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Thermosyphoning freeze protection - Page 3

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Posted by daestrom on July 14, 2003, 9:30 pm
 


So the riser piping is bare and the return is insulated?  Otherwise the
riser and return would both pick up the same amount of heat.  In fact with
return cooler than riser, it would pick up more heat and reduce the delta T
for natural circulation??

When the water is warmer than the house, this will radiate heat to the house
and only absorb heat from the house when the hot-water is cooling off?

daestrom



Posted by Nick Pine on July 15, 2003, 9:57 am
 


The circulation can happen in either direction, and thermally equal legs
seem to work more reliably than unequal ones, according to Steve Baer's
experience, with no significant metastability problems or delays. There's
always enough "noise" to start a flow going, electrically-speaking.
 

We might use a cold water tempering tank (eg a well pressure tank or
4"x10' of PVC pipe) or a heat tape with a thermostat.


Btu/h. Sigh.

Nick


Posted by andy on July 15, 2003, 6:18 pm
 
Yes, presently beating myself on the head with a hammer. I can't
beleive I have succumbed to this completely false lazy jargon.

I tend to stick to Watts in everyday life..but this is still no excuse
for stating misleading information. My apologies.

Posted by daestrom on July 15, 2003, 10:39 pm
 

While internal circulation within the riser and return may well happen, that
wouldn't create any flow through the header box.  To get circulation through
the header  you need to somehow keep the riser warmer than the return.  The
storage tank will do this for a while unless some cold supply water is
admitted to replace warm water drawn off by the homeowner.  Stripping
insulation from the riser and leaving insulation on the return could also
work if enough of the vertical run is in the warm house (and the house stays
warm ;-).  But warming the water halfway up the riser is not as effective
because it has less elevation to 'work' through.

A more accurate calculation for the density of water in this range is
((-8.5010e-5*T+4.6829e-3)*T+62.384) where T is in degrees F and the results
are in lbm/ft^3.  This shows that with warmer water (say 80 - 100 degF) the
change per degree is more so you can get more driving head for a given delta
T. (this is what I used in these calculation).

If the return goes down the dip tube inside the storage tank, say a distance
of 5 feet, and the top of the tank is appreciably warmer than the riser (say
100 degrees while the riser is 80 for an average dip-tube temp of 90), then
this little section of pipe will create a dp of about  0.539035 lbf/ft^2,
but in the wrong direction.  If the rest of the elevation above the tank is
a total of 16 feet, then the return leg will have to be about 76 degF, *just
to balance the dip tube*.  Given Nick's calculations, it might take another
3 degrees to develop the desired flow.  So the riser might have to be kept
about 7 degrees warmer than the return to keep up the flow. (or put another
way, the return would have to be 7 degrees cooler).  And that has to be over
most of its length!!!

This would be a case for putting the return into the bottom of the tank
through a fitting at the bottom of the tank.  This way the return will *not*
be warmed that last five feet and you won't lose that driving head.  With
the dip tube, even though it ends at the bottom of the tank, it is warmed
that last five feet or so and that hurts the overall natural circulation.
'Course, this all depends on how much you're going to need that driving head
:-)

The same calculation of the dip tube issue using 70 at the top of the tank
and 50 at the bottom shows a disaster.  The dp caused by an average dip tube
temperature of 60 and riser of 50 is 0.23657 lbf/ft^2.  Sound better right??
But, even with the return cooled down to 40.44 and the riser at 50, no
natural circulation would occur (16 feet of elevation with 50 and 40.44 degF
gives a delta P of only 0.2237 lbf/ft^2).  Thermosyphoning would try and
reverse, but your check-valve would stop all flow.

If the tank stratifies a lot more than 20 degrees, lets see, maybe 120 on
the top and 60 on the bottom??  The dp caused by a 5 ft dip tube with
average 90 degrees and riser of 60 degrees is 1.4581 lbf/ft^2.  And again,
with the riser at 60, no natural circulation can occur.  With the top of the
tank at 120, the most stratification you can have and just balance the dp
would be about 54 degrees from top to bottom (120 on top and 65 on bottom).

Like I said before, warn the homeowner *not* to use any water when the power
is out.  Cold supply will cool the bottom of the tank quickly and stall the
thermosyphon.

As you may guess, in a former career, I had a lot of experience with natural
circulation systems :-)  Sorry for rambling on, but when I get a problem
laid out in Excel, I can't help but tweak all the variables looking for what
makes a difference :-)

Turns out the dip tube is one of those things you wouldn't have thought of,
but has an impact.  Might be better off returning into a bottom fitting.
This would add another 5 feet or so of elevation as well.

How many 'heat-pipes' are you thinking about brazing onto the header?  As
far as conduction of heat out passed the header's insulation are they like a
hollow metal pipe?  What diameter & wall thickness?  Might be interesting to
see how much heat might be conducted away from the header at night through
the heat pipe's metal.

daestrom



Posted by Nick Pine on July 16, 2003, 9:24 am
 


Seems unlikely, without a return path through another pipe.


Then again...
 

I could imagine a little flow. If thermosyphoning can occur within
a single pipe, why not within the header pipe as well? But that seems
like a matter of hope, vs a more pro-active and predictable design.


Or vice-versa, assuming no check valve (as in the Aug 2003 HP system.)


Not exactly something to count on either.


This might work better with both pipes bare.


Agreed...


I agree that it doesn't seem like a good idea to depend on such warnings,
but a cold tank might be OK, if the supply and return pipes are warmer...

Another freezing scenario: it's cold outdoors, and the kids have exhausted
the solar hot water, leaving 38 F water in the tank bottom, and then the
power fails. If the house heating fails at the same time, we drain the pipes
and go somewhere else, eg Florida. But suppose the house is still warm, and
we are all sitting around the woodstove with candles... We might avoid
collector freezing, if the pipes inside the house have enough conductance
to warm house air, to a) supply the header heat loss to the outdoors and b)
warm the water that flows out of the tank to a thermosyphoning temperature.

Roughly speaking, suppose the pipes have a conductance of G Btu/h-F to
70 F house air and the pipe water is about 50 F and the header has 1 Btu/h-F
of conductance to -10 F outdoor air. Then the house air has to supply
(50-(-10))1 = 60 Btu/h to the header. Suppose the flow in the pipe loop
Q = 1000dT lb/h, which supplies 1000dT^2 Btu/h of heat to the header.
Then dT = sqrt(60/1000) = 0.245 F, which makes Q = 245 lb/h, which means
we have to heat 245 lb/h from 38 to 50 F. (70-50)G = 60+(50-38)245 = 3000
makes G = 150 Btu/h-F, eg 40' of fin tube pipe :-) Not too practical,
altho this might work with slightly cooler pipe water. Then again, we'd
also like it to work with a slightly cooler house.

Most of that heat (98%) is needed to warm the cold water that comes out
of the tank. Without that requirement, G = 3 Btu/h-F, eg 10' of bare pipe
and 0.5 gpm of flow. Is there a way to add some sort of bypass pipe between
the supply and return pipes to complete the loop at the bottom and avoid
having to heat the cold tank water? It might tend to short out the pumped
solar loop in normal times. Perhaps this could work with a normally-open
check valve that closes when the pump runs but allows thermosyphoning
through the bypass pipe with the pump off.

Nick


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