I have hard numbers and questions......

concretefire · August 2012

I seem to be having a really hard time understanding this solar thing. The over all picture I understand. It's the math that's killing me.

So here are my numbers (per Kill-A-Watt) for my computer only.
Volt 118, Amp 2.01 ~ 3.00 (depending on how much power my processor is using) 157 Watts at idle closer to 230 watts when being heavily used. 239 Va 60HZ .65 PF 7.57 KWH in 71:12 time.

So my question is based on these numbers, WHAT equipment would I need to run this computer???? Allowing for the usual 3 days without much sun thing...... someone please help me because if I can get my head around running 1 thing then I think I will have it. Thank you.

inetdog · August 2012

Re: I have hard numbers and questions......

concretefire wrote: »

I seem to be having a really hard time understanding this solar thing. The over all picture I understand. It's the math that's killing me.

So here are my numbers (per Kill-A-Watt) for my computer only.
Volt 118, Amp 2.01 ~ 3.00 (depending on how much power my processor is using) 157 Watts at idle closer to 230 watts when being heavily used. 239 Va 60HZ .65 PF 7.57 KWH in 71:12 time.

So my question is based on these numbers, WHAT equipment would I need to run this computer???? Allowing for the usual 3 days without much sun thing...... someone please help me because if I can get my head around running 1 thing then I think I will have it. Thank you.

First question: How many hours per day will the computer be running? It will be an enormous help if you can sleep or hibernate it when not in use.

157 watts time 24 hours is 3800 watt-hours, which would require a substantial (greater than 1000 watt) PV array just for the computer, along with a battery bank of over 22,000 watt-hours. At 24 volts, that battery bank would have to be over 900 amp-hours if my arithmetic is correct.
If you cut your active use of the computer down to only six hours per day, the whole thing becomes a lot more manageable.

niel · August 2012

Re: I have hard numbers and questions......

higher power usages are also possible when accessing or using the cd or dvd drives. i would not go by the idle as few use a computer in idle or you might just as well shut it off. i do shut mine off after use and on one other pc i physically shut it off with a power strip after it has successfully shut down to wipe out the phantom load a pc will draw even when off.

nobody can even guess how long you will use your pc and you can measure the kwh for yourself with a killawatt meter. do this over a week and divide by 7 to get your average kwh/day.

Photowhit · August 2012

Re: I have hard numbers and questions......

Well he gave us a pretty clear usage for 3 days "...7.57 KWH in 71:12 time. .." while we all would suggest conservation as that's a huge load, perhaps it's a server than needs to be up and running 24-7.

So... we all do things a bit differently, figure storage, so at a max 3 days with out charging, drawing the batter down to 50%, 7570 watt hours x 1.2 (inverter effecencies) = 9084 KWh x 2 (since you are only drawing down to 50%) 18168 KWh or at 24 volts (18168/24) 757Amp hour battery bank.

inetdog · August 2012

Re: I have hard numbers and questions......

niel wrote: »

the phantom load a pc will draw even when off.

Nobody can even guess how long you will use your pc and you can measure the kwh for yourself with a killawatt meter. do this over a week and divide by 7 to get your average kwh/day.

Expanding on the phantom load issue:
Don't forget that what you are measuring is the input to the AC power supply of the computer (as seen by the power factor not equal 1, it will be pulling some VA from an inverter even when the electronics of the PC are shut down completely. A pure waste to you.) To that you add any "standby" load that the PC itself may be taking from the power supply even when "off" to handle such things as 'wake up on timer' or 'wake up on oncoming messge'. Are there one or more fans which which are running whenever the computer is on, whether it is being used or not?

I am guessing that you are talking about a desktop computer, not a laptop, and they are generally not designed for high efficiency as a laptop where someone will be watching the battery life and publishing it would be. On a laptop, the power brick will be pulling power from AC even when the laptop itself is shut down and unplugged from the brick.

concretefire · August 2012

Re: I have hard numbers and questions......

Let's assume I'm running this PC at least 12 hrs a day.

BB. · August 2012

Re: I have hard numbers and questions......

C-Fire,

I will start with what you posted here first.

concretefire wrote: »

So here are my numbers (per Kill-A-Watt) for my computer only.
Volt 118,

Amp 2.01 ~ 3.00 (depending on how much power my processor is using)

157 Watts at idle closer to 230 watts when being heavily used.

239 Va 60HZ

0.65 PF

7.57 KWH

71:12 (71.2 hour measuring period)

Volts--118 Volts nominal (usually between ~106 to 132 VAC). That is perfectly normal, and remember that this is a sine wave that goes from zero to ~+167 volts back to zero, and to -167 volts (peak) back to zero--60 times a second (Hz).

