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a firmament greenhouse proposal - Page 4

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Posted by Toby Anderson on July 28, 2004, 6:50 pm
 
Dear Solar Heads,

Any fisherman knows, that you're more likely to get sunburned in your
boat than on dry land.

A lot of the Energy from the sunlight will be reflected --- first by
the glass/plastic, and then by the water. A good way of estimating how
much energy is reflected back into the atmosphere, off of the glass,
is by using the data from the 30 year Aveages for hundreds of US
cities, from  ‘Solar Radiation Manual' from the Government's
‘Renewable Resource Data Center' (RREL) at:

http://rredc.nrel.gov/solar/old_data/nsrdb/bluebook/state.html

Since the water in our ‘water roof' is horizontal (otherwise the water
will run off) we need to look at the Horizontal portion of the Sun's
Energy.

 For Las Vegas, Nevada:

The Horizontal Global "AVERAGE INCIDENT SOLAR RADIATION
(Btu/sq.ft./day) in July, is: 2490 Btu/sq.ft./day.

While the:

The Horizontal unshaded "AVERAGE TRANSMITTED SOLAR RADIATION
(Btu/sq.ft./day) FOR ---DOUBLE GLAZING----," is: 1830 Btu/sq.ft./day.

This later number, 1830, is for a window in the roof of your house,
while the former number, 2490, is just for the ambient outside air.

This means that 2490 – 1830 =  660 Btu/sq.ft./day. was reflected, off
of the glass, back into the atmosphere. That is: 660/2490, i.e.  26.5%

What is happening here, is that about 25% of the 2490 Btu/sqft/day is
reflected off the glass (leaving 1830), and then about another 25%
percent of this, i.e. 0.25*1830= 460, is reflected off of the water,
leaving 1830-460 = 1370. Thus, about only 50% (1370/2490) of the sun's
light actually enters the water.

Now, this light is effective for about 13 to 15 hours each day, and
hence, the 1370 Btu/sqft/day translates to about 100 Btu/sqft/hour

From Lambert's law, my calculations above, show, that 37% of these
100Btu/hr are absorbed into the water,

Eabs_h20_sun = 0.37 * 100 = 37 Btu/sqft/hr

Eg_sun

According to Lambert's law, as just shown, 37% of the energy from the
sun's light entering the water, i.e. 100 Btu/sqft/hr, is absorbed by
the water. Thus, 63% will enter the greenhouse, i.e.

Eg_sun = 0.63 * 100 = 63 Btu/hr

That means, by having a water-roof, only about 1/3 (63/200= 0.32) of
the sun's energy actually makes it into the greenhouse itself!!!

Toby

Posted by nicksanspam on July 28, 2004, 8:09 pm
 


No. The 660 also includes radiation absorbed by the glass.


Why on earth would you say that?


Are the following "your calculations above"?


No. e^(-0.01524) = + 0.985.

As for the rest, perhaps we have "garbage in--garbage out."

Nick


Posted by Toby Anderson on July 30, 2004, 12:04 am
 nicksanspam@ece.villanova.edu wrote in message

How much?

Please provide some evidence for your statement. I suggest that 'most'
of that 660 Btu is reflected. Why so? Well, here is my evidence:

The following article, on the properties of glass from:

http://www.squ1.com/index.php?http://www.squ1.com/concepts/glass-properties.html

 suggests that glass DOES NOT absorb the sun's radiation because it is
shortwave. It says:

"As shown above, most of the energy from the Sun is short-wave, which
passes straight through the glass."

Here's a more complete quote:

"To illustrate this, imagine you happened to be carrying around a
large pane of glass while wandering around in the desert (bear with me
on this one). In the middle of the day, if you held the glass between
you and the searingly hot sun, you would find that it doesn't make a
very effective shade - in fact it makes almost no difference at all.
Then at night, you build a roaring campfire to protect you from the
rapidly falling temperature. As you are enjoying the warmth of the
radiated heat, you hold the pane of glass between you and the flames -
does it make a difference ? As shown above, most of the energy from
the Sun is short-wave, which passes straight through the glass. "

Toby wrote:

Nick replied:

Snell's law. Are you familiar with it?

