Recently, it was discussed whether the finite resistivity of the alumina reactor tube at elevated temperatures matters in any way for LENR reactions. It does not. You can completely forget about the currents through the reactor tube.
Take a look in a good reference book like "CRC Handbook of chemistry and physics"
If you keep in mind that excess heat was already observed at 800°C and if you look at specific resistivities at these temperatures you will find:
Specific resistivity of Alumina is: rho_Al = 100000 Ohm * m
Specific resistivity of Kanthal: rho_K = 1.03*1.45e-6 Ohm * m
(The latter according to http://kanthal.com/en/products…istance-wire/kanthal-a-1/)
For example: If you insert the geometric parameters of @me356 's last run, you will get
R_Al = 100 MOhm
R_K = 10 Ohm
If you consider the current divider, and keep in mind that the total current was about 2 to 3 amps at 800°C
there will be a current of I_K = 2.5 amp * 100 MOhm / ( 100 MOhm + 10 Ohm) = 2.49999975 A flowing through the Kanthal wire
and consequently a current of I = 0.00000025 A flowing through the tube. Even if you assume that the resistivity of Ni LiAlH4 is negligible.
Completely forget about the conductivity of the alumina tube.