Amps--Now that gets a little tricky. You are measuring a 118 volt 60 Hz sine wave--AC (alternating current) voltage. And as such, the relationship between voltage and current sine waves is not obvious. If this was driving a resistive load (electric heater or filament lamp are two examples), the current would also be a sine wave "in phase" with the voltage sine wave. Basically, this is the most efficient use of AC power.

If the sine wave "lags" (or follows by XX degrees --0 to -90 degrees for an inductive circuit), the current wave form does not track the voltage wave form, but is delayed for part of a cycle (remember 360 degrees in a cycle). This means that while we have "full amount of current", since it not "in phase" with the AC wave form, we don't get 100% of the expected "work" from the voltage * current equals power equation.

The "difference" is the Cosine of the phase angle. So, the "real" power equation is (Cosine XX degrees = Power Factor; power factor ranges from 1.0 "good" to 0.5 "pretty bad" to 0.0 "no power transfer at all):

Power = Voltage * Current * Cosine of phase = Voltage * Current * Power Factor

The above are "vectors" if you remember your physics.... Or a simple example.

You pull on a rope tied to a car. If you are in font of the car, 100% of your effort pulling the rope moves the card forward (Cos 0 = 1.0). If you stand off 60 degrees to the side, only 1/2 of your pull moves the car forward (Cos 60 = 0.5). Stand to the side, and you don't pull the car forwards at all (Cos 90 = 0.0).

Now, what does all that mean... You have two numbers you need to worry about... The traditional Power:

Power = Voltage * Current * Power Factor

Which is the electricity you draw for from the utility in Watts (or kW), or the power from your solar panels, or the power from your battery. Multiply by time (hours) and that is the actual amount of electricity yo use (watt*hours, kWH, etc.). This is the power/energy used to move the car forward when pulling on the rope.

The other is VA (kVA, etc.), which is used to size the wiring, fuses, generator, transformers, inverters, etc. This is the equivalent of sizing the rope itself in the above example. If you stand to the side of the car and you want to pull it up hill--standing to the side you would have to pull harder and the rope would have to be stronger.

There are other details... Not all loads are "linear" or pull sine wave shaped current. Computer power supplies are known for pulling a whole bunch of current just at the peaks of the sine wave voltage (+ and - peaks). That also can give us "power factor" (PF = 0.65 in your measurements).

So my question is based on these numbers, WHAT equipment would I need to run this computer???? Allowing for the usual 3 days without much sun thing...... someone please help me because if I can get my head around running 1 thing then I think I will have it. Thank you.

OK--now lets say you want to run the computer from a backup battery bank and AC inverter.

So, the minimum size of the inverter will need to be 239 VA minimum (or higher during heavy loads) because of the PF=0.65 ...

The average amount of power from the battery bank is around:

7.57 kWH / 71.2 hours = 0.106 kW = 106 W (based on your usage over ~3 days)

Your daily power usage would be:

106 Watts * 24 hours per day = 2,544 Watt*Hours = 2.544 kWH per day

Now--That is probably a good average for that computer... Assuming it is off/sleeping for around 12 hours per day. That is the nice thing about a Kill-a-Watt meter--It measures total power used based on your usage pattern. No fancy averaging/estimating 157 watts for 6 hours, 230 watts for 6 hours and zero watts for 12 hours, etc...

To put that into perspective... That is about 2x the average power usage of todays "Energy Star" rated 21 cuft full sized refrigerator/freezer (around 404 kWH per year or ~1.1 kWH per day).

For most off-grid folks, the refrigerator is the #1 power use.

If you were going off the grid with a battery based solar power system, You should be aiming to around ~3.3 kWH per day for a very efficient off grid home with "normal" appliances (fridge, lights, tv, well pump, laptop computer, clothes washing machine, etc.).

That one computer is drawing about 2/3rds of a suggested power budget.

Power usage is highly personal, so if you cannot substitute a 30 watt laptop computer for the 230 watt desktop--that is fine. We will work with you to generate the power needed.

But--as a rough rule of thumb, a pure off grid system in the US will cost around $1 to $2+ per kWH (capital costs, installation, replacement batteries, inverter, charge controllers, etc. assuming 20+ year system life). Or round $2.50 to $5.00 per day just for that one computer.

In the end, it is almost always cheaper to conserve power than it is to generate it.

One thing you did not discuss was if you wanted battery bank emergency power, full off grid solar power, or to reduce power bills using Grid Tied solar power...

