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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 ohm- cm
Atomic weight of iron = 55.85
Valency of Iron = 2
Faraday's constant = 96540 Coulombs
The material removal rate (in g/s) is
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- 0.3471
- 3.471
- 34.71
- 347.1
Correct Option: A
We know, R = | |
A |
where, ρ = resistivity
= special resistance of electrolyte
= 2Ω
R =resistance of electrolyte between electrodes
l = interelectrode gap = 0.2mm
A = electrodes cross-sectional area = 20 × 20 mm2
∴ R = (2 × 10) × | = 0.01Ω | |
20 × 20 |
∴ Current, I = | = | = 1200 A | ||
R | 0.01 |
Faraday of electricity, E = | |
v |
where, E = equivalent weight
M = molecular weight
v = valency
M.R.R (gm/sec) = | × | ||
F | v |
= | × | = 0.3471 | ||
96540 | 2 |