Heat exchanger design/usage is a whole science in and of itself. But here
are a couple of points:
1) The metal walls in heat exchangers are *not* the major resistance to heat
transfer. The metal is a good conductor (duh!), but the laminar layer of
air/water right next to the metal is not as good.
Ah, hence the wrinkles to provide turbulence...
2) With that in mind, to get the most heat transfer for a given sized heat
exchanger, you want to minimize the resistance/thickness of the two laminar
films (remember, one on each side of the metal wall). The easiest thing to
do this is increase the local velocity of the air/water. But as you
suspect, there is a point where a further increase in flow doesn't reduce
the laminar layer much more at all.
But, I think I can do a "Here's how good I can get without spending more
money than this will save" and get _somewhere_. If I do it right, I
should be able to add heat with this arrangement in the winter as well
(outside wood furnace / heat exchanger).
3) And while the differential pressure needed to move water goes up with
flow-rate squared, the *power* needed goes up with flow-rate cubed. So
doubling the flow through the system requires *eight* times the power.
Ah, so it's mostly about the pump, which for this project is a fixed
parameter anyway. So, do it & see what my numbers are, then...
4) A 'rule-of-thumb' we use in industrial heat-exchangers is to target water
flow for something like 4 to 7 feet/second flow through the actual
heat-exchanger tubes. This reduces the laminar layer pretty far, and
doesn't require too much pumping power. The trick is to figure out just
what the cross-section of the tube is, and the number of tubes that are in
parallel.
I've got 4 or 5 more radiators just like this one; one leaks, and I have
a bandsaw. I think I can get a cross section directly, then.
5) As one increases the flow through the heat exchanger, the temperature
rise/drop of the water/air will decrease. This is because although the
amount of heat transferred *increases*, the mass flowrate increases faster
(the amount of heat Q=<massflowrate>*<heatcapacity>*(Tin - Tout) ) This
sometimes confuses folks since the outlet temperature approaches the inlet
temperature and some people suppose this means that *less* heat is being
transferred.
Right, I figured out how that works when my a-coil froze up a month ago.
It was choked with dust from the underside, or choked enough to be a
problem. So, flow was down, delta-T was up, and the only thing cool in
the house whas the ball of ice around the a-coil. Lots of differential,
very little cooling.
6) Counter-flow arrangement is best, but most practical designs are
'cross-flow'. Parallel flow is worst (and easily converted to
counter-flow).
A car radiator is cross-flow, yes?
In conclusion, no you will never increase flow the point where heat transfer
does *down*. But extreme velocities can cause erosion problems and pumping
power requirements can be enormous.
So, here's what I'm thinking. I have the pump pushing at a certain flow
rate. I have a resistance which is currrently 7 loops of tubing through
various sections of various floors, which act as resistances in parallel.
That gives me a flow, after the slabs, of a constant value. I'll
measure that (easy enough given a bucket and a clock), and compare it to
the cross sections of my radiators to see if it gets the velocity we're
after. If we're in-range, bolt the sucker to the cold air inlet, plumb
it in (add another air filter...) and let the furnace blower suck air
through it. Temperature delta on the water side will tell me how much
heat I'm moving, since I know the flow, and temperature delta on the air
side tells me what good it's really doing.
Thanks for the non-trivial amount of time and effort you put into this
for me.
Dave Hinz
Some of the most widely used are a compromise
