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I was poking around to check on solar PV prices today as the remodel project gets underway (post-garage/office disaster discovery, new architectural and engineering drawings starting now) and found this page:

http://www.dmsolar.com/solar-gridtie-system.html

which has three complete systems (just-under-9 to about-10.5 kW peak, panels + inverter + mounts + wire + lightning arrestor + etc) for $16.7k to $19.4k, i.e. about $1.85/watt. That's a seriously good price. The just-under-9 kW system would make about 36 kWh/day for me here (yearly average, lots more in summer than winter of course).

[Edit: correction, apparently the SLC area gets 5, not 4, peak sun hours per day. That 8.9 kW system makes more than 44 kWh/day, possibly more than I need even with car-charging.]

My electric rate is now (after an increase) about $.09/kWh average so that's just under $3.25/day or about $1180 a year. Over 20 years that's about $23600 in electric energy that I can buy up front for $16738 (plus installation, of course). Discounting at 3% that's actually slightly better than break-even (though of course installation will make it "slightly worse", but utility rate increases in the future—which will happen, I just don't know enough to estimate them—will bring it back in line or better).

[Edit: using 44 kWh/day I get 16060 kWh/yr or about $1445/yr or $28900 over 20 years. I think I am going to do this!]

Anyway, worth a look.

 

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My system is 7.3 KW, split into two inverters, one with about 3 KW facing south and the other about 4 KW facing South-East. In Northern California, it generated 9938 kwh in 2011 ... So, your estimates for making 44 kwh/day from an 8.9 KW system may be a tad high. ...

Yes, it's going to depend on a number of things. I have no idea where you are in northern-Calif; the amount of sun you get in the SFBayArea varies wildly based on location. Here in SLC we have a lot of sun in the summer, so even though we are a little further north I think we'll get as many sun hours as one would in (say) Concord or Walnut Creek, which is a lot more than one would get in the Berkeley hills or much of SF.

If we get the same as you do, and if my roof (which is basically flat so that there's easy panel orientation) gives me slightly better panel orientation, I could get close to 40 kWh/day. (27.227 / 7.3 * 8.9 = 33.195, more or less.)

Rocky Mountain Power has no obnoxious feed-in tariffs that I can find, and does net metering over the course of a year with power generated credited at the normal retail rate. We have some rate hikes baked in for the next few years but the amounts are still "under discussion". My best guess is that $.09/kWh is going to $.11/kWh in the summer within two years, and then upwards from there. (Also: we already have summer vs winter rates, which is new as of a couple of years ago. I used .09 as an average, it's .08 in winter and .10 in summer.)

[Edit: here's a table with maps: http://www.bigfrogmountain.com/SunHoursPerDay.html which suggests that SFBayArea gets roughly 4.5 peak sun hours, vs 5.0 for most of Utah, specifically 5.26 for Salt Lake City itself. I downloaded a Utah slice from the NREL data and they also have 5.3 for best-fixed-alignment, 5.0 for off-by-15-degree, and just over 7 if you use single-axis tracking, which I doubt I will do but 7 sure is a lot better than 5...]

 

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Don't forget that there is a 30% tax credit too. In the US.

Oo, I thought this had been terminated, but maybe it's still active.

http://www.absak.com/library/solar-power-tax-credits

If I get a $5k (30% of $17k, more or less) solar PV credit plus a $7500 EV credit plus roughly $1500 in tax deductions on the (new) mortgage (I don't have one now but plan to take one out for the home remodel), that's about $14k off my 2012 Fed taxes (nominally returned in 2013 of course but I can probably spread it out over the year in terms of forking over fewer dollars each quarter ... I'll have to talk to some tax people here when I do the 2011 taxes).

This also means that the installed system, which I've been thinking of as "costing" roughly $20k, might cost closer to $15k. Depending on what happens with retail energy rates here, if the system produces $25k or more of energy over 20 years, that's a decent enough return.

Just for fun (?) I made up a spreadsheet: https://docs.google.com/spreadsheet/ccc?key=0AucA52IiMClUdERBbkc4MmJfZHhJc1dvb0VtOFNNd1E

Even assuming the utility only raises rates 3%/yr and I can somehow earn a steady 8%/yr on the cost of the solar PV system, I break even in the 18th year. I used a "derate factor" of 0.9 (winter/spring/fall) and 0.8 (summer) to adjust for not actually getting all the energy out of the panels that the NREL red book would imply. Increasing the derate factor obviously makes things worse. Increasing the utility inflation rate (or making "summer" price extend a little longer, which it really does) makes things better, and reducing the "safe" rate of return from 8% to (say) 3% makes things hugely better.

 

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Just for fun you made a spreadsheet?

Spreadsheets can't be fun?

 

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Note: I updated my spreadsheet after finding an NREL page that suggested a derate factor of 0.77. After doing that, I get about 36.2 kWh/day (but of course much higher in summer and lower in winter), which matches pretty well with what Dave_Car_Guy is observing: take 9938, multiply by 8.9/7.3, multiply again by 5.0/4.5, final result is 13462, vs about 13222 in the spreadsheet.

Had to lower return-on-cash to 6.5% to get breakeven to remain in Year 19. Treasuries are producing about 2% so I feel pretty reasonable about that. Also, I only did calculations to 20 years, but I believe the system will keep working fine for 30 or more.

 

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- With solars panels you're also helping the planet a bit.
- With solar panels you are more or less 'off the grid' and more independent.
- With solar panels you are protected against steep price increases.

Well, yes, I'm just trying to prove that it makes financial sense with the spreadsheet.

As I said to someone on line last night, the real point is ... well, here in Utah, we have almost all coal power. So I imagine I'll get the inevitable questions about the Karma, and when I say "I'm driving on electricity hence not dumping crap into our air here" someone will say "yeah but you're just using coal", and then I can be all smug and say "no, I have solar PV on my roof!"

 

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How much power is needed for a full charge?

Technically, it's "energy", not "power". (Energy is a "stock" like a pool full of water, power is a "flow" like the water draining out of, or refilling the pool. Or, mathematically, energy = ? (power) dt; power = d(energy)/dt.)

Thus, to answer both questions: 3.3 kW of power for about 6 hours, or about 20 kWh of energy. There's a little bit of loss in the charging cycle so that loading the batteries with 20 kWh can take up to 22 kWh at the wall-socket-plug.

Whether one will charge from "zero" to "full" every time is an important question. In my case I expect most of my charges will be starting from at least 70% full anyway, so no more than about 7 kWh/day most of the time.

 

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Sort of on this point, does the Karma support Level 3 charging -- something faster than 6 hours?

No. And in fact, even with Level 2 it could be done in 3 hours, if it would draw 6.6 kW instead of 3.3, but it won't do that either. I don't know if the battery can handle this rate (it seems likely to me that it can, I just don't know) but the charger electronics have to negotiate this through the J1772 connector first, and they don't now.

 

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Only 16-17 kWh (80-85%) of the 20+ kWh is being used, in order to prolong battery life.

Oops, I was using a base size of 24 kWh or so (from, obviously faulty, memory ).

The main point—how much energy is required to charge the battery depends on how much energy you draw out of the battery before recharging—remains, though. If your battery range is roughly 40 miles and you usually drive about 20, you will only do a "half recharge". (Actual ranges seem to be 40 to 45+ miles, though I expect lower numbers here in the winter when it's 20 degrees F outside. 40 and 20 here are mainly for illustration and the obviousness of "half".)