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...
Robert Gammon wrote:

... physics clearly tells us that

1. Metal conducts heat FAR more efficiently than plastics

But the conductivity of the pipe wall is only a minor factor in fluid heat
transfer. In almost all situations, the conductivity of the film layer
*next* to the wall is the dominant factor. Just look at the R values for
two conventional water films versus that of 1/16" of Cu or 3/16" of plastic.
When conducting heat through a wall, the two films and wall material are in
series so it is appropriate to just sum the R values. (we'll neglect the
calculation accounting for the wall being cylindrical and just *assume* flat
plates)

Forced convection water films R values range from 0.02 m^2-K/W to as low as
0.0001 m^2-K/W. For a flow of about 2 m/s through a 3 cm pipe, we get a
Reynolds number of about 6.4e4. For water around 20C, that gives us a
Nusselt number of about 350, and a heat transfer coefficient of about 7000
W/m^2-K (or an R value of 1.44e-4). Cu has an R value of about 0.0025
m-K/W, or about 5.0e-6 m^2-K/W for a 2mm thick layer.

So the total R value for heat transfer across a water-water heat exchanger
tube might run about 1.44e-4 + 5.0e-6 + 1.44 e-4 = 2.93e-4 m^2-K/W

If the PEX has a conductivity of only 1/10th that of copper, and is three
times thicker, we would have about 1.44e-4 + 1.5e-4 + 1.44e-4 = 4.38e-4
m^2-K/W. Worse, true. But still about 67% that of the Cu.

And that is with rather optimal surface conditions and relatively high flow
(~2.1 m/s is a common 'rule of thumb' design flow rate, it balances between
poor film coefficients and excessive erosion).


2. Water falls in a thin vertical film FAR faster than water that is
flowing horizontally in a pipe.


But the flow through a flooded horizontal pipe means a much thicker film
layer. The novelty of the GFX design is that the water film formed by
having a small flow rate of say 2 gpm flowing over the inside surface of a
3" diameter pipe. This means the total thickness layer in the GFX flow is
about the same or *less* than the boundary layer thickness in conventional
pipe flow. So the average thickness between the bulk of the water and the
pipe wall is about 1/2 that of the flow layer. This reduces one of those
two film coefficients by an order of 2. This could be...

1.44e-4 + 1.0e-5 + 7.2e-5 = 2.26e-4 m^2-K/W (assuming twice the thickness
of Cu since it is double wall design).

With the high velocity of the water film on the drain side, overall heat
transfer could even be a bit better than this.

Flow in a horizontal pipe could be done in two ways. Flood the pipe
completely. But then you have issues of venting both sides of the drain
line, and the bore of the pipe would result in very low velocities and
correspondingly poor film coefficients. Or leave the pipe only partially
filed (like most current drain lines) and then you only have a tiny surface
area coming in contact with the drain water.

While not the *best* possible performance, like many designs it compromises
between getting better heat transfer coefficient, material costs, ease of
maintenance and installation.

The GFX does well with its small surface...

OK, GFX doesn't help with heat recovery for a bath, but great for hot
showers, dishwashing, clothes washing.

...60% is not "great," IMO.


For a total surface area of just (4 in)*pi *60 in /144 = 5.24 ft^2, 60% is
pretty 'great'. How much surface area does your setup require?

Here's what physics tells us on page 3.4 of the 1993 ASHRAE HOF:

1. E = (Thi-Tho)/(Thi-Tci) when Ch = Cmin and
= (Tco-Tci)/(Thi-Tci) when Ch = Cmin, where

Ch = hot fluid capacity rate, Btu/h-F
Cc = cold fluid capacity rate, Btu/h-F
Cmin = smaller of the two rates
Th = terminal temp of hot fluid (F). Subscript i indicates
entering condition; o indicates leaving condition.
Tc = terminal temp of cold fluid (F)...

2. Number of Exchanger Heat Transfer Units NTU = AUavg/Cmin.

3. Capacity rate ratio Z = Cmin/Cmax.

Generally, the heat transfer effectiveness can be expressed for a given
exchanger as a function of NTU and Z: E = f(NTU,Z,flow arrangement).
The effectiveness is independent of the temps in the exchanger.

For any exchanger with Z = 0 (where one fluid undergoes a phase change,
eg in a condenser or evaporator), E = 1-e^(-NTU).

For parallel flow exchangers, E = [1-e^(-NTU(1+Z))]/(1+Z).

For counterflow exchangers, E = [1-e^(-NTU(1-Z))]/[(1-Z(e^(-NTU(1-Z))],
= NTU/(NTU+1), when Z = 1.

For instance, if we use 50 gallons per day of hot water in short bursts
and Cmin = Cmax = 50x8.33/24h = 17.4 Btu/h-F and A = 78.5 ft^2 (a $60
300' piece of 1" polyethylene pipe with a 50 year guarantee) and U = 10
Btu/h-F-ft^2 (with slow-moving greywater and crud outside and slow-moving
fresh water inside), NTU = 78.5x10/17.4 = 45.2, and E = 0.98.


We've been through this before Nick. You can't calculate Cmin or Cmax using
a 24 hour 'average' flow rate. When water is flowing, (say 16 lbm/minute or
960 lbm/hr), your Cmin=Cmax = 960 Btu/h-F.

So *while* the water is flowing, you might see NTU=78.5*10/960 = 0.818. And
*that* would give you about E = 0.45.

By using an average flow rate that includes long periods when there is no
flow at all, you make it seem as though the heat exchanger is much longer
than just 300'. If you want to get your kind of performance with the
existing surface area and U, you would need to reduce the flow to 0.034 gpm
and keep it there all day/night. To get your kind of performance at 2 gpm,
you would need about 55 times longer tubing (~3 miles).

When you look at it that way, GFX's 0.60 performance in a 60" tall package
starts to look pretty good.

daestrom