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Is Efficiency dangerous ?

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Posted by Rob Dekker on August 10, 2005, 3:40 am
 
Hi,

I am thinking about building or buying a solar water heater system,
but when making some efficiency calculations to determine how large
the collector should be, I run into an interesting problem.

I want to be able to heat 50 gallons of water/day to 60 C or so with solar.
Lets call that Tsys (system temperature).
Naturally, the system should not boil over in summer when I'm on vacation,
or don't use hot water for a prolonged period, so I guess enough heat should
radiate so that in full California summer sun, the system does not
get above 95 C. That would be Tmax.
So at Tmax, the system should radiate as much energy as the collector receives.

Since heat is radiated at T^4 (Stefan-Bolzmann law),  if we thus collect Tmax^4
heat, the
efficiency drops to 0% at Tmax. And this is what we want, otherwise
the system boils over.

At Tsys, there is thus Tmax^4 (solar input) - Tsys^4 (radiation) heat available
for usage.
Since solar input is Tmax^4, and usefull heat available is Tmax^4 - Tsys^4,
the efficiency of the system is   (Tmax - Tsys / Tmax)^4.

For Tmax C (368K) and Tsys` C (333K), the efficiency would have to
be as low as 0.008%. That's absurt. There must be another way to get rid of
unused heat.

So is there some sort of special design needed to cool down the system when it's
not used (like run water through the collector over night if heat was not used
during the day) ?

Or is there an 'optimal' efficiency of solar water heaters which prevent them
from
boiling over when not used ?

Rob



Posted by Robert Morien on August 10, 2005, 5:29 am
 


Active system and your temperature/pressure valve on the storage tank.

For prolonged periods of non-use, cover with a tarp.

Posted by nicksanspam on August 10, 2005, 11:57 am
 

That's one way to prevent overheating...


That's one way to prevent overheating...


If radiation were the only heat loss...


If a^4-b^4 = (a-b)^4
 

...(368^4-333^4)/368^4 = 1-(333/368)^4 = 0.33.


Several...


That would use electrical energy. We can let the collector run dry, with
components that can take the heat, or use a steeper tilt and overhang.

Nick


Posted by Rob Dekker on August 10, 2005, 8:35 pm
 
Thanks guys !

I knew there was something wrong with my calcs. Thanks for pointing that out.
So if the efficiency of the collector is in the range of 35%, it won't overheat,
by radiation of the collector alone. Not even considering other heat losses
elsewhere in the system. And the temp/pressure valve will take care of the rest,
just in case.

So what is the typical overall efficiency of a solar water heater system ?
And are convection losses more important than radiation losses ?

Thanks gain

Rob



Posted by Cosmopolite on August 10, 2005, 5:44 pm
 Rob Dekker wrote:


Tmax^4 heat, the

available for usage.

unused heat.

it's

during the day) ?

from

I'm with the other Robert, cover it up.

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