vacuum tube solar collector CPC absorber temperature 1

[Ciaci]

Hi all,
I am trying to demonstrate that a vacuum tube solar collector CPC have not caused the fire on the roof of wood structure. So I am trying evaluate the thermal condition of the plant, in particular the temperature of the absorber part of the collector. I found the dry bulb air temperature (it was 6°C) and the radiation on a tilt surface that is 120 W/m^2 (it was early morning).

Since the trasmission coefficient multiplicated for the absorption coefficient is 0,7, the absorbed radiation energy is 0,7*120=84 W/m^2.

Then I thought to use the Stefan-Boltzman low,

Eass = (5,67*10^-8)*(Tabsorber^4-Tair^4)

Where
Eass is absorbed radiation energy
Tabsorber is the temperature of the absorbe
Tair is the already measured themeperature of the air

So i found the Tabsorber since Eass and Tair were already known.

Do you think that this method could be right? I personally have some doubt since I have not considered the temperature of the glass collector face, and I have not considered the convection.
Do you have any advice? Have you ever heard of fire caused by thermal collectors?
Thank you so much

 

[racookpe1978]

Tracking collector or fixed?

Photograph the installation, and photograph where the fire occurred.
WHEN did it occur? Midday? End of afternoon? At night?

 

[Ciaci]

It was a fixed collector and 9.30 a.m. of the 21 december (Italy). I have already analyzed the fire patterns of the fire scenario, and I am pretty sure about the fact that it was not the collector but another source, i.e. a domestic chimney really bad insulated and placed near wood shreaded insulation of the roof.
I was just trying to make a mathematical demonstration, in order to demonstrate that the temperature was low.

 

[Ciaci]

Maybe I am trying to make something of impossible and there are too many other variable to know in order to find the temperature of the collector. What do you think about it?

 

[racookpe1978]

OK. Assume mid-Italy, latitude = 42 north (near Rome), 21 December at 9:30 local solar time.

My default spreadsheet is for a horizontal flat-plate collector, so it won't be an exact match to you inclined (sloped) collector facing due south, but it will give you some idea.

At 9:30 local solar time, the collector is seeing very little solar energy (lots of absorption in the atmosphere, solar elevation still very low above the horizon, collector and pipes and glass still cool from being in the winter air overnight, etc. At 14:30 in the afternoon, it would have been soaking up heat all day, and be at it near-maximum temperature! (Even though the solar energy present is the same at 9:30 in the morning as at 2:30 in the afternoon, the temperature of everything would be far hotter.)

At 9:30 am on 21 Dec at latitude 42 the sun is very low: 16 degrees above the horizon.
On a clear day, a 1m x 1m plate at 9:30 am receives
500 watts/m^2 on a plate perpendicular to the sun but only
138 watts/m^2 on a horizontal plate.
Sun is only 16 degrees above the horizon, air mass = 3.546

To compare heat loads, on June 22 at 9:30 am that same 1m x 1m plate receives
923 watts/m^2 on a perpendicular plate to the sun but only
745 watts/m^2 on a horizontal plate.
Sun is at 53.8 degrees above the horizon, air mass = 1.238

At noon on June 22, that same 1m x 1m plate receives
973 watts/m^2 on a perpendicular plate to the sun but only
922 watts/m^2 on a horizontal plate.
The sun at noon is 71.5 degrees above the horizon, air mass = only 1.054

So if the solar collector didn't burn up in the noon sun on 22 June, it would not burn up on a colder day on 21 Dec.

 

[Ciaci]

Thank you very much racookpe for your very professional help.

 

[MintJulep]

, the absorbed radiation energy is 0,7*120=84 W/m^2.

Check your units. You need to multiply by the area of the collector to get the correct units for the Eass term.

Your approach will provide a higher than actual temperature, because it neglects all heat rejected from the collector to anything else, including whatever it was that caught fire.

I think you need to do a complete heat balance for the entire system, including all modes of heat transfer.

The opposing expert witness will be able to poke all sorts of holes in this approach, noting that most of the coefficients that you use will likely be assumptions.

How long had the collector been installed? Was it there in the summer? Much better argument would be that the roof didn't catch on fire in the summer when the collector was there and it was much hotter outside. The roof only caught on fire when there was hot gas in the chimney.

 

[Ciaci]

You are right MintJulep, my thought requires too much assumption. The collector was installed the 10th of July, so it worked in July and August.