Power maths
Kris
I have a 50w device (0.05 kwh) and I'd like to know what UPS I need to power it for 16 hours.
UPS seem to be rated as 1000VA 230v which I don't understand how to equate this to 0.05kwh @ 240 over 16 hours
Kris
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Lachlan Horne
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ada
On May 17, 3:31 pm, Lachlan Horne <wil5oncle...@gmail.com> wrote:
> You need 0.05kW for 16 hours: 16 * 0.05 = 0.8kWh. As for your UPS rating,
> nothing in what you've quoted mentions time, it's only power. The UPS is
> rated for 1000VA which is effectively 1kW so you are well within its
> capability, but the units you have don't suggest how long it will last for.
load /DC supply, because of the notion of 'power factor'.
most UPSes are rated by the size of the output inverter, and then have
batteries sufficient for (typically) 2-15 minutes of runtime at that
rated output power. (see https://www.apc.com/products/runtime_for_extendedruntime.cfm?upsfamily=165
for example.)
is this something that you want to run off batteries for 16 hours, or
do you want to be able to survive a 16 hour outage with continued
service?
kris
I think it's weird that they wouldn't state a nominal value for how long it could provide power. that really boggles me.
Not doubting you Ada just i would have thought it would be a factor a shopper might want to know.
anyway, yes now that i'm not on the phone, i can type betterer* and give you a bit more background on what i want to do
I'm investigating powering a device 24/7 using as little 'expensive' electricity as possible.
I have an 'off peak' meter which provides my water heater with electricity from 11pm to 7am at 7c per kWh
I normally pay 26c per kWh at all other times.
The device I'd like to power sucks 50w constantly (measured from the 240v plug)
What i would like to do is:
have a battery power the device
have a solar panel charge the battery (or maintain the charge and power the device if the battery is full)
have the off peak power charge the battery to full IF the battery does not have 12 hours of juice left.
the device is very stable in its use of 50w
the solar panel i'm looking to get is an ebay 120w folding solar panel caravan kit. It comes with a battery charging regulator but i'm unsure how well this would work if i'm constantly drawing 50w as its trying to charge.
It is important that the device remains powered, more important than having the battery charge by solar alone.
So:
is this doable with off the shelf stuff (or simple circuit building). Ie do i just go and buy a UPS unit
In your opinion am i better off simply buying a big battery and have it charge at night (forget about the solar since at 7c K/w it would be a long time to pay off the investment of the solar panel)
What size battery do i need? I'm struggling to understand 800w per hour at 240 from a 12v source
as i understand it, batteries are described as amps per hour. in that a 1ah battery will provide me with X voltage at 1 amp for 1 hour and then be empty. Or it can provide X voltage and 0.5 amp for 2 hours etc.
but i need 800w to provide enough juice to run 50w for 16 hours
I = p/v
so
I = 800/240
so 3.3Amps total
or i need a battery that is 3.3Ah
but wait, there's more. Lead acid/gel and the other one all don't like being drained past 50%, it will severely shorten their life.
So I need a 7AHh (rounding) battery?
but that is a 12v battery, not a 240v ac battery.
though since this thing uses a computer power supply i think i can just remove the AC part of it and provide 12v and 5v through a BEC
Kris
*betterer, yes its a word. You can look it up in the krisionary.
Aaron Power
Kris,
You are getting yourself confused by the units.
Your device is drawing 50W. It is running off 240V. Therefore it is drawing about (50 / 240) or 0.2 Amps (I say “about” because of the power factor that Ada already mentioned – but let’s assume a resistive load for the moment).
This is simply worked out from P=IV (or Power equals Current multiplied by Voltage).
So we now know how much it draws from the supply (both “Power” and “Current”) at every instant in time. Both Power and Current are instantaneous values – there is no concept of “time” with either value.
Now you want to know how long it will last. To do that, just multiply the above values by the time you are interested in – in this case 16 hours.
