Posted by *Nick Pine* on August 16, 2003, 10:44 am

*>: Where I live near Phila, NREL says 1020 Btu/ft^2 of sun falls on level *

*>: ground on an average 56.4 F October day...*

*>Is that 1020 Btu/ft^2 a typo?*

No. That's the amount of sun on a typical day.

*>About 100 watts per square foot is the maximum possible,*

As I recall, 311 Btu/h-ft^2 is the max in the Sahara at noon.

We are more likely to see 250.

*>Also please check that publication to see if it gives BTU*

*>as a per hour number,*

I don't see that in NREL's Solar Radiation Data Manual for Buildings.

*>the BTU is a quantity, while sunshine is a rate of energy transfer,*

It is? :-) "Penelope dear, let's go bask in the rate of energy transfer..."

*>and is either watts, or BTU per hour.*

Or Btu per square foot per day.

Nick

Posted by *Nick Pine* on August 16, 2003, 1:17 pm

*>>: Where I live near Phila, NREL says 1020 Btu/ft^2 of sun falls on level *

*>>: ground on an average 56.4 F October day...*

*>>Is that 1020 Btu/ft^2 a typo?*

No. That's the amount of sun on an average October day. It's a useful way

to specify solar heat in calculations that use Ohm's law for heatflow over

24 hours. If a thing gathers 1020 Btu per day of heat which warms it to

a constant temp T and it loses heat through a thermal resistance R to an

ambient temp Ta over 24 hours, T = Ta + 1020/(24R).

For example, a pool with an R1 cover with 80% solar transmission

would gain 0.8x1020 = 816 Btu/ft^2 on an average October day and

lose 24h(T-56.4)1ft^2/R1, which makes T = 56.4+816/(24x1) = 90.4 F.

In October, where I live near Phila, with an average humidity ratio w = 0.007,

the vapor pressure of water in air Pa = 29.921/(0.62198/w+1) = 0.333 "Hg, so

at pool temp T (F), with no wind, we might lose Qc = 2(T-56.4) Btu/h-ft^2 by

convection. Bowen's equation says we'd lose Qe = 200(Pw-Pa) by evaporation,

where Pw = e^(17.863-9621/(460+T))... 24h(Qc+Qe)+1020 = 0 makes T = 59.3 F

for a well-stirred uncovered pool.

Nick

Posted by *Nick Pine* on August 16, 2003, 5:57 pm

*>:>: Where I live near Phila, NREL says 1020 Btu/ft^2 of sun falls on level *

*>:>: ground on an average 56.4 F October day...*

*>: *

*>:>Is that 1020 Btu/ft^2 a typo?*

*>:*

*>: No. That's the amount of sun on a typical day.*

*> Ok, then 1020 BTU/ft^2/day would have saved me from asking. *

Then again, you mighta read this:

*>:>: on an average 56.4 F October day...*

*>:>About 100 watts per square foot is the maximum possible,*

*>: *

*>: As I recall, 311 Btu/h-ft^2 is the max in the Sahara at noon.*

*>: We are more likely to see 250.*

*> Which would only be about 80 watts, but the*

*>hours per day in the Sahara is probably more like 8.*

But this is a peak rate. Perhaps you failed to read this:

*>: the max in the Sahara at noon.*

*>I don't know your latitude, but at greater than 30 degrees North,*

*>more sun falls on a vertical surface than a horizontal surface*

*>in November, December and January. *

NREL says 450, 360 and 190 Btu/ft^2 fall on a north wall on an average day

in Nov, Dec and Jan in 40 N Phila. East and west walls get 450, 370, and

430... 680, 530 and 620 fall on a horizontal surface. Barrow, AK (N 71)

walls and roofs receive 0 Btu/ft^2-day in Dec and Jan. You've stated your

point (whatever it was) imprecisely.

*>: "Penelope dear, let's go bask in the rate of energy transfer..." *

*>: *

*>:>and is either watts, or BTU per hour.*

*>: *

*>: Or Btu per square foot per day.*

*> Ok, but that seems like an odd way to say it,*

Not if you care about the amount of sun on an average day.

*>I think isolation is usually given in watts/meter^2.*

InsOlation is...

*> And cloud cover and day length vary a lot.*

That's why an average day number is useful.

Nick

Posted by *Joe Fischer* on August 17, 2003, 12:01 am

In alt.solar.photovoltaic

: NREL says 450, 360 and 190 Btu/ft^2 fall on a north wall

: on an average day in Nov, Dec and Jan in 40 N Phila.

I can't disagree, but the poor Jan number is due

to statistical cloud cover for that city, and may not

apply to all of the 40th parallel.

This seems more relevant to photovoltaics than

to outdoor pool heating.

I don't think statistical cloud cover should

carry a lot of weight in designing a system.

: East and west walls get 450, 370, and

: 430... 680, 530 and 620 fall on a horizontal surface. Barrow, AK (N 71)

: walls and roofs receive 0 Btu/ft^2-day in Dec and Jan. You've stated your

: point (whatever it was) imprecisely.

If you are going to bring up Alaska, obviously. :-)

It seems that people living where there is less

sun are the ones always asking about solar energy.

I was going by some graphs in a 1955 book

of the ASHAE on sun angle, I will look at it again

and see if they consider cloud cover.

Joe Fischer

--

3

>: Where I live near Phila, NREL says 1020 Btu/ft^2 of sun falls on level>: ground on an average 56.4 F October day...>Is that 1020 Btu/ft^2 a typo?