Nick replied:

I greatly respect your knowledge, but please stop being so mean, or I
will not respond to your postings.

Toby

Posted by nicksanspam on July 30, 2004, 10:15 am
 

Maybe 10%, after subtracting a 16% (4% per surface) Fresnel reflection.


This one? :-)


Gustav Robert Kirchoff (1824-1887) said all radiation hitting a surface
must be transmitted, absorbed, or reflected. Radiation obeyed. T+A+R=1.


Sure. Why on earth would you guess 25% for the water? Recall that water
and glass have about the same index, so a water-glass surface will have
little reflection.


You might consider not posting nonsense :-)

Nick


Posted by Toby Anderson on July 30, 2004, 9:14 pm
 Nick,

http://rredc.nrel.gov/solar/old_data/nsrdb/bluebook/state.html
-------------------------------------------------------------------------------The
Horizontal Global "AVERAGE INCIDENT SOLAR RADIATION
(Btu/sq.ft./day) in July, is: 2490 Btu/sq.ft./day.

The Horizontal unshaded "AVERAGE TRANSMITTED SOLAR RADIATION
(Btu/sq.ft./day) FOR ---DOUBLE GLAZING----," is: 1830 Btu/sq.ft./day.
------------------------------------------------------------------------------

Toby wrote:

In regards to how much energy the glass absorbs, Nick wrote:
"Maybe 10%, after subtracting a 16% (4% per surface) Fresnel
reflection."

WHere did you get these numbers from?

I don't think 10% is absorbed by the glass because if it did, the
glass
would be HOT, and the article I quoted in a previous post, said glass
does not get hot from the sun's radiation.

Even if you are correct, it doesn't really matter as far as my
firmament
 greenhouse is concerned. The point is, that we both agree thta 26.5%
 of the sun's energy does not make it through the  glass... This makes
sense, as the above website has this definition about the above data:

"Transmitted solar radiation.  
          The solar radiation transmitted into a living space"

THe sun's energy will get reflected off of 3 surfaces:
-  the glass on top
- the water's surface
- the glass on the bottom.

The sun's energy will get absorbed by the same 3 media. Let's look
at all these factors....


Hence, we both agree that 73.5% of the  energy transmitted through
the glass.  
------ 73.5% through the glass on top

Then, when this light hits the water, another 16% (Nick's View) or
26.5% (Toby's View) is reflected again. Leaving 84% (Nick) or
73.5% (Toby) to enter the water.
------ 73.5% into the water (Toby)
------ 84% into the water (Nick)

In my earlier post, I  estimated, using Lambert's Law., 37% of this
 is absorbed into the water , hence, 63% of this leaves the water.
------63% through the water.

The sun, then, hits the glass on the bottom.. where it is reflected
 again: (16%-Nick, 2 6.5% - Toby), but Nick has 10% of this being
absorbed by the glass.
----- 73.5%  through the glass on the bottom into the greenhouse.

Now, taking all those percentages together, we get the following
energy entering the greenhouse:

Toby's theory: 0.735 * 0.735 *  0.63 * 0.735 * Esun = 0.25 * Esun
----- 0.25 * 2490 = 623 BTU/sqft/day
Nick's theory:   0.735 * 0.840 *  0.63 * 0.735 * Esun = 0.285 * Esun
----- 0.285 * 2490 = 710 BTU/sqft/day

Toby and Nick are not that far off.

Over athe 14 hours the sun shines each day, there is only about:
------ 623/14 = 44 BTU/sqft/hr
of energy from the sunlight entering the firmament greenhouse

Compare this to a standard greenhouse with no water, just
glass on the roof, in which 73.5% of the Sun's energy enters it:

0.735 * 2490 / 14 = 131 Btu/sqft/hr

About 3 times (131/44=2.97) the amount of energy enters the
standard greenhouse as opposed to the firmament, water-
roofed greenhouse.

Toby

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