Power multiplied by time gives you energy*. Energy can be measured in Watt-Hours (that is Watts *times* Hours – not Watts *per* Hour) or in Watt-Seconds, which are also known as Joules. Most science and engineering will talk about energy in Joules, but we buy our electricity in kW-Hr. You are still buying energy, not power.
So over the 16 hour period your device will consume (50 * 16) or 800 Watt-Hours, which can also be written as 0.8 kW-Hr. It could also be written as (50 * 16 * 60 * 60) or 2.88 Mega Joules.
We also know that over 16 hours it will draw (0.2 * 16) or 3.2 Amp-Hours. So assuming everything had perfect efficiency, your device could be run from a 3.2 Amp-Hr battery.
In reality there is a lot of conversions going on here – you are converting from AC to DC; 240 V to 12 V; electrical energy to chemical energy; etc. There will be some loss in each step of the process.
When you are working out what size UPS you need there are 2 factors to consider:-
1. How much power is drawn (instantaneous);
2. How much energy in total can be delivered.
Your load is only 50W, so the first condition would be easily met by an off-the-shelf 1000VA UPS. For the moment let’s assume the UPS is rated for 10 minutes of run time at full load (this is a fairly typical figure for small desktop UPS’es – they are really only designed to run a PC for long enough to shut it down cleanly).
If the UPS can run 1000W (again assuming resistive load**) for 10 minutes, then it could run 50W (1/20th of 1000W) for 20 times as long. So your device could be powered for around 200 minutes, or 3 hours & 20 minutes. Not long enough for your application. And again, in reality things do not always scale linearly, so the real time if you tested this would probably be anywhere from 2 hours to 4 hours – but hopefully you get the picture.
So if a 1000VA UPS won’t cut it, you obviously need a bigger UPS, right? NO! (or at least not necessarily). You could buy a bigger UPS, say 5000VA, still with a run time of 10 minutes at full load. This would then run your 50W load for (theoretically) 1000 minutes, or just over 16 ½ hours. But if you did this you would be paying more money for the bigger electronics need to supply 5000VA of (instantaneous) power, as well as for the extra energy (batteries) that you need to run the load for the full 16 hours.
The better option is to look for a UPS that can supply at least 2 to 4 times your instantaneous power (let’s say a minimum of 200VA), but with bigger, or expandable, battery capacity.
Hope that helps clear up some of the confusion.
Regards,
Aaron.
* Mathematically, it is the integral of power over time ( S P dt )
** I’ve been very sloppy here with equating VA to Watts. They both refer to volts multiplied by amps, but when you are working out Watts you only consider the current that is “in phase” with the voltage. It gets complicated, but basically if you draw a right-angled triangle the current used in calculating the VA rating would be the hypotenuse of the triangle. The current used in calculating the Watt rating would be one of the other (shorter) sides. Thus the VA rating is always higher. So a 1000VA UPS is usually only capable of running around 800W of computer equipment. Since manufactures like to quote the bigger numbers on their glossy brochures, UPS’es are usually sized in VA.
And one last thing – it has been a while since I worked in this space, but there used to be fairly strict restrictions on what you could connect to your off-peak meter. Even getting to the connections can be tricky as they are usually sealed by you energy supplier.
--
James Laird
> What size battery do i need? I'm struggling to understand 800w per
> hour at 240 from a 12v source
work out energy, you need to add it up over time - so here, 50W for 16 hours of
the day means you need 800Wh (no "per hour").
Power P = V * I.
So for 50W, at 12V, you'll need to supply 4.2 amps.
To supply that for 16 hours with a (nominally) 12V battery you will need
a battery with rating of at least 16*4 = 64 Ah.
A car battery will be on the order of 40-80Ah, and if you don't want to
discharge it past halfway you'll need one closer to 120Ah. Which is big,
expensive, or requires heavy (choose at least two).
So a solar panel looks pretty good for keeping the battery size down, as
long as you can fall back on pricy mains power if your juice is
insufficient eg. cloudy day.
