Hybrid Car – More Fun with Less Gas

Re: Air to Water heat transfer

register ::  Login Password  :: Lost Password?
Posted by nicksanspam on October 30, 2004, 2:42 pm
 
<richard> wrote:


Plates and pipes sound expensive. Used auto radiators are nice, but
their electric fans are inefficient and require a 12 V supply.
I'd use 120 or 240V fans.


There may be easier ways to heat a pool. Paint it black. Then add
a clear cover, when not in use. Then paint the skirt black, and
flood it whenever it reaches 120 F. Or use two layers of
poly film on a flat platform.


Space heating with sunspaces or thermosyphoning air heaters can be.
Water heating can pay off year-round.

daestrom@NO_SPAM_HEREtwcny.rr.com writes:


About $ and 5 Btu/h-F per linear foot.


A 35 C (95 F) attic might give (95-75)5x16 = 1600 Btu/h to 16' of fin
tube with 80 F pool water inside.


A $5 used auto radiator or a $50 SHW2347 MagicAire heat exchanger might
collect (95-75)800 = 16K Btu/h for 80 F water or (95-55)800 = 32K Btu/h
for 55 F well water. Being more compact than fin-tube, it's easier to
freeze-protect with a pump at night, and collecting condensation is easier.

Another alternative: collect heat in a 4' wide 10 cent/ft^2 drain-down
poly film duct on a horizontal platform (eg 10 cent/ft^2 welded-wire
fence or PV panels under a reflective north wall) under the ridge.


With A ft^2 of dark opaque R1 roof in full sun (250 Btu/h-ft^2) on a 70 F
still day and another A ft^2 on the shady side, collecting heat from T (F)
attic air into 80 F water with an 800 Btu/h-F heat exchanger, we might have
something like this, viewed in a fixed font like Courier:

      south roof         north roof
                     T
    0.5/A  X   1/A   |   1/A  0.5/A
70---www-------www---*---www---www---70
     ---   |         |
|---|-->|--          w
     ---             w 1/800   X is the outside surface of the south roof
    250A             w
    Btu/h            |
                    --- 80
                    ---
                     |
                     -

If A = 32'x20' = 640 ft^2, we have

                     T
    0.5/640    1/640 |  1/640 0.5/640
70---www-------www---*---www---www---70
     ---   |         |
|---|-->|--          w
     ---             w 1/800
    160K             w
    Btu/h            |
                    --- 80
                    ---
                     |
                     -

which simplifies to this

                     T
    0.5/640    1/640 |  1/427        1/640+0.5/640 = 1/427
70---www-------www---*---www---70
     ---   |         |
|---|-->|--          w
     ---             w 1/800
    160K             w
    Btu/h            |
                    --- 80
                    ---
                     |
                     -
and this

                     T
    0.5/640   1/640  |  1/427
   --www-------www---*---www---70
  |                  |
 --- 195             w                 70+160Kx0.5/640 = 195 F
 ---                 w 1/800
  |                  w
  -                  |
                    --- 80
                    ---
                     |
                     -
and this
                     T
          1/427      |  1/427
   --------www-------*---www---70
  |       ----->     Y
 --- 195    I        w                 0.5/640+1/640 = 1/427
 ---                 w 1/800
  |                  w                 Disconnecting the heat exchanger
  -                  |                 at Y, I = (195-70)/(1/427+1/427)
                    --- 80                     = 26.7K Btu/h.
                    ---
                     |
                     -
and this

          1/853
   --------www-------            
  |                  |
 --- 132.5           w                 70+26.7K/427 = 132.5
 ---                 w 1/800
  |                  w                 So (132.5-80)(1/853+1/800)
  -                  |                 = 21.7K Btu/h (6.35 kW) flows
                    --- 80             into water, if I did that right.
                    ---
                     |
                     -

Insulating the north roof and replacing the south roof with transparent
polycarbonate would raise the (14%) efficiency, especially in wintertime.

