solar Radiation on an object 1

[Russ0448]

I'm at a loss.
I have an object that is subject to solar radiation. I'm trying to understand what a good assumption for the heat flux would be. I have been told that a reasonable assumption would be 1000W/m^2. The surface of the object is painted white with (e=. 96 and Alpha=. 21). The surface temperature of the object is estimated at about 115F to 120F. The removal of heat is only going to be free convection and radiation back of the surface. The surrounding temperature is estimated at 95F.
I guess my question is : Is the 1000W/m^2 a reasonable estimation of the total (diffuse as well as direct) irradiation per unit area?
Also should I take into consideration the Atmospheric radiation associated with CO2 and H2O. I have reason to believe that I should, but am unsure of how to do so.

Any help would be much appreciated.
Thanks
Russ

 

[CountZero]

1000 W/m^2 would be a gross assumption. No, at max, there is 700 W/m^2 at the surface of the earth. In outer space, just above the earth's atmosphere the solar flux is 1367 +/- 2- W/m^2. There are 5 major factors which reduce solar flux. Rayleigh scattering, aersol extinction, ozone absorption, water vapour absorption, and permanent gas absortpion. There are many ways to calculate solar flux, but most are either simplistic in their methods or specific to a very small region. I'll give you the most commonly used method by engineers; i.e. the ASHRAE CLEARSKY MODEL.

I'm assuming a horizontal surface on the surface of the earth. Sin B = cos(l)cos(h)cos(d)+sin(l)sin(d)
B = solar altitude and where, l = latitude of point on surface of earth, h = hour angle (i.e. angle of sun away from solar noon)always positive no matter if it's before or after solar noon, and d = sun's declination (for June 21 d = 23.45 deg)
the max. solar flux occurs at solar noon, so then
Bnoon = 90 - [l-d]absolute bars [in units of degrees]
and cos (theta) = sin B (for horizontal surface)
where (theta) = angle of incidence

To calculate direct and diffuse radiation use the following equations.

G(direct) = ConstantX{346.6/exp(0.185/sinB)X cos (theta)
assume Constant = 1, this is the fallback of the ASHRAE model.
G(diffuse) = 0.137(346.3/exp(0.185/sinB))
So then, Gtotal = G(dir) + G(dif) both of these are in units of BTU/hr-ft^2, this can easily be converted into SI if you so choose. These calues, it should be noted, are for June 21.

One last problem is calculating the hour angle. Here is an example to aid your calculations.
determine local solar time(LST) to 11:00 AM Central DAylight savings time (CDST) on February 21 at 95 deg W longtitude. CST = CDST - 1 hour = 11:00 - 1hr = 10:00 AM
CST is at 90 deg W longitude so then 95 - 90 deg = 5 deg
5 deg X 4min/deg (sun's movement) = 20 mins
LCT (local central time) = CST - 20 mins = 9:40 AM
need table to know equation of time (could neglect this in your calc) = -13.9 mins.
LST = LCT + eq. of Time = ~9:26 AM
so then the hour angle (using LST) is 12:00 - 9:26AM =
2 hrs and 34 mins or 2.567 hrs.
h = 2.567 hrs X (15 deg/hr) = 38.505 deg

There ya go, hope that helps. Send a reply if you need anything else.

 

[IRstuff]

MIL-STD-810F, Method 505.4, Solar Radiation, Figure 505.4-1 shows a 99-percentile peak diurnal solar radiation of 1120 W/m^2.



TTFN

 

[Russ0448]

Thanks for the help,
One other thing I should add is that the structure under investigation is in Spain at an altitude of around 5000 to 6000 Feet above sea level. This may limit the amount of Rayleigh scattering, aerosol extinction, ozone absorption, water vapor absorption, and permanent gas absorption.(being at a higher elevation) However this may be of a negligible amount. In any case Thanks for your comment.

Russ

 

[CountZero]

Well, the first 3 will not be effected. i.e. Rayleigh scattering, aerosol extinction, and ozone absorption since they are mostly effected above the 6000 ft elevation you described. However, because of reduced pressure at that higher elevation, you should see a decrease in the amount of water vapor absorption and permanent gas absorption.

You know, if you check out the Journal "Solar Energy" they probably will have a numerical method of determining the solar flux in a place like spain up high in elevation.

To figure out the 5 major factors which reduce solar flux, you must integrate over a certain wavelength bandwidth. I forget what it is, but thermal radiation goes from mid IR to low blue.

Check out "A HYBRID MODEL FOR ESTIMATING GLOBAL SOLAR RADIATION" by Lang, K., el. Al. They have a method for determining the 5 types of radiation damping processes. It's very regionalized (i.e. Japan), but Japan has many mountains and is near the ocean just like spain is.

 

[Tunalover]

Russ0448-
I would ignore the participatory effects of the atmosphere; worst case is with no effects anyway.
As IRStuff suggested, using 1120W/m^2 is good if you wish to assume the solar radiation of a worst-case desert climate. If your object is heavy, then you should take a time average of the solar "hump" in MIL-STD-810F, Method 505.4 using the heating and cooling time constants of the object. Otherwise you will be overstating the solar load. If you are cooling at an altitude, you need to derate the convection coefficients for the thinner air.

I can send you a paper and spreadsheet model for a similar analysis but you would first need to sign a non-disclosure agreement; I wrote the paper for a publication but it hasn't been published yet.
Tunalover