Hybrid Car – More Fun with Less Gas

An Alberta bungalow - Page 5

register ::  Login Password  :: Lost Password?
Posted by wmbjkREMOVE on September 13, 2007, 2:22 pm
 
On Wed, 12 Sep 2007 18:45:50 -0800, Anthony Matonak


They too have everything they "require", except for water pressure, a
toaster.... hey, wait a minute! Aren't all the things the Ghinius
doesn't require, the very same items he couldn't power with his 1kWh
per day even if his life depended on it? What a coinkydink!

BTW, George is a propheshinel riter, not to mention editor,
screenwriter, pundit, poet, welder, hydraulic expert, power
consultant, teacher, swordsman, geostationary polar orbit wiz, baby
anticipater, and exultant deezyner of magic mass. So that apparent
grammar slip you mentioned couldn't possibly have been his fault. His
computer is surely infected with the dreaded Nitwit-C virus, which
targets incipient messertations and then utilizes artificial
intelligence to subtly change prose in order to make the author appear
to be a slow-witted 5th grader. Don't be fooled, George's talent is
undeniable! http://www.citlink.net/~wmbjk/tbfduwisdumb.htm

Wayne

Posted by nicksanspam on September 13, 2007, 10:05 am
 

With constant weak sun at 48KBtu/h/480ft^2 inside the sunspace, 64 ft^2
of 140 F bare solar collectors or Big Fins would absorb 6400 Btu/h and
lose about (140-T)64x1.5 Btu/h, so we'd have something like this:

  42.2K-6.4K = 35.8K Btu/h                                            
        ---                                              1/278        
   |---|-->|---------------- T     -----------------------www--- T
        ---             |         | 14+35.8K/278 = 143F
                        |    -   ---
   1/480x0.58 = 1/278   |    -    -
                        |    -    |
14 F ---www-------------          -

and this:
                        
               1/278    T       1/800
     -----------www--------------www-----
    | 143 F             |                | 140
   ---                  w               ---
    -                   w 1/96           -   radiator
    |     6400 Btu/h    w                |
    -        ---        |                -
         |--|-->|-------|  
             ---        | 140
                       ---    
                        -  collector
                        |
                        -

which is equivalent to this:
                        
               1/278    T = 141   1/896
     -----------www----------------www-----
    | 143 F             I-->               | 140
   ---       (143-140)/(1/278+1/800)      ---
    -             = 589 Btu/h              -            
    |                                      |
    -                                      -
                              
so the radiator and collector would gain 6400+589 = 6989 Btu/h, 30% more
than the radiator alone. But sun isn't constant. It might be 50% diffuse
in December, so we might heat water whenever the house needs are met and
the collector output is at least 140 F in the first two phases above,
as well as the third. This can also work well with poor collectors, eg
copper pipes banged into grooves in thin brown aluminum coil stock,
since their heat losses to sunspace air become radiator gains.

Engineer Nathan Hurst will talk about his Mazda radiator experiments in
Australia at the Pennsylvania Renewable Energy Festival on 9/22-23/07,
http://www.paenergyfest.com .

Nick


Posted by nicksanspam on September 15, 2007, 4:10 pm
 
With 50% diffuse sun over 6 hours per day, we'd have 751Btu/ft^2/2 = 375
for 1.5 hours at 250 Btu/h-ft^2 and 375 for 4.5 at 83 Btu/h-ft^2, so
phase 1 in diffuse sun would become:

        26.2K Btu/h                                            
        ---                                              1/278        
   |---|-->|---------------- T     -----------------------www--- T
        ---             |         | 14+26.2K/278 = 108F
                        |    -   ---
   1/480x0.58 = 1/278   |    -    -
                        |    -    |
14 F ---www-------------          -
                          
