"Martin Sykes" wrote in message
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"IMM" wrote in message
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"IMM" wrote in message
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"RichardS" noaccess@invalid wrote in message
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"IMM" wrote in message
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I could use 400 squ foot of flat plate collector on the roof,
twice
the
area, and produce the same volume of solar hot water as the
Thermomax
solar
collectors which takes up half as much square footage.

The area is "very" important in this instant. Is that clear?

Yes, but efficiency is independent of that IMM.

Thats tantamount to saying a 5oKw boiler is twice as efficient
as
a
25Jw
one.

It is not. Solar collectors produce hot water. Some produce more
than
others for the same area, hence more efficient for a given area.

No.

Efficiency is the ratio of converted power out to power in .

The area doesn't come into it.

In this case it does. Area is the most important factor as it is
limited
on
a roof. Solar panel X can be more efficient (ratio of converted
power
out
to power in) than panel Y. But panel X may take up four times the
area
of
panel Y. It means eff all if the area is not taken into account.
For
a
given area which is the most efficient? Area, area, area.

Also, with solar panels the input doesn't really matter as you don't pay
for
it. The output per square foot, or metre (hot water generated), is what
matters. The efficiency of a boiler comes into the "ratio of converted
power out to power in", and is important as you pay for the fuel.






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Technically Richard's right. You keep switching your definitions.

Quote: Solar panel X can be more efficient (ratio of converted power out
to
power in) than panel Y snip For a given area which is the most
efficient?
Answer: Obviously panel X. You said so yourself.

Quote: panel X may take up four times the area of panel Y
OK. So you could swap each panel X for four Y's in the same area. This
effectively gives you a large panel Y with the same area as a panel X and
all that matters is the relative efficiency.

But practically you are right as well - area is very important assuming
the
area cannot be tiled by panels because it is too small. But, the
efficiency
is still the same. Surely all you have to do is:

1. For each panel type, work out how many (if any) you can fit on the
roof.
2. Multiply by the efficiency

Whichever gets the highest score will give you the most energy output from
your roof.

Qiute correct. And you did not need a mention of the spurious and
non-existent concept of "efficiency per square foot".

Franz