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August 17th, 2015, 06:42 PM   #1
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From: aberbaizan

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Practical Math - For Solar Power

Can someone help me out with an expression to help me find the best time/price cost for choosing solar power?

There is a graph with two curves. One showing the domestic usage. The other curve is the available electricity from the solar panels.

They both have peaks. Panels one peak, usage two peaks. The peaks are offset. The usage peaks when the panels do not. The panels peak when there's no usage.

Hence in the absence of any way of saving the electricity we're wasting a large chunk of the electricity we've bought with the panels.

Now the electricity we pay for may be, say 40cents kWh.

'Unused' electricity produced by the panels over and beyond what's required by the house can be fed back into the grid but is only credited at say 10c per kWh.

So production which offsets use gains 40c where purely excess production gains only 10.

There's one more factor: the base cost of the system. That could be, say, $6000 for a 3kW system. It could be paid in advance or paid over some years. The lifetimes of such systems are about 25years I think. I think the actual figures are irrelevent to you mathematicians, aren't they? It is all a matter of proportion. But I chuck them in anyway, they are all I have to offer. I would like some mathematical rationale for all of this. Something that says when it would be profitable and when not. And perhaps quantifies the profit. Could anyone do anything like that? Here is an image of the curves in question. The blue line represents home usage of electricity. Peaking before the workday and after. The yellow line represents output from the solar panels, peaking right in the middle of the day when they are least required. The almost perfect match between the panel production and the after work usage is clear. The whole problem being the offset. Attached Images  electricityusageimage.jpg (34.4 KB, 4 views) Last edited by abrogard; August 17th, 2015 at 06:45 PM.  August 18th, 2015, 06:57 AM #2 Global Moderator Joined: Dec 2006 Posts: 20,931 Thanks: 2207 If one is using the power to heat the home, the benefit will occur mainly during the colder months of the year. The power can also be used to provide hot water that is stored in a well-insulated tank.  August 18th, 2015, 08:42 PM #3 Newbie Joined: Aug 2015 From: aberbaizan Posts: 3 Thanks: 0 Well, yes, thanks for that. Not really a mathematical treatment of the subject, though, is it? Nor anything anyone wouldn't be aware of. Perhaps I should go into it a bit more. Seems to me there's a couple or three main things: 1. It is basically a question of NPV, I guess, net present value, as we're seeking to minimise cost. Adds up to the same thing? 2. That question can be looked at in a couple of different ways and all ways in between. i.e. NPV of buying a system and amortizing it over say 3 years - a commonly quoted period hereabouts for a system, 40 months they'll often say. And the other way: the NPV of a system continuing to perform long after it has been paid for. 3. The above systems are basically very different inasmuch as a major feature here is that a large system is largely uneconomic from first principles (cost principles) because your excess capacity is only bought by the Utility Provider for a fraction of what you must pay. For instance electricity cost for me is 35c/kWh we can say, for argument. So when I produce electricity I need then I save 35c/kWh. But when I produce electricity beyond what I need then I only get back 10c/kWh. This is virtually a dead loss when set against amortizing the cost of a system. But in the future, when the system is operating and already paid for then the larger capacity can absorb increased use, is available when costworthy batteries come along and the larger envelope in any case offsets a larger proportion of 'peak use'. (though that bit applies to basic amortization, too, doesn't it). System costs I've seen quoted are: 3kW$5k - $6k 4kWh$6.5k - $7.5k 5kWh$7.5k - $8.5k 10kWh$13k - \$16k That's all up. Installed. Includes inverter cost and allowance for carbon tax rebates. 4. The biggest constant factor I see in there is the proportion of output that can be utilized. The output is a Sine wave, I take it. The peak of that wave occurs in the middle of the day when the electricity is not required. Built in inefficiency. What is the size of that factor? That's my rough and ready non-mathematician look at it. From consideration of it all how would I decide what size system to buy to cover needs economically and how would I structure the financing - pay up front or pay it off? I thought mathematicians might have some views on that. Last edited by abrogard; August 18th, 2015 at 08:45 PM.

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