# CHAPTER 4.3: ELECTROLYTIC AND ELECTRO-METALLURGICAL PROCESSES¶

## Example 4.3.1, Page number 747-748¶

In [1]:
import math

#Variable declaration
l = 20.0      #Length of shaft(cm)
d = 10.0      #Diameter of shaft(cm)
thick = 1.5   #Layer of nickel(mm)
J = 195.0     #Current density(A/sq.m)
n_I = 0.92    #Current efficiency
g = 8.9       #Specific gravity of nickel

#Calculation
Wt = math.pi*l*d*thick/10*g*10**-3    #Weight of nickel to be deposited(kg)
ece_nickel = 1.0954                   #Electro-chemical equivalent of nickel(kg/1000 Ah)
Q_I = Wt*1000/(ece_nickel*n_I)        #Quantity of electricity required(Ah)
time = Q_I/(math.pi*l*d*10**-4*J)     #Time taken(hours)

#Result
print('Quantity of electricity = %.f Ah' %Q_I)
print('Time taken for the process = %.f hours' %time)

Quantity of electricity = 832 Ah
Time taken for the process = 68 hours


## Example 4.3.2, Page number 748¶

In [1]:
#Variable declaration
no_cells = 600.0    #Number of cells employed for copper refining
I = 4000.0          #Current(A)
V = 0.3             #Voltage per cell(V)
hour = 90.0         #Time of plant operation(hours)
ece_cu = 1.1844     #Electro-chemical equivalent of copper(kg/1000 Ah)

#Calculation
Ah_week  = I*hour                           #Ah per week per cell
Ah_year = Ah_week*52                        #Ah per year per cell
Wt = no_cells*ece_cu*Ah_year/(1000*10**3)   #Weight of copper refined per year(tonnes)
energy = V*I*no_cells*hour*52/1000          #Energy consumed(kWh)
consumption = energy/Wt                     #Consumption(kWh/tonne)

#Result
print('Annual output of refined copper = %.f tonnes' %Wt)
print('Energy consumption = %.1f kWh/tonne' %consumption)
print('\nNOTE: ERROR: Substitution & calculation mistake in the textbook solution')

Annual output of refined copper = 13303 tonnes
Energy consumption = 253.3 kWh/tonne

NOTE: ERROR: Substitution & calculation mistake in the textbook solution


## Example 4.3.3, Page number 748¶

In [1]:
#Variable declaration
hour = 24.0         #Time(hour)
I = 3500.0          #Average current(A)
n = 0.9             #Current efficiency
valency = 3.0       #Aluminium valency
w = 27.0            #Atomic weight of aluminium
ece_Ag = 107.98     #Electro-chemical equivalent of silver
Wt_dep = 0.00111    #Silver deposition by one coulomb(gm)

#Calculation
chemical_eq_Al = w/valency                    #Chemical equivalent of aluminium
eme_Al  = Wt_dep/ece_Ag*chemical_eq_Al        #Electro-chemical equivalent of aluminium(gm/coulomb)
Wt_Al_liberated = I*hour*3600*n*eme_Al/1000   #Weight of aluminium liberated(Kg)

#Result
print('Weight of aluminium produced from aluminium oxide = %.1f kg' %Wt_Al_liberated)

Weight of aluminium produced from aluminium oxide = 25.2 kg