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re: poor man's co-generator
10 jan 2001
georges   wrote:

>i'm going to install a 1.5 kw 24vdc gasoline powered generator/charger...

it might also make 6 kw (20k btu/h) of heat. not water cooled, i spose.

>i've been thinking of making a heat exchanger for the exhaust gas...

sounds like a good idea. would this be 50% of the heat output (vs 25% for
a water-cooled engine?) intelligen's water-cooled 11 hp diesel had a 93%
cogen efficiency. some of that came from a plate-type exhaust-gas-to-water
heat exchanger about the size of a softball. they ran the cooled (150 f)
exhaust through an empty 55 gallon drum to remove particulates. they might
have put some water in the drum... you could multiply displacement by rpm
to find the exhaust cfm (50?), counting cycles and cylinders.

> the simplest implementation, i was going to exhaust the generator
>through an insulated bucket of water...

how will you keep it full of water without much back pressure? maybe float
the bubbler on top? coupling to the vibrating engine (a stainless steel
natural gas supply flex hose?) and removing the exhaust gas from the bucket
without leaks is also a challenge, especially inside a house.

>and put a coil in the bucket. the coil would provide heat transfer from
>the warm water in the bucket to the solar thermal circuit...

another heat exchanger lowers efficiency, and it has to be large, for a
large heat transfer rate. you might use a smaller coil with a lower heat
transfer rate and an insulated 55 gallon drum instead of a bucket(?) or
circulate the exhaust water itself and replace it often. drill a small
hole in the bucket and let a float valve keep it full?

>i could get more complicated and run the exhaust through a separate coil,
>but i like the idea of a bubbler as a noise abatement method
pollution abatement too...

>nick... how would you do the heat transfer calculation here?

not easily. the chapter on boiling in the 1998 2nd edition of schaum's
outline on heat transfer by donald pitts and leighton sissom (a very 
nice book for $14.95) says
   as in simple convection, a heat transfer coefficient h is used to
   relate the heat flux to the temperature differential between the
   heating surface [they are thinking about a hot plate or a hot wire,
   which is less complicated and maybe less efficient than a bubble]
   and the saturated liquid.

                             q = ha(ts-tsat)             (9.1)

   however, since phase change processes involve changes in density,
   viscosity, specific heat, and thermal conductivity of the fluid, while
   the fluid's latent heat is either liberated (condensation) or absorbed
   (vaporization), the heat transfer coefficient for boiling and condensation
   is much more complicated than that for single-phase convective processes.
   because of this, most engineering calculations involving boiling and
   condensation are made from empirical correlations...

they distinguish 6 types of boiling which happen as the hot-surface-to-
boiling "excess temperature" difference increases, free convection (i, ~1 c),
individual bubbles (ii, <~6 c), bubble column (iii, <~30 c), unstable film 
(iv, <~100 c "bubbles form so rapidly that they blanket the heating surface,
preventing fresh fluid from moving in to take their place. the increased 
resistance of this film reduces the heat flux, and the heat transfer
decreases with increasing temperature differential. because the film
intermittently collapses and reappears, this regime is very unstable."),
stable film (v, <~300 c), and "radiation affects film" (vi, >~300 c.)

     | free surface|    nucleate boiling   |     film boiling...
     | evaporation |                       |

     | i           ii          iii         iv         v        vi
     | free        individual  bubble      unstable   stable   radiation
     | convection  bubbles     columns     film       film     affects
     |                                                        .film
10^6 |                                 .  .b                 .
     |                              .        *              .
10^5 |                            .            *           .
     |                           .                *      .
10^4 |                          .                     c
     |                         a
10^3 |                       .
     |                     .
10^2 |                   .
  ^  |                 .
  |  |               .
w/m^2|            .        fig 9.1 (use courier font)
     dt (c)-->    1           6            30         100     300

their fig 9.1 shows q/a as roughly 10^4 to 10^6 w/m^2 for bubble and stable
and unstable film boiling. the 1973 itt ref data for radio engineers book
discusses "a tube with a specially-designed anode immersed in a boiler...
when power is dissipated on the anode, the water boils and the steam is
conducted upward through an insulating pipe to a condenser. the condensate
is then gravity fed back to the boiler... the anode dissipation should not 
exceed 135 w/cm^2 [1350 kw/m^2] of external anode surface, because at this
point, often referred to as the "leidenfrost" or "calefaction" point, the
surface becomes completely covered with a sheath of vapor and the thermal
conductivity between the anode and the cooling liquid drops to 30 w/cm^2 
[300 kw/m^2], with resultant overheating of the anode." sounds like unstable
film boiling. maybe you don't have to worry much about that, since you are
injecting a stream of bubbles vs putting a hot surface into the water. 