With lots of airflow, we can charge the house and ceiling from 60 to 70 F
in -16.2ln((70-108)/(60-108)) = 3.8 hours and charge the ceiling from 70
to 80 in another -1.8ln((80-108)/(70-108)) = 0.5 hours, leaving phase 3
with 1.7 hours at an average 0.8x250x480x1.5/1.7 = 84.7K Btu/h, like this:

  84.7K-11.3K-5.8K = 67.7K Btu/h                                            
        ---                                              1/278        
   |---|-->|---------------- T     -----------------------www--- T
        ---             |         | 14+67.7K/278 = 258F
                        |    -   ---
                        |    -    -
       1/278            |    -    |
14 F ---www-------------          -

and this:
                        
               1/278    T       1/800
     -----------www--------------www-----
    | 258 F             |                | 140
   ---                  w               ---
    -                   w 1/96           -   radiator
    |    11.3K Btu/h    w                |
    -        ---        |                -
         |--|-->|-------|  
             ---        | 140
                       ---    
                        -  collector
                        |
                        -

which is equivalent to this:
                        
               1/278    T = 175   1/896
     -----------www----------------www-----
    | 258 F             I-->               | 140
   ---       (258-140)/(1/278+1/800)      ---
    -             = 24.2K Btu/h            -            
    |                                      |
    -                                      -

so the radiator and collector would gain (11.3K+24.2K)1.7h = 60K Btu/day,
doubling the 6989x4.2h = 29.4K gain in constant sun. We could refine this
with a simple simulation using an Energy Plus Alberta hourly weather data
file for a typical meteorological year.

Engineer Nathan Hurst (with his new PhD :-) will talk about his Mazda
radiator experiments in Australia at the Pennsylvania Renewable Energy
Festival on 9/22-23/07, http://www.paenergyfest.com .

Nick


Posted by nicksanspam on September 12, 2007, 6:00 pm
 An auto radiator in a vertical duct that returns air to the sunspace without
mixing with room air could heat water for an unpressurized tank on sunny days
and heat the house on cloudy days...

Nathan Hurst wrote: "It took me and my brother about 15 minutes to remove
one from a Mazda - just a bunch of plate screws and stud bolts then pop off
the hose, avoid the toxic green water and you're set." Does this take longer
in the dark? :-)

A radiator mechanic said '86-'96 Tauruses are easy, but Accords of any year
are easier, with 2 plastic posts that slide into holes beneath and 2 posts
that slide into holes in plates with 2 screws above. I just bought a '95
Mitsubishi 2.0 Eclipse radiator with its electric fans for $5. It looks
like an Accord. Butch from Reed's auto salvage near Lancaster, PA removed
it from a working car and guaranteed to replace it if it leaks. It's about
19"x28". The '83 and '87 Accords on his lot had metal vs plastic tanks which
might corrode galvanically, even with an ACI-100 0.5% sodium silicate
aluminum chemical corrosion inhibitor solutiton from D. W. Davies. (The LD50
for this solution is about 70 quarts, so the ICC would rate it safe for use
with a single wall heat exchanger, eg a $0 1"x300' 13-gallon pressurized
PE pipe coil in the 140 F tank to make hot water for showers.)

We might have something like this, conceptually, viewed in a fixed font:

                 up         2'                  up
     ----------------------------          -----------
    |         r          s|      |       r|a          |
    |        fa          d|      |       o|.d  room   |
    |         d 2155A    a|      |       o| .a  air   |
    |    <== ai room <== m| 2'   |       m|  .m ==>   | 2'
    |         a fan      p|      |       f|30 .p      |
    |        nt          e|      |       a|    .e     |
    |         o          r|  s   |       n|degr?.r    |
    |         r-----------|  u   |        |-----------| west
    |              4'?    |  n   |        |           |
    |                     |  s   |        |           |
    |                     |  p   | 20'    |           |
    |                     |  a   |        |           |
    |                     |  c   | s      |           |
    |                     |  e   | o      |           |
    |                     |      | u      |           |
   a|                     |s     | t      |           |
   d| room       sunspace |d     | h      |           |
   a|  air            air |a     |        |  adamper  |
   m| <==             ==> |m     |        |           |
   p|                     |p     |        |  sdamper  |
   e|                     |e     |        |           |
   r|                     |r     |        |           |
     ----------------------------          -----------
                                                12'
               room fan
     ----------------------------     Drawing not to scale.
   s|         r adamper  s|s     |        
   d|        fa    |     d|d     |
   a|         d    v     a|a     |
   m|    <== ai      <== m|m 4'  | 12' south
   p|         a   up     p|p     |
   e|        nt          e|e     |
   r|         o          r|r     |
     ---------r------------------
                 west

