On Oct 8, 2011, at 9:58 PM, Alan Fletcher wrote:
Let's see ... The total length of that section looks to be about
10 cm.
Let's apply your "resistor" calculation.
As a first approximation, consider only the shortest path from the
thermistor to the fluid.
Vin = 100 (Voltage :: Temperature) Steam
Vout = 30 : Output of heat exchanger.
The "resistance" is proportional to the length of brass between the
thermistor and the heat source.
For the steam output .. the closest it gets to the thermistor is
about 5 cm (half the total length)
For the heat exchanger it's the thickness of the tube .. say 0.2 cm
On this we may disagree significantly. Take a look at the photos
kindly provided by Enzo:
http://www.redmatica.com/media/Thermo1.jpg
http://www.redmatica.com/media/Thermo2.jpg
The central brass fitting is very thick. Given the hose ID is about
1.5 cm I would guess over a cm thick. It appears the thermocouple was
placed not far from it.
The intermediate section looks to be at least 0.75 cm thick
From the location of the tape, and the protruding thermocouple, in:
http://www.redmatica.com/media/Thermo2.jpg
it looks like the thermocouple may have been taped to the
intermediate section and oriented axially toward the large steel nut.
We have a loop of
Vin -- Rin -- Rout -- Vout (Kirchoff)
and V = IR (good old ohm)
This is the equivalent of a simple voltage divider. I would have
drawn it like so:
Tw -- Rin -- Tout - Rout -- Thot
where Tw is the *actual* secondary loop (cooling water) exit
temperature, Thot is the actual steam/water temperature at the
primary circuit entry to the heat exchanger, and Tout is the
thermocouple location. Tout is the equivalent of a voltage probe,
and Tw and Thot are equivalent to a low and high voltage
respectively. Rin and Rout are the equivalent to resistors of course.
The temperature Tout is thus given by:
Tout = Tw + Rin/(Rin+Rout)*(Thot-Tw)
and the error Terr is:
Terr = Tout - Tw = Rin/(Rin+Rout)*(Thot-Tw)
Which is essentially your formula below. The analog of current need
not be calculated.
At any rate, we seem to be in good agreement on this part. Now I
wonder if it is even that relevant.
If you look at:
http://www.redmatica.com/media/Thermo1.jpg
You can see the thermocouple wire is still taped at its base. You
can see it probably extends to about the location of the steel nut.
It appears it rested right on the middle sized fitting, which looks
to be at least 0.75 cm thick. However, I wonder how relevant that
is. If you look at:
http://www.redmatica.com/media/Thermo2.jpg
you can see the piece of tape is back a bit from the tip of the
thermocouple. It thus can not direct significant force on the
thermocouple. Besides, the thermocouple is round, the brass fitting
is round, so the surface contact should not be very good. If this is
all true then the temperature the thermocouple is sensing is largely
the air temperature at that location under the silicon wool.
Since Vin and Vout oppose, V = Vin - Vout = 100 - 30 = 70
Rin and Rout are in series, and their resistance is proportional to
distance ... arbitrarily 1 ohm/cm.
(The actual resistivity/thermal conduction would just cancel out).
R = Rin + Rout = 5 + 0.2 = 5.2
I = V/R = 70 / (Rin + Rout) = 13.5 (I'm rounding all values off a
spreadsheet)
Then we can calculate the voltage (temp) drop across Rout -- Vdrop
-- which is the error due to heat conduction from the steam input.
Vdrop = I * Rout = 70 * Rout / ( Rin + Rout) = 70 * 0.2 / (5 +
0.2 ) = 2.7
This is essentially the same method I posted earlier, except I did
not calculate an analog to current.
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Here it is again: "At the heat exchanger side of things, a similar
formula applies, but the water does not even have to be 100°C, merely
hot enough to obtain a small delta T to the Tout temperature. If we
designate Thot to be the temperature of the water arriving at the
steam/hot water entry port, then there is some composite thermal
resistance R1 from the Tout water to the Tout thermocouple, and a
similar thermal resistance R2 to the Thot water/steam, then the
thermocouple will be at a temperature of 24°C + (R2/(R1+R2)*100°C. To
get an 8°C difference all is needed is for r=(R2/(R1+R2)) to satisfy:
r * (100°C-24°C) = 8°C
r = 8/76 = 0.1
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In your equation above I would use 0.75 instead of 0.2 here, based on
the photos. This gives an r of 0.13 and:
Vdrop = 70 * 0.75 / (5 + 0.75 ) = 9.1 °C
which is in the ball park. However we don't know the effect of the
large fitting on the air temperature under the silicon wool blanket
there, or the magnitude of the effect of air temperature on the
thermocouple there.
Since the measured drop across the heat exchanger was about 6 C,
that's a bit close for comfort.
I suspect that if you actually did a 2D or 3D FEM calculation would
come out a LOT smaller.
I don't know, because, due to the nut protruding on top and the
narrowing of the cross section, I wonder just how tight the silicon
wool was in that vicinity.
If only there had been good calibration sessions.
----- Original Message -----
Attached is a jpg of the fitting for the hot end of the Rossi heat
exchanger. The finger points to where the Tout themocouple was
located. The other side of this big brass fitting was the entry
point for the steam/water from the E-cat.
You can see white streak marks on the tape both sides of the
fitting. I wonder if those are "footprints" of the thermocouples
used.
Best regards,
Horace Heffner
[image/jpeg:Tout.jpg]
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
