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re: yoghurt maker
19 feb 1999
jean p nance  wrote:
>there are lots of ways to keep yogurt at the requiared temperature
>which is somewhere between 100 and 115 degrees f...

indeed. a friend makes frugal yogurt. his recipe (as i understand it) is:
heat a gallon of milk to just under boiling for about 15 minutes, add some
powdered milk and gelatin to make it thicker, cool it to 115 f, add some
active yogurt as a starter, and maintain it at 110 f +/- 2 f for 6 hours.

the milk might cool on the stove, then be poured into containers in
a machine with a small electric heater and thermostat, eg an aquarium
with a heater. this process might be improved...

how fast will the milk cool to 115 f on the stove, if initially at
200 f in an 8" diameter x 8" tall shiny pot with a lid in a 70 f room?
the pot has about 2 ft^2 of surface, and the thermal conductance of
the still air film around it might be about 1.5 btu/h-f, and water
has a heat capacity of 1 btu/f-lb or 8 btu/f-gallon, so the natural
time constant rc = 2ft^2/1.5btu/h-f-ft^2x8btu/f = 10.7 hours, and the
milk might cool until 115 = 70+(200-70)exp(-t/10.7h), ie t = 12.1 hours.

taking the lid off or blowing air on the pot or putting the milk in
smaller containers or putting the pot in cool water would make this happen
faster. i measured an r-value of 1/30 cooling some water in my kitchen
this way. putting the pot in a large tub of 60 f water would make
rc = 1/30/2x8 = 0.133 hours or 8 minutes, so 115 = 60+(200-60)exp(-t/8m)
when t = 7.5 minutes :-)

but why waste the heat? why not use the cooling water as thermal mass
in a passive yogurt machine, say a 10x10x6" tall (id) box with 3"
styrofoam walls and top made with 2 layers of 1.5" foam and some 4" 
deck screws and silicone caulk or a plastic garbage bag liner? adding
x pounds of 60 f water to the box will make an average temperature of
112 f = (60x+200(8))/(x+8), if x = 13.54 pounds, which will just about
fill up the box, with a volume of 0.347 ft^3.

the milk won't cool so fast this way, because as it cools, the water
warms significantly, and the temperature difference and heatflow
decrease until they both become 112 f. how long do we have to wait
until the milk becomes 115 f, so we can add the starter? here's an
electrical analog:

           s     r
  v1  ---- \----www----  v2       the switch s is closed at t=0.
     |        ------>  |   
     | c1 = 8    i     | c2 = 13.54
    ---               ---
    ---               ---
     |                 |
     _                 _

we want r small to speed this up, so let's put the hot milk right into
1 quart plastic yogurt containers (i tried this with boiling water, and
they didn't melt, when surrounded by air.) the containers are about
4.5 inches in diameter and 5.5" tall, with a surface of about 0.65 ft^2,
not counting the lid, so r might be about 0.65/4/30 = 0.0054 for 4 of
them, so rc1 = 8r = 0.043 hours or 2.6 minutes, so they would cool to
115 f in -2.6ln((115-60)/(200-60)) = 2.4 minuntes, if c2 were huge.

but it isn't, and every time 1 btu of heat flows into c2 it raises v2
by 1/c2 degrees f. we can write v2(t) = v2(0) + 1/c2 integral i(t) dt.
every time 1 btu flows out of c1 its temperature drops by 1/c1 degrees f,
so v1(t) = v1(0) - 1/c1 integral i(t) dt. and i(t) = (v1(t)-v2(t))/r,
and di(t)/dt = -(1/rc1+1/rc2)i(t). let's call 1/(1/rc1+1/rc2) tau.
it's about 1.6 minutes, in this case. 

then di(t)/i(t) = -dt/tau, so integrating, ln(i(t)) = -t/tau + a, if
i did that right, where ln is the natural log function. at time 0,
a = ln(i(0)) = ln(v1(0)-v2(0))/r). raising both sides of the equation
for ln(i(t)) to the power of e ("e-to-the-x-ing," the inverse of the
natural log on a $20 calculator), i(t) = (v1(0)-v2(0))/r exp(-t/tau).
so we have an initial heat flow that is eventually squashed to zero
by our old friendly decaying exponential function (quickly, with
this 1.6 minute version.)

integrating again to find out how long we have to wait to add the starter,

v1(t) = v1(0)-1/c1 integral i(t) dt = v1(0)-i(0)/c1 integral exp(-t/tau) dt, 
      = v1(0)-i(0)tau/c1(exp(-t/tau)-1) = v1(0)-c2/(c1+c2)(exp(-t/tau)-1)
      = v1(0)-c2/(c1+c2)(v1(0)-v2(0))+c2/(c1+c2)(v1(0)-v2(0))exp(-t/tau)
      = 200  -13.54/(8+13.54)(200-60)+13.54/(8+13.54)(200-60)exp(-t/1.6m).
      = 112 + (200-112)exp(-t/1.6m) (whew.) 

so, another familiar and simple formula, a final value (112 f, after
a fairly long time) plus an initial difference (200-112) eventually 
squashed to zero by the same quickly decaying exponential. we have
to wait -1.6ln(3/88) = 5.4 minutes before adding the starter. 

what will the temperature be inside the box after 6 hours? we need
to use the log mean area of the inside and outside of a box this small
with such thick insulation to calculate its thermal resistance. the
inside has area 200+2(20)6 = 440 in^2 with ln 6.09 and the outside
has area 256+2(32)12 = 1024 in^2 with ln 6.93, and the average ln
is 6.51, corresponding to an area of 671 in^2 or 4.66 ft^2. so
rc = r15/4.66ft^2(8+13.54) = 69.3 hours, and after 6 hours, the
inside temperature should be 70+(112-70)exp(-6/69.3) = 108.5 f.

if we initially heat the milk in a pot in a pan of water on a
woodstove, we don't need any electrical power at all for this
precise yogurtmaking...

a simulation-check:

10 c1=8
20 c2=13.54
30 r=.0054
40 tau=1/(1/(r*c1)+1/(r*c2))
50 v1=200
60 v2=60
70 dt=.00001
80 for t=0 to .09174 step dt
90 i=(v1-v2)/r*dt
100 v1=v1-i/c1
110 v2=v2+i/c2
120 v1h=112+(200-112)*exp(-(t+dt)/tau)
130 tf=t
140 next
160 print tf,v1,v1h

    9.173259e-02                114.9862      115.0009


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