electricity calculation

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eric b 125

Guest
i'm getting some lights on monday that total 884 watts. 2X250 MH, 4X96 PC, 8 X moonlights. either way, i'm looking at my electric bill here, trying to get an estimate on monthly costs but i'm not sure what number i'm supposed to use. it looks like all of them add up to my bill.
Supply: 7.1560 cents/kWh
Transmission: 1.3767 cents/ kWh
Distribution: 4.2313 cents/ kWh
i will probably do a 12 hr photoperiod: 2, 8, 2. anyone that can help me with the formula would be greatly appreciated.
 
E

eric b 125

Guest
it also says on my bill: "for comparison to other electric suppliers--your supply/generation and transmission Price to Compare is 8.53 cents/ kWh.
 
xcali1985

xcali1985

Active Member
Most people with MH notice an increase of about 50$ on their bill. I cant do the math to figure it out also you need to take into account the amount of time you have the lights on. The good thing though is you don't need to run your lights as long with MH as you do with other forms.
 
posiden

posiden

Active Member
You owe me at least one beer for this.

Watts / by 1000= kilowatts
Kilowatts x hours run= kilowatt hours
Kilowatt hours x cost per kilowatt hour= your bill.
384 watts= your Pc lights
500 watts= your halides.
384watts/1000= .384 kilowatts
500watts/1000= .5 kilowatts
.5 kilowatts x 8 hours= 4 kilowatt hours
.384 kilowatts x 12 hours= 4.608 kilowatt hours.
4 + 4.608 kilowatt hours= 8.608 kilowatt hours in a day.
Now you can do it by the month or the year here.
8.608 KH x 31 days= 266.848 KH a month
8.608 KH x 365 day= 3141.92 KH per year.
So monthly, I will figure the most expensive number you posted as I don't know which one you pay. It is always the highest number tho right.

266.848 KH x 7.156 cents= 1909.564 cents
1909.564 cents/100cents= $19.09564 Dollars a month.
I will leave you to the yearly. Though it is the simplest on to do.
 
scsinet

scsinet

Active Member
Don't forget to count ballast overhead. The 500w and 384 watts are just what the lamps draw. The ballasts are not 100% efficient, in fact in most cases, they add about 20% of additional current draw.
Don't bother changing the calculations, just add 20% to posiden's total... so
$19.10 * 1.20 = $22.92/mo.
 
calbert0

calbert0

Member
Originally Posted by Posiden
http:///forum/post/3160361
You owe me at least one beer for this.

Watts / by 1000= kilowatts
Kilowatts x hours run= kilowatt hours
Kilowatt hours x cost per kilowatt hour= your bill.
384 watts= your Pc lights
500 watts= your halides.
384watts/1000= .384 kilowatts
500watts/1000= .5 kilowatts
.5 kilowatts x 8 hours= 4 kilowatt hours
.384 kilowatts x 12 hours= 4.608 kilowatt hours.
4 + 4.608 kilowatt hours= 8.608 kilowatt hours in a day.
Now you can do it by the month or the year here.
8.608 KH x 31 days= 266.848 KH a month
8.608 KH x 365 day= 3141.92 KH per year.
So monthly, I will figure the most expensive number you posted as I don't know which one you pay. It is always the highest number tho right.

266.848 KH x 7.156 cents= 1909.564 cents
1909.564 cents/100cents= $19.09564 Dollars a month.
I will leave you to the yearly. Though it is the simplest on to do.

sorry Poseidon but i think your wrong there.
First of all, before you go doing a bunch of math, you should first define a kilowatt hour.
Definition: The amount of energy expended (or dissipated) if work is done at a constant rate of one thousand watts for one hour.
and beside
Watt / 1000 = miliwatt
Watt * 1000 = killowat
think about it, a gram times 1,000 is a kilogram not a miligram.
So if he is using 384 watt and 500 watt lights for say 12 hours a day than... 884 x 12 = 10,608 or 10.86 KWH of energy. or $.075 x 10.86 = $.81 a day
But after looking at it i guess both of our methods would have produced the same answer if i used different photo periods for the lights rather than having them both on for 12 hours.
Funny how you can do things two different ways and they both work
 
posiden

posiden

Active Member
Originally Posted by calbert0
http:///forum/post/3160372
sorry Poseidon but i think your wrong there.
First of all, before you go doing a bunch of math, you should first define a kilowatt hour.
Definition: The amount of energy expended (or dissipated) if work is done at a constant rate of one thousand watts for one hour.
and beside
Watts / 1000 = miliwatts
Watts * 1000 = killowats
think about it, a gram times 1,000 is a kilogram not a miligram.
So if he is using 384 watt and 500 watt lights for say 12 hours a day than... 884 x 12 = 10,608 or 10.86 KWH of energy. or $.075 x 10.86 = $.81 a day

Maybe you would be a good sport and explane your theroy a bit more. Where did you come up with the .075??
Also the way you have it, 884x12=10608. All you did was figure out the total number of watts, as if there were 12..... 884 watt pieces of equipment.
 
