formula for converting AC Killawatt reading over to DC panel wattage?
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Urbandialect
Solar Expert Posts: 107 ✭✭✭✭
http://www.lowes.com/pd_322553155844460_0__?storeId=10151&Ntt=killawatt&UserSearch=killawatt&productId=3191393&N=0&catalogId=10051&langId=1
^ just bought this
Once i get the numbers, how do i convert that over to DC panel wattage? Lets say my Computer is pulling 136 watts (AC) what is that in DC? How many panel watts will I need to cover that? and what about the MSW inverter? http://www.justcardvdplayers.com/vectorvec050c1500wattpowerinverter.aspx < this is what i'm using, how much am I loosing in the DC to AC conversion w/the inverter that I will need to factor into the equations?
Thank you
^ just bought this
Once i get the numbers, how do i convert that over to DC panel wattage? Lets say my Computer is pulling 136 watts (AC) what is that in DC? How many panel watts will I need to cover that? and what about the MSW inverter? http://www.justcardvdplayers.com/vectorvec050c1500wattpowerinverter.aspx < this is what i'm using, how much am I loosing in the DC to AC conversion w/the inverter that I will need to factor into the equations?
Thank you
Comments

Re: formula for converting AC Killawatt reading over to DC panel wattage?
Direct conversion to panel Watts isn't actually possible. Panels produce power only during sunlight; usage can be any time, and you have to factor in the hours.
But here's the Icarus Formula that can give you a pretty good estimate of overall conversion:
Panel Watts * hours of good sun / 2 = AC Watt hours.
Otherwise you need the Watt hours the device uses plus the Watt hours the inverter uses factored by the inverter's efficiency divided by the nominal system Voltage to get a rough idea of how many Amp hours you need. Since most batteries should not be discharged below 50% you multiply that by at least 2 to get the minimum battery bank size. That number would be rounded up to the nearest available batteries. The size the battery bank ends up being determines how much PV you need to recharge it. 5%  13% of the total Amp hour capacity is recommended for the peak current, times the system charging Voltage (according to battery maker's specs) less panel efficiency (which will vary with the particular install site).
Yes, I can make it even more complicated than that! Did I mention wire losses? 
Re: formula for converting AC Killawatt reading over to DC panel wattage?
A 136 watts sounds like a desktop computer... My laptop is around 2030 watts...
And, it depends on how many hours a day you want to run your computer.
Say you want 10 hours a day, 136 watts average power, and you get 4 hours or better of full sun equivalent per day 9 months of the year.
Assuming average system efficiency of 0.52 (very close to the 1/2 losses): 136 watts * 10 hours * 1/0.52 system eff * 1/4 hours of sun = 654 watts of solar panels minimum
 136 watts * 10 hours * 1/12 volts * 1/0.85 eff * 2 days * 1/0.50 max discharge = 533 AH battery bank
 533 AH * 14.5 volts charging * 1/0.77 derating * 0.10 rate of charge = 1,004 watts...
BillNear San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset 
Re: formula for converting AC Killawatt reading over to DC panel wattage?Cariboocoot wrote: »Direct conversion to panel Watts isn't actually possible. Panels produce power only during sunlight; usage can be any time, and you have to factor in the hours.
But here's the Icarus Formula that can give you a pretty good estimate of overall conversion:
Panel Watts * hours of good sun / 2 = AC Watt hours.
Otherwise you need the Watt hours the device uses plus the Watt hours the inverter uses factored by the inverter's efficiency divided by the nominal system Voltage to get a rough idea of how many Amp hours you need. Since most batteries should not be discharged below 50% you multiply that by at least 2 to get the minimum battery bank size. That number would be rounded up to the nearest available batteries. The size the battery bank ends up being determines how much PV you need to recharge it. 5%  13% of the total Amp hour capacity is recommended for the peak current, times the system charging Voltage (according to battery maker's specs) less panel efficiency (which will vary with the particular install site).
Yes, I can make it even more complicated than that! Did I mention wire losses?
What the ****? How about I come back w/the Kill a watt numbers tomorrow of the devices i'm trying to run, then we figure it out from there. 
Re: formula for converting AC Killawatt reading over to DC panel wattage?A 136 watts sounds like a desktop computer... My laptop is around 2030 watts...
And, it depends on how many hours a day you want to run your computer.
Say you want 10 hours a day, 136 watts average power, and you get 4 hours or better of full sun equivalent per day 9 months of the year.
Assuming average system efficiency of 0.52 (very close to the 1/2 losses): 136 watts * 10 hours * 1/0.52 system eff * 1/4 hours of sun = 654 watts of solar panels minimum
 136 watts * 10 hours * 1/12 volts * 1/0.85 eff * 2 days * 1/0.50 max discharge = 533 AH battery bank
 533 AH * 14.5 volts charging * 1/0.77 derating * 0.10 rate of charge = 1,004 watts...
Bill
Yes, i'm running a desktop, with a surround sound system hooked up to it, haven't got the killawatts reading off it yet, I'm almost finished with my 600 watt array, and you know I have 2 battery banks, 6 6volt amg batteries rated at 200amp hours, and 6 6volt T105s at 225amp hours... but what's shocking to me is that it takes almost 1kw of DC to run and maintain 150 watts of AC.. mindblowing 
Re: formula for converting AC Killawatt reading over to DC panel wattage?
