Electrolysis energy/power problem

[vulgarian]

Im working through an assy about electrowinning copper and done some calcs that im not too happy with. Im not too used to working with amps coulombs, KWh, etc so maybe this is the problem. Any comments on my workings would be most useful cos i gotta get it in, along with a load of other stuff!

 

I need to calculate the energy/power required to electrowin 270 tonnes of copper per day from a sulphuric acid solution.

 

Using faraday's laws of electrolysis:

 

63.54 * I * 24*3600 = 270,000,000

2 * 96,500

=>

270,000,000 * 2 * 96500 = I

63.54 * 24*3600

 

That makes 9.5 million amps per day!

 

The PD between electrodes is 2V so to find power / energy:

 

(9500000 * 2) / 24 = 791 KWh

791 / 3600 = 219W

 

I think the current calc is correct but sure im getting pickled with the power and energy calculations cos im sure 2 light bulbs do not use enough power to create 270tonnes of copper in a day!

 

another way i worked it is from trial data:

 

2A for 24 hours yielded 50g of copper.

so 270,000,000 / 50 = 5400000

multiplying the 2A by 5400000 gives 10800000A

(the diff is because the 9.5MA calc is theoretical)

 

Im just really unsure about how to get from 9.5MA/day at 2V to wattage and KWh per day.

 

help me and i'll shower you with blessings, rescue cats from trees and smile at strangers that pass my way!

[John Cuthber]

Well, I think th e9.5Ma is about right. It's not amps per day though, it's just amps. That current has to flow to get 270T/day of copper.

Since the cell voltage is 2V the easy way to calculate the power is to use the power = current times voltage and so you get

2 V times 9500000 A = 19000000 W

That's 19 megawatts of power. Rather more than a couple of light bulbs.

Your experimental data would, siimilarly give a little over 20 MW as the power required.