Hybrid Car – More Fun with Less Gas

heat transfer plates - Page 2

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Posted by John on December 10, 2003, 2:58 am
 
Check Ebay using  "bead roller".  Yeah, I guess I was off on the price.
Going on EBay for about 115.00.  You can look at the pics and get ideas for
home made axles.  Think of Grandma's old washing machine.  Two parallel
shafts with locking collars to popsitioncollar



Posted by John on December 10, 2003, 1:18 pm
 
You can get a design idea at EBay.  Use "Bead Roller" for keyword search.
You will see some in the $0 to $20 range. Harbor Freight sold a cheapie
but as you noted, it is discontinued.  The tool doesn't look difficult to
fabricate.  A pair of shafts, few collar locks and gear set, make a handle
and off you go.

John




Posted by <farview on December 10, 2003, 4:27 pm
 Duane,

From looking at the pictures, building a bead roller for 8 to 16 inch wide
for light gauge aluminum will be feasible once I get my slab floor done and
can set up my welder and drill press.  I'm trying to pin down a good source
for roll flashing or sheet.  Most roll flashing runs 0.010" to 0.012" for
$.50/sf.  This may be a bit too thin for effective heat transfer.  I'd like
to use 0.024" but having trouble locating it.  A local metal supplier can
get me 0.032" sheets for $/sf (cheaper if I buy a lot).

Considerations for home-brew plates are the heat transfer rate for different
thicknesses (Nick - what do you think?) and expansion noises (100% silicon
sealant/adhesive might cure that i guess).
Chuck



Posted by Nick Pine on December 11, 2003, 10:04 am
  

I seem to recall we can order 2'x50'x0.027" Al coils for $.50/sf
from Home Depot, painted brown on one side and white on the other.
What's the cost of 10' of 1/2" copper pipe? And a jig to nicely
bend the coil around one or two pipes in a U-shape and then unbend
the wings to make a solar collector or heat distributor? The two-pipe
coil might look like this, before unbending, viewed in a fixed font:

coilcoileecoilcoil   ee are the coil edges
lpipe        pipec
ioclioclioclioclio  

After unbending, one pipe's worth might look like this:

|      L       |                  L = 12/2-Pi0.625/2-0.25 = 4.77", ie 0.4'?
---------------   ---------------
         0.25"?| | bolts?              
          --- .   .
        0.625".   .
          ---   .


If the distance from the pipe to the edge of the fin is a bit too large.


Al has a thermal conductivity k = 128 Btu/h-ft-F. If the slow-moving airfilm
conductance from both sides of the plate is 1.5 Btu/h-F-ft^2, we can model
1 linear foot of L-foot-wide half-fin like this:

              70 F
              |
              w      
              w Rc = 1/(2x1.5L) = 0.333/L
              w
              |
pipe ---www---
        Rf = L/(2Ak) = L/(2x0.027/12x128) = 1.74L for half of the half-fin,

So the resistance from an L = 0.5' half-fin to air is 1.74/2+0.666
= 1.54 F-h/Btu. If the pipe is 100 F, 2(100-70)/1.54 = 39 Btu/h
flows out of both sides of the fin.

Here's a more accurate model:

              70 F      70 F
              |         |
              w         w
              w Rc      w Rc = 1/(2x1.5L/2) = 0.666/L = 1.33
              w         w
              |         |
pipe ---www---*---www---
        Rf        Rf = L/(3Ak) = L/(3x0.027/12x128) = 1.16L = 0.58,

which simplifies to

              70 F
              |  
              w  
              w 1.33
              w  
              |  
pipe ---www---*---www--- 70, and
        0.58      1.91

pipe ---www--- 70, so 2(100-70)/1.37 = 44 Btu/h would flow out of this pipe.
        1.37  

Adding another section:

              70 F      70 F      70 F
              |         |         |
              w         w         w
              w Rc      w Rc      w Rc = 1/(2x1.5L/3) = 1/L = 2
              w         w         w
              |         |         |
pipe ---www---*---www---*---www---
        Rf        Rf        Rf = L/(4Ak) = L/(4x0.027/12x128) = 0.868L = 0.434,
...

pipe ---www--- 70, so 2(100-70)/1.30 = 46 Btu/h would flow out of this pipe.
        1.30  

Continuing, we might have 39, 44, 46, 47, 47.5, 47.75...

Pages 26 and 27 of the 1998 Schaum's Outline on Heat Transfer say a "very
long" (case 1) L' rectangular half-fin in Ti air with base temp Tb would
have q = kAnThetab Btu/h flowing through it, where k = 128 and A = 1'x0.027/12
= 0.00225 and n = sqrt(2h/(kt)) = sqrt(2x1.5/(128x0.027/12) = 3.23 and Thetab
= Tb - Ti = 30, so q = 128x0.00225x3.23x30 = 27.9, and 2q = 56 Btu/h-F flows
out of the pipe...

A "finite length" (case 2) fin with q = kAnThetabtanh(nL) = 27.9tanh(1.62)
= 25.8 would have 2q = 51.6 Btu/h flowing out of the pipe.  

A $ foot of fin-tube pipe would move about (100-70)5 = 150 Btu/h, or more,
with a column of hot air above it, increasing its airspeed.

Nick


Posted by Nick Pine on December 11, 2003, 3:31 pm
  

Lowes sells a 2'x50'x0.018" brown and white coil (sku 17376, Napco
(704) 473-7868) for $3.78. Home Depot's (sku 125467, Amerimax
(717) 299 3711 p/n 69124) is $3.50. It meets AAMA standard 1402-86.
Will this stay crimped hard around a 1/2" copper pipe, with no bolts?


$.68 for type N from Lowes, or $.98 at Home Depot.

Pages 26 and 27 of the 1998 Schaum's Outline on Heat Transfer say a "very
long" (case 1) rectangular half-fin in Ti air with base temp Tb would move
q = kAnThetab Btu/h, where k = 128 Btu/h-ft-F and A = 1'x0.018/12 = 0.0015
ft^2 and n = sqrt(2h/(kt))= sqrt(2x1.5/(128x0.018/12) = 3.95 and Thetab
= Tb-Ti = 100-70 = 30 F, so q = 128x0.0015x3.95x30 = 22.8, and 2q = 45.5
Btu/h-F would flow from pipe to air.

A "finite length" (case 2) L = 0.4' half-fin with q = kAnThetabtanh(nL)
= 22.8tanh(1.58) = 20.9 would have 2q = 41.9 Btu/h flowing out of the pipe.  
Tanh(x) = (e^x-e^-x)/(e^x+e^-x) is the hyberbolic tangent "tanch x."

A much thicker fin might move (100-70)2x0.4x2x1.5 = 72 Btu/h out of the pipe.

A $ foot of fin-tube pipe would move about (100-70)5 = 150 Btu/h, or maybe
more, with a column of hot air above it to increase its airspeed. Fin-tube
may be cheaper for heat distribution, but inefficient as a solar collector.

Nick


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