The reactor used in the Lugano test was equipped with 3 heater coils in an AC 3-phase balanced delta configuration (see Figure 4 of the Lugano test report (LTR) "Observation of abundant heat production from a reactor device and of isotopic changes in the fuel").
There had been discussions, wether the power consumption values in the LTR are plausible using usual wires for the heater coils or if these values would require a miracle material.
Some calculations (published by discussion members) showed large differences between the single coil resistance of the dummy reactor, when compared to values calculated for the active E-Cat.
The lack of information about voltages, resistances etc. in the LTR, makes it not possible to calculate exact values for the active E-Cat. But it is possible to probe, wether the values given in the LTR might be plausible.
This task requires some knowledge of the conditions in 3-phase AC balanced delta configurations. So, there is a shot summery:
For 3 Phase AC and a balanced Delta Configuration the total power equals 3 times the power of a single phase.
The total current is sqrt(3) times the current of a single phase.
The total voltage is sqrt(3) times the voltage of a single phase (e.g. 400V total gives 230V single phase) and the total resistance is the resistance of a single phase divided by sqrt(3).
Calculations of heater coil resistances
Dummy reactor at 450°C:
From the LTR one can take the values of 486 W input - 7 W joule heating (line-losses subtracted) = 479 W with 19.7 A current for a single phase.
An exact calculation of the single coil resistance is possible:
19.7A * sqrt(3) = 34.12 A (total current = single phase current * sqrt(3))
479 W / 34.12 A = 14.04 V (total voltage = total power / total current)
(The 14.04 V are for example comparable to the 400V line voltage. The voltage of a single line 230 V e.g. is calculated 400 V devided by sqrt(3)).
14.04 V / sqrt(3) = 8.11 V (single phase voltage = total voltage / sqrt(3))
8.11 V / 19.7 A = 0.41 Ohms (single coil resistance = single phase voltage / single phase current)
Probe:
8.11 V * 19.7 A = 159.77 W (single phase power = single phase voltage * single phase current)
159.77 W * 3 = 479.03 W (total power = single phase power * 3)
For a single heater coil at 450°C and a net input power of 479 W a resistance of 0.41 Ohms is calculated (the total resistance equals to 0.41 Ohms / sqrt(3) = 0.237 Ohms - further needed for verification using web-calculator).
Active E-Cat at 1260°C:
Consumption is 815.86 Watt subtracting the joule heating of 37.77 W one get a net consumption of the 3 heater coils in total of 778.09 W.
There is no exact data for the coil currents available. The report mentions 40 - 50 Amps for the total of the 3 coils.
But it is possible to prove whether the info in the report might be be plausible or not.
Let someone for example assume a total current of 43A.
The current for a single heater coil equals to:
43 A / sqrt(3) = 24.83 A (single phase current = total current / sqrt(3))
The calculation steps as for the dummy reactor:
778.09 W / 43 A = 18.09 V (total voltage = total power / total current)
18.09 V / sqrt(3) = 10.45 V (single phase voltage = total voltage / sqrt(3))
10.45 V / 24.83 A = 0.42 Ohms (single coil resistance = single phase voltage / single phase current)
Probe:
10.45 V * 24.83 A = 259.47 W (single phase power = single phase voltage * single phase current)
259.47 W * 3 = 778.41 W (total power = single phase power * 3)
For a single heater coil at 1260°C and a net power input of 778.09W a resistance of 0.42 Ohms is calculated (the total resistance equals to 0.42 Ohms / sqrt(3) = 0.242 Ohms - further needed for verification using web-calculator).
Conclusions:
The information in the LTR is plausible, at least according to the input power consumption with respect to real existing and usual heater wire materials.
My calculations show a single coil resistance of 0.41 Ohms at 450°C and a single coil resistance of 0.42 Ohms at 1260°C. (The 0.42 Ohms for the active E-Cat at 1260° were calculated under the assumption, that the total input current is 43 A - other assumptions (40-50A) and values are possible).
Real existing and usual wire materials could have been used.
With best wishes
Tom
P.S.
If you doubt my calculations, then you may check these using the 3-phase delta calculator at:
https://www.watlow.com/Resourc…neering-Tools/Calculators
Be sure you have selected "3-Phase Delta (Balanced Load)" in the calculator selection box!
Then enter only two values:
Phase Current (lp)> 19.7
Watts (W)> 479
Leave the other fields empty and press calculate.
The applet shows phase voltage 8.1049... and Ohms 0.2375... . The resistance in Ohms is that of the total circuit, for a single coil and multiplied by sqrt(3) that gives 0.411141... .
Press RESET
Enter only two example values for the active E-Cat:
Phase Current (lp)> 24.83
Watts (W)> 778.09
Leave other fields empty and press calculate.
The applet shows phase voltage 10.4455... and Ohms 0.24288... . The resistance in Ohms is that of the total circuit, for a single coil and multiplied by sqrt(3) that gives 0.420668... .