# Collector Material... - Page 2

Posted by Sonideft on April 28, 2004, 8:03 am

Here is a re-post of my reply for all to see.

Hi Duane:

You are right that the 74% can be misleading. It is only an indicator of how
well the solar fins, collector box, glazing and insulation will operate as a
"set" in a completed solar collector. The de-rating curve that was also in
the document I referenced actually shows the reduced performance based on
the temperature differential of the collector fluid and the outside air
temperature.
h = 0.738 - 5.247(Ti - Ta)/G

Perhaps better measurements to point out would have been the solar fin's
solar absorptivity of 95 % and its infrared emissivity of 25%. This is based
on solar
fins that are painted. We also use solar fins that have a Anodic-Cobalt
treated surface with slightly lower absorptivity but much lower emissivity
(92% and 15% respectively).

Regards,

Stephen Cumminger

Posted by nicksanspam on April 29, 2004, 8:05 am

with 0.0667 kg/s of inlet water at Ti (C), and (Ti-Ta)/G in Cm^2/W...

Duane writes:

How would you answer this question, with a 50 C average water temp?

Nick

Posted by Sonideft on April 30, 2004, 1:44 am
Based on Richfield, Missouri with an average (yearly) ground-level radiation
of 5300 Watts/m^2/day, the efficiency equation becomes 0.738-5.247*(60)/5300
= 67.8%. The following link has some info on how the efficiency ratings are
determined, if anyone is interested.
http://www.fsec.ucf.edu/solar/testcert/collectr/gp6ops.htm

--

Regards,

Stephen Cumminger
Email: stephen.cumminger@heatwithsolar.com
www.heatwithsolar.com

Posted by nicksanspam on April 30, 2004, 7:08 am

Watts per day? :-)

http://www.fsec.ucf.edu/solar/testcert/collectr/gp6ops.htm

You've ducked Duane's question again Stephen, and this time you got
the numbers AND the units AND the equation wrong, while coming up
with an overly optimistic answer. You may have a brilliant career
as a politician.

Can you find someone at the factory who actually understands Duane's
question and post a correct answer here?

Posted by Sonideft on April 30, 2004, 5:42 pm
I think that I left out an important part. My apologies upfront. The subject
of collector efficiency can be a bit confusing, so I'll attempt to explain
it in a bit more detail.

Collector efficiency = Solar Energy Collected / Solar Energy Available. This
is what I am referring to when I talk about efficiency. I trust we can all
agree with this starting point.

The simplified equation for the amount of energy you can collect is:

Q = FR*G - FrUL ( Ti- Ta);  This first term is the solar gain and the second
represents the heat losses of the collector to the outside air.

FR = 0.74 (for the collectors I mentioned). This number has no units of
measurement.
G = Solar radiation available. This can be in several different units of
measure. Watts/m^2 is common for instantaneous or average hourly figures.
Since Watts = Joules/second, you can see that these units really describe
the rate/speed that energy that can be collected. For solar system sizing
purposes, the total amount of solar energy per day that is available can be
very useful. This can be expressed in kilo-watt hours or Joules. 1 kwhr =
1000*60*60 Joules.
FrUL = 5.247 W/m^c/Degree C (for the collectors I mentioned).  This
represents the rate of heat lost to the outside air from the heated fluid
circulating through the collector. Again, the rate of heat loss * time =
energy lost.
(Ti - Ta) = the temperature difference between the average fluid temperature
in the collector and the outside air. We use the average daytime temperature
in our calculations as that is when the fluid would be circulating.

Putting this all together for a typical day in Richfield, Missouri that has
average total Solar Radiation of 6.4 Kwhr/m^2 in May with an average daytime
temperature of  10.6 C and 10 hours of good Solar Collection time (fluid
circulation):

Q = 0.74 * 6400 Watts-hour/m^2*3600sec/hour  - 5.247 Watts/m^2/C*(50 C -
10.6 C) * 10 hours*3600sec/hour   =  9607 Kilo-Joules
(1 hour = 3600 seconds, 1 Watt = 1 Joule/sec)

Solar Energy Available = 6400 Watt-hours * 3600 sec/hour = 23040 Kilo -
Joules

Collector Efficiency then = 9607 / 23040 = 41.6%
This assumes that the input fluid temperature is close to 50 C at all times.
That may be true for some applications, but not for typical domestic hot
water (DHW) applications. In those cases the fluid temperature would be very
low in the morning and high by the end of the day. That would make the
average fluid temperature to be closer to 30 C, resulting in an efficiency
of 57.6%. Different applications can get different performance out of the
same collectors. This is why the ratings for FR and FrUL are so important in
comparing different Solar Collectors.

In summary, thanks Nick for picking up my error.

--

Regards,

Stephen Cumminger,Email: stephen.cumminger@heatwithsolar.com
www.heatwithsolar.com

•
• Subject
• Author
• Date
 Re: Collector Material... Cosmopolite 04-05-2004
 Re: Collector Material... Sonideft 04-26-2004
 Re: Collector Material... Duane C. Johnso... 04-27-2004
 Re: Collector Material... Sonideft 04-28-2004
 Re: Collector Material... nicksanspam 04-29-2004
 Re: Collector Material... Sonideft 04-30-2004
 Re: Collector Material... nicksanspam 04-30-2004
 Re: Collector Material... Sonideft 04-30-2004
 Re: Collector Material... nicksanspam 05-01-2004
 Re: Collector Material... nicksanspam 05-04-2004
 Re: Collector Material... Sonideft 05-06-2004
 Re: Collector Material... Nick Pine 05-06-2004