Nick


Posted by Toby Anderson on November 1, 2004, 5:32 pm
 
nicksanspam@ece.villanova.edu wrote

MOst of the sun's heat is carried on the Infrared wavelength which
only penetrates about 3 feet into the water. So, if your pool is
deeper than 2 feet, painting the bottom black won't help.

Toby

Posted by nicksanspam on November 1, 2004, 7:07 pm
 

It appears you have erred again, my good man.

HC Bryant and Ian Colbeck's paper, "A Solar Pond for London,"
(Solar Energy, v. 19, p 321-322), says the fraction of incident
energy remaining after passing through x meters of water (x varies
from 1 cm to 10 m) is 0.73-0.08ln(x), eg 0.73 after 1 meter.

Nick


Posted by daestrom on November 1, 2004, 10:47 pm
 

Actually, water is pretty transparent from 400 to 700 nm wavelengths.  And
the sun radiates about 50% of its total energy in this energy band.  So
quite a bit of the sun's energy penetrates pretty well.

http://hyperphysics.phy-astr.gsu.edu/hbase/chemical/watabs.html

But, who *really* wants a black pool to swim in??

daestrom



Posted by Toby Anderson on November 2, 2004, 7:57 pm
 
Hi Daestrom,

Infrared light, the light I was talking about, has wavelengths ranging
from 1 mm - 750 nm, which is  greater than 400-700nm you referenced.

I got my information, partically from:

The following article 'LIGHT PENETRATION IN WATER'
(http://www.utoronto.ca/env/jah/lim/lim02f99.htm ) says this:

"The absorption of water (in %) is very high in the infrared portion
(long wavelengths) and results in rapid heating of water by incident
light. Approximately 53% of total light energy is transformed into
heat in the first meter of water."

Consider also:
http://www.lander.edu/rsfox/415lightLec.html

-------------------------------------------
color  penetration (m)
UV 70-80
blue >100
green 80
yellow 70
orange 17
red 4
IR <1
-----------------------------------------

The article goes on to give the Lambert's law formula:

-----------------------------------------
"• Light attenuation with depth increases exponentially according to
Lambert's Law:
Iz = Ioe ^(-nz)
ln Io - ln Iz = nz
where n is the extinction coefficient and Io is the irradiance at the
surface and Iz is irradiance at a particular depth z. usually
taken at 1 m, i.e. meters intervals below Io and n is the extinction
coefficient.
...
• The most common way of expressing the transmission or absorption of
light in water was developed by Birge:
% Transmission - 100(Iz/I0) = 100e ^(-n)
% Absorption - 100*(Io-Iz)/Io = 100(1-e ^(-n))"
-----------------------------------------

The 'extinction coeficient' varies according to wavelength. Here is a
table of the Infrared wavelengths, from the same website mentioned
above:

-----------------------------------------
  Table 41 Extinction coefficients for light in the visible and near
visible spectrum (Hutchinson, 1957).

Wave-   Extinction
length  Coefficient
(nm)   N
------------
820 2.42
800 2.24
780 2.31
760 2.45
740 2.16
720 (IR)1.04
700 .598
-----------------------------------------

Here are my calculations for the amount of IR energy absorbed in 1
meter of water, for 3 different extinction co-efficients.

%absorption (at n=1.04) = 100 * (1 – e ^(-1.04*1)) = 65%
%absorption (at n=2.4) = 100 * (1 – e ^(-2.4*1)) = 91%
%absorption (at n=2) = 100 * (1 – e ^(-2*1)) = 86%

Here is a table that I made for Infrared, Red, and Green light at
various depths:

Table BB of how much energy from the light is absorbed into the water
for Infrared (this wavelenth causes the most heat) and ‘Red' (also
called scarlet) wavelengths.
Column 1 = light color
Column 2 = 3in (0.0762m)of water    
Column 3 = 13ft (4m) of water    
Column 4 = 1 meter of water    
Column 5 = 51 feet
Infrared    14%    99.97%    86%    
Red        3.4%    84%        
Green            3.9%        40%

Toby

This Thread
Bookmark this thread:
 
 
 
 
 
 
  •  
  • Subject
  • Author
  • Date
please rate this thread