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

*> Hi Sonideft;*

*> > www.heatwithsolar.com/products/G_Series_tech.pdf*

*> > The end result is a collector with an efficiency FR value*

*> > of 74%, which is pretty good for a flat-plate collector.*

*> That's a bit misleading.*

*> Your 74% efficiency number, while technically true, is not*

*> very useful as the temperature rise is zero degrees C.*

*> What is the efficiency of your collector*

*> at a more useful temperature rise?*

*> Let's say the temperature rise is 60C (108F):*

*> 50C (122F) average water temperature.*

*> -10C ( 14F) ambient temperature.*

*> Duane*

*> -- *

*> Home of the $5 Solar Tracker Receiver*

*> http://www.redrok.com/electron.htm#led3X [*]*

*> Powered by \ \ \ //|*

*> Thermonuclear Solar Energy from the Sun / |*

*> Energy (the SUN) \ \ \ / / |*

*> Red Rock Energy \ \ / / |*

*> Duane C. Johnson Designer \ \ / \ / |*

*> 1825 Florence St Heliostat,Control,& Mounts |*

*> White Bear Lake, Minnesota === \ / \ |*

*> USA 55110-3364 === \ |*

*> (651)426-4766 use Courier New Font \ |*

*> redrok@redrok.com (my email: address) \ |*

*> http://www.redrok.com (Web site) ==*

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

*>You are right that the 74% can be misleading... The de-rating curve...*

*>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*

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

Duane writes:

*>> What is the efficiency of your collector*

*>> at a more useful temperature rise?*

*>> Let's say the temperature rise is 60C (108F):*

*>> 50C (122F) average water temperature.*

*>> -10C ( 14F) ambient temperature.*

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

*> >You are right that the 74% can be misleading... The de-rating curve...*

*> >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*

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

*> Duane writes:*

*> >> What is the efficiency of your collector*

*> >> at a more useful temperature rise?*

*> >> Let's say the temperature rise is 60C (108F):*

*> >> 50C (122F) average water temperature.*

*> >> -10C ( 14F) ambient temperature.*

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

*> Nick*

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

*>Based on Richfield, Missouri with an average (yearly) ground-level radiation*

*>of 5300 Watts/m^2/day...*

Watts per 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

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?

*>> >You are right that the 74% can be misleading... The de-rating curve...*

*>> >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*

*>>*

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

*>>*

*>> Duane writes:*

*>>*

*>> >> What is the efficiency of your collector*

*>> >> at a more useful temperature rise?*

*>> >> Let's say the temperature rise is 60C (108F):*

*>> >> 50C (122F) average water temperature.*

*>> >> -10C ( 14F) ambient temperature.*

*>>*

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

*>>*

*>> Nick*

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

*> >Based on Richfield, Missouri with an average (yearly) ground-level*

radiation

*> >of 5300 Watts/m^2/day...*

*> Watts per 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 *

*> 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?*

*> >> >You are right that the 74% can be misleading... The de-rating curve...*

*> >> >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*

*> >>*

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

*> >>*

*> >> Duane writes:*

*> >>*

*> >> >> What is the efficiency of your collector*

*> >> >> at a more useful temperature rise?*

*> >> >> Let's say the temperature rise is 60C (108F):*

*> >> >> 50C (122F) average water temperature.*

*> >> >> -10C ( 14F) ambient temperature.*

*> >>*

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

*> >>*

*> >> Nick*

> Hi Sonideft;> > www.heatwithsolar.com/products/G_Series_tech.pdf> > The end result is a collector with an efficiency FR value> > of 74%, which is pretty good for a flat-plate collector.> That's a bit misleading.> Your 74% efficiency number, while technically true, is not> very useful as the temperature rise is zero degrees C.> What is the efficiency of your collector> at a more useful temperature rise?> Let's say the temperature rise is 60C (108F):> 50C (122F) average water temperature.> -10C ( 14F) ambient temperature.> Duane> --> Home of the $5 Solar Tracker Receiver> http://www.redrok.com/electron.htm#led3X [*]> Powered by \ \ \ //|> Thermonuclear Solar Energy from the Sun / |> Energy (the SUN) \ \ \ / / |> Red Rock Energy \ \ / / |> Duane C. Johnson Designer \ \ / \ / |> 1825 Florence St Heliostat,Control,& Mounts |> White Bear Lake, Minnesota === \ / \ |> USA 55110-3364 === \ |> (651)426-4766 use Courier New Font \ |> redrok@redrok.com (my email: address) \ |> http://www.redrok.com (Web site) ==