Question

Electrochemical machining is performed to remove material from an iron surface of 20 mm × 20 mm under the following conditions:
Inter electrode gap = 0.2 mm
Supply voltage (DC) = 12 V
Specific resistance of electrolyte = 2 $\Omega$ cm
Atomic weight of iron = 55.85
Valency of Iron = 2
Faraday's constant = 96540 Coulombs
The material removal rate (in g/s) is

• A. 0.3471
• B. 3.471
• C. 34.71
• D. 347.1

Answer 1

The correct option is A 0.3471

Electrochemical machining (ECM):
Inter electrode gap, y = 0.2 mm
Supply voltage (DC). V = 12 V
Specific resistance of electrolyte,
$\text{1/K}=2\Omega$ cm
Atomic weight of iron A = 55.85
Valency of iron: Z = 2
Faraday's constant: F = 96540 coulombs

Material is removed from iron surface of 20 mm × 20 mm
${\text{Surface area: a = 20 × 20 mm}}^2$
$=2\times 2c{m}^2$

Rate of mass removal in the form
$m=\frac{AI}{ZF}$
Current density with this gap is
$J=\frac{K(V-\Delta V)}{y}$

$\text{Assume total over voltage to be 0 volt}(\Delta V=0)$

$J=\frac{KV}{y}=\frac{V}{y\left(\frac{1}{K}\right)}$

$=\frac{12V}{\left(\frac{0.2}{10}cm\right)\left(2\Omega cm\right)}=300Amp/c{m}^2$

Currrent: I = (Current Density) × (Surface Area)
$=30Amp/c{m}^2\times (2\times 2c{m}^2)$

= 1200 Amp

Rate of mass removal is
$m=\frac{AI}{ZF}=\frac{\left(55.85\right)\left(1200\right)}{2\times 96540}=0.3471gm/s$