I might have to try that too. I used to go to the park long time ago but not recently, and use my calculator. Would be interesting to go and work on a PC board on a picnic table on a nice day too. I picked up a portable soldering iron recently, i never 'believed' in them but it works pretty well (butane).

 

[KeepItSimpleStupid]

Aside: While I was using my laptop I lost my wireless network and I smelled smoke. It came from an APC Back-ups UPS, BUT it wasn't connected to my QNAP NAS. The BACK-UPS has super simple serial signaling, but the QNAP does have USB in. I was thinking at one point to try to add an Ethernet to serial converter to the NAS, but I never tried.

I have no idea what smoked yet. I might be in the market for a UPS. I would like to hide it in the rafters. The UPS powers a DSL modem via POE, has 2 switches, an access point, a Skype interface, a USB serial server, a router and a repeater. It will also support an automation server and a wireless alarm receiver and a NAS. The AP and router will get replaced by the repeater reconfigured at some point. I might add a 24 port switch and it might be POE.

Any suggestions for a UPS that can inform the QNAP NAS of an impending failure?

 

[KeepItSimpleStupid]

Mr Al:

I have this https://www.masterappliance.com/heat-tool-products/butane-powered/ultratorches/ultratorch-ut-100si butane soldering iron. I've had it a LONG time. It's been factory rebuilt twice. Yes, they repair them.

As for soldering in the park. I plug the inverter in and generally keep track of the time and run a 100 ft extension cord out the window.
I can easily get withing 100' of a picnic bench under the tall trees.

 

[MrAl]

Hi,

That looks like a nice iron. They appear to make lots of different tips for it too for soldering. That could be good. I dont think they make more than one type of tip for the Dremel.

Wow you are really into the park scene
I guess you dont use your butane iron out there.

 

[PG1995]

This is continuation of discussion from post #31 above.

Q1:
So, let me summarize the C-rate by updating the original data. Do you agree with the updated data?

Q2:
Did you have this graph data in mind?

Anyway, I believe that terminal voltage is not a good indicator of battery's capacity as I have been told in the past. We need to calculate Ah capacity of a battery by measuring the current going in/out of the battery and not rely on the terminal voltage.

On the other hand, I do agree that it looks like most UPS systems use terminal voltage (i.e. closed circuit voltage) to turn off the system so that battery(ies) aren't discharged 'too much' and that 'too much' point need not be 50% depth of discharge. For instance, we can choose the voltage points intersected by the dashed line like this to turn off the UPS system (assuming the system is not used at more than 1C capacity). In other words, as soon as voltage drops below the intersected value the system is turned off. We can see that for each curve the UPS system will be turned off for different Ah rating. Do you agree?

Q3:
Suppose you are going to design a 1000W UPS system. You can either make it a 12V system or 24V system. In a 12V system, you could connect two or more 12V batteries in parallel and not in series, and in a 24V system you could connect two 12V batteries in series. I believe that given the choice, one should design a 24V system rather than a 12V one because a lead-acid battery's capacity is greatly affected by the current drawn and connecting two 12V batteries could provide the same amount of power as two 12V batteries in parallel at half the current compared to a 12V system. So, in a way it 'enhances' overall capacity. What do you suggest? Thank you.

Regards
PG

 

[KeepItSimpleStupid]

Q1: Y
Q2: Y, Y
Q3: Y

Don't forget temperature effects and battery temperature monitoring. Battery balancing is probably easier to do in a series stack and there are IC's to do so. So called "battery gas guages" are available too and battery end of life estimators as well. I just ordered a refurbished APC SMT1500 Smart UPS from one vendor with a network management card/console cable from the bay,

 

[Tony Stewart]

Yes this is the SLA end voltage per cell.
I believe 1.85 is for ~10% SOC and 1.60V for 0% SOC so the average difference in Watts or Amps should be 10% for these extremes.

 

[PG1995]

Hi

Lead-acid batteries come in different Ah rating. Like a 12V battery could be rated 20Ah, 60Ah, 175Ah, etc. A 12V battery consist of six individual cells connected in series. What determineS the Ah rating of a lead-acid battery? I believe that it's size of cells. Likewise, the watt rating of solar panels of similar voltage rating differ by the size of individual cells. Here, you can see that 12V, 100W panel is double the length of 12V, 50W panel but both consist of 36 cells. Thanks.

Regards
PG