1. Which of the following is true for ECG?
a) Higher the grinding wheel rpm, higher is the MRR
b) Higher the current density, faster the removal rate
c) Stronger the electrolyte, poor surface finish
d) Higher the hardness of the workpiece, lower the current density required
Explanation: The removal rate for ECG is governed by the current density, just as in ECM: as with ECM, the higher the current density, the faster the removal rate and better the resulting surface finish.
2. Which of the following is true about ECG?
a) Lower MRRs when grinding hard, heat sensitive materials
b) Machining of soft metals like aluminium can be easily done
c) Difficult to machine materials with high hardness like tungsten carbide
d) Burr-free sharpening of needles
Explanation: Applications of ECG are as follows;
(1) Single largest use for ECG is in the manufacturing and remanufacturing of turbine blades and vanes for aircraft turbine engines
(2) Grinding of tungsten carbide tool inserts
(3) Re-profiling worn locomotive traction motor gears
(4) Burr-free sharpening of hypodermic needles
(5) Grinding of surgical needles, other thin wall tubes, and fragile parts
(6) Machining of fragile or very hard and tough material – honey comb, thin walled tubes and skins
(7) High MRR’s when grinding hard, tough, stringy, work-hardenable or heat sensitive materials.
3. ECM can also be called as un-controlled anodic dissolution.
a) True
b) False
Explanation: ECM can be thought of a controlled anodic dissolution at atomic level of the workpiece that is electrically conductive by a shaped tool due to flow of high current at relatively low potential difference through an electrolyte which is quite often water based neutral salt solution.
4. For ECM of steel which is used as the electrolyte?
a) Kerosene
b) NaCl
c) Deionised water
d) HNO3
Explanation: During ECM, there will be reactions occurring at the electrodes i.e. at the anode or workpiece and at the cathode or the tool along with within the electrolyte. For electrochemical machining of steel, generally a neutral salt solution of sodium chloride (NaCl) is taken as the electrolyte.
5. MRR is ECM depends on _____________
a) hardness of work material
b) atomic weight of work material
c) thermal conductivity of work material
d) ductility of work material
Explanation: In ECM, material removal takes place due to an atomic dissolution of work material. Electrochemical dissolution is governed by Faraday’s laws. Also, for ECM, MRR= IA/(Fρv), where I= current, ρ= density of the material, A= atomic weight, v= valency, F= faraday’s constant.
6. For which of the following ECM cannot be undertaken?
a) Steel
b) Nickel based superalloy
c) Aluminium oxide
d) Titanium alloy
Explanation: ECM can machine any electrically conductive work material irrespective of their hardness, strength or even thermal properties. In ECM, material is removed from the workpiece by oxidizing it. Aluminium oxide cannot be oxidized further; therefore it cannot be machined by ECM.
7. Commercial ECM is carried out at a combination of ___________
a) low voltage high current
b) low current low voltage
c) high current high voltage
d) high voltage only
Explanation: As we know, in ECM, MRR= IA/(Fρv), where I= current. Therefore, the amount of electrochemical dissolution or deposition is proportional to the amount of charge passed through the electrochemical cell, i.e.
m∝Q, where m = mass of material dissolved or deposited
Q = amount of charge passed.
8. In ECM of pure iron a material removal rate of 600 mm3/min is required. What will be the current requirement?
a) 157A
b) 183.6A
c) 247.8A
d) 268.8A
Explanation: MRR= IA/(Fρv)
MRR = 600 mm3/min = 600/60 mm3/s = 10 mm3/s = 10×10-3cc/s
A= 56,
v= 2,
F= 96500 coulomb, and
ρ = 7.8 gm/cc
Therefore, I= (96500*10*10-3*7.8*2)/56
I= 268.8A.
9. In ECM operation of pure iron an equilibrium gap (h) of 2 mm is to be kept. What will be the supply voltage (v), if the total overvoltage is 2.5 Volts. The resistivity(r) of the electrolyte is 50 Ω-mm and the set feed rate (f) is 0.25 mm/min.
a) 5v
b) 7.8v
c) 11.3v
d) 13.2v
Explanation: h= c/f
Where, c= (v- overvoltage)*A/(F*ρ*r*v)
Therefore, c= (v-2.5)*56/(96500*7.8*10-3*50*2)
c= (v-2.5)/1344.1
Now, h= c/f
2= [(v-2.5)/1344.1]/[0.25/60] Therefore, v= 13.2 volts.
10. Electrolyte used in ECM should have _____________
a) high specific heat
b) lower resistance to film formation on the workpiece
c) higher viscosity
d) corrosive nature
Explanation: Desirable properties of the electrolytes are as follows:
• High electrical conductivity – for easy ionization
• Low viscosity – for easy flow
• High specific heat – to carry more heat
• Chemical stability – to be chemically neutral or
does not disintegrate during the reaction
• Resistance to formation of passivating film on
the workpiece surface
• Non corrosiveness and non-toxicity
• Inexpensiveness and easy availability.