with 3 modes:

1. To heat the tank, pull sunspace air through the radiator with its fan
and push it back into the sunspace below. One-way lightweight plastic film
convection sdampers over vertical hardware cloth grates prevent reverse
thermosyphoning at night.

2. To heat the room, push air through the upper adamper and radiator with
a room fan and back into the room through a lower adamper near the bottom
of the vertical duct. The upper sdamper prevents room air from flowing
through the sunspace at night.

3. To do both, run both fans. It might be OK if the room fan (2470 cfm at
full speed with 90 watts) backdrafts the upper sdamper to prevent sunspace
airflow while it is running, although it seems better to avoid that.

Will the radiator fan draw room air in through the upper adamper and push
it out the lower adamper and heat the room in summertime? Lasko's $0 2155A
fan has 3 speeds and it's reversible, so we could run it backwards slowly
when the radiator fan runs to backdraft the adamper, or motorize it, but
it would be nice to avoid those complications.

If the radiator fan draws 1000 cfm through a 2'x4' sdamper orifice, one
chimney formula says 1000 cfm = 16.6x2'x4'Asqrt(HdT), so HdT = 57, eg
H = 7' and dT = 8.1 F, with dP = 7(1-(460+70)/(460+78.1)) = 0.105 psf,
or 0.63 lb for 6 ft^2, with an opening moment of 0.63 ft-lb for a hinge
at the top of a 3'-wide by 2'-tall rigid adamper.

Hem-Fir and Atlas Energy Shield double-foil-faced polyiso board weigh about
28 and 2 lb/ft^3, so a 2'x3' 2" foil-faced foamboard damper with a 2' flat
1.46 lb 2x3 along the top edge 2" away from the hinge would have a closing
moment of (2+0.75)/12x1.46lb = 0.33 ft-lb for the 2x3 + (6+1)/12x2lb = 1.17
for the foamboard, over twice the radiator fan's opening moment.

The room fan might push 2000 cfm with a 1/8" H20 static pressure (or more,
with the radiator fan helping), ie about 0.125/12x62.33 = 0.65 psf, or 3.9
pounds on a 6 ft^2 damper, with an opening moment of 3.9 ft-lb. Nobody's
measured a 2155A fan curve, but we might use Q = 60sqrt(2x32.174xAFt/rho)
cfm, where A is the air outlet area in ft^2, Ft is thrust in pounds, and
rho is the air density, about 0.075 lb/ft^3. We might attach the fan to
a 2' cardboard cube with a diagonal damper, hinged at the top, and hang it
from 2 long ropes and estimate thrust by the weight of the fan and the box
and its sideways deflection when the fan runs, and measure the pressure
difference with a $0 Magnehelic gauge. Or estimate Q with a pressure and
a sharp-edged orifice area, as in the chimney calc above.

For example, if the fan and box weigh 20 lb and the outlet area A = 2 ft^2
and the ceiling-outlet center distance is 7' (84") and we measure a 1.5"
horizontal deflection at the outlet center, Ft = 1.5x20/84 = 0.36 pounds,
and Q = 1757sqrt(2ft^2x0.36) = 1485 cfm.

I'll be talking about this at the PA Renewable Energy Festival on 9/22/07,
http://paenergyfest.com .

Nick


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