GT solar power is the cheapest, most cost effective, and most reliable of any home based alternative power system today. But, of course, you have to have utility power at your home, and the utility has to support Net Metering/Grid Tied solar power.

-Bill

Photowhit · August 2012

Re: I have hard numbers and questions......

BB. wrote: »

GT solar power is the cheapest, most cost effective, and most reliable of any home based alternative power system today.

I'll give you Cheapest, and I'll give you Cost effective, but since grid tied require the grid to be up and running, I can count my off grid down time in minutes, comared to days for the grid in the last 5 years,...

niel · August 2012

Re: I have hard numbers and questions......

BB. wrote: »

Volts--118 Volts nominal (usually between ~106 to 132 VAC). That is perfectly normal, and remember that this is a sign wave that goes from zero to ~+167 volts back to zero, and to -167 volts (peak) back to zero--60 times a second (Hz).

-Bill

i think we can say that these kind of waves aren't desirable.;):p love that blooper.8)

~~~~~~~~~ now this is a sign of sine waves.:D

BB. · August 2012

Re: I have hard numbers and questions......

You mean that somebody reads these posts.

-Bill

vtmaps · August 2012

Re: I have hard numbers and questions......

The sine qua non of a spell checker (rather than a proofreader) is a sign wave. --vtMaps

concretefire · August 2012

Re: I have hard numbers and questions......

Bill, thank you. Everyone else as well but i'm still confused. I DID NOT ASK MY QUESTION CORRECTLY!!

Ok, so yes, I can do the math of 7.57kwh / (time) of 71:12 which = .106 = 106 watt
Which in turn = •106 Watts * 24 hours per day = 2,544 Watt*Hours = 2.544 kWH per day

Ok, those parts I understand.

Here is what I DO NOT understand. I do not understand EXACTLY what equipment I would need to buy in order to produce that amount of power.....and have the standard 3 day reserve in case of no sun.

I don't know if I should buy like an 100 watt solar panel or a 1,000,000 watt solar panel. I don't understand how that math works.

I image I could do it with a 100 watt solar panel, BUT, it would take me days or weeks to fill up a batter bank for what I would need just to run that ONE computer.

Furthermore, I don't understand how you convert that power into amp hours to get an idea of how many batteries or amp hours I would need.

Thirdly, I don't understand the inverter. I mean I know what it does, but if I'm going to get an inverter, and ASSUMING I will be adding more panels and more gadgets to my 'solar setup',,,,,wouldn't I just buy the biggest, baddest inverter out there so it could handle anything?

Example of #1..... say I bought a 135 watt solar panel.....ok...??...so what? I don't understand the math. How much power will I actually GET out of that panel, per day, assuming normal/good conditions??? I honestly have no idea.

Then, that power goes to a 'controller' before it ever get's to the batter bank. So is there power lost at that controller before it gets to the battery bank?

Then, how do you do the math to convert that power into REAL Amp hours that just got put into the battery bank per day, assuming good/normal conditions for solar.

Then you're losing more when it runs from the battery thru the inverter....

That's what I don't understand. I don't understand the basic math of solar, starting with the solar panel itself. (or panels)

Please help.

ywhic · August 2012

Re: I have hard numbers and questions......

Avg panel is 77% being able to get to the controller.. avgs AC side of things is rumored to be 52-62% of the panel number..

135w = 103.95 watts to the charge controller.. and 75.2 watts (based on the 52%) getting to the other side of the inverter (aka 110-120 AC)

These numbers are based on wire loss and inverter eff and the controller usage losses..

I'm thinking/hoping I will be at like 62% of my 870 watts worth of panels.. 539.4 watts getting thru to the 110-120 AC side..

BB. · August 2012

Re: I have hard numbers and questions......

OK--We can go to the next level in sizing.

Using PV Watts for Charlotte N.C., fixed array tilted to latitude (35 degrees from horizontal) and 0.52 derating for off grid system end to end efficiency, your "hours of noontime sun per day by month would be:

Month    Solar Radiation
(kWh/m 2/day)
1      4.04     
2      4.46     
3      5.30     
4      5.89     
5      5.79     
6      5.77     
7      5.55     
8      5.59     
9      5.26     
10      5.23     
11      4.25     
12      3.74     
Year      5.07

If you toss the bottom three months (assume you make up with genset during bad weather), the lowest month would be February at 4.46 hours of sun per day. So, to generate 2.544 kWH per day average in February, you would need a minimum array of:

2,544 WH per day * 1/0.52 system derating * 1/4.46 hours per day of sun = 1,097 Watt array minimum

So February would be your transition month. You may need some generator support during bad weather (you should always have a "buffer" if you really intend to generate 100% of your power--like 1/75% or 1/66% for "base loads" and use a genset or cut back on optional loads during bad weather (as needed).