You could then have a smaller battery that would hold you over through
the evening - the time between sundown and cheap electricity, and then
again until dawn after charging it off the mains.
This is starting to sound complicated though, and definitely beyond what
you'll find a simple UPS for. You might be able to find a solar charge
manager that has an external power input that it uses when available -
then control the mains PSU with a) a timer for nighttime plus b) an
override for low-battery trouble conditions.
Cheerio,
James
Aaron Power
Sorry – when I said 3.2 Amp-Hr battery, that is at 240V, so you would need a string of 20 by 12V batteries (impractical) or a single 12V battery 20 times as big (64 Amp-Hr). (I knew something didn’t quite sound right, but was concentrating on something else)
Aaron.
ada
On May 17, 7:22 pm, kris <k...@sleepingplanet.com> wrote:
> I'm investigating powering a device 24/7 using as little 'expensive'
> electricity as possible.
> I have an 'off peak' meter which provides my water heater with
> electricity from 11pm to 7am at 7c per kWh
> I normally pay 26c per kWh at all other times.
> have a solar panel charge the battery (or maintain the charge and power
> the device if the battery is full)
> have the off peak power charge the battery to full IF the battery does
> not have 12 hours of juice left.
> charge at night (forget about the solar since at 7c K/w it would be a
> long time to pay off the investment of the solar panel)
- this device will consume 2.8c per night (offpeak) and 20.8c per day
(peak), running it just off the mains. It will consume 8.4c per day
if you had a perfect battery and charged it overnight - or a saving of
15.2c per day. Consider what the lifecycle of your system is and
whether having it run off cheap electricity is worthwhile.
- if you have a 12V device, I would not use a UPS - the inefficiencies
from inverting and then downconverting again (IMO) far outweigh the
simplicity of having an OTS solution.
- having a solar system with small batteries and being willing to use
occasional expensive electricity might be better than using lots of
batteries.
I don't know what the typical cycle life of deep cycle (VRLA/AGM) SLAs
is, but the last time I looked, they ran about 20c/Whr - and you need
in the order of 800Whrs of usable capacity (with say a fudge factor of
2) costing around $300. it would take approx 5 years to pay that off.
Kris
Ada, I think you hit the nail on the head. This current project will depreciate in value at 60% per year so I was trying to maximise the gain as quickly as possible but at this scale I think you are right and I should not use the battery idea.
It will be slightly more expensive to run as it will run on the 26c tariff unless i have some sort of make before break switching mechanism to move it over to the off peak tariff when available.
interestingly (i was speaking to some ex sparkys at work) off peak doesn't come on at a set time each night. it can have up to 2 hour variance
and yes i also heard that the electricity company are picky about what i run from the off peak meter. water heater obviously. pool filters are also allowed but im trying to find agl's actual policy. its clear as mud.
kris
kaindub
I just did a quick calculation comparing running your device from 'expensive' mains, versus powering it up from 'cheap' mains and running on an UPS.
I assumed that there was 100% onversion efficiencies
You will save the whole of 2.85 cents (not dollars) per day
Now if you get a conversion efficiency of 50% (pretty reasonable considering the numbers of elements you will need) the saving is less than 1.5 cents per day
How long is it going to take to recoup the cost of the battery, UPS charger etc? Years or decades?
Don't kid yourself. Mains power even at the peak rate is very cheap comoared to other forms of electricity. Why has renewable energy not been taken up more? Because of its costs compared to coal fired electricity.
