*>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.*

It seems to me that you are the person who is confused :-)

*>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.*

I'd say "power," vs energy. Collector efficiency is an instantaneous

parameter, hot water output power divided by solar input power, with

a particular flow rate and inlet temperature.

*>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...*

*>G = Solar radiation available. This can be in several different units of*

*>measure. Watts/m^2 is common for instantaneous or average hourly figures.*

And collector efficiency calculations.

*>For solar system sizing purposes, the total amount of solar energy per day*

*>that is available can be very useful...*

That's a different subject.

*>FrUL = 5.247 W/m^c/Degree C (for the collectors I mentioned).*

FrUL = 5.247 W/m^2/Degree C?

*>(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.*

Ti is the INLET (i for inlet) vs the average fluid temperature,

which would be higher.

*>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):*

Duane didn't ask about Richfield, and the solar intensity on the panel

and water and air temps can vary a lot over 10 hours, which makes this

kind of calculation meaningless:

*>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*

Very mysterious... I can't imagine what was going on in your brain when

your fingers typed that. The units don't even work. Where did the m^2 go?

You might say the average solar power is 6.4kWh/m^2/10hours = 640 W/m^2

(altho it isn't because the sun angle and intensity vary over the day)

and estimate the efficiency as 0.738-5.247*(50-10.6)/640 = 0.41, but

that would be a very gross approximation.

*>Solar Energy Available = 6400 Watt-hours * 3600 sec/hour = 23040 KiloJoules*

That part looks OK, for a square meter.

*>Collector Efficiency then = 9607 / 23040 = 41.6%*

Very mysterious.

*>This assumes that the input fluid temperature is close to 50 C at all times.*

Duane didn't ask about an input fluid temperature. He asked about an

average fluid temperature. To answer his question, you might do some sort

of calculation involving the fluid flow rate and the collector area to

find the outlet temperature, then average the inlet and outlet temps.

*>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.*

With an undersized heat storage tank.

*>That would make the average fluid temperature to be closer to 30 C,*

*>resulting in an efficiency of 57.6%.*

But your efficiency equation uses an inlet vs average fluid temp.

You've ducked Duane's question a third time, Stephen.

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?

*>> >> 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 50C average water temp?*

Say water goes in at temp Ti, which is less than 50C, at the flow rate

specified on your data sheet, and the sun on the panel is 1000 W/m^2...

Nick

*>(Ti - Ta) = the temperature difference between the average fluid temperature*

*>in the collector and the outside air...*

That's apparently how Europeans spec collector efficiency,

but your Ti is the inlet vs average temp.

*>Putting this all together for a typical day in Richfield, Missouri...*

Duane didn't ask about Richfield...

*>This assumes that the input fluid temperature is close to 50 C at all times.*

Duane asked about an average vs input fluid temperature. To answer his

question, you might do some sort of calculation involving the fluid flow

rate and the collector area to find the outlet temperature, then average

the inlet and outlet temps.

*>Can you find someone at the factory who actually understands*

*>Duane's question and post a correct answer here?*

Peter Allen understands Duane's question.

*>> >> 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 50C average water temp?*

Say water goes in at temp Ti, which is less than 50C, at the flow rate

specified on the data sheet, and the sun on the panel is 1000 W/m^2...

With 0.0687 kg/s of water flow and 2.78 m^2 of solar aperture and

efficiency E = 0.738-5.247(Ti-10)/1000 = 0.6855-0.005247Ti (1) and

(Ti+To)/2 = 50 (2), To = Ti+2.78x1000E/(0.0687kg/sx4.19kJ/kg-c)

= Ti+9.66E (3). Substituting To = 100-Ti (2) on the left side of (3)

and (1) on the right side makes 100-Ti = Ti+9.66(0.6855-0.005247Ti),

so Ti = 47.91 C and E = 0.434, for a 43.4% collection efficiency.

Peter says (2) isn't quite accurate because the water gains more heat

in the first half of the collector than in the second half, since it

is cooler in the first half and loses less heat to the outdoors.

Nick

Sounds like Peter and I were very close in our estimates; 43% vs. 41%. Not

bad.

Regards,

Stephen Cumminger

Email: stephen.cumminger@heatwithsolar.com

www.heatwithsolar.com

*> >(Ti - Ta) = the temperature difference between the average fluid*

temperature

*> >in the collector and the outside air...*

*> That's apparently how Europeans spec collector efficiency,*

*> but your Ti is the inlet vs average temp.*

*> >Putting this all together for a typical day in Richfield, Missouri...*

*> Duane didn't ask about Richfield...*

*> >This assumes that the input fluid temperature is close to 50 C at all*

times.

*> Duane asked about an average vs input fluid temperature. To answer his*

*> question, you might do some sort of calculation involving the fluid flow*

*> rate and the collector area to find the outlet temperature, then average*

*> the inlet and outlet temps.*

*> >Can you find someone at the factory who actually understands*

*> >Duane's question and post a correct answer here?*

*> Peter Allen understands Duane's question.*

*> >> >> 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 50C average water temp?*

*> Say water goes in at temp Ti, which is less than 50C, at the flow rate*

*> specified on the data sheet, and the sun on the panel is 1000 W/m^2...*

*> With 0.0687 kg/s of water flow and 2.78 m^2 of solar aperture and*

*> efficiency E = 0.738-5.247(Ti-10)/1000 = 0.6855-0.005247Ti (1) and*

*> (Ti+To)/2 = 50 (2), To = Ti+2.78x1000E/(0.0687kg/sx4.19kJ/kg-c)*

*> = Ti+9.66E (3). Substituting To = 100-Ti (2) on the left side of (3)*

*> and (1) on the right side makes 100-Ti = Ti+9.66(0.6855-0.005247Ti),*

*> so Ti = 47.91 C and E = 0.434, for a 43.4% collection efficiency.*

*> Peter says (2) isn't quite accurate because the water gains more heat*

*> in the first half of the collector than in the second half, since it*

*> is cooler in the first half and loses less heat to the outdoors.*

*> Nick*

>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.