Next, sizing the battery bank. Nominally, we recommend 1-3 days of "no sun" with 50% maximum discharge (for longer battery life). A good sized system is 2 days, but I will do 3 here (note, will do a 48 volt battery bank as you want a "large" battery bank--but a 24 volt (or even 12 volt) battery bank will cost about the same--it is just how the cells are arranged in series or parallel/larger AH capacity cells):

2,544 WH per day * 1/0.85 inverter losses * 1/48 volt battery bank * 3 days backup * 1/0.50 max discharge = 374 AH @ 48 volt battery bank

Now, because we also recommend a 5% to 13% rate of charge for a battery bank rule of thumb (for many reasons), a large battery bank can "force" you to have a larger solar array to meet the "minimum/nominal" charge requirements:

374 AH * 59 volts charging * 1/0.77 panel+charger deratings * 0.05 = 1,433 Watt array minimum
374 AH * 59 volts charging * 1/0.77 panel+charger deratings * 0.10 = 2,866 Watt array nominal
374 AH * 59 volts charging * 1/0.77 panel+charger deratings * 0.13 = 3,725 Watt array "maximum cost effective"

So, even though you only need a 1,097 watt array, because of the "larger" battery bank (3 days instead of 2, or even 1), the minimum recommended array size would be 1,433 Watts. And ~2,866 Watt array would be recommended if this was using flooded cells and used every day off grid to keep the battery well conditioned.

You can put upwards of a 3,725 watt array on the battery bank--A remote battery temperature sensor would be highly recommended (actually, always recommend the option if available). Any larger would, more or less, be a waste of array unless you have special needs (say pumping lots of water during the day for irrigation).

Of course, with the larger array, you will be generating a lot more power (well pumping during the day, draw the battery bank down to 2 days+50% maximum discharge, etc.):

2,866 Watts * 0.52 system eff * 4.46 hours of sun per day February = 6,647 WH = 6.65 kWH per day February

Anyway--That is what a "rule of thumb" design would look like based on your estimated power needs and battery bank size--And my guess where the system will be installed.

Is this what you were looking for?

-Bill

BB. · August 2012

Re: I have hard numbers and questions......

After we "agree" on your power needs. We can go into the rest of the equipment sizing/selection.

If we do it now--it will get a bit more confusing (different size loads/panels/batteries require different brands/models of equipment).

Also, with off grid and power usage in general--We aim for "balanced" system design. Too big of battery bank is a waste of money and requires a larger solar array/genset to keep happy. And when you need to replace the battery bank every 5-8 years or so--That 2x larger battery bank will cost 2x as much to replace. It is true that a larger battery bank may last longer--But from what I have seen, a 2x battery bank may last 2.2x longer--And if there is an "oops" (inverter/pump left on when array is shutdown/family on vacation), that battery bank will hurt your pocket book a lot more too.

Too large of AC inverter can draw 20-40+ watts just on its own to "turn on" with no loads. For example, you can get a very nice 12 volt 300 watt TSW inverter that will draw ~6 watts with no load, and even shutdown (automatically--"search mode") and draw much less power "in standby" when AC loads are turned off.

Same thing with generators... A 1,600 watt generator may run 4-10 hours on 1 gallon of gasoline. A 6kW genset may run 1-2 hours on a gallon of gasoline.

If your loads are typically less than 1,600 watts--the smaller genset will cost you a lot less in fuel usage (sometimes, it is better to have two generators... A small one for average loads, and a second/larger genset for shop/tools and emergency backup for the smaller one).

Anyway--the whole idea of balance design and doing "extreme" conservation with your loads/choices of loads. It can easily swing the costs by 2x or more one way or the other depending on your eventual design decisions.

-Bill

rollandelliott · August 2012

Re: I have hard numbers and questions......

well I'll be the old nag to remind you that conserving is the firs step. Get yourself the least power full pc that will do the job. an intel, atom net top would be my suggestion. you'll be using 1/8th of the power right off the bat!

concretefire · August 2012

Re: I have hard numbers and questions......

Everyone, yes, I agree that conservation is of course the #1 thing to consider. That's the cheapest thing to do. The example of my PC was just to get some kind of numbers going so I could try to wrap my head around what it would take to run that beast. (It's a Quad Core

) There's no way I would run that off grid. Lol. However, the numbers are still the numbers.... meaning insert your appliance here......PC, Toaster, Microwave, whatever....