this schaum's outline continues with an estimate of peak stable film heat
transfer (point b of fig 9.1):

    q/a max = 0.149rvhfg[s(rl-rv)ggc/rv^2]^1/4 = 1059 kw/m^2, where

            rv is the steam density, roughly 0.4 kg/m^3,
            hfg is water's enthalpy of vaporization, 2.27x10^6 j/kg,
	    s is the surface tension, 0.063 n/m for (say) 80 c water,
              (0.00528(1-0.0013t) = 0.00407 lb/ft for t = 176 f (80 c))
	    rl is the 80 c water density, 974 kg/m^3,
            g is gravitational acceleration, 9.81 m/s^2, and 
	    gc is 1 kg-m/n-s^2 (a unit fudge.)

they also estimate the minimum stable film heat transfer (point c):

    q/a min = 0.09rvfhfge^1/2f^1/4 = 13 kw/m^2, where

            e = [g(rl-rvf)/(rl+rvf)] = 9.8 and
	    f = [gcs/(g(rl-rvf)] = 6.6x10^-6,

and the excess temperature at which this occurs:

    dt = 0.127rvfhfg/kvfe^2/3f^1/2[uf/(g((rl-rv)]^1/3 = 44.9 c, where

            uf is the viscosity of 80 c water, 354.6 kg/m-s.

they continue...

    film boiling (regimes iv, v, and vi)

    horizontal tube. from a study of conduction through the film... l.a.
    bromley [chem. eng. prog. 46: 221 (1950)] proposed that the boiling
    heat transfer coefficient in these regimes be... h = hc(hc/h)^1/3+hr,
    where hc = 0.62[kvf^3rvf(rl-rvf)g(hfg+0.4cpvfdt/(duvfdt)]^0.25 and
    hr = se(ts^4-tsat^4)/(ts-tsat) and h = hc+0.75hr (~5%) if hr/hc < 1.

if the emissivity e = 1, with stefan-boltzman constant s = 5.67x10^-8
w/m^2-k^4 and ts = 755 k (900 f) and tsat = 373 k, hr = 45.4 w/m^2-k.

d is the outside tube diameter, say 1/2" (0.0134 m) copper tubing.
the vapor properties at mean film temp tf = (ts+tsat)/2 = 564 k are
kvf = 0.0379 w/m-k and rvf = 0.4 kg/m^3 and uvf = 18.84x10^-6 kg/m-s
and cpvf = 1997 j/kg-k. dt = 755-373 = 382 k makes hc = 169 w/m^2-k. 

so hr/hc < 1 and h = 169+0.75(45.4) = 203.3, inside the cube root, and
h = 169(169/203.3)^1/3+45.4 = 204 w/m^2-k and q/a = 204x382 = 78 kw/m^2.

for nucleate boiling (types ii and iii), they give rhosenow's 1952 formula
q/a = ulhfgij, where ulhfg = 804.7x10^6, i = sqrt(g(rl-rv)/(gcs)) = 389.2,
and j = (cl(ts-tsat)/(hfgprcsf))^3, ie the specific heat of water,
cl = 4196.4j/kg-k, times the excess temperature (dt = ts-tsat = 10 c?) ie
the hot surface temp minus the boiling temp, divided by that ol' enthalpy
of vaporization again (2.27x10^6j/kg) and the prandtl number of 80 c water,
pr = 2.22 (with exponent 1 vs 1.7) and the "surface fluid constant" csf,
"a function of surface roughness (number of nucleating sites) and the angle
of contact between the bubble and the heating surface, [which] ranges from
0.0025 to 0.015." csf = 0.01 makes j = 0.577 and q/a = 181 gw/m^2. wow...
somebody made a mistake. well, they also say heat tranfer increases with
the cube of the temperature difference for nucleate boiling.

so the itt book says 300-1350 kw/m^2, and these calcs show 13, 78, and
1060 kw/m^2. the average is 560 kw/m^2 (ignoring the 181 gw :-) so you
might be able to move 6 kw with 6/560 = 0.0107 m^2 or 0.115 ft^2 of hot
surface. a 1/16" bubble has 4pi(1/32/12)^2 = 0.00008522 ft^2 of surface,
so you might use 0.115/0.00008522 = 1352.29 bubbles, or 1353 to be sure.

i'd suggest you buy 25' of soft copper tubing with 3.3 ft^2 of surface and
solder a cap on the end and bend it into a 4 turn 2' od spiral and drill
a 1/8" hole every half-inch on the bottom and suspend it a foot below the
water line in a steel 55 gallon drum with a clamp-on top and 3 nested drum
liners to keep it from rusting, and let us know how it works. 

>i assume the back pressure effects on the motor are minimal ( ~1 psi
>for every 2 foot height of the bucket)

mssrs pitts and sissom also address that, saying a balance of forces for
a spherical vapor bubble of radius r in a liquid with surface tension s
and vapor pressure pv and liquid pressure pl has pir^2(pv-pl) = 2pirs,
so pv = pl+2s/r, ie smaller bubbles have higher pressures; s = 0.00407
lb/ft for 80 c water, so 1/8" bubbles under 1' of water would have pv
= 64+0.00407/(1/16/12) = 64.78 psf or 0.450 psi, plus some pneumatic flow
resistance at something like 50cfm/(25x12x2xpi(1/8/2/12)^2) = 977 fpm
(about 10 mph.)


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