4 ms·
That's fine, but a bit of mathematics... Roof area of an average new house: maybe 100 m^2? Peak solar flux: 1 kW/m^2 Take into account the existence of night
by hugh3 16y ago
That's fine, but a bit of mathematics...
Roof area of an average new house: maybe 100 m^2?
Peak solar flux: 1 kW/m^2
Take into account the existence of night, cloudy days, and the fact you're probably not at the equator: divide that by four.
Efficiency of these solar cells: they have a "goal" of 10%, so let's optimistically assume they make it.
So we're collecting, on average, 0.11001000/4 =2.5 kW.
Actually, y'know, when I started this calculation I was anticipating something ridiculously small, but that's a not-insignificant fraction of your usage. Am I missing a factor somewhere?
- MikeCapone 16y agoIt's especially "not insignificant" if you have batteries and can store the excess power generated during the times of day when you aren't using much power. Most houses are empty a large fraction of most week days.
- deleted 16y ago[deleted]
- ars 16y agoUsing batteries is not sustainable, they just don't last long enough and use too many resources to make. Use the power grid as your battery.
- fraserharris 16y agoIf you are storing the energy in Li-ion batteries, you'll have an added 10 - 20% loss.
- davidj 16y agothe numbers work out pretty well when you disregard the reality . Only in fantasy land the sun is always at peak, technology that claims to be 10% efficient is actually 10%, DC-AC conversion and storage is 100% efficient, and if you disregard the government subsidies and the energy needed to manufacture.
- bsk 16y agoDo you work at BP or Exxon Mobil? :D
- pjscott 16y agoI know you're joking, but a big problem in discussing energy sources is that a lot of people assume that anybody who tries to explain the difficult engineering issues must be a paid shill for a big energy company. It's kind of irritating, even if it is both false and an ad hominem fallacy.
- davidj 16y agoBP is one of the largest researcher and manufacture of solar panels.
- hugh3 16y agoWell, I set out to debunk the economics of it, but my back of the envelope calculations gave better numbers than I was expecting. You're right, though, lots of factors I have failed to take into account. Another big one not mentioned yet: ever noticed how dirty your car looks after being left outside for a couple of weeks? Well have fun washing your roof as often as you wash your car.
- joeag 16y agoActually oil, gas and coal get more than 10X (in dollar amounts) the subsidy that renewable energy and technologies get. Granted, it is because the quantities are much higher, but it's not a "subsidized vs not subsidized" argument.
- eru 16y agoNet subsidies? I know that in Europe there tends to be a huge tax on petrol.
- pjscott 16y ago> Take into account the existence of night, cloudy days, and the fact you're probably not at the equator: divide that by four. That's still somewhat optimistic; about 10--20% of peak is more realistic, based on what we've seen with existing solar installations. It's closer to 20% in sunny places, and closer to 10% in places with less sun. However, that still leaves us with a respectable amount of electricity. Good for them. Of course, the panels themselves are not the whole cost. There's other electrical equipment needed to make that energy usable, for example, and someone has to install and maintain the solar panels. I think the next big source of money in this market, after they get the cost of manufacturing the panels down far enough, is streamlining installation and maintenance. Perhaps someone could make durable "solar shingles" that are relatively easy to put on a house, or some such thing. Longer-term, solar photovoltaics look like they could provide a pretty hefty amount of energy cheaply, if we solve a lot of engineering issues. Where they fall down is producing power reliably, close to where it's needed. One way of dealing with this is to have the solar panels in space, but obviously that's pretty tricky. Another way of dealing with it is to use something else to provide the base load of your electrical grid -- nuclear could handle this with cheap green aplomb -- and use the solar power for things which can easily soak up cheap intermittent power, such as charging the batteries of electric or plug-in hybrid cars. Theoretically, it could also be used to power interesting energy-hungry chemical processes, like ammonia synthesis or aluminum refining. Making this financially viable is another big engineering challenge, but definitely a worthwhile one if it works out.
- joeag 16y agoSolar Shingles by Dow - http://www.dow.com/news/corporate/2009/20091005b.htm http://www.dow.com/news/corporate/2009/20091005b.htm Note: these are not cost competitive with "regular" CSI panels - they are mostly for people who put a premium on appearance.
- ars 16y agoI would divide by a lot more than four. Just because of day and night divide by three. (2 for day night, plus some extra due to twilight). Then for upper latitudes divide by 2 again, and then finally divide by 3 because only part of the roof will face the sun. So your factor should be 18 not 4, i.e. your numbers are 4.5 times larger than they should be. BTW your multiplication symbols turned the number inside it italic.
- mseebach 16y agoYour math isn't horribly wrong. In fact, Wikipedia mentions ~1kWh/day as achievable. That's why cheap, efficient mass-produced solar cells are so damn attractive. They just have the minor hitch of not, in fact, existing. http://en.wikipedia.org/wiki/Photovoltaic_array#Performance http://en.wikipedia.org/wiki/Photovoltaic_array#Performance
- hugh3 16y agoPet peeve: I don't like the unit "kWh", since it's a unit of energy, divided by a unit of time, multiplied by a different unit of time, to give a unit of energy. Why don't they just give it in Joules? kWh/day is even worse -- it's a unit of power made by dividing a unit of energy by a unit of time, multiplying it by a second unit of time, and dividing it by a third unit of time! And now if you'll excuse me, I'm going to go for a jog at 0.11 meter days per hour per second.
- Vivtek 16y agoThey don't give it in Joules because I have no way of understanding how many Joules it takes to run my 17W bulb for two hours per day - but I do understand rather easily how many kWh/day that comes to. So while kWh/day may be confusing from the standpoint of relating it to the amount of coal burned, in point of fact it is the natural unit to use when discussing consumption.
- eru 16y ago17W * 2 hours = 17 W * 2 * 3600s = 0.122400 MJ. For comparison 1kWh = 3.6 MJ.
- joeag 16y agoIn US units one square foot of roof can support 10 watts. So a kilowatt takes 100 sf of roof space. In much of the US you would generate 1,400 to 1,600 "AC sun hours" per year, so a 1 Kw system generates 1,400 to 1,600 kwhrs per year. If your grid electricity costs $0.20 per kwhr, 1 kwh then produces $280 to $320 worth of electricity per year.