The lower the numbers, the lower the cost of equipment. I get that, but the bottom line is the numbers are still the numbers. You just have to be wise when you make purchases, which I understand. The example of my pc was just a rough draft in my head and Bill broke it down very well I think.

Bill, thank you, and yes most of that is what I was looking for. I am going back to digest it some more and do the calculations along with yours to reinforce what you are saying.

Bill, I stated I did not know what a solar panel would produce or give me power wise....so let me ask you this question to see if I'm on the right track.
(assuming 100% efficiency, no losses, perfect world)

a 200 watt panel would produce 1000 watt hours of power if I had 5 hours of perfect sun. Is that correct?
But in the real world it would be 200 w = 770 watt hours (same 5 hour time frame) given the .77 efficiency factor. Is that correct?

icarus · August 2012

Re: I have hard numbers and questions......

My rule of off grid is this (which actually works well in the real world) take the name plate rating of the PV, divide that number in halve to account for all cumulative system losses, then multiply that number by 4 to account for The AVERAGE hours of GOOD sun one can reasonably expect, per day, over the course of the yerar,, sort of factoring in season, clouds etc. Unless you are in some place like the desert south west you will seldom get to average more than 4 over the course of the year.

So, 200/2=100*4=400 WH/day out the inverter.

We have 400 watts of panel, routinely use 5-800 WH/day, and seldom have to run a generator. On a ideal day, we can generate 1.5 kwh/day, but the 400/2*4 works pretty well, day in day out,, which fer all is what you want.

Tony

BB. · August 2012

Re: I have hard numbers and questions......

More or less, the equation for power from a solar panel will be:

200 watts * 0.77 panel+controller derating * 0.80 battery derating * 0.85 invert losses * 5 hours of sun per day = 524 WH at the AC plug

Note, that flooded cell lead acid batteries run around 80-90% efficiency (depending on how you use/charge them and their age). AGM will run from ~90% to 98% efficiency.

Inverters will run from ~80% to 94% depending on quality, loading, etc...

The 52% or 50% rule of thumb is conservative and most people will meet those efficiency levels.

There are other thing too--For example, if you pump water/wash clothes, etc. during the day--Then you can bypass the 0.80 battery losses. If you do that with a DC pump, you can skip the inverter losses too...

-Bill

concretefire · August 2012

Re: I have hard numbers and questions......

Bill, funny you should mention a DC pump. I just posted a new ? in the water section. Could you have a look and advise? Also to icarus, that is a very good rule of thumb and even I can remember that. So I can get all technical if I wanna but a good rule of thumb is nice and simple. Thank you.

concretefire · August 2012

Re: I have hard numbers and questions......

icarus wrote: »

My rule of off grid is this (which actually works well in the real world) take the name plate rating of the PV, divide that number in halve to account for all cumulative system losses, then multiply that number by 4 to account for The AVERAGE hours of GOOD sun one can reasonably expect, per day, over the course of the yerar,, sort of factoring in season, clouds etc. Unless you are in some place like the desert south west you will seldom get to average more than 4 over the course of the year.

So, 200/2=100*4=400 WH/day out the inverter.

We have 400 watts of panel, routinely use 5-800 WH/day, and seldom have to run a generator. On a ideal day, we can generate 1.5 kwh/day, but the 400/2*4 works pretty well, day in day out,, which fer all is what you want.

Tony

Ok, I really like this formula. It's quite simple and seems fairly accurate according to what I have read. Another guy said take the power you need per day and double it....that's the amount of solar array you will need....then double that # again, and that is the amount of your battery bank you will need.

So by that if you use a 1000 kw per day you would need 2000 kw of solar array (200w panel x 5hrs sun (in his example) = 1000 kw. X 2. Meaning TWO, 200 watt solar panels.) Then you would need a 4000 watt battery bank.....

But that is where I get lost. The battery bank. I can do the math as far as power but how do you convert the watt hours needed (4000) into Amp Hours..???...to let you know what sized (and what kind) of batteries to use....???

That's where I get lost. Even in your example.

BB. · August 2012

Re: I have hard numbers and questions......

Be careful, you are using kw when you should have typed wh (watt**hour).

Divide watt or wh by voltage to get amps or amphours.

-Bill

concretefire · August 2012

Re: I have hard numbers and questions......

BB. wrote: »

Be careful, you are using kw when you should have typed wh (watt**hour).

Divide watt or wh by voltage to get amps or amphours.

-Bill

oops.... good point. Thank you. 1000 kw a day would require one helluva a solar array. My bad. Sweet forumula for the amhours Bill. Thank you.

I have hard numbers and questions......

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