Robert
Kris
its hardly comparing apple with apples. For 800wh i would say a solar/ups/battery rig will be much more cost efficient than me building my own coal fired power plant. the cost savings are because all of you are subsidising the cost of building and running a coal fired power plant in order to supply me with a tiny percentage of its output (for a tiny % of the cost).
i spent 3k on a 1.5 solar system on my roof, it has reduced my electricity bill by 25-30% saving me around $100 a quarter (around 3 years before it pays for itself).
so at the current rate it will take a while to pay off. but given that 80% output is guaranteed for 20 years this means that solar renewable is 'cheaper'
also are you saying that hydro generation is not as cost effective as coal fired?
what really bugs me is that any electricity i don't use from my solar array gets 'purchased' by agl at a shitty rate of 7c kwh yet they charge me 26c!
300% profit. nice work if you can get it.
kris
ada
On May 18, 9:26 am, Kris <k...@sleepingplanet.com> wrote:
> i spent 3k on a 1.5 solar system on my roof, it has reduced my electricity bill by 25-30% saving me around $100 a quarter (around 3 years before it pays for itself).
or 7.5 years to pay for itself. also, was that 3k with the solar
panel rebate?
> so at the current rate it will take a while to pay off. but given that 80% output is guaranteed for 20 years this means that solar renewable is 'cheaper'
the cost of electricity-from-coal is going to be lower than 7c/kWh.
> also are you saying that hydro generation is not as cost effective as coal fired?
> what really bugs me is that any electricity i don't use from my solar array gets 'purchased' by agl at a shitty rate of 7c kwh yet they charge me 26c!
> 300% profit. nice work if you can get it.
on the spot market when you're selling back in? who takes the risk of
not being able to sell the electricity?
why not charge batteries off your solar array and run a DC bus around
your house to power your low-power equipment? if you do the numbers
right you can probably save money with that.
power electrical is a complex topic and I don't pretend to know the
ins and outs of it, but to attempt to oversimplify it like this is
fundamentally cognitively flawed.
kris
The hydro price per mwh vs coal i was surprised at. On further reading
the raw costs do show that coal is cheaper, but in other countries with
different subsidies and equity pricing systems (cts) it's not, which is
where i had got the idea that hydro was chaper than coal.
I'm unsure how much it costs to build a hydro plant vs coal plant
(probably a lot more for hydro if you have to build a dam).
http://www.energy.ca.gov/2007publications/CEC-200-2007-011/CEC-200-2007-011-SD.PDF
I still feel I'm valid in stating that buying solar at 7c and selling
'grid' at 26c is excessive profit taking.
even with peak generation shorting and spot prices etc.
though of course the rebate system before that with nearly 67c paid for
electricity worth a third of that was not very sensible either.
I have though about the DC ring, pretty much all that equipment would be
in the garage and my room so it wouldn't be that much wiring, but the
speed at which things are becoming more energy efficient i would worry
that the capital outlay would never pay for itself.
Kris
Iain Chalmers
>
> I still feel I'm valid in stating that buying solar at 7c and selling 'grid'
> at 26c is excessive profit taking.
I think you'll fond that's a completely typical markup to retail. How
much do you suppose the factory in China who makes those $26 kettles
on the shelf in your local retailer gets? I think they'd be perfectly
happy with $7 of that...
</hat>
big
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That's the only way to be sure." - Ellen Ripley
Luke Weston
On Saturday, 18 May 2013 09:26:11 UTC+10, kris wrote:
what really bugs me is that any electricity i don't use from my solar array gets 'purchased' by agl at a shitty rate of 7c kwh yet they charge me 26c!
It seems reasonable to me. What you're buying from the grid is energy with a very-close-to-100% availability factor, but what you're selling into the grid is energy with a relatively low availability factor, maybe ~20% or so. And they're paying you more than 20% of the cost, so you're doing alright.
And do the people running the big coal-fired generators (or gas turbines, or wind farms, whatever) get paid all of that 26c/kWh? No! Obviously those major generators only get paid a portion of that. So why should it be different for you?
Arik Baratz
It seems reasonable to me. What you're buying from the grid is energy with a very-close-to-100% availability factor, but what you're selling into the grid is energy with a relatively low availability factor, maybe ~20% or so. And they're paying you more than 20% of the cost, so you're doing alright.