E

eric b 125

Guest
sounds good dude, next time you're in town drinks are on me. youre the man.
 
posiden

posiden

Active Member
Originally Posted by Eric B 125
http:///forum/post/3160418
sounds good dude, next time you're in town drinks are on me. youre the man.

That wont be any time soon. Your a long ways away.
I was just razzing you.
Thanks though. I want to see some pics of this thing you are building. I have read a few posts by you on it. Will you do a thead on it?
 
posiden

posiden

Active Member
Originally Posted by SCSInet
http:///forum/post/3160369
Don't forget to count ballast overhead. The 500w and 384 watts are just what the lamps draw. The ballasts are not 100% efficient, in fact in most cases, they add about 20% of additional current draw.
Don't bother changing the calculations, just add 20% to posiden's total... so
$19.10 * 1.20 = $22.92/mo.

Thank you for the addition. I was unaware of this 20% ish addition for the ballasts.
I is now.
 
calbert0

calbert0

Member
Originally Posted by Posiden
http:///forum/post/3160388
Maybe you would be a good sport and explane your theroy a bit more. Where did you come up with the .075??
Also the way you have it, 884x12=10608. All you did was figure out the total number of watts, as if there were 12..... 884 watt pieces of equipment.
$.075 is the same as saying 7.5 cents
what is wrong with adding up the wattage being constantly used and multiplying by time to get the total watts used for the day. thats how i got the 10,608
then i divided by 1,000 to get my KWH usage per day
i then multiplied by the price of the KWH which was 7.5 cents or $.075 dollars.
 
E

eric b 125

Guest
posiden: i most definately will do a re-build thread on it when the time comes. i really regretted not doing a build thread when the tank was first coming together. things were coming along really smoothly and then BAM... life came at me quick. my car got broken into and they got me real good, relieving me of things that my insurance wont pay for. but what can ya do other than keep moving your feet and try to do the next right thing. either way, i'm getting a piece of equipment here and there, and by the end of the year i'm expecting to have everything in place for a killer re-build.
 
johnr2604

johnr2604

Member
Originally Posted by Posiden
http:///forum/post/3160426
Thank you for the addition. I was unaware of this 20% ish addition for the ballasts.
I is now.

Yep, SCSI is correct. I have to Look cause its been a while, but I Think the National electric code adds 25% when figuring for service calculations.
The 20% would probably be more like what its actually using
 
posiden

posiden

Active Member
Originally Posted by Eric B 125
http:///forum/post/3161351
posiden: i most definately will do a re-build thread on it when the time comes. i really regretted not doing a build thread when the tank was first coming together. things were coming along really smoothly and then BAM... life came at me quick. my car got broken into and they got me real good, relieving me of things that my insurance wont pay for. but what can ya do other than keep moving your feet and try to do the next right thing. either way, i'm getting a piece of equipment here and there, and by the end of the year i'm expecting to have everything in place for a killer re-build.
I will be looking for it.

Originally Posted by johnr2604
http:///forum/post/3161358
Yep, SCSI is correct. I have to Look cause its been a while, but I Think the National electric code adds 25% when figuring for service calculations.
The 20% would probably be more like what its actually using
Thank you.
 
calbert0

calbert0

Member
Originally Posted by Posiden
http:///forum/post/3160388
Maybe you would be a good sport and explane your theroy a bit more. Where did you come up with the .075??
Also the way you have it, 884x12=10608. All you did was figure out the total number of watts, as if there were 12..... 884 watt pieces of equipment.

Originally Posted by calbert0
http:///forum/post/3161219
$.075 is the same as saying 7.5 cents
what is wrong with adding up the wattage being constantly used and multiplying by time (12) to get the total watts used for the day. thats how i got the 10,608
then i divided by 1,000 to get my KWH usage per day
i then multiplied by the price of the KWH which was 7.5 cents or $.075 dollars.
Why wouldnt this work?
 
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