Hi Friends,
We all know that production of aluminium from calcined alumina in Aluminium is a power intensive process. On an average, the efficient aluminium producers are consuming about 13.5 MWh to 15.5 MWh of electrical power per tonne of aluminium. Most of the technology suppliers are putting efforts to minimise the specific electrical power consumption. But it appears that we have to travel a long way to bring down the electrical power requirement close to theoretical power requirement of about 9.53 MWh per tonne of aluminium.
For arriving at the theoretical power requirement, I have simply followed the Faraday's law of electrolysis as step by step described here under:
We all know that production of aluminium from calcined alumina in Aluminium is a power intensive process. On an average, the efficient aluminium producers are consuming about 13.5 MWh to 15.5 MWh of electrical power per tonne of aluminium. Most of the technology suppliers are putting efforts to minimise the specific electrical power consumption. But it appears that we have to travel a long way to bring down the electrical power requirement close to theoretical power requirement of about 9.53 MWh per tonne of aluminium.
For arriving at the theoretical power requirement, I have simply followed the Faraday's law of electrolysis as step by step described here under:
- Atomic mass of aluminium = 27 gram / mole,
- Valency of aluminium = 3,
- Electro-chemical equivalent of aluminum, z = 27/3 = 9 grams.
- The mass of aluminium deposited on electrode on paasage of one coulomb of charge is termed as Electrochemical equivalent.
- For production of 1 tonne of aluminium per hour, m = 1 tonne = 1000,000 grams, t = 1 hour = 60 x 60 seconds = 3600 sec., z = 9.
- Let i amperes of current is required.
- Hence as per Faraday's Law of Electrolysis, m = z.i.t/F
- or, 1000,000 = 9 x i x 3600/96500
- Therefore, i = (1000, 000x 96500) /(9 x 3600) = 2978395 amperes.
- Let the voltage across electrodes = 3.2 v.
- Hence electrical power required = i x v /1000,000 MWh = (2978358 x 3.2) / 1000, 000 = 9.53 MWh/t of aluminium metal.
- Taking 10% extra for inefficiencies, power required = 1.10 x 9.53 = 10.5 MWh/ tonne of aluminium.
The above calculation signifies that there is tremendous scope for reduction in electrical power consumption at least by 25 to 30% from present level in aluminium smelters.
Regards.
Kunwar Rajendra