tinsnano
Industrial
- Jan 3, 2015
- 62
Dear all
I want your suggestion on the following method of chlor-alkin production as shown on the attached picture. Imagine a pump pushing brine through a pipe to a high velocity and transverse intense magnetic field to the pipe. according to Faraday's law:- moving ions in the presence of magnetic field with the right orientation will experience a force according to the right hand rule. Therefore, the sodium ions will concentrate on the upper side of the pipe as the brine solution pumped. On the other hand chlorine ions will concentrate on the bottom side of the pipe, then there will be a partition which dived the pipe in to upper and lower parts so the upper pipe part will contains more of sodium ions and the lower pipe part/partition will have concentrated chlorine ions.
As the this journey continue these two ions will encounter two electrodes which are externally shorted and hence the the sodium ions gain electrons and the chlorine lose electrons with a complete path for the electrons along the externally shorted wire.
The sodium ions are now sodium atoms so they react with nearby water molecule and forms sodium hydroxide and hydrogen while at the bottom side the chlorine ions lose electrons and forms chlorine molecules. As the two liquors get out of their corresponding pipes the gases also liberated. In general we will have sodium hydroxide solution and hydrogen at the top and chlorine and depleted brine at the bottom.
please see the attached picture
Thanks
I want your suggestion on the following method of chlor-alkin production as shown on the attached picture. Imagine a pump pushing brine through a pipe to a high velocity and transverse intense magnetic field to the pipe. according to Faraday's law:- moving ions in the presence of magnetic field with the right orientation will experience a force according to the right hand rule. Therefore, the sodium ions will concentrate on the upper side of the pipe as the brine solution pumped. On the other hand chlorine ions will concentrate on the bottom side of the pipe, then there will be a partition which dived the pipe in to upper and lower parts so the upper pipe part will contains more of sodium ions and the lower pipe part/partition will have concentrated chlorine ions.
As the this journey continue these two ions will encounter two electrodes which are externally shorted and hence the the sodium ions gain electrons and the chlorine lose electrons with a complete path for the electrons along the externally shorted wire.
The sodium ions are now sodium atoms so they react with nearby water molecule and forms sodium hydroxide and hydrogen while at the bottom side the chlorine ions lose electrons and forms chlorine molecules. As the two liquors get out of their corresponding pipes the gases also liberated. In general we will have sodium hydroxide solution and hydrogen at the top and chlorine and depleted brine at the bottom.
please see the attached picture
Thanks