While I think youR 15% conversion is not quite accrete, a 50% is closer.
Consider this, PV will typically only put out ~80 of rated into a battery under ideal, so there is a 20% loss. Add to that, it takes ~120 ah of power to replace 100 ah of draw, so there is another ~20% loss. Add in inverter efficiency, charge controller loses, wiring loses, and bingo, you are near 50%.
On top of that, you haven't even thought about the opportunity loses when the PV is dialed back because the battery is close to full. This could account for another 1020% depending on battery/PV configuration.
Thees are just a few of the reasons why we always suggest going grid tie if you have the grid available. As we say, off gird PV comes at twice the price, and half the efficiency, leading to power that is 4 times as expensive per useable KWH.
Tony 
Re: formula for converting AC Killawatt reading over to DC panel wattage?Urbandialect wrote: »what's shocking to me is that it takes almost 1kw of DC to run and maintain 150 watts of AC.. mindblowing
Hahaha Reality begins to sink in. You have no idea how many people think all they have to do is stick a $150.00 panel on the roof and all their energy problems are over.
Living off grid demands a whole different way of thinking/lifestyle. Unless you have extremely deep and full pockets, there is one rule to live by off grid, and that rule consists of three words: Conserve, conserve, conserve! As has been said by others so many times on this form, it's far easier and cheaper to conserve than to waste power. Unfortunately, most of us have been brought up to see excess energy consumption as normal, not recognizing it as waste.
Good luck. It can be done, we did it. 
Re: formula for converting AC Killawatt reading over to DC panel wattage?A 136 watts sounds like a desktop computer... My laptop is around 2030 watts...
And, it depends on how many hours a day you want to run your computer.
Say you want 10 hours a day, 136 watts average power, and you get 4 hours or better of full sun equivalent per day 9 months of the year.
Assuming average system efficiency of 0.52 (very close to the 1/2 losses): 136 watts * 10 hours * 1/0.52 system eff * 1/4 hours of sun = 654 watts of solar panels minimum
 136 watts * 10 hours * 1/12 volts * 1/0.85 eff * 2 days * 1/0.50 max discharge = 533 AH battery bank
 533 AH * 14.5 volts charging * 1/0.77 derating * 0.10 rate of charge = 1,004 watts...
Bill
so without doing the double check part of it, just the math from the first part.
here is how it seems to break down to get a ball park estimate on what to buy batteries and panels to get started based upon ones needs
for 10 hours a day usage of ac watts
and
assuming 85% efficiency inverter
for every 1w ac usage need 4.8w of panels
for every 1w ac used need 4 ah 12v battery 
Re: formula for converting AC Killawatt reading over to DC panel wattage?Urbandialect wrote: »but what's shocking to me is that it takes almost 1kw of DC to run and maintain 150 watts of AC.. mindblowing
You are mistaking power (watts) for energy (kilowatthours). Watts is a RATE of energy use. Kilowatthours are an AMOUNT of energy.
150 watts for 10 hours is 1.5 kilowatthours of energy. That needs to be stored in your battery. A battery that size needs a certain minimum solar array just to charge it. And, as mentioned, there are many losses going into and out of the battery.
btw, if you only use the computer while the sun shines, you can get away with a much smaller system.Urbandialect wrote: »I'm almost finished with my 600 watt array, and you know I have 2 battery banks, 6 6volt amg batteries rated at 200amp hours, and 6 6volt T105s at 225amp hours
Please elaborate... I hope all those batteries are not all connected together... you will ruin all those batteries pretty quickly if they are wired together to make one battery bank.
Even if they are two separate banks, each of those banks is a far from optimal design... especially the AGM batteries. The problem is that parallel batteries are unstable... when charging them it is difficult to get the charging current to divide up evenly among the parallel paths. AGM batteries have lower resistance, so the resistance of the wiring exacerbates the problem, relative to the flooded batteries.
short discussion here: http://www.windsun.com/ForumVB/showthread.php?14674
vtMaps4 X 235watt Samsung, Midnite ePanel, Outback VFX3524 FM60 & mate, 4 Interstate L16, trimetric, Honda eu2000i 
Re: formula for converting AC Killawatt reading over to DC panel wattage?FarmerBlue wrote: »so without doing the double check part of it, just the math from the first part.
here is how it seems to break down to get a ball park estimate on what to buy batteries and panels to get started based upon ones needs
for 10 hours a day usage of ac watts
and
assuming 85% efficiency inverter
for every 1w ac usage need 4.8w of panels
for every 1w ac used need 4 ah 12v battery
Well, no because Watts isn't what you're after; Watt hours is. You can't really equate it directly. The batteries provide the power to run the inverter and anything connected to that; the panels recharge the batteries.
1 Watt hour AC @ 85% efficiency becomes 1.18 Watt hours DC. The inverter consumes power to make that which may be 6 Watts for a small inverter or 20 Watts for a large one depending on what the total maximum power needs are. Thus small loads on large inverters are very inefficient.
That 1.18 Watt hours on 12 Volts is 0.098 Amp hours. You could call it 1.0 and say you need 4.0 Amp hours @ 12 Volts, relying on 25% DOD. But ... time of use and size of load demand once again comes in to skew the results.
That is another way to end up with a disappointing system: trying to shortcut between AC Watts and PV panel size and/or battery capacity.
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