Posted by DavidIa on October 27, 2008, 8:39 pm
Help me understand what capacity a PV system would need to be to power
my house. Here is what I get from the electric company:
Days Consumption Units
28 1953 KWH
30 3180 KWH
32 4493 KWH
30 4130 KWH
33 4493 KWH
29 2201 KWH
29 1367 KWH
33 1289 KWH
30 1092 KWH
33 1731 KWH
31 1562 KWH
31 1264 KWH
29 2929 KWH
30 4033 KWH
Would a 6kwh system be sufficient? Do I take the consumption/days/24
to calculate the per hour need?
Thx
D
Posted by Ron Rosenfeld on October 27, 2008, 10:36 pm
wrote:
>Help me understand what capacity a PV system would need to be to power
>my house. Here is what I get from the electric company:
>Days Consumption Units
>28 1953 KWH
>30 3180 KWH
>32 4493 KWH
>30 4130 KWH
>33 4493 KWH
>29 2201 KWH
>29 1367 KWH
>33 1289 KWH
>30 1092 KWH
>33 1731 KWH
>31 1562 KWH
>31 1264 KWH
>29 2929 KWH
>30 4033 KWH
>Would a 6kwh system be sufficient? Do I take the consumption/days/24
>to calculate the per hour need?
>Thx
>D
You don't provide sufficient information. Since a PV system requires the
SUN to generate power, you need to know how much sunlight you will get at
the location and orientation of your PV panels.
If you are in the US, you can use the PVWATTS (google for the URL)
calculator to estimate electricity production for different sized arrays.
The very basic calculations go something like:
Days Total: 428
Consumption Total: 35,717 kWh
Your Average daily consumption: 83.45 kWh
For a grid-connected system, you would then divide that value by the hours
of FULL sunlight per day, averaged over a year. That is considerably less
than the time from sunrise to sunset, by the way. And it may have to be
adjusted if you cannot optimally orient your array.
Assume for this exercise that you have 5 hrs of full sun equivalent (also
called ESH for effective sun hours, or insolation) per day.
83.45 / 5 = 16.69 kWh.
But the manner in which PV arrays are rated does not usually reflect "real
world" electricity generation. There will also be losses in your inverter.
So your hypothetical system my have to be sized as a 20+ kWh array to
replace all of your generated electricity. The cost to do that would
probably be on the order of 160,000 USD.
You may want to consider other alternatives to reduce your electric bill.
--ron
Posted by Neon John on October 28, 2008, 1:41 am
wrote:
>wrote:
>>Help me understand what capacity a PV system would need to be to power
>>my house. Here is what I get from the electric company:
>>Days Consumption Units
>>28 1953 KWH
>>30 3180 KWH
>>32 4493 KWH
>>30 4130 KWH
>>33 4493 KWH
>>29 2201 KWH
>>29 1367 KWH
>>33 1289 KWH
>>30 1092 KWH
>>33 1731 KWH
>>31 1562 KWH
>>31 1264 KWH
>>29 2929 KWH
>>30 4033 KWH
>Assume for this exercise that you have 5 hrs of full sun equivalent (also
>called ESH for effective sun hours, or insolation) per day.
>83.45 / 5 = 16.69 kWh.
Actually, if you cancel your units properly, that's 16.7 kilowatts and not
kilowatt-hours.
>But the manner in which PV arrays are rated does not usually reflect "real
>world" electricity generation. There will also be losses in your inverter.
>So your hypothetical system my have to be sized as a 20+ kWh array to
>replace all of your generated electricity. The cost to do that would
>probably be on the order of 160,000 USD.
Again, it's 20 kilowatts and not kWh.
Of course, with that kind of consumption variability, I'd never even remotely
consider recommending a system large enough to pick up all the peak load.
Without any more information at hand than what was given, I'd probably go for
something in the 5kW range. Enough to handle all the load on the low
consumption months, most of the load on the medium consumption months and then
I'd find out how to trim the high consumption months.
>You may want to consider other alternatives to reduce your electric bill.
Yes, I agree. Further I have some questions. Your usage varies over a 4:1
ratio, especially if the numbers are normalized to constant days.
Assuming this is a domestic situation, what are you doing to vary your usage
that much? That would be the very first thing I'd want to investigate. You
didn't put any months next to the readings but still, I don't see a seasonal
pattern there.
If this is a heating application such as a pool or hot tub then it would
foolish to spend money to turn radiant energy (sunlight) into electricity,
only to turn it back into heat. You'd be MUCH better off collecting the solar
energy, including the infrared that solar cells don't do anything with, with
solar-to-water collectors and using that flow to heat your pool or tub
directly. Solar water heater panels cost a tiny fraction of solar electric
panels and can easily be home-built if you're the type to do that.
If you're comfort heating your house then again, solar heat collectors are a
much more sensible way to go.
Please do tell us what causes these huge consumption swings. Of course, if
the load involves lots of big ole grow lights, never mind..... :-) That would
be a great application for solar light ducts. Home Depot has 'em.
John
--
John De Armond
See my website for my current email address
http://www.neon-john.com
http://www.johndearmond.com <-- best little blog on the net!
Tellico Plains, Occupied TN
No one can be right all of the time but I'm getting close.
Posted by Eeyore on October 28, 2008, 6:37 am
Neon John wrote:
> Ron Rosenfeld wrote:
> >DavidIa wrote:
> >
> >>Help me understand what capacity a PV system would need to be to power
> >>my house. Here is what I get from the electric company:
> >>Days Consumption Units
> >>28 1953 KWH
> >>30 3180 KWH
> >>32 4493 KWH
> >>30 4130 KWH
> >>33 4493 KWH
> >>29 2201 KWH
> >>29 1367 KWH
> >>33 1289 KWH
> >>30 1092 KWH
> >>33 1731 KWH
> >>31 1562 KWH
> >>31 1264 KWH
> >>29 2929 KWH
> >>30 4033 KWH
Are the high kWh hours due to A/C or heating ?
> >Assume for this exercise that you have 5 hrs of full sun equivalent (also
> >called ESH for effective sun hours, or insolation) per day.
> >
> >83.45 / 5 = 16.69 kWh.
> Actually, if you cancel your units properly, that's 16.7 kilowatts and not
> kilowatt-hours.
And 5 hours peak equivalent is a LOT. Certainly won't get that in New England.
And
doubly not in winter.
Oh he's in Texas. At least that won't be a problem then.
Graham
Posted by Ron Rosenfeld on October 28, 2008, 7:30 am
>Again, it's 20 kilowatts and not kWh.
You're absolutely correct. And I do know better. But sometimes, typing
late at night, errors occur.
--ron
>my house. Here is what I get from the electric company:
>Days Consumption Units
>28 1953 KWH
>30 3180 KWH
>32 4493 KWH
>30 4130 KWH
>33 4493 KWH
>29 2201 KWH
>29 1367 KWH
>33 1289 KWH
>30 1092 KWH
>33 1731 KWH
>31 1562 KWH
>31 1264 KWH
>29 2929 KWH
>30 4033 KWH
>Would a 6kwh system be sufficient? Do I take the consumption/days/24
>to calculate the per hour need?
>Thx
>D