How to do it and what you need to know.
1. Where do you live. This means the latitude of your home. this is needed because the sun only shines straight down on homes between the tropics of Cancer and capricorn at Midsummer and midwinter respectively. At my latitude ( 41 degrees North) the best angle of the sun at Midsummer is xxxx degrees.
2. the angle of your roof. If solar panels are flat on your roof, then this affects the best angle that the sun can have on your roof. This can be calculated by counting the number of bricks horizontally and then the number of bricks vertically between the eave and the peak, and then measuring these distances against the bricks at a lower height, and calculating the tangent of the angle.
3. is the orientation of the peak of the roof. Ideally it should be East-West so that the roof faces south. As the sun processes across the sky it will be at the best angle at Midday in this instance. If the roof is at a different angle - My roof faces SW by W. Then the sun strikes the roof later in the day.
It is a double integral to calculate the amount of incident sunlight on your solar panels over the year. It also depends on the efficiency of the panels at every angle.
There is another double integral of the depth of the atmosphere at every angle to take into consideration.
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Two quotes received. One for a 4kw system and the other for a 2.6 kw system as that is what he thinks will be sufficient. However, the question is- when will it give this power? At Midsummer at noon? If I play that game I would say that I want a system that will deliver 700watts during most of the day, say from 8AM to 4PM in midwinter. 700 watts seems to be the high regular sort of usage I see on the power meter, with peaks up to 2kw when the kettle is in use.
These quotes have very basic calculations to determine the values, but are based on charts of radiation that contain some of this information.
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Accepted the quote for the 2.6KW system using Microinverters and 10 panels.
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Assuming I wanted to have two more panels added to this system. Is it viable?
from a DIY Solar PV site I found figures for panels.
Assuming a panel + inverter costs £350, including the trivial cost of installation with the rest of the system. What is payback on that £700?
Negative - That £700 could earn me 2%? in the bank.
2 panels generates maximum 270w x 2 =540watts.
Cost of generation income = 14.38p/Kwhr + export of 4.77pKwhr for 50% of generated electricity (= 2.38pKwhr) = total revenue of 16.76pKwhr.
equation is 540w x h = 700£ / (£0.1676/Kwhr)
h = 700 £Kwhr / (£0.1676 x 540w) = 7734 hr
(Ultimate guestimate) Assuming that you only get this power for 4 hours a day, this gives 1933 days for payback =
5.3 years.
