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 We have a loop of Vin -- Rin -- Rout -- Vout (Kirchoff) and V = IR (good old ohm) 